A lighting system consists of two long metal rods with a potential difference maintained between them. Identical lamps can be connected between the rods as required.
\nThe following data are available for the lamps when at their working temperature.
\n\n
Lamp specifications 24 V, 5.0 W
\nPower supply emf 24 V
\nPower supply maximum current 8.0 A
\nLength of each rod 12.5 m
\nResistivity of rod metal 7.2 × 10–7 Ω m
\nA step-down transformer is used to transfer energy to the two rods. The primary coil of this transformer is connected to an alternating mains supply that has an emf of root mean square (rms) magnitude 240 V. The transformer is 95 % efficient.
\nEach rod is to have a resistance no greater than 0.10 Ω. Calculate, in m, the minimum radius of each rod. Give your answer to an appropriate number of significant figures.
\nCalculate the maximum number of lamps that can be connected between the rods. Neglect the resistance of the rods.
\nOne advantage of this system is that if one lamp fails then the other lamps in the circuit remain lit. Outline one other electrical advantage of this system compared to one in which the lamps are connected in series.
\nOutline how eddy currents reduce transformer efficiency.
\nDetermine the peak current in the primary coil when operating with the maximum number of lamps.
\nALTERNATIVE 1:
\nOR ✔
\nr = 5.352 × 10−3 ✔
\n5.4 × 10−3 «m» ✔
\n\n
For MP2 accept any SF
\nFor MP3 accept only 2 SF
\nFor MP3 accept ANY answer given to 2 SF
\n\n
ALTERNATIVE 2:
\n✔
\nr = 5.352 × 10−3 ✔
\n5.4 × 10−3 «m» ✔
\n\n
For MP2 accept any SF
\nFor MP3 accept only 2 SF
\nFor MP3 accept ANY answer given to 2 SF
\ncurrent in lamp = «= 0.21» «A»
\nOR
\nn = 24 × ✔
\n\n
so «38.4 and therefore» 38 lamps ✔
\n\n
Do not award ECF from MP1
\n\n
when adding more lamps in parallel the brightness stays the same ✔
\nwhen adding more lamps in parallel the pd across each remains the same/at the operating value/24 V ✔
\nwhen adding more lamps in parallel the current through each remains the same ✔
\nlamps can be controlled independently ✔
\nthe pd across each bulb is larger in parallel ✔
\nthe current in each bulb is greater in parallel ✔
\nlamps will be brighter in parallel than in series ✔
\nIn parallel the pd across the lamps will be the operating value/24 V ✔
\n\n
Accept converse arguments for adding lamps in series:
\nwhen adding more lamps in series the brightness decreases
\nwhen adding more lamps in series the pd decreases
\nwhen adding more lamps in series the current decreases
\nlamps can’t be controlled independently
\nthe pd across each bulb is smaller in series
\nthe current in each bulb is smaller in series
\n\n
in series the pd across the lamps will less than the operating value/24 V
\nDo not accept statements that only compare the overall resistance of the combination of bulbs.
\n«as flux linkage change occurs in core, induced emfs appear so» current is induced ✔
\ninduced currents give rise to resistive forces ✔
\neddy currents cause thermal energy losses «in conducting core» ✔
\npower dissipated by eddy currents is drawn from the primary coil/reduces power delivered to the secondary ✔
\npower = 190 OR 192 «W» ✔
\nrequired power «200 or 202 W» ✔
\nso OR 0.84 «A rms» ✔
\npeak current = « OR » = 1.2/1.3 «A» ✔
\nIon-thrust engines can power spacecraft. In this type of engine, ions are created in a chamber and expelled from the spacecraft. The spacecraft is in outer space when the propulsion system is turned on. The spacecraft starts from rest.
\nThe mass of ions ejected each second is 6.6 × 10–6 kg and the speed of each ion is 5.2 × 104 m s–1. The initial total mass of the spacecraft and its fuel is 740 kg. Assume that the ions travel away from the spacecraft parallel to its direction of motion.
\nAn initial mass of 60 kg of fuel is in the spacecraft for a journey to a planet. Half of the fuel will be required to slow down the spacecraft before arrival at the destination planet.
\nIn practice, the ions leave the spacecraft at a range of angles as shown.
\nDetermine the initial acceleration of the spacecraft.
\n(i) Estimate the maximum speed of the spacecraft.
\n(ii) Outline why the answer to (i) is an estimate.
\nOutline why scientists sometimes use estimates in making calculations.
\nOutline why the ions are likely to spread out.
\nExplain what effect, if any, this spreading of the ions has on the acceleration of the spacecraft.
\nchange in momentum each second = 6.6 × 10−6 × 5.2 × 104 «= 3.4 × 10−1 kg m s−1» ✔
\nacceleration = « =» 4.6 × 10−4 «m s−2» ✔
\n(i) ALTERNATIVE 1:
\n(considering the acceleration of the spacecraft)
\ntime for acceleration = = «4.6 × 106» «s» ✔
\nmax speed = «answer to (a) × 4.6 × 106 =» 2.1 × 103 «m s−1» ✔
\n\n
ALTERNATIVE 2:
\n(considering the conservation of momentum)
\n(momentum of 30 kg of fuel ions = change of momentum of spacecraft)
\n30 × 5.2 × 104 = 710 × max speed ✔
\nmax speed = 2.2 × 103 «m s−1» ✔
\n\n
(ii) as fuel is consumed total mass changes/decreases so acceleration changes/increases
OR
external forces (such as gravitational) can act on the spacecraft so acceleration isn’t constant ✔
problem may be too complicated for exact treatment ✔
\nto make equations/calculations simpler ✔
\nwhen precision of the calculations is not important ✔
\nsome quantities in the problem may not be known exactly ✔
\nions have same (sign of) charge ✔
\nions repel each other ✔
\nthe forces between the ions do not affect the force on the spacecraft. ✔
\nthere is no effect on the acceleration of the spacecraft. ✔
\nA chicken’s egg of mass 58 g is dropped onto grass from a height of 1.1 m. Assume that air resistance is negligible and that the egg does not bounce or break.
\nDefine impulse.
\nShow that the kinetic energy of the egg just before impact is about 0.6 J.
\nThe egg comes to rest in a time of 55 ms. Determine the magnitude of the average decelerating force that the ground exerts on the egg.
\nExplain why the egg is likely to break when dropped onto concrete from the same height.
\nforce × time
\nOR
\nchange in momentum ✔
\nEk = mgh = 0.058 × 9.81 ×1.1 = 0.63 J ✔
\nAllow use of g = 10 m s−2 (which gives 0.64 «J»)
\nSubstitution and at least 2 SF must be shown
\nALTERNATIVE 1:
\ninitial momentum = mv = «= 0.27 kg m s−1»
\nOR
\nmv = «= 0.27 kg m s−1» ✔
\nforce = « =» ✔
\n4.9 «N» ✔
\nF − mg = 4.9 so F = 5.5 «N» ✔
\n\n
ALTERNATIVE 2:
\n«Ek = mv2 = 0.63 J» v = 4.7 m s−1 ✔
\nacceleration = « =» = «85 m s−2» ✔
\n4.9 «N» ✔
\nF − mg = 4.9 so F= 5.5 «N» ✔
\n\n
Accept negative acceleration and force.
\nALTERNATIVE 1:
\nconcrete reduces the stopping time/distance ✔
\nimpulse/change in momentum same so force greater
\nOR
\nwork done same so force greater ✔
\n\n
ALTERNATIVE 2:
\nconcrete reduces the stopping time ✔
\ndeceleration is greater so force is greater ✔
\n\n
Allow reverse argument for grass.
\nA pipe is open at both ends. A first-harmonic standing wave is set up in the pipe. The diagram shows the variation of displacement of air molecules in the pipe with distance along the pipe at time t = 0. The frequency of the first harmonic is f.
\nA transmitter of electromagnetic waves is next to a long straight vertical wall that acts as a plane mirror to the waves. An observer on a boat detects the waves both directly and as an image from the other side of the wall. The diagram shows one ray from the transmitter reflected at the wall and the position of the image.
\nSketch, on the diagram, the variation of displacement of the air molecules with distance along the pipe when t = .
\nAn air molecule is situated at point X in the pipe at t = 0. Describe the motion of this air molecule during one complete cycle of the standing wave beginning from t = 0.
\nThe speed of sound c for longitudinal waves in air is given by
\n\n
where ρ is the density of the air and K is a constant.
\nA student measures f to be 120 Hz when the length of the pipe is 1.4 m. The density of the air in the pipe is 1.3 kg m–3. Determine the value of K for air. State your answer with the appropriate fundamental (SI) unit.
\nDemonstrate, using a second ray, that the image appears to come from the position indicated.
\nOutline why the observer detects a series of increases and decreases in the intensity of the received signal as the boat moves along the line XY.
\nhorizontal line shown in centre of pipe ✔
\n«air molecule» moves to the right and then back to the left ✔
\nreturns to X/original position ✔
\nwavelength = 2 × 1.4 «= 2.8 m» ✔
\nc = «f λ =» 120 × 2.8 «= 340 m s−1» ✔
\nK = «ρc2 = 1.3 × 3402 =» 1.5 × 105 ✔
\nkg m–1 s–2 ✔
\nconstruction showing formation of image ✔
\nAnother straight line/ray from image through the wall with line/ray from intersection at wall back to transmitter. Reflected ray must intersect boat.
\ninterference pattern is observed
\nOR
\ninterference/superposition mentioned ✔
\n
maximum when two waves occur in phase/path difference is nλ
OR
\nminimum when two waves occur 180° out of phase/path difference is (n + ½)λ ✔
\nThe diagram shows the position of the principal lines in the visible spectrum of atomic hydrogen and some of the corresponding energy levels of the hydrogen atom.
\nA low-pressure hydrogen discharge lamp contains a small amount of deuterium gas in addition to the hydrogen gas. The deuterium spectrum contains a red line with a wavelength very close to that of the hydrogen red line. The wavelengths for the principal lines in the visible spectra of deuterium and hydrogen are given in the table.
\nLight from the discharge lamp is normally incident on a diffraction grating.
\nDetermine the energy of a photon of blue light (435nm) emitted in the hydrogen spectrum.
\nIdentify, with an arrow labelled B on the diagram, the transition in the hydrogen spectrum that gives rise to the photon with the energy in (a)(i).
\nExplain your answer to (a)(ii).
\nThe light illuminates a width of 3.5 mm of the grating. The deuterium and hydrogen red lines can just be resolved in the second-order spectrum of the diffraction grating. Show that the grating spacing of the diffraction grating is about 2 × 10–6 m.
\nCalculate the angle between the first-order line of the red light in the hydrogen spectrum and the second-order line of the violet light in the hydrogen spectrum.
\nThe light source is changed so that white light is incident on the diffraction grating. Outline the appearance of the diffraction pattern formed with white light.
\nidentifies λ = 435 nm ✔
\nE = « =» ✔
\n4.6 ×10−19 «J» ✔
\n–0.605 OR –0.870 OR –1.36 to –5.44 AND arrow pointing downwards ✔
\nArrow MUST match calculation in (a)(i)
\nAllow ECF from (a)(i)
\nDifference in energy levels is equal to the energy of the photon ✔
\nDownward arrow as energy is lost by hydrogen/energy is given out in the photon/the electron falls from a higher energy level to a lower one ✔
\n«lines» ✔
\nso spacing is «= 1.9 × 10−6 m» ✔
\n\n
Allow use of either wavelength or the mean value
\nMust see at least 2 SF for a bald correct answer
\n2 × 4.1 × 10−7 = 1.9 × 10−6 sin θv seen
\nOR
\n6.6 × 10−7 = 1.9 × 10−6 sin θr seen ✔
\n\n
θv = 24 − 26 «°»
\nOR
\nθr = 19 − 20 «°» ✔
\n\n
Δθ = 5 − 6 «°» ✔
\n\n
For MP3 answer must follow from answers in MP2
\nFor MP3 do not allow ECF from incorrect angles
\ncentre of pattern is white ✔
\ncoloured fringes are formed ✔
\nblue/violet edge of order is closer to centre of pattern
\nOR
\nred edge of order is furthest from centre of pattern ✔
\nthe greater the order the wider the pattern ✔
\nthere are gaps between «first and second order» spectra ✔
\nis formed when a nucleus of deuterium () collides with a nucleus of . The radius of a deuterium nucleus is 1.5 fm.
\nState how the density of a nucleus varies with the number of nucleons in the nucleus.
\nShow that the nuclear radius of phosphorus-31 () is about 4 fm.
\nState the maximum distance between the centres of the nuclei for which the production of is likely to occur.
\nDetermine, in J, the minimum initial kinetic energy that the deuterium nucleus must have in order to produce . Assume that the phosphorus nucleus is stationary throughout the interaction and that only electrostatic forces act.
\nundergoes beta-minus (β–) decay. Explain why the energy gained by the emitted beta particles in this decay is not the same for every beta particle.
\nState what is meant by decay constant.
\nIn a fresh pure sample of the activity of the sample is 24 Bq. After one week the activity has become 17 Bq. Calculate, in s–1, the decay constant of .
\nit is constant ✔
\nR = «m» ✔
\nMust see working and answer to at least 2SF
\nseparation for interaction = 5.3 or 5.5 «fm» ✔
\nenergy required = ✔
\n= 6.5 / 6.6 ×10−13 OR 6.3 ×10−13 «J» ✔
\n\n
Allow ecf from (b)(i)
\n«electron» antineutrino also emitted ✔
\nenergy split between electron and «anti»neutrino ✔
\nprobability of decay of a nucleus ✔
\nOR
\nthe fraction of the number of nuclei that decay
\nin one/the next second
\nOR
\nper unit time ✔
\n1 week = 6.05 × 105 «s»
\n17 = ✔
\n5.7 × 10−7 «s–1» ✔
Award [2 max] if answer is not in seconds
\nIf answer not in seconds and no unit quoted award [1 max] for correct substitution into equation (MP2)
\nA small electric motor is used with a 12 mF capacitor and a battery in a school experiment.
\nWhen the switch is connected to X, the capacitor is charged using the battery. When the switch is connected to Y, the capacitor fully discharges through the electric motor that raises a small mass.
\nThe battery has an emf of 7.5 V. Determine the charge that flows through the motor when the mass is raised.
\nThe motor can transfer one-third of the electrical energy stored in the capacitor into gravitational potential energy of the mass. Determine the maximum height through which a mass of 45 g can be raised.
\nAn additional identical capacitor is connected in series with the first capacitor and the charging and discharging processes are repeated. Comment on the effect this change has on the height and time taken to raise the 45 g mass.
\ncharge stored on capacitor = 12 × 10−3 × 7.5 = 0.09 «C» ✔
\nenergy stored in capacitor «CV2 or QV =» × 12 × 10−3 × 7.52 «= 0.338 J» ✔
\nheight = «» 0.25/0.26 «m»
\n\n
Allow use of g = 10 m s−2 which gives 0.25 «m»
\nC halved ✔
\nso energy stored is halved/reduced so rises «less than» half height ✔
\ndischarge time/raise time less as RC halved/reduced ✔
\n\n
Allow 6 mF
\nIn an investigation to measure the acceleration of free fall a rod is suspended horizontally by two vertical strings of equal length. The strings are a distance d apart.
\nWhen the rod is displaced by a small angle and then released, simple harmonic oscillations take place in a horizontal plane.
\nThe theoretical prediction for the period of oscillation T is given by the following equation
\n\n
where c is a known numerical constant.
\nIn one experiment d was varied. The graph shows the plotted values of T against . Error bars are negligibly small.
\nState the unit of c.
\nA student records the time for 20 oscillations of the rod. Explain how this procedure leads to a more precise measurement of the time for one oscillation T.
\nDraw the line of best fit for these data.
\nSuggest whether the data are consistent with the theoretical prediction.
\nThe numerical value of the constant c in SI units is 1.67. Determine g, using the graph.
\n✔
\n\n
Accept other power of tens multiples of , eg: .
\nmeasured uncertainties «for one oscillation and for 20 oscillations» are the same/similar/OWTTE
\nOR
\n% uncertainty is less for 20 oscillations than for one ✔
\n\n
dividing «by 20» / finding mean reduces the random error ✔
\nStraight line touching at least 3 points drawn across the range ✔
\nIt is not required to extend the line to pass through the origin.
\ntheory predicts proportional relation «, slope = Td = = constant » ✔
\nthe graph is «straight» line through the origin ✔
\ncorrectly determines gradient using points where ΔT≥1.5s
\nOR
\ncorrectly selects a single data point with T≥1.5s ✔
\n\n
manipulation with formula, any new and correct expression to enable g to be determined ✔
\nCalculation of g ✔
\nWith g in range 8.6 and 10.7 «m s−2» ✔
\n\n
Allow range 0.51 to 0.57.
\nThere is a proposal to place a satellite in orbit around planet Mars.
\nThe satellite is to have an orbital time T equal to the length of a day on Mars. It can be shown that
\nT2 = kR3
\nwhere R is the orbital radius of the satellite and k is a constant.
\nThe ratio = 1.5.
\nOutline what is meant by gravitational field strength at a point.
\nNewton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting Mars.
\nMars has a mass of 6.4 × 1023 kg. Show that, for Mars, k is about 9 × 10–13 s2 m–3.
\nThe time taken for Mars to revolve on its axis is 8.9 × 104 s. Calculate, in m s–1, the orbital speed of the satellite.
\n\n
Show that the intensity of solar radiation at the orbit of Mars is about 600 W m–2.
\nDetermine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.
\nThe atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the mean temperature of Earth is strongly affected by gases in its atmosphere but that of Mars is not.
\nforce per unit mass ✔
\nacting on a small/test/point mass «placed at the point in the field» ✔
\nMars is spherical/a sphere «and of uniform density so behaves as a point mass» ✔
\nsatellite has a much smaller mass/diameter/size than Mars «so approximates to a point mass» ✔
\n« hence» . Also
\nOR
\nhence ✔
\n\n
uses either of the above to get
\nOR
\nuses ✔
\n\n
k = 9.2 × 10−13 / 9.3 × 10−13
\n\n
\n
Unit not required
\n\n
R = 2.04 × 107 «m» ✔
\n\n
v = «» 1.4 × 103 «m s–1»
\nOR
\nv = «» 1.4 × 103 «m s–1» ✔
\nuse of «1.36 × 103 × » ✔
\n604 «W m–2» ✔
\nuse of for mean intensity ✔
\ntemperature/K = «» 230 ✔
\nreference to greenhouse gas/effect ✔
\nrecognize the link between molecular density/concentration and pressure ✔
\nlow pressure means too few molecules to produce a significant heating effect
\nOR
\nlow pressure means too little radiation re-radiated back to Mars ✔
\n\n
The greenhouse effect can be described, it doesn’t have to be named
\nA beam of electrons each of de Broglie wavelength 2.4 × 10–15 m is incident on a thin film of silicon-30 . The variation in the electron intensity of the beam with scattering angle is shown.
\nUse the graph to show that the nuclear radius of silicon-30 is about 4 fm.
\nEstimate, using the result from (a)(i), the nuclear radius of thorium-232 .
\nSuggest one reason why a beam of electrons is better for investigating the size of a nucleus than a beam of alpha particles of the same energy.
\nOutline why deviations from Rutherford scattering are observed when high-energy alpha particles are incident on nuclei.
\nread off between 17 and 19 «deg» ✔
\ncorrect use of d = = 7.8 × 10−15 «m» ✔
\nso radius = «fm» = 3.9 «fm» ✔
\nAward ecf for wrong angle in MP1.
\nAnswer for MP3 must show at least 2 sf.
\nRTh = Rsi or substitution ✔
\n7.4 «fm» ✔
electron wavelength shorter than alpha particles (thus increased resolution)
OR
electron is not subject to strong nuclear force ✔
\n
nuclear forces act ✔
\nnuclear recoil occurs ✔
\nsignificant penetration into nucleus / probing internal structure of individual nucleons ✔
\nincident particles are relativistic ✔
\nThis question was left blank by many candidates and many of those who attempted it chose an angle that when used with the correct equation gave an answer close to the given answer of 4 fm. Very few selected the correct angle, calculated the correct diameter, and divided by two to get the correct radius.
\nThis question was also left blank by many candidates. Many who did answer simply used the ratio of the of the mass numbers of the two elements and failed to take the cube root of the ratio. It should be noted that the question specifically stated that candidates were expected to use the result from 2ai, and not just simply guess at the new radius.
\nThis question was very poorly answered with the vast majority of candidates simply listing differences between alpha particles and electrons (electrons have less mass, electrons have less charge, etc) rather than considering why high speed electrons would be better for studying the nucleus.
\nCandidates struggled with this question. The vast majority of responses were descriptions of Rutherford scattering with no connection made to the deviations when high-energy alpha particles are used. Many of the candidates who did appreciate that this was a different situation from the traditional experiment made vague comments about the alpha particles “hitting” the nucleus.
\nA beam of microwaves is incident normally on a pair of identical narrow slits S1 and S2.
\nWhen a microwave receiver is initially placed at W which is equidistant from the slits, a maximum in intensity is observed. The receiver is then moved towards Z along a line parallel to the slits. Intensity maxima are observed at X and Y with one minimum between them. W, X and Y are consecutive maxima.
\nExplain why intensity maxima are observed at X and Y.
\nThe distance from S1 to Y is 1.243 m and the distance from S2 to Y is 1.181 m.
\nDetermine the frequency of the microwaves.
\nOutline one reason why the maxima observed at W, X and Y will have different intensities from each other.
\nThe microwaves emitted by the transmitter are horizontally polarized. The microwave receiver contains a polarizing filter. When the receiver is at position W it detects a maximum intensity.
\nThe receiver is then rotated through 180° about the horizontal dotted line passing through the microwave transmitter. Sketch a graph on the axes provided to show the variation of received intensity with rotation angle.
\ntwo waves superpose/mention of superposition/mention of «constructive» interference ✔
\nthey arrive in phase/there is a path length difference of an integer number of wavelengths ✔
\npath difference = 0.062 «m»✔
\nso wavelength = 0.031 «m»✔
\nfrequency = 9.7 × 109 «Hz»✔
\nAward [2 max] for 4.8 x 109 Hz.
\nintensity is modulated by a single slit diffraction envelope OR
\nintensity varies with distance OR points are different distances from the slits ✔
cos2 variation shown ✔
\nwith zero at 90° (by eye) ✔
\nAward [1 max] for an inverted curve with maximum at 90°.
\nMany candidates were able to discuss the interference that is taking place in this question, but few were able to fully describe the path length difference. That said, the quality of responses on this type of question seems to have improved over the last few examination sessions with very few candidates simply discussing the crests and troughs of waves.
\nMany candidates struggled with this question. Few were able to calculate a proper path length difference, and then use that to calculate the wavelength and frequency. Many candidates went down blind paths of trying various equations from the data booklet, and some seemed to believe that the wavelength is just the reciprocal of the frequency.
\nThis is one of many questions on this paper where candidates wrote vague answers that did not clearly connect to physics concepts or include key information. There were many overly simplistic answers like “they are farther away” without specifying what they are farther away from. Candidates should be reminded that their responses should go beyond the obvious and include some evidence of deeper understanding.
\nThis question was generally well answered, with many candidates at least recognizing that the intensity would decrease to zero at 90 degrees. Many struggled with the exact shape of the graph, though, and some drew a graph that extended below zero showing a lack of understanding of what was being graphed.
\nThe moon Phobos moves around the planet Mars in a circular orbit.
\nOutline the origin of the force that acts on Phobos.
\nOutline why this force does no work on Phobos.
\nThe orbital period T of a moon orbiting a planet of mass M is given by
\n\n
where R is the average distance between the centre of the planet and the centre of the moon.
\nShow that
\nThe following data for the Mars–Phobos system and the Earth–Moon system are available:
\nMass of Earth = 5.97 × 1024 kg
\nThe Earth–Moon distance is 41 times the Mars–Phobos distance.
\nThe orbital period of the Moon is 86 times the orbital period of Phobos.
\nCalculate, in kg, the mass of Mars.
\nThe graph shows the variation of the gravitational potential between the Earth and Moon with distance from the centre of the Earth. The distance from the Earth is expressed as a fraction of the total distance between the centre of the Earth and the centre of the Moon.
\nDetermine, using the graph, the mass of the Moon.
\ngravitational attraction/force/field «of the planet/Mars» ✔
\nDo not accept “gravity”.
\nthe force/field and the velocity/displacement are at 90° to each other OR
\nthere is no change in GPE of the moon/Phobos ✔
\nALTERNATE 1
\n«using fundamental equations»
\nuse of Universal gravitational force/acceleration/orbital velocity equations ✔
\nequating to centripetal force or acceleration. ✔
\nrearranges to get ✔
\nALTERNATE 2
\n«starting with »
\nsubstitution of proper equation for T from orbital motion equations ✔
\nsubstitution of proper equation for M OR R from orbital motion equations ✔
\nrearranges to get ✔
\nor other consistent re-arrangement ✔
\n6.4 × 1023 «kg» ✔
\n\n
read off separation at maximum potential 0.9 ✔
\nequating of gravitational field strength of earth and moon at that location OR ✔
7.4 × 1022 «kg» ✔
\nAllow ECF from MP1
\nThis was generally well answered, although some candidates simply used the vague term “gravity” rather than specifying that it is a gravitational force or a gravitational field. Candidates need to be reminded about using proper physics terms and not more general, “every day” terms on the exam.
\nSome candidates connected the idea that the gravitational force is perpendicular to the velocity (and hence the displacement) for the mark. It was also allowed to discuss that there is no change in gravitational potential energy, so therefore no work was being done. It was not acceptable to simply state that the net displacement over one full orbit is zero. Unfortunately, some candidates suggested that there is no net force on the moon so there is no work done, or that the moon is so much smaller so no work could be done on it.
\nThis was another “show that” derivation. Many candidates attempted to work with universal gravitation equations, either from memory or the data booklet, to perform this derivation. The variety of correct solution paths was quite impressive, and many candidates who attempted this question were able to receive some marks. Candidates should be reminded on “show that” questions that it is never allowed to work backwards from the given answer. Some candidates also made up equations (such as T = 2𝝿r) to force the derivation to work out.
\nThis question was challenging for candidates. The candidates who started down the correct path of using the given derived value from 5bi often simply forgot that the multiplication factors had to be squared and cubed as well as the variables.
\nThis question was left blank by many candidates, and very few who attempted it were able to successfully recognize that the gravitational fields of the Earth and Moon balance at 0.9r and then use the proper equation to calculate the mass of the Moon.
\nLiquid oxygen at its boiling point is stored in an insulated tank. Gaseous oxygen is produced from the tank when required using an electrical heater placed in the liquid.
\nThe following data are available.
\nMass of 1.0 mol of oxygen = 32 g
\nSpecific latent heat of vaporization of oxygen = 2.1 × 105 J kg–1
\nAn oxygen flow rate of 0.25 mol s–1 is needed.
\nDistinguish between the internal energy of the oxygen at the boiling point when it is in its liquid phase and when it is in its gas phase.
\nCalculate, in kW, the heater power required.
\nCalculate the volume of the oxygen produced in one second when it is allowed to expand to a pressure of 0.11 MPa and to reach a temperature of –13 °C.
\nState one assumption of the kinetic model of an ideal gas that does not apply to oxygen.
\nInternal energy is the sum of all the PEs and KEs of the molecules (of the oxygen) ✔
\nPE of molecules in gaseous state is zero ✔
\n(At boiling point) average KE of molecules in gas and liquid is the same ✔
\ngases have a higher internal energy ✔
\n\n
Molecules/particles/atoms must be included once, if not, award [1 max]
\nALTERNATIVE 1:
\nflow rate of oxygen = 8 «g s−1» ✔
\n«2.1 × 105 × 8 × 10−3» = 1.7 «kW» ✔
\n\n
ALTERNATIVE 2:
\nQ = «0.25 × 32 × 10−3 × 2.1 × 105 =» 1680 «J» ✔
\npower = «1680 W =» 1.7 «kW» ✔
\nT = 260 «K» ✔
\nV = «» 4.9 × 10−3 «m3» ✔
\nideal gas has point objects ✔
\nno intermolecular forces ✔
\nnon liquefaction ✔
\nideal gas assumes monatomic particles ✔
\nthe collisions between particles are elastic ✔
\n\n
Allow the opposite statements if they are clearly made about oxygen eg oxygen/this can be liquified
\nA student makes a parallel-plate capacitor of capacitance 68 nF from aluminium foil and plastic film by inserting one sheet of plastic film between two sheets of aluminium foil.
\nThe aluminium foil and the plastic film are 450 mm wide.
\nThe plastic film has a thickness of 55 μm and a permittivity of 2.5 × 10−11 C2 N–1 m–2.
\nThe student uses a switch to charge and discharge the capacitor using the circuit shown. The ammeter is ideal.
\nThe emf of the battery is 12 V.
\nCalculate the total length of aluminium foil that the student will require.
\nThe plastic film begins to conduct when the electric field strength in it exceeds 1.5 MN C–1. Calculate the maximum charge that can be stored on the capacitor.
\nThe resistor R in the circuit has a resistance of 1.2 kΩ. Calculate the time taken for the charge on the capacitor to fall to 50 % of its fully charged value.
\nThe ammeter is replaced by a coil. Explain why there will be an induced emf in the coil while the capacitor is discharging.
\nSuggest one change to the discharge circuit, apart from changes to the coil, that will increase the maximum induced emf in the coil.
\nlength = ✔
\n= 0.33 «m» ✔
\nso 0.66/0.67 «m» «as two lengths required» ✔
\n1.5 × 106 × 55 × 10-6 = 83 «V» ✔
\nq «= CV»= 5.6 × 10-6 «C»✔
\n\n
t = «−» 1200 × 6.8 × 10−8 × ln0.5 ✔
\n5.7 × 10−5 «s» ✔
\nOR
\nuse of t = RC × ln2 ✔
\n1200 × 6.8 × 10−8 × 0.693 ✔
\n5.7 × 10−5 «s» ✔
\nmention of Faraday’s law ✔
\nindicating that changing current in discharge circuit leads to change in flux in coil/change in magnetic field «and induced emf» ✔
\ndecrease/reduce ✔
\nresistance (R) OR capacitance (C) ✔
\nMany candidates were able to use the proper equation to calculate the length of one piece of aluminum foil for the first two marks, but very few doubled the length for the final mark.
\nThis question was challenging for many candidates. While some candidates were able to use proper equations for capacitors to determine the charge some of the candidates attempted to use electrostatic equations for the electric field around a point charge to solve this problem.
\nThis question was also challenging for many candidates, with not an insignificant number leaving it blank. The candidates who did attempt it generally set up a correct equation, but ran into some simple calculation and power of ten errors. Some candidates attempted to solve the equation using basic circuit equations, which did not receive any marks.
\nThis is an explain question, so there was an expectation for a fairly detailed response. Many candidates missed the fact that the discharging capacitor is causing the current in the coil to change in time, and that this is what is inducing the emf in the coil. Many simply stated that the current created a magnetic field with not complete explanation of induction.
\nCandidates who recognized that something about the discharge circuit (not the charging circuit) needed to be changed generally suggested that something had to change with the resistance or capacitance. It should be noted that even though this was the last question on the exam, it was attempted at a higher rate than many of the other questions on the exam.
\nA small metal pendulum bob is suspended at rest from a fixed point with a length of thread of negligible mass. Air resistance is negligible.
\nThe pendulum begins to oscillate. Assume that the motion of the system is simple harmonic, and in one vertical plane.
\nThe graph shows the variation of kinetic energy of the pendulum bob with time.
\nWhen the 75 g bob is moving horizontally at 0.80 m s–1, it collides with a small stationary object also of mass 75 g. The object and the bob stick together.
\nCalculate, in m, the length of the thread. State your answer to an appropriate number of significant figures.
\nLabel on the graph with the letter X a point where the speed of the pendulum is half that of its initial speed.
\nThe mass of the pendulum bob is 75 g. Show that the maximum speed of the bob is about 0.7 m s–1.
\nCalculate the speed of the combined masses immediately after the collision.
\nShow that the collision is inelastic.
\nSketch, on the axes, a graph to show the variation of gravitational potential energy with time for the bob and the object after the collision. The data from the graph used in (a) is shown as a dashed line for reference.
\nThe speed after the collision of the bob and the object was measured using a sensor. This sensor emits a sound of frequency f and this sound is reflected from the moving bob. The sound is then detected by the sensor as frequency f′.
\nExplain why f and f′ are different.
\nidentifies T as 2.25 s ✔
\nL = 1.26 m ✔
\n1.3 / 1.26 «m» ✔
\nAccept any number of s.f. for MP2.
\nAccept any answer with 2 or 3 s.f. for MP3.
\nX labels any point on the curve where EK of maximum/5 mJ ✔
\nmv2 = 20 × 10−3 seen OR × 7.5 × 10-2 × v2 = 20 × 10-3 ✔
\n0.73 «m s−1» ✔
\nMust see at least 2 s.f. for MP2.
\n0.40 «m s-1» ✔
\ninitial energy 24 mJ and final energy 12 mJ ✔
\nenergy is lost/unequal /change in energy is 12 mJ ✔
\ninelastic collisions occur when energy is lost ✔
\ngraph with same period but inverted ✔
\namplitude one half of the original/two boxes throughout (by eye) ✔
\nmention of Doppler effect ✔
\nthere is a change in the wavelength of the reflected wave ✔
\nbecause the wave speed is constant, there is a change in frequency ✔
\nThis question was well approached by candidates. The noteworthy mistakes were not reading the correct period of the pendulum from the graph, and some simple calculation and mathematical errors. This question also had one mark for writing an answer with the correct number of significant digits. Candidates should be aware to look for significant digit question on the exam and can write any number with correct number of significant digits for the mark.
\nThis question was well answered. This is a “show that” question so candidates needed to clearly show the correct calculation and write an answer with at least one significant digit more than the given answer. Many candidates failed to appreciate that the energy was given in mJ and the mass was in grams, and that these values needed to be converted before substitution.
\nCandidates fell into some broad categories on this question. This was a “show that” question, so there was an expectation of a mathematical argument. Many were able to successfully show that the initial and final kinetic energies were different and connect this to the concept of inelastic collisions. Some candidates tried to connect conservation of momentum unsuccessfully, and some simply wrote an extended response about the nature of inelastic collisions and noted that the bobs stuck together without any calculations. This approach was awarded zero marks.
\nMany candidates drew graphs that received one mark for either recognizing the phase difference between the gravitational potential energy and the kinetic energy, or for recognizing that the total energy was half the original energy. Few candidates had both features for both marks.
\nThis question was essentially about the Doppler effect, and therefore candidates were expected to give a good explanation for why there is a frequency difference. As with all explain questions, the candidates were required to go beyond the given information. Very few candidates earned marks beyond just recognizing that this was an example of the Doppler effect. Some did discuss the change in wavelength caused by the relative motion of the bob, although some candidates chose very vague descriptions like “the waves are all squished up” rather than using proper physics terms. Some candidates simply wrote and explained the equation from the data booklet, which did not receive marks. It should be noted that this was a three mark question, and yet some candidates attempted to answer it with a single sentence.
\nA student blows across the top of a cylinder that contains water. A first-harmonic standing sound wave is produced in the air of the cylinder. More water is then added to the cylinder. The student blows so that a first-harmonic standing wave is produced with a different frequency.
\nWhat is the nature of the displacement in the air at the water surface and the change in frequency when the water is added?
\nD
\nA third-harmonic standing wave of wavelength 0.80 m is set up on a string fixed at both ends. Two points on the wave are separated by a distance of 0.60 m. What is a possible phase difference between the two points on the wave?
\nA.
\nB.
\nC.
\nD.
\nC
\nThis question had a low discrimination index with response D the most popular and an even spread between the other 3 answers. A third-harmonic standing wave of wavelength 0.8m must be on a string of length 1.2m giving 3 loops of 0.4m each. Depending on where the initial point is chosen, two points separated by 0.6m will either be in adjacent loops e.g. at 0.1m and 0.7 m with a phase difference of π or in the two end loops e.g at 0.3 m and 0.9m with a phase difference of 0. So for a standing wave there are only two possible answers, π (response C) or 0 (not included in these responses).
\nA particle with a charge ne is accelerated through a potential difference V.
What is the magnitude of the work done on the particle?
\nA.
B.
C.
D.
\nB
\nA train approaches a station and sounds a horn of constant frequency and constant intensity. An observer waiting at the station detects a frequency fobs and an intensity Iobs. What are the changes, if any, in Iobs and fobs as the train slows down?
\nD
\nAn unusual way of considering the Doppler effect, this had a very low discrimination index with the most popular answer A when D was correct. It is likely the candidates have confused what the train is producing – a constant intensity sound – and what the observer hears, Io, where the intensity is going to increase as the train approaches. This immediately eliminates options A and C.
\nIn an experiment to determine the resistivity of a material, a student measures the resistance of several wires made from the pure material. The wires have the same length but different diameters.
Which quantities should the student plot on the -axis and the -axis of a graph to obtain a straight line?
\nC
\nThree resistors of resistance 1.0 Ω, 6.0 Ω and 6.0 Ω are connected as shown. The voltmeter is ideal and the cell has an emf of 12 V with negligible internal resistance.
\nWhat is the reading on the voltmeter?
A. 3.0 V
B. 4.0 V
C. 8.0 V
D. 9.0 V
\nD
\nMost candidates at both levels gave option A as the correct response instead of D. This would indicate that they have misread the diagram thinking the voltmeter was across the 1.0Ω resistor not the parallel combination.
\nTwo stars are viewed with a telescope using a green filter. The images of the stars are just resolved. What is the change, if any, to the angular separation of the images of the stars and to the resolution of the images when the green filter is replaced by a violet filter?
\nA
\nAnother question with a low discrimination index and candidates choosing all 4 responses with B the most popular. Remembering that angular separation is dependent on the stars position in space relative to each other so unlikely to have been changed by a coloured filter would have helped to eliminate A and D.
\nThe input to a diode bridge rectification circuit is sinusoidal with a time period of 20 ms.
\nWhich graph shows the variation with time t of the output voltage Vout between X and Y?
\nA
\nThree identical capacitors are connected in series. The total capacitance of the arrangement is mF. The three capacitors are then connected in parallel. What is the capacitance of the parallel arrangement?
\nA. mF
B. 1 mF
C. 3 mF
D. 81 mF
\nB
\nA transformer with 600 turns in the primary coil is used to change an alternating root mean square (rms) potential difference of 240 Vrms to 12 Vrms.
When connected to the secondary coil, a lamp labelled “120 W, 12 V” lights normally. The current in the primary coil is 0.60 A when the lamp is lit.
\nWhat are the number of secondary turns and the efficiency of the transformer?
\nD
\nA circular coil of wire moves through a region of uniform magnetic field directed out of the page.
\nWhat is the direction of the induced conventional current in the coil for the marked positions?
\nC
\nAn electron is fixed in position in a uniform electric field. What is the position for which the electrical potential energy of the electron is greatest?
\nD
\nA proton of velocity v enters a region of electric and magnetic fields. The proton is not deflected. An electron and an alpha particle enter the same region with velocity v. Which is correct about the paths of the electron and the alpha particle?
\nD
\nA particle of mass 0.02 kg moves in a horizontal circle of diameter 1 m with an angular velocity of 3 rad s-1.
What is the magnitude and direction of the force responsible for this motion?
\nD
\nA radioactive nuclide with atomic number Z undergoes a process of beta-plus (β+) decay. What is the atomic number for the nuclide produced and what is another particle emitted during the decay?
\nA
\nThe meson contains an up () quark. What is the quark structure of the meson?
A.
B.
C.
D.
\nC
\nThree conservation laws in nuclear reactions are
I. conservation of charge
\nII. conservation of baryon number
\nIII. conservation of lepton number.
\nThe reaction
\n\n
is proposed.
\nWhich conservation laws are violated in the proposed reaction?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nC
\nA neutron collides head-on with a stationary atom in the moderator of a nuclear power station. The kinetic energy of the neutron changes as a result. There is also a change in the probability that this neutron can cause nuclear fission.
What are these changes?
\nB
\nThe orbital radius of the Earth around the Sun is 1.5 times that of Venus. What is the intensity of solar radiation at the orbital radius of Venus?
A. 0.6 kW m-2
B. 0.9 kW m-2
C. 2 kW m-2
D. 3 kW m-2
\nD
\nThis had a low discrimination index at both SL and HL and although the correct answer was the most popular, all options gained high support. Candidates should be reminded that they have a data booklet and become familiar with its contents before the exam.
\nA coil is rotated in a uniform magnetic field. An alternating emf is induced in the coil. What is a possible phase relationship between the magnetic flux through the coil and the induced emf in the coil when the variations of both quantities are plotted with time?
\nB
\nPhotons of a certain frequency incident on a metal surface cause the emission of electrons from the surface. The intensity of the light is constant and the frequency of photons is increased. What is the effect, if any, on the number of emitted electrons and the energy of emitted electrons?
\nB
\nA low discrimination index with the majority of candidates choosing option D when B is correct. Students tend to link the intensity of light to the number of photons but forget that it is the energy (per unit time per unit area) of the light so if the photon energy increases (frequency increases) then the number of photons must decrease.
\nA capacitor of capacitance 1.0 μF stores a charge of 15 μC. The capacitor is discharged through a 25 Ω resistor. What is the maximum current in the resistor?
\nA. 0.60 mA
B. 1.7 mA
C. 0.60 A
D. 1.7 A
\nC
\nA diode bridge rectification circuit is constructed as shown. An alternating potential difference is applied between M and N.
\nThree statements about circuits are
\nI. when diode P conducts, Q does not conduct
II. when diode S conducts, neither P nor R conducts
III. the direction of conventional current in the resistor is from left to right.
Which statements are correct for this circuit?
\n\n
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nA
\nAn electron of low energy is enclosed within a high potential barrier. What is the process by which the electron can escape?
\nA. Quantum tunneling
B. Energy–mass conversion
C. Diffraction
D. Barrier climbing
\nA
\nA beam of monochromatic radiation is made up of photons each of momentum . The intensity of the beam is doubled without changing frequency. What is the momentum of each photon after the change?
\nA.
\nB.
\nC.
\nD.
\nB
\nThree observations of the behaviour of electrons are
\nI. electron emission as a result of the photoelectric effect
II. electron diffraction as an electron interacts with an atom
III. emission of radio waves as a result of electrons oscillating in a conductor.
Which observations are evidence that the electron behaves as a particle?
\n\n
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nB
\nA pure sample of a radioactive nuclide contains N0 atoms at time t = 0. At time t, there are N atoms of the nuclide remaining in the sample. The half-life of the nuclide is .
\nWhat is the decay rate of this sample proportional to?
\n\n
A. N
\nB. N0 – N
\nC. t
\nD.
\nA
\nThe graph shows the variation with time t of the horizontal force F exerted on a tennis ball by a racket.
\nThe tennis ball was stationary at the instant when it was hit. The mass of the tennis ball is 5.8 × 10–2 kg. The area under the curve is 0.84 N s.
\nCalculate the speed of the ball as it leaves the racket.
\nShow that the average force exerted on the ball by the racket is about 50 N.
\nDetermine, with reference to the work done by the average force, the horizontal distance travelled by the ball while it was in contact with the racket.
\nDraw a graph to show the variation with t of the horizontal speed v of the ball while it was in contact with the racket. Numbers are not required on the axes.
\nlinks 0.84 to Δp ✔
\n«» 14.5 «m s–1»✔
NOTE: Award [2] for bald correct answer
\nuse of Δt = «(28 – 12) × 10–3 =» 16 × 10–3 «s» ✔
\n=«» OR 53 «N» ✔
\nNOTE: Accept a time interval from 14 to 16 ms
Allow ECF from incorrect time interval
Ek = × 5.8 × 10–2 × 14.52 ✔
\nEk = W ✔
\ns = «» 0.12 « m » ✔
\n\n
Allow ECF from (a) and (b)
\nAllow ECF from MP1
\nAward [2] max for a calculation without reference to work done, eg: average velocity × time
\ngraph must show increasing speed from an initial of zero all the time ✔
overall correct curvature ✔
The air in a kitchen has pressure 1.0 × 105 Pa and temperature 22°C. A refrigerator of internal volume 0.36 m3 is installed in the kitchen.
\nThe refrigerator door is closed. The air in the refrigerator is cooled to 5.0°C and the number of air molecules in the refrigerator stays the same.
\nWith the door open the air in the refrigerator is initially at the same temperature and pressure as the air in the kitchen. Calculate the number of molecules of air in the refrigerator.
\nDetermine the pressure of the air inside the refrigerator.
\nThe door of the refrigerator has an area of 0.72 m2. Show that the minimum force needed to open the refrigerator door is about 4 kN.
\nComment on the magnitude of the force in (b)(ii).
\nOR ✔
\n✔
\nNOTE: Allow [1 max] for substitution with T in Celsius.
Allow [1 max] for a final answer of n = 14.7 or 15
Award [2] for bald correct answer.
use of = constant OR OR ✔
« Pa »✔
NOTE: Allow ECF from (a)
Award [2] for bald correct answer
✔
\nOR 4.3 × 103 « N »✔
\nNOTE: Allow ECF from (b)(i)
Allow ECF from MP1
force is «very» large ✔
\nthere must be a mechanism that makes this force smaller
OR
assumption used to calculate the force/pressure is unrealistic ✔
The solid line in the graph shows the variation with distance of the displacement of a travelling wave at t = 0. The dotted line shows the wave 0.20 ms later. The period of the wave is longer than 0.20 ms.
\nOne end of a string is attached to an oscillator and the other is fixed to a wall. When the frequency of the oscillator is 360 Hz the standing wave shown is formed on the string.
\nPoint X (not shown) is a point on the string at a distance of 10 cm from the oscillator.
\nCalculate, in m s–1, the speed for this wave.
\nCalculate, in Hz, the frequency for this wave.
\nThe graph also shows the displacement of two particles, P and Q, in the medium at t = 0. State and explain which particle has the larger magnitude of acceleration at t = 0.
\nState the number of all other points on the string that have the same amplitude and phase as X.
\nThe frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string.
\nv = «» 250 «m s–1»✔
\nλ = 0.30 «m» ✔
= «» 830 «Hz» ✔
NOTE: Allow ECF from (a)(i)
Allow ECF from wrong wavelength for MP2
Q ✔
acceleration is proportional to displacement «and Q has larger displacement» ✔
3 «points» ✔
\nfirst harmonic mode drawn ✔
\nNOTE: Allow if only one curve drawn, either solid or dashed.
\nA proton is moving in a region of uniform magnetic field. The magnetic field is directed into the plane of the paper. The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton.
\nThe speed of the proton is 2.0 × 106 m s–1 and the magnetic field strength B is 0.35 T.
\nExplain why the path of the proton is a circle.
\nShow that the radius of the path is about 6 cm.
\nCalculate the time for one complete revolution.
\nExplain why the kinetic energy of the proton is constant.
\nmagnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔
force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔
force is centripetal/towards the centre ✔
\nNOTE: Accept reference to acceleration instead of force
\n✔
\nOR 0.060 « m »
\nNOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures
\n✔
\n« s » ✔
\nNOTE: Award [2] for bald correct answer
\nALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔
NOTE: Award [2] for a reference to work done is zero hence Ek remains constant
\n
ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔
NOTE: Accept mention of speed or velocity indistinctly in MP2
\n
An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.
\n\n
Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 108 N C–1.
\nCalculate the magnitude of the initial acceleration of the electron.
\nDescribe the subsequent motion of the electron.
\n✔
\nOR ✔
\nNOTE: Ignore any negative sign.
\nOR ✔
\n✔
\nNOTE: Ignore any negative sign.
Award [1] for a calculation leading to
Award [2] for bald correct answer
\n
the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔
NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase
\nWind is incident on the blades of a wind turbine. The radius of the blades is 12 m. The following data are available for the air immediately before and after impact with the blades.
\nDetermine the maximum power that can be extracted from the wind by this turbine.
\nSuggest why the answer in (a) is a maximum.
\nOR ✔
\n «in incoming beam» = 1.4×105 «W»
OR
«in outgoing beam» = 1.9×104 «W» ✔
subtracts both P to obtain 1.2×105 «W» ✔
\nNOTE: Condone use of a wrong area or use of circumference in MP1.
Allow ECF from MP2.
Award [1] max for any attempt to use the formula for wind power which cubes the difference of velocities
Award [3] for a bald correct answer
\n
because some power is lost due to inefficiencies in the system/transfers to the surroundings ✔
\nNOTE: Accept power or energy indistinctly
\nA stationary nucleus of uranium-238 undergoes alpha decay to form thorium-234.
\nThe following data are available.
\nEnergy released in decay 4.27 MeV
Binding energy per nucleon for helium 7.07 MeV
Binding energy per nucleon for thorium 7.60 MeV
Radioactive decay is said to be “random” and “spontaneous”. Outline what is meant by each of these terms.
\nRandom:
\nSpontaneous:
\nCalculate the binding energy per nucleon for uranium-238.
\nCalculate the ratio .
\nrandom:
it cannot be predicted which nucleus will decay
OR
it cannot be predicted when a nucleus will decay ✔
NOTE: OWTTE
\n
spontaneous:
the decay cannot be influenced/modified in any way ✔
NOTE: OWTTE
234 × 7.6 OR 4 × 7.07 ✔
\nBEU =« 234 × 7.6 + 4 × 7.07 – 4.27 =» « MeV » ✔
\n« MeV » ✔
\n NOTE: Allow ECF from MP2
Award [3] for bald correct answer
Allow conversion to J, final answer is 1.2 × 10–12
states or applies conservation of momentum ✔
\nratio is «» 58.5 ✔
NOTE: Award [2] for bald correct answer
\nIn a classical model of the singly-ionized helium atom, a single electron orbits the nucleus in a circular orbit of radius r.
\nThe Bohr model for hydrogen can be applied to the singly-ionized helium atom. In this model the radius , in m, of the orbit of the electron is given by where is a positive integer.
\nShow that the speed of the electron with mass , is given by .
\nHence, deduce that the total energy of the electron is given by .
\nIn this model the electron loses energy by emitting electromagnetic waves. Describe the predicted effect of this emission on the orbital radius of the electron.
\nShow that the de Broglie wavelength of the electron in the state is m.
The formula for the de Broglie wavelength of a particle is .
\nEstimate for , the ratio .
\nState your answer to one significant figure.
\nThe description of the electron is different in the Schrodinger theory than in the Bohr model. Compare and contrast the description of the electron according to the Bohr model and to the Schrodinger theory.
\nequating centripetal to electrical force to get result ✔
\nuses (a)(i) to state OR states ✔
\nadds « » to get the result ✔
\nthe total energy decreases
OR
by reference to ✔
the radius must also decrease ✔
\nNOTE: Award [0] for an answer concluding that radius increases
\nwith ✔
\nOR ✔
\n ✔
NOTE: Allow ECF from (b)(i)
\nreference to fixed orbits/specific radii OR quantized angular momentum in Bohr model ✔
electron described by a wavefunction/as a wave in Schrödinger model OR as particle in Bohr model ✔
reference to «same» energy levels in both models ✔
reference to «relationship between wavefunction and» probability «of finding an electron in a point» in Schrödinger model ✔
\nX has a capacitance of 18 μF. X is charged so that the one plate has a charge of 48 μC. X is then connected to an uncharged capacitor Y and a resistor via an open switch S.
\nThe capacitance of Y is 12 μF. S is now closed.
\nCalculate, in J, the energy stored in X with the switch S open.
\nCalculate the final charge on X and the final charge on Y.
\nCalculate the final total energy, in J, stored in X and Y.
\nSuggest why the answers to (a) and (b)(ii) are different.
\nOR ✔
\n✔
\nALTERNATIVE 1
✔
✔
\nsolving to get ✔
\n\n
ALTERNATIVE 2
\n✔
\n✔
\n✔
\n\n
NOTE: Award [3] for bald correct answer
\nALTERNATIVE 1
\n✔
\n✔
\n\n
ALTERNATIVE 2
\n✔
\n✔
\n\n
NOTE: Allow ECF from (b)(i)
Award [2] for bald correct answer
Award [1] max as ECF to a calculation using only one charge
charge moves/current flows «in the circuit» ✔
thermal losses «in the resistor and connecting wires» ✔
NOTE: Accept heat losses for MP2
\nThe lens of an optical system is coated with a thin film of magnesium fluoride of thickness d. Monochromatic light of wavelength 656 nm in air is incident on the lens. The angle of incidence is θ. Two reflected rays, X and Y, are shown.
\nThe following refractive indices are available.
\nAir = 1.00
Magnesium fluoride = 1.38
Lens = 1.58
The thickness of the magnesium fluoride film is d. For the case of normal incidence (θ = 0),
\nLight from a point source is incident on the pupil of the eye of an observer. The diameter of the pupil is 2.8 mm.
\nPredict whether reflected ray X undergoes a phase change.
\nstate, in terms of d, the path difference between the reflected rays X and Y.
\ncalculate the smallest value of d that will result in destructive interference between ray X and ray Y.
\ndiscuss a practical advantage of this arrangement.
\nDraw, on the axes, the variation with diffraction angle of the intensity of light incident on the retina of the observer.
\nEstimate, in rad, the smallest angular separation of two distinct point sources of light of wavelength 656 nm that can be resolved by the eye of this observer.
\nthere is a phase change ✔
of OR as it is reflected off a medium of higher refractive index ✔
2d ✔
NOTE: Accept 2dn
✔
\n✔
\n\n
NOTE: Award [2] for bald correct answer
\nreflection from «front surface of» lens eliminated/reduced
OR
energy reaching sensor increased ✔
at one wavelength ✔
NOTE: Accept reference to reduction of glare for MP1
\nstandard single slit diffraction pattern with correct overall shape ✔
secondary maxima of right size ✔
\n\n
NOTE: Secondary maximum not to exceed 1/5th of maximum intensity
Ignore width of maxima
\nuse of ✔
\n«rad» ✔
\n\n
NOTE: Award [2] for bald correct answer
\nMonochromatic light of very low intensity is incident on a metal surface. The light causes the emission of electrons almost instantaneously. Explain how this observation
\nIn an experiment to demonstrate the photoelectric effect, light of wavelength 480 nm is incident on a metal surface.
\nThe graph shows the variation of the current in the ammeter with the potential of the cathode.
\ndoes not support the wave nature of light.
\ndoes support the photon nature of light.
\nCalculate, in eV, the work function of the metal surface.
\nThe intensity of the light incident on the surface is reduced by half without changing the wavelength. Draw, on the graph, the variation of the current with potential after this change.
\n«low intensity light would» transfer energy to the electron at a low rate/slowly ✔
time would be required for the electron «to absorb the required energy» to escape/be emitted ✔
NOTE: OWTTE
\n«in the photon theory of light» the electron interacts with a single photon ✔
and absorbs all the energy OR and can leave the metal immediately ✔
NOTE: Reference to photon-electron collision scores MP1
\n✔
\n✔
\n✔
\nNOTE: Allow reading from the graph of leading to an answer of 1.2 «eV».
\nsimilar curve lower than original ✔
with same horizontal intercept ✔
\nA student investigates how the period T of a simple pendulum varies with the maximum speed v of the pendulum’s bob by releasing the pendulum from rest from different initial angles. A graph of the variation of T with v is plotted.
\nSuggest, by reference to the graph, why it is unlikely that the relationship between T and v is linear.
\nDetermine the fractional uncertainty in v when T = 2.115 s, correct to one significant figure.
\nThe student hypothesizes that the relationship between T and v is T = a + bv2, where a and b are constants. To verify this hypothesis a graph showing the variation of T with v2 is plotted. The graph shows the data and the line of best fit.
\nDetermine b, giving an appropriate unit for b.
\nThe lines of the minimum and maximum gradient are shown.
\nEstimate the absolute uncertainty in a.
\na straight line cannot be drawn through all error bars
OR
the graph/line of best fit is /curved/not straight/parabolic etc.
OR
graph has increasing/variable gradient ✔
NOTE: Do not allow “a line cannot be drawn through all error bars” without specifying “straight”.
\nAND ✔
\n«»0.04 ✔
\nNOTE: Accept 4 %
\nuse of 2 correct points on the line with Δv2 > 2 ✔
\nb in range 0.012 to 0.013 ✔
\ns3 m–2 ✔
\n«s» ±0.001 «s» AND «s» ±0.001 «s» ✔
\n«» 0.003 «s» ✔
\nThe resistance R of a wire of length L can be measured using the circuit shown.
\nIn one experiment the wire has a uniform diameter of d = 0.500 mm. The graph shows data obtained for the variation of R with L.
\nThe gradient of the line of best fit is 6.30 Ω m–1.
\nEstimate the resistivity of the material of the wire. Give your answer to an appropriate number of significant figures.
\nExplain, by reference to the power dissipated in the wire, the advantage of the fixed resistor connected in series with the wire for the measurement of R.
\nThe experiment is repeated using a wire made of the same material but of a larger diameter than the wire in part (a). On the axes in part (a), draw the graph for this second experiment.
\nevidence of use of ρ = given gradient × wire area
OR
substitution of values from a single data point with wire area ✔
✔
\nNOTE: Check POT is correct.
MP2 must be correct to exactly 3 s.f.
measurement should be performed at a constant temperature
OR
resistance of wire changes with temperature ✔
series resistance prevents the wire from overheating
OR
reduces power dissipated in the wire ✔
by reducing voltage across/current through the wire ✔
\nANY straight line going through the origin if extrapolated ✔
ANY straight line below existing line with smaller gradient ✔
A train is moving across a bridge with a speed v = 0.40c. Observer A is at rest in the train. Observer B is at rest with respect to the bridge.
\nThe length of the bridge LB according to observer B is 2.0 km.
\nAccording to observer B, two lamps at opposite ends of the bridge are turned on simultaneously as observer A crosses the bridge. Event X is the lamp at one end of the bridge turning on. Event Y is the lamp at the other end of the bridge turning on.
\nEvents X and Y are shown on the spacetime diagram. The space and time axes of the reference frame for observer B are and ct. The line labelled ct' is the worldline of observer A.
\nCalculate, for observer A, the length LA of the bridge
\nCalculate, for observer A, the time taken to cross the bridge.
\nOutline why LB is the proper length of the bridge.
\nDraw, on the spacetime diagram, the space axis for the reference frame of observer A. Label this axis '.
\n\n
Demonstrate using the diagram which lamp, according to observer A, was turned on first.
\nDemonstrate, using the diagram, which lamp observer A observes to light first.
\nDetermine the time, according to observer A, between X and Y.
\n✔
\n1.8 «km» ✔
\n\n
ALTERNATIVE 1
\ntime = ✔
\n1.5 × 10–5 «s» ✔
\n\n
ALTERNATIVE 2
\n«s» ✔
\n«s» ✔
\nLB is the length/measurement «by observer B» made in the reference frame in which the bridge is at rest ✔
NOTE: Idea of rest frame or frame in which bridge is not moving is required.
\nx′ axis drawn with correct gradient of 0.4 ✔
\n\n
NOTE: Line must be 1 square below Y, allow ±0.5 square.
Allow line drawn without a ruler.
\nlines parallel to the x′ axis through X and Y intersecting the worldline ct′ at points shown ✔
so Y/lamp at the end of the bridge turned on first ✔
\nNOTE: Allow lines drawn without a ruler
Do not allow MP2 without supporting argument or correct diagram.
light worldlines at 45° from X AND Y intersecting the worldline ct′ ✔
so light from lamp X is observed first ✔
\nNOTE: Allow lines drawn without a ruler.
Do not allow MP2 without supporting argument or correct diagram.
ALTERNATIVE 1
\n✔
\n= «–»2.9 × 10–6 «s» ✔
\n\n
ALTERNATIVE 2
\nequating spacetime intervals between X and Y
relies on realization that eg:
\n✔
\n«s» ✔
\n\n
ALTERNATIVE 3
use of diagram from answer to 4(c)(ii) (1 small square = 200 m)
\ncounts 4.5 to 5 small squares (allow 900 – 1000 m) between events for A seen on B’s ct axis ✔
\n«s» ✔
\nA flywheel is made of a solid disk with a mass M of 5.00 kg mounted on a small radial axle. The mass of the axle is negligible. The radius R of the disk is 6.00 cm and the radius r of the axle is 1.20 cm.
A string of negligible thickness is wound around the axle. The string is pulled by an electric motor that exerts a vertical tension force T on the flywheel. The diagram shows the forces acting on the flywheel. W is the weight and N is the normal reaction force from the support of the flywheel.
\nThe moment of inertia of the flywheel about the axis is .
\nThe flywheel is initially at rest. At time t = 0 the motor is switched on and a time-varying tension force acts on the flywheel. The torque exerted on the flywheel by the tension force in the string varies with t as shown on the graph.
\nAt t = 5.00 s the string becomes fully unwound and it disconnects from the flywheel. The flywheel remains spinning around the axle.
\nState the torque provided by the force W about the axis of the flywheel.
\nIdentify the physical quantity represented by the area under the graph.
\nShow that the angular velocity of the flywheel at t = 5.00 s is 200 rad s–1.
\nCalculate the maximum tension in the string.
\nThe flywheel is in translational equilibrium. Distinguish between translational equilibrium and rotational equilibrium.
\nAt t = 5.00 s the flywheel is spinning with angular velocity 200 rad s–1. The support bearings exert a constant frictional torque on the axle. The flywheel comes to rest after 8.00 × 103 revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.
\nzero ✔
\n«change in» angular momentum ✔
\nNOTE: Allow angular impulse.
\nuse of L = lω = area under graph = 1.80 «kg m2 s–1» ✔
rearranges «to give ω= area/I» 1.80 = 0.5 × 5.00 × 0.0602 × ω ✔
«to get ω = 200 rad s–1 »
\n✔
\ntranslational equilibrium is when the sum of all the forces on a body is zero ✔
rotational equilibrium is when the sum of all the torques on a body is zero ✔
\nALTERNATIVE 1
\n✔
\n✔
\ntorque = ✔
\n\n
ALTERNATIVE 2
\nchange in kinetic energy ✔
\nidentifies work done = change in KE ✔
\ntorque = ✔
\nA long straight current-carrying wire is at rest in a laboratory. A negatively-charged particle P outside the wire moves parallel to the current with constant velocity v relative to the laboratory.
\nIn the reference frame of the laboratory the particle P experiences a repulsive force away from the wire.
\nOne of the two postulates of special relativity states that the speed of light in a vacuum is the same for all observers in inertial reference frames. State the other postulate of special relativity.
\nState the nature of the force on the particle P in the reference frame of the laboratory.
\nDeduce, using your answer to part (a), the nature of the force that acts on the particle P in the rest frame of P.
\nExplain how the force in part (b)(ii) arises.
\nThe velocity of P is 0.30c relative to the laboratory. A second particle Q moves at a velocity of 0.80c relative to the laboratory.
\nCalculate the speed of Q relative to P.
\nlaws of physics are the same for all observers
OR
laws of physics are the same in all «inertial» frames ✔
NOTE: OWTTE
\nmagnetic ✔
\n«from 3a»
force must still be repulsive ✔
for P there is no magnetic force AND force is electric/electrostatic
OR
since P is at rest the force is electric/electrostatic ✔
protons and electrons in the wire move with different velocities «relative to P»
OR
speed of electrons is greater ✔
«for P» the density of protons and electrons in wire will be different «due to length contraction»
OR
«for P» the wire appears to have negative charge «due to length contraction» ✔
«hence electric force arises»
NOTE: Do not award mark for mention of length contraction without details.
\n✔
\n✔
\nNOTE: Accept 0.89c if all negative values used. Accept –0.89c even though speed is required.
\nAn ideal gas consisting of 0.300 mol undergoes a process ABCD. AB is an adiabatic expansion from the initial volume VA to the volume 1.5 VA. BC is an isothermal compression. The pressures at C and D are the same as at A.
\nThe following data are available.
\nPressure at A = 250 kPa
Volume at C = 3.50 × 10–3 m3
Volume at D = 2.00 × 10–3 m3
The gas at C is further compressed to D at a constant pressure. During this compression the temperature decreases by 150 K.
\nFor the compression CD,
\nShow that the pressure at B is about 130 kPa.
\nCalculate the ratio .
\ndetermine the thermal energy removed from the system.
\nexplain why the entropy of the gas decreases.
\nstate and explain whether the second law of thermodynamics is violated.
\n✔
\n= 127 kPa ✔
\n1.31 ✔
\nALTERNATIVE 1
\nwork done ✔
\nchange in internal energy
OR
✔
thermal energy removed
OR
✔
\n
ALTERNATIVE 2
\n✔
\nthermal energy removed ✔
\n✔
\nALTERNATIVE 1
\n«from b(i)» is negative ✔
\nAND is negative ✔
\n\n
ALTERNATIVE 2
\nT and/or V decreases ✔
\nless disorder/more order «so S decreases» ✔
\n\n
ALTERNATIVE 3
\nT decreases ✔
\n✔
\n\n
NOTE: Answer given, look for a valid reason that S decreases.
\nnot violated ✔
\nthe entropy of the surroundings must have increased
OR
the overall entropy of the system and the surroundings is the same or increased ✔
The diagram, drawn to scale, shows an object O placed in front of a converging mirror. The focal point of the mirror is labelled F.
\nA planar wavefront of white light, labelled A, is incident on a converging lens. Point P is on the surface of the lens and the principal axis. The blue component of the transmitted wavefront, labelled B, is passing through point P.
\nConstruct a ray diagram in order to locate the position of the image formed by the mirror. Label the image .
\nEstimate the linear magnification of the image.
\nDescribe two features of the image.
\nSketch, on the diagram, the wavefront of red light passing through point P. Label this wavefront R.
\nExplain chromatic aberration, with reference to your diagram in (b)(i).
\nAn achromatic doublet reduces the effect of chromatic aberration. Describe an achromatic doublet.
\ncorrectly draws any 2 of the 4 conventional rays from the object tip ✔
correctly extends reflections to form virtual upright image in approximate position shown ✔
NOTE: No ECF for incorrect rays in MP1.
Award [0] for rays of converging lens or diverging mirror.
1.5 ✔
\nNOTE: For “correct” image position in (a)(i) allow 1.3 to 1.7
\nAny two of:
virtual OR upright OR larger than the object ✔
“circular” wave front through P: symmetric about the principal axis AND of greater radius than B ✔
\nred and blue wave fronts have different curvature/radius
OR
red and blue waves are refracted differently/have different speeds ✔
so different colors have different foci/do not focus to one point
OR
so image is multi-coloured/blurred ✔
NOTE: MP1 is for the reason for the aberration, MP2 is for the effect.
\nmention combination of converging and diverging lenses ✔
of different refractive index/material ✔
NOTE: Achromatic doublet is in the question, so no marks for mentioning this.
\nA substance in the gas state has a density about times less than when it is in the liquid state. The diameter of a molecule is . What is the best estimate of the average distance between molecules in the gas state?
\nA.
\nB.
\nC.
\nD.
\nB
\nThis question gives good discrimination at HL but less so at SL. Teacher comments felt that the question was too mathematical but it can be noted that it asks for an estimation of the average distance which is related to the cube root of the volume and 1000 is 103. At both levels option D proved a popular alternative suggesting that candidates were forgetting the cube root.
\nA bicycle of mass comes to rest from speed using the back brake. The brake has a specific heat capacity of and a mass . Half of the kinetic energy is absorbed by the brake.
\nWhat is the change in temperature of the brake?
\nA.
\nB.
\nC.
\nD.
\nA
\nAn object moves with simple harmonic motion. The acceleration of the object is
\nA. constant.
\nB. always directed away from the centre of the oscillation.
\nC. a maximum at the centre of the oscillation.
\nD. a maximum at the extremes of the oscillation.
\nD
\nA travelling wave has a frequency of . The closest distance between two points on the wave that have a phase difference of is . What is the speed of the wave?
\nA.
\nB.
\nC.
\nD.
\nC
\nWhat changes occur to the frequency and wavelength of monochromatic light when it travels from glass to air?
\nB
\nThe air in a pipe, open at both ends, vibrates in the second harmonic mode.
\nWhat is the phase difference between the motion of a particle at P and the motion of a particle at Q?
\nA.
\nB.
\nC.
\nD.
\nC
\nFour resistors of each are connected as shown.
\nWhat is the effective resistance between P and Q?
\nA.
\nB.
\nC.
\nD.
\nB
\nThis has a very low discrimination index. It is suspected that students did not realise that PQ has 2 branches in parallel and many chose D, 4 ohm, the value of a single resistor.
\nMass is attached to one end of a string. The string is passed through a hollow tube and mass is attached to the other end. Friction between the tube and string is negligible.
\nMass travels at constant speed in a horizontal circle of radius . What is mass ?
\nA.
\nB.
\nC.
\nD.
\nD
\nPlanet X has a gravitational field strength of at its surface. Planet Y has the same density as X but three times the radius of X. What is the gravitational field strength at the surface of Y?
\nA.
\nB.
\nC.
\nD.
\nC
\nA metal wire has free charge carriers per unit volume. The charge on the carrier is . What additional quantity is needed to determine the current per unit area in the wire?
\nA. Cross-sectional area of the wire
\nB. Drift speed of charge carriers
\nC. Potential difference across the wire
\nD. Resistivity of the metal
\nB
\nWhat are the principal roles of a moderator and of a control rod in a thermal nuclear reactor?
\nC
\nA nuclear power station contains an alternating current generator. What energy transfer is performed by the generator?
\nA. Electrical to kinetic
\nB. Kinetic to electrical
\nC. Nuclear to kinetic
\nD. Nuclear to electrical
\nB
\nAn electric motor raises an object of weight through a vertical distance of in . The current in the electric motor is at a potential difference of . What is the efficiency of the electric motor?
\nA.
\nB.
\nC.
\nD.
\nC
\nThe average temperature of the surface of a planet is five times greater than the average temperature of the surface of its moon. The emissivities of the planet and the moon are the same. The average intensity radiated by the planet is . What is the average intensity radiated by its moon?
\nA.
\nB.
\nC.
\nD.
\nC
\nA current in a wire lies between the poles of a magnet. What is the direction of the electromagnetic force on the wire?
\n\n
A
\nMany chose option C, the opposite of the correct response. As always in electromagnetism questions, students need to consider carefully which hand and which fingers to use. We do tend to say that there is little that needs to be memorised in physics, this is probably one of them.
\nA car is driven from rest along a straight horizontal road. The car engine exerts a constant driving force. Friction and air resistance are negligible. How does the power developed by the engine change with the distance travelled?
\nA. Power does not change.
\nB. Power decreases linearly.
\nC. Power increases linearly.
\nD. Power increases non-linearly.
\nD
\nLowish discrimination with C the most popular choice. It was felt that candidates normally analyse in terms of the time taken whereas this question refers to the distance travelled so with a constant driving force the velocity increases linearly with time but non linearly with distance.
\nWhich graph shows the variation of activity with time for a radioactive nuclide?
\nD
\nWhat is not an assumption of the kinetic model of an ideal gas?
\nA. Attractive forces between molecules are negligible.
\nB. Collision duration is negligible compared with time between collisions.
\nC. Molecules suffer negligible momentum change during wall collisions.
\nD. Molecular volume is negligible compared with gas volume.
\nC
\nEven though both difficulty and discrimination index are acceptable a significant number of candidates chose incorrect options B or D. The examiners appreciated that this was a challenging question which required some thought partly because it asked what is not an assumption. Candidates need to be aware that although questions are normally phased in a positive sense there will occasionally be ones like this and they need to hold the idea of 'not' when looking at the possible answers. A useful strategy is to look for correct assumptions and when those are identified there should be just one left — the required answer.
\nWavefronts travel from air to medium Q as shown.
\nWhat is the refractive index of Q?
\nA.
\nB.
\nC.
\nD.
\nB
\nThis has a negative discrimination index and the majority of candidates chose option C which would be correct if we were considering rays but the question asks about wavefronts. It must be stressed that it is important to read the question carefully and not skim over the introductory stem. In this type of question students are advised to draw the rays on the diagram perpendicular to the wavefronts to make it easier to work out which angles to use.
\nTwo containers X and Y are maintained at the same temperature. X has volume and Y has volume . They both hold an ideal gas. The pressure in X is and the pressure in Y is . The containers are then joined by a tube of negligible volume. What is the final pressure in the containers?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhat statement about alpha particles, beta particles and gamma radiation is true?
\nA. Gamma radiation always travels faster than beta particles in a vacuum.
\nB. In air, beta particles produce more ions per unit length travelled than alpha particles.
\nC. Alpha particles are always emitted when beta particles are emitted.
\nD. Alpha particles are deflected in the same direction as beta particles in a magnetic field.
\nA
\nA cell of electromotive force (emf) and zero internal resistance is in the circuit shown.
\nWhat is correct for loop WXYUW?
\nA.
\nB.
\nC.
\nD.
\nD
\nThere are a high number of blanks responses indicating that some candidates decided to leave it until the end. It perhaps looked complicated but pleasingly over half did get the correct answer with choices reasonably evenly balanced between the three wrong answers suggesting that these were guesses.
\nFour of the energy states for an atom are shown. Transition between any two states is possible.
\nWhat is the shortest wavelength of radiation that can be emitted from these four states?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhat is the relationship between the resistivity of a uniform wire, the radius of the wire and the length of the wire when its resistance is constant?
\nA.
\nB.
\nC.
\nD.
\nD
\nIt was pointed out on the G2s that the proportional symbol is incorrect. The high number of correct responses indicates that it did not disadvantage the students and will be corrected for publication.
\nThe Feynman diagram shows some of the changes in a proton–proton collision.
\nWhat is the equation for this collision?
A.
\nB.
\nC.
\nD.
\nA
\nThere were some teacher comments that this was not a complete Feynman diagram however the stem does say that the diagram shows some of the changes and is intended to make the question easier by not complicating with particles that do not change. Students should be made aware that they can expect to see diagrams like this in the future as partial diagrams do tend to make the situation simpler for students to solve.
\nA power station generates of power at a potential difference of . The energy is transmitted through cables of total resistance .
\nWhat is the power loss in the cables?
\nA.
\nB.
\nC.
\nD.
\nB
\nAn electrical power supply has an internal resistance. It supplies a direct current to an external circuit for a time . What is the electromotive force (emf) of the power supply?
\nA.
\nB.
\nC.
\nD.
\nA
\nThree historical developments in physics were
\nI. wave–particle duality
II. the kinetic model
III. the equivalence of mass and energy.
Which of these represented a paradigm shift in scientific thinking?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nA body is held in translational equilibrium by three coplanar forces of magnitude , and . Three statements about these forces are
\nI. all forces are perpendicular to each other
II. the forces cannot act in the same direction
III. the vector sum of the forces is equal to zero.
Which statements are true?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nThe mass of nuclear fuel in a nuclear reactor decreases at the rate of every hour. The overall reaction process has an efficiency of . What is the maximum power output of the reactor?
\nA.
\nB.
\nC.
\nD.
\nA
\nThe discrimination index was below the desired 0.2 with a high number of blank responses and many candidates choosing each of the options. This is a question requiring consideration of units using 10-6 for mg to kg and also remembering to allow for efficiency.
\nPower is dissipated in a resistor of resistance when there is a direct current in the resistor.
\nWhat is the average power dissipation in a resistance when the alternating root-mean-square (rms) current in the resistor is ?
\nA.
\nB.
\nC.
\nD.
\nB
\nA simple pendulum and a mass–spring system oscillate with the same time period. The mass of the pendulum bob and the mass on the spring are initially identical. The masses are halved.
\nWhat is when the masses have been changed?
\nA.
\nB.
\nC.
\nD.
\nC
\nLight is incident on a diffraction grating. The wavelengths of two spectral lines of the light differ by and have a mean wavelength of . The spectral lines are just resolved in the fourth order of the grating. What is the minimum number of grating lines that were illuminated?
\nA.
\nB.
\nC.
\nD.
\nB
\nWhite light is incident normally on separate diffraction gratings X and Y. Y has a greater number of lines per metre than X. Three statements about differences between X and Y are
\nI. adjacent slits in the gratings are further apart for X than for Y
II. the angle between red and blue light in a spectral order is greater in X than in Y
III. the total number of visible orders is greater for X than for Y.
Which statements are correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nMany candidates chose incorrect options. They need to be aware that there will usually be questions of this style and they need to practice them. The pattern of the answers is always the same and the best strategy is to try to identify a wrong answer which will then help to eliminate incorrect combinations.
\nTwo satellites W and X have the same mass. They have circular orbits around the same planet. W is closer to the surface than X. What quantity is smaller for W than for X?
\nA. Gravitational force from the planet
\nB. Angular velocity
\nC. Orbital speed
\nD. Orbital period
\nD
\nA rectangular coil rotates at a constant angular velocity. At the instant shown, the plane of the coil is at right angles to the line . A uniform magnetic field acts in the direction .
\nWhat rotation of the coil about a specified axis will produce the graph of electromotive force (emf) against time ?
\nA. Through about
\nB. Through about
\nC. Through about
\nD. Through about
\nC
\nA capacitor of capacitance has initial charge . The capacitor is discharged through a resistor of resistance . The potential difference across the capacitor varies with time.
\nWhat is true for this capacitor?
\nA. After time the potential difference across the capacitor is halved.
\nB. The capacitor discharges more quickly when the resistance is changed to .
\nC. The rate of change of charge on the capacitor is proportional to .
\nD. The time for the capacitor to lose half its charge is .
\nC
\nP and S are two points on a gravitational equipotential surface around a planet. Q and R are two points on a different gravitational equipotential surface at a greater distance from the planet.
\nThe greatest work done by the gravitational force is when moving a mass from
\nA. P to S.
\nB. Q to R.
\nC. R to P.
\nD. S to R.
\nC
\nMonochromatic light is incident on a metal surface and electrons are released. The intensity of the incident light is increased. What changes, if any, occur to the rate of emission of electrons and to the kinetic energy of the emitted electrons?
\nD
\nA device sends an impulse of electrical energy to maintain a regular heartbeat in a person. The device is powered by an alternating current (ac) supply connected to a step-up transformer that charges a capacitor of capacitance 30 μF.
\n
The voltage across the primary coil of the transformer is 220 V. The number of turns on the secondary coil is 15 times greater than the number of turns on the primary coil.
\nThe switch is moved to position B.
\nExplain the role of the diode in the circuit when the switch is at position A.
\nShow that the maximum energy stored by the capacitor is about 160 J.
\nCalculate the maximum charge Q0 stored in the capacitor.
\nIdentify, using the label + on the diagram, the polarity of the capacitor.
\nDescribe what happens to the energy stored in the capacitor when the switch is moved to position B.
\nShow that the charge remaining in the capacitor after a time equal to one time constant of the circuit will be 0.37 Q0.
\nThe graph shows the variation with time of the charge in the capacitor as it is being discharged through the heart.
\nDetermine the electrical resistance of the closed circuit with the switch in position B.
\nIn practice, two electrodes connect the heart to the circuit. These electrodes introduce an additional capacitance.
\nExplain the effect of the electrode capacitance on the discharge time.
\nto charge a capacitor current must be direct ✓
\ndiode will only allow current to flow in one direction
\nOR
\nthe diode provides half wave rectification ✓
\n\n
✓
\n\nOR
\n163 «J» ✓
\nAllow use of 220 V as an RMS value to calculate Vs = 467 V and E = 327 J for full marks if appropriate work is provided.
Answer must be to 3 or more sf or working shown for MP2
✓
\n\n
Allow ECF from (b)(i) (Q = 30 μF x V)
\nlabels + on the lower side of the capacitor ✓
\n\n
the energy stored in the capacitor is delivered to the resistor/heart ✓
\n\n
✓
\n\n
ALTERNATIVE 1
\nreads from the graph ✓
\nso ✓
\nALTERNATIVE 2
\nreads a correct value from the graph for and ✓
\nso ✓
\n«the capacitors are in parallel hence» capacitances are added / more charge is stored
OR
Ceq is larger
OR
electrode capacitor charges and discharges ✓
«therefore» discharge takes longer/increases ✓
\nThe graph shows the variation of electric field strength with distance from a point charge.
\nThe shaded area X is the area under the graph between two separations and from the charge.
\nWhat is X?
\nA. The electric field average between and
\nB. The electric potential difference between and
\nC. The work done in moving a charge from to
\nD. The work done in moving a charge from to
\nB
\nThe diameter of a nucleus of a particular nuclide X is . What is the nucleon number of X?
\nA.
\nB.
\nC.
\nD.
\nC
\nWhy are high voltages and low currents used when electricity is transmitted over long distances?
\nA. Cables can be closer to the ground.
\nB. Electrons have a greater drift speed.
\nC. Energy losses are reduced.
\nD. Resistance of the power lines is reduced.
\nC
\nA photon has a wavelength . What are the energy and momentum of the photon?
\nA
\nThe Rutherford-Geiger-Marsden experiment shows that
\nA. alpha particles do not obey Coulomb’s law.
\nB. there is a fixed nuclear radius for each nucleus.
\nC. a large proportion of alpha particles are undeflected.
\nD. the Bohr model of the hydrogen atom is confirmed.
\nC
\nThis has a low discrimination index but it was felt that perhaps as it was the last question students were guessing the answer especially those choosing option A.
\nA painting is protected behind a transparent glass sheet of refractive index nglass. A coating of thickness w is added to the glass sheet to reduce reflection. The refractive index of the coating ncoating is such that nglass > ncoating > 1.
\nThe diagram illustrates rays normally incident on the coating. Incident angles on the diagram are drawn away from the normal for clarity.
\nState the phase change when a ray is reflected at B.
\nExplain the condition for w that eliminates reflection for a particular light wavelength in air .
\nState the Rayleigh criterion for resolution.
\nThe painting contains a pattern of red dots with a spacing of 3 mm. Assume the wavelength of red light is 700 nm. The average diameter of the pupil of a human eye is 4 mm. Calculate the maximum possible distance at which these red dots are distinguished.
\n✓
\n\n
«to eliminate reflection» destructive interference is required ✓
\nphase change is the same at both boundaries / no relative phase change due to reflections ✓
\ntherefore
\nOR
\n\nOR
\n ✓
central maximum of one diffraction pattern lies over the central/first minimum of the other diffraction pattern ✓
\n✓
\n✓
\nA spaceship is travelling at , away from Earth. It launches a probe away from Earth, at relative to the spaceship. An observer on the probe measures the length of the probe to be .
\nThe Lorentz transformations assume that the speed of light is constant. Outline what the Galilean transformations assume.
\nDeduce the length of the probe as measured by an observer in the spaceship.
\nExplain which of the lengths is the proper length.
\nCalculate the speed of the probe in terms of , relative to Earth.
\nconstancy of time
OR
speed of light > c is possible ✓
\n
OWTTE
\n ✓
length = ✓
\n\n
Allow length in the range to .
\nAllow ECF from wrong
\nAward [2] marks for a bald correct answer in the range indicated above.
\n / measurement made on the probe ✓
the measurement made by an observer at rest in the frame of the probe ✓
\n ✓
✓
\n\n
Allow all negative signs for velocities
Award [2] marks for a bald correct answer
\nAlthough the expected answers was the constancy of time, the markscheme allowed references to the speed of light not being constant, as this was a common answer, deriving from the stem used in the question.
\nVery well answered.
\n\"In the same frame\" does not highlight the need to be \"at rest\" in that frame, and was the most frequent wrong answer, although a vast majority scored full marks here.
\nVery well answered.
\nThe graph shows the variation of speed v of an object with time t.
\nWhich graph shows how the distance s travelled by the object varies with t?
\nB
\nA company delivers packages to customers using a small unmanned aircraft. Rotating horizontal blades exert a force on the surrounding air. The air above the aircraft is initially stationary.
\nThe air is propelled vertically downwards with speed . The aircraft hovers motionless above the ground. A package is suspended from the aircraft on a string. The mass of the aircraft is and the combined mass of the package and string is . The mass of air pushed downwards by the blades in one second is .
\nState the value of the resultant force on the aircraft when hovering.
\nOutline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.
\nDetermine . State your answer to an appropriate number of significant figures.
\nThe package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.
\nzero ✓
\nBlades exert a downward force on the air ✓
\n
air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓
Downward direction required for MP1.
«lift force/change of momentum in one second» ✓
\n✓
\nAND answer expressed to sf only ✓
\nAllow from .
\nvertical force = lift force – weight OR OR ✓
\nacceleration✓
\nTwo boxes in contact are pushed along a floor with a force F. The boxes move at a constant speed. Box X has a mass m and box Y has a mass 2m.
\nWhat is the resultant force acting on Y?
A. 0
B.
C. F
D. 2F
A
\nAn elevator (lift) and its load have a total mass of 750 kg and accelerate vertically downwards at 2.0 m s–2.
\nWhat is the tension in the elevator cable?
\n
A. 1.5 kN
B. 6.0 kN
C. 7.5 kN
D. 9.0 kN
B
\nThe Rotor is an amusement park ride that can be modelled as a vertical cylinder of inner radius rotating about its axis. When the cylinder rotates sufficiently fast, the floor drops out and the passengers stay motionless against the inner surface of the cylinder. The diagram shows a person taking the Rotor ride. The floor of the Rotor has been lowered away from the person.
\nDraw and label the free-body diagram for the person.
\n\n
The person must not slide down the wall. Show that the minimum angular velocity of the cylinder for this situation is
\n\nwhere is the coefficient of static friction between the person and the cylinder.
\nThe coefficient of static friction between the person and the cylinder is . The radius of the cylinder is . The cylinder makes revolutions per minute. Deduce whether the person will slide down the inner surface of the cylinder.
\narrow downwards labelled weight/W/ and arrow upwards labelled friction/ ✓
\narrow horizontally to the left labelled «normal» reaction/ ✓
\n
Ignore point of application of the forces but do not allow arrows that do not touch the object.
Do not allow horizontal force to be labelled ‘centripetal’ or .
\nSee AND ✓
\n«substituting for N» ✓
\nALTERNATIVE 1
\nminimum required angular velocity ✓
\nactual angular velocity ✓
\nactual angular velocity is greater than the minimum, so the person does not slide ✓
\n\n
ALTERNATIVE 2
\nMinimum friction force ✓
\nActual friction force ✓
\nActual friction force is greater than the minimum frictional force so the person does not slide ✓
\n\n
Allow from .
\nA graph shows the variation of force acting on an object moving in a straight line with distance moved by the object. Which area represents the work done on the object during its motion from P to Q?
\nA. X
B. Y
C. Y + Z
D. X + Y + Z
C
\nRadioactive uranium-238 produces a series of decays ending with a stable nuclide of lead. The nuclides in the series decay by either alpha (α) or beta-minus (β−) processes.
\nThe graph shows the variation with the nucleon number A of the binding energy per nucleon.
\nUranium-238 decays into a nuclide of thorium-234 (Th).
\n
Write down the complete equation for this radioactive decay.
Thallium-206 decays into lead-206 .
\nIdentify the quark changes for this decay.
\n\n
Outline why high temperatures are required for fusion to occur
\n\n
Outline, with reference to the graph, why energy is released both in fusion and in fission.
\n\n
Uranium-235 () is used as a nuclear fuel. The fission of uranium-235 can produce krypton-89 and barium-144.
\nDetermine, in MeV and using the graph, the energy released by this fission.
\n\n
✓
\nAllow He for alpha.
\nudd→uud
OR
down quark changes to up quark ✓
links temperature to kinetic energy/speed of particles ✓
\nenergy required to overcome «Coulomb» electrostatic repulsion ✓
\n«energy is released when» binding energy per nucleon increases
\nany use of (value from graph) x (number of nucleons) ✓
\n✓
\nOn a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger.
\n
The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with a frequency of 195 Hz. The sounding length of the string is 62 cm.
\nOutline how a standing wave is produced on the string.
\nShow that the speed of the wave on the string is about 240 m s−1.
\nSketch a graph to show how the acceleration of point P varies with its displacement from the rest position.
\n
Calculate, in m s−1, the maximum velocity of vibration of point P when it is vibrating with a frequency of 195 Hz.
\nCalculate, in terms of g, the maximum acceleration of P.
\nEstimate the displacement needed to double the energy of the string.
\nThe string is made to vibrate in its third harmonic. State the distance between consecutive nodes.
\n«travelling» wave moves along the length of the string and reflects «at fixed end» ✓
\nsuperposition/interference of incident and reflected waves ✓
\nthe superposition of the reflections is reinforced only for certain wavelengths ✓
\n✓
\n✓
\nAnswer must be to 3 or more sf or working shown for MP2.
\nstraight line through origin with negative gradient ✓
\nmax velocity occurs at x = 0 ✓
\n✓
\n✓
\n✓
\nuse of ✓
\n✓
\n✓
\nRadioactive uranium-238 produces a series of decays ending with a stable nuclide of lead. The nuclides in the series decay by either alpha (α) or beta-minus (β−) processes.
\nThe graph shows the variation with the nucleon number A of the binding energy per nucleon.
\nUranium-238 decays into a nuclide of thorium-234 (Th).
\n
Write down the complete equation for this radioactive decay.
Thallium-206 decays into lead-206 .
\nIdentify the quark changes for this decay.
\nThe half-life of uranium-238 is about 4.5 × 109 years. The half-life of thallium-206 is about 4.2 minutes.
\nCompare and contrast the methods to measure these half-lives.
\nOutline why high temperatures are required for fusion to occur.
\n\n
Outline, with reference to the graph, why energy is released both in fusion and in fission.
\n\n
Uranium-235 is used as a nuclear fuel. The fission of uranium-235 can produce krypton-89 and barium-144.
\nDetermine, in MeV and using the graph, the energy released by this fission.
\n✓
\nAllow He for alpha.
\nudd→uud
OR
down quark changes to up quark ✓
measure «radio»activity/«radioactive» decay/A for either
OR
take measurements with a Geiger counter. ✓
for Uranium measure number/N of radioactive atoms/OWTTE ✓
\nfor Thalium measure «rate of» change in activity over time. ✓
\ncorrect connection for either Uranium or Thalium to determine half life ✓
\nlinks temperature to kinetic energy/speed of particles ✓
\nenergy required to overcome «Coulomb» electrostatic repulsion ✓
\n«energy is released when» binding energy per nucleon increases
\nany use of (value from graph) x (number of nucleons) ✓
\n«235 × 7.6 – (89 × 8.6 + 144 × 8.2) =» 160 «MeV» ✓
\nA car travelling at a constant velocity covers a distance of 100 m in 5.0 s. The thrust of the engine is 1.5 kN. What is the power of the car?
\nA. 0.75 kW
B. 3.0 kW
C. 7.5 kW
D. 30 kW
D
\nAn inelastic collision occurs between two bodies in the absence of external forces.
\nWhat must be true about the total momentum of the two bodies and the total kinetic energy of the two bodies during this interaction?
\nA. Only momentum is conserved.
B. Only kinetic energy is conserved.
C. Both momentum and kinetic energy are conserved.
D. Neither momentum nor kinetic energy are conserved.
A
\nConservation of energy and conservation of momentum are two examples of conservation laws.
\nOutline the significance of conservation laws for physics.
\nWhen a pi meson π- (du̅) and a proton (uud) collide, a possible outcome is a sigma baryon Σ0 (uds) and a kaon meson Κ0 (ds̅).
\n
Apply three conservation laws to show that this interaction is possible.
they express fundamental principles of nature ✓
\nallow to model situations ✓
\nallow to calculate unknown variables ✓
\nallow to predict possible outcomes ✓
\nallow to predict missing quantities/particles ✓
\nallow comparison of different system states ✓
\nthree correct conservation laws listed ✓
\nat least one conservation law correctly demonstrated ✓
\nall three conservation laws correctly demonstrated ✓
\nA liquid is initially at its freezing point. Energy is removed at a uniform rate from the liquid until it freezes completely.
Which graph shows how the temperature T of the liquid varies with the energy Q removed from the liquid?
A
\nA thin-walled cylinder of weight W, open at both ends, rests on a flat surface. The cylinder has a height L, an average radius R and a thickness x where R is much greater than x.
\nWhat is the pressure exerted by the cylinder walls on the flat surface?
\nA.
\nB.
\nC.
\nD.
\nA
\nIn an electric circuit used to investigate the photoelectric effect, the voltage is varied until the reading in the ammeter is zero. The stopping voltage that produces this reading is 1.40 V.
\nDescribe the photoelectric effect.
\nShow that the maximum velocity of the photoelectrons is .
\nThe photoelectrons are emitted from a sodium surface. Sodium has a work function of 2.3 eV.
\nCalculate the wavelength of the radiation incident on the sodium. State an appropriate unit for your answer.
\nelectrons are ejected from the surface of a metal ✓
\nafter gaining energy from photons/electromagnetic radiation ✓
\nthere is a minimum «threshold» energy/frequency
OR
maximum «threshold» wavelength ✓
«» and manipulation to get ✓
\nOR ✓
\nMust see either complete substitution or calculation to at least 3 s.f. for MP2
\n✓
\n✓
\n✓
\n
Must see an appropriate unit to award MP3.
A fixed mass of an ideal gas in a closed container with a movable piston initially occupies a volume V. The position of the piston is changed, so that the mean kinetic energy of the particles in the gas is doubled and the pressure remains constant.
\nWhat is the new volume of the gas?
\nA.
\nB.
\nC. 2V
\nD. 4V
\nC
\nA vertical tube, open at both ends, is completely immersed in a container of water. A loudspeaker above the container connected to a signal generator emits sound. As the tube is raised the loudness of the sound heard reaches a maximum because a standing wave has formed in the tube.
\nDescribe two ways in which standing waves differ from travelling waves.
\nOutline how a standing wave forms in the tube.
\nThe tube is raised until the loudness of the sound reaches a maximum for a second time.
\nDraw, on the following diagram, the position of the nodes in the tube when the second maximum is heard.
\nBetween the first and second positions of maximum loudness, the tube is raised through 0.37 m. The speed of sound in the air in the tube is 320 m s−1. Determine the frequency of the sound emitted by the loudspeaker.
\n\n
energy is not propagated by standing waves ✓
\namplitude constant for travelling waves OR amplitude varies with position for standing waves OR standing waves have nodes/antinodes ✓
\nphase varies with position for travelling waves OR phase constant inter-node for standing waves ✓
\ntravelling waves can have any wavelength OR standing waves have discrete wavelengths ✓
\n
OWTTE
«sound» wave «travels down tube and» is reflected ✓
\nincident and reflected wave superpose/combine/interfere ✓
\n
OWTTE
Do not award MP1 if the reflection is quoted at the walls/container
\nnodes shown at water surface AND way up tube (by eye) ✓
\n
Accept drawing of displacement diagram for correct harmonic without nodes specifically identified.
Award [0] if waveform is shown below the water surface
\n✓
\n✓
\n
Allow ECF from MP1
The graph shows the variation with temperature T of the pressure P of a fixed mass of helium gas trapped in a container with a fixed volume of 1.0 × 10−3 m3.
\nDeduce whether helium behaves as an ideal gas over the temperature range 250 K to 500 K.
\nHelium has a molar mass of 4.0 g. Calculate the mass of gas in the container.
\nA second container, of the same volume as the original container, contains twice as many helium atoms. The graph of the variation of P with T is determined for the gas in the second container.
\nPredict how the graph for the second container will differ from the graph for the first container.
\n«He behaves as ideal gas if» «at constant V» ✓
\nuses two points to show that ✓
\n
MP1 can also be described as
\n
✓
\n✓
\n
Allow any correct data point to be used.
Allow ECF from MP1
\nrecognizes that pressure will double ✓
\ngraph will be steeper OR gradient will be larger ✓
\ngraph will still go through the origin ✓
\n
MP1 can be expressed as e.g.“” OR “will double”.
Accept for MP1.
\nA particle undergoes simple harmonic motion (SHM). The graph shows the variation of velocity v of the particle with time t.
\nWhat is the variation with time of the acceleration a of the particle?
\nA
\nWhat statement about X-rays and ultraviolet radiation is correct?
\nA. X-rays travel faster in a vacuum than ultraviolet waves.
\nB. X-rays have a higher frequency than ultraviolet waves.
\nC. X-rays cannot be diffracted unlike ultraviolet waves.
\nD. Microwaves lie between X-rays and ultraviolet in the electromagnetic spectrum.
\nB
\nA vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.
\nThe centre of the ball, still carrying a charge of 1.2 × 10−6 C, is now placed 0.40 m from a point charge Q. The charge on the ball acts as a point charge at the centre of the ball.
\nP is the point on the line joining the charges where the electric field strength is zero. The distance PQ is 0.22 m.
\nThe charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation
\n.
\nDemonstrate that the units of the quantities in this equation are consistent.
\nThe thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg.
\nDetermine the horizontal force that acts on the ball.
\nThe charge on the ball is 1.2 × 10−6 C. Determine σ.
\nThe thread breaks. Explain the initial subsequent motion of the ball.
\nCalculate the charge on Q. State your answer to an appropriate number of significant figures.
\nOutline, without calculation, whether or not the electric potential at P is zero.
\nidentifies units of as ✓
\nseen and reduced to ✓
\n\n
Accept any analysis (eg dimensional) that yields answer correctly
\nhorizontal force on ball ✓
\n✓
\n✓
\n
Allow g = 10 N kg−1
Award [3] marks for a bald correct answer.
\nAward [1max] for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.
\n✓
\n✓
\n
Allow ECF from the calculated F in (b)(i)
Award [2] for a bald correct answer.
\n\n
horizontal/repulsive force and vertical force/pull of gravity act on the ball ✓
\nso ball has constant acceleration/constant net force ✓
\nmotion is in a straight line ✓
\nat 30° to vertical away from wall/along original line of thread ✓
\n✓
\n✓
\n2sf ✓
\n
Do not award MP2 if charge is negative
Any answer given to 2 sig figs scores MP3
\n\n
work must be done to move a «positive» charge from infinity to P «as both charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value ✓
therefore, point P is at a positive / non-zero potential ✓
\n
Award [0] for bald answer that P has non-zero potential
A photovoltaic cell is supplying energy to an external circuit. The photovoltaic cell can be modelled as a practical electrical cell with internal resistance.
\nThe intensity of solar radiation incident on the photovoltaic cell at a particular time is at a maximum for the place where the cell is positioned.
\nThe following data are available for this particular time:
\n Operating current = 0.90 A
Output potential difference to external circuit = 14.5 V
Output emf of photovoltaic cell = 21.0 V
Area of panel = 350 mm × 450 mm
Explain why the output potential difference to the external circuit and the output emf of the photovoltaic cell are different.
\n\n
Calculate the internal resistance of the photovoltaic cell for the maximum intensity condition using the model for the cell.
\n\n
The maximum intensity of sunlight incident on the photovoltaic cell at the place on the Earth’s surface is 680 W m−2.
\nA measure of the efficiency of a photovoltaic cell is the ratio
\n\nDetermine the efficiency of this photovoltaic cell when the intensity incident upon it is at a maximum.
\nState two reasons why future energy demands will be increasingly reliant on sources such as photovoltaic cells.
\nthere is a potential difference across the internal resistance
OR
there is energy/power dissipated in the internal resistance ✓
when there is current «in the cell»/as charge flows «through the cell» ✓
\n
Allow full credit for answer based on
ALTERNATIVE 1
pd dropped across cell ✓
internal resistance ✓
\n✓
\n
ALTERNATIVE 2
so ✓
\n✓
\n✓
\n
Alternative solutions are possible
Award [3] marks for a bald correct answer
\npower arriving at cell = 680 x 0.35 x 0.45 = «107 W» ✓
\npower in external circuit = 14.5 x 0.9 = «13.1 W» ✓
\nefficiency = 0.12 OR 12 % ✓
\n
Award [3] marks for a bald correct answer
Allow ECF for MP3
\n«energy from Sun/photovoltaic cells» is renewable
OR
non-renewable are running out ✓
non-polluting/clean ✓
\nno greenhouse gases
OR
does not contribute to global warming/climate change ✓
OWTTE
Do not allow economic aspects (e.g. free energy)
\nTwo pulses are travelling towards each other.
\nWhat is a possible pulse shape when the pulses overlap?
\nA
\nThe primary coil of a transformer is connected to a 110 V alternating current (ac) supply. The secondary coil of the transformer is connected to a 15 V garden lighting system that consists of 8 lamps connected in parallel. Each lamp is rated at 35 W when working at its normal brightness. Root mean square (rms) values are used throughout this question.
\nThe primary coil has 3300 turns. Calculate the number of turns on the secondary coil.
\nDetermine the total resistance of the lamps when they are working normally.
\nCalculate the current in the primary of the transformer assuming that it is ideal.
\nFlux leakage is one reason why a transformer may not be ideal. Explain the effect of flux leakage on the transformer.
\nA pendulum with a metal bob comes to rest after 200 swings. The same pendulum, released from the same position, now swings at 90° to the direction of a strong magnetic field and comes to rest after 20 swings.
\n
Explain why the pendulum comes to rest after a smaller number of swings.
\n✓
\n\n
ALTERNATIVE 1
\ncalculates total current ✓
\nresistance ✓
\n
ALTERNATIVE 2
calculates total power ✓
\nresistance ✓
\n
ALTERNATIVE 3
calculates individual resistance ✓
\nresistance ✓
\ntotal power required = 280 «W»
OR
uses factor
OR
total current = 18.7 « A» ✓
current = 2.5 OR 2.6 «A» ✓
\n
Award [2] marks for a bald correct answer.
Allow ECF from (a)(ii).
\n\n
the secondary coil does not enclose all flux «lines from core» ✓
\ninduced emf in secondary
OR
power transferred to the secondary
OR
efficiency is less than expected ✓
Award [0] for references to eddy currents/heating of the core as the reason.
Award MP2 if no reason stated.
\n\n
bob cuts mag field lines
OR
there is a change in flux linkage ✓
induced emf across bob ✓
\nleading to eddy/induced current in bob ✓
\neddy/induced current produces a magnetic field that opposes «direction of» motion ✓
\nforce due to the induced magnetic field decelerates bob ✓
\ndamping of pendulum increases/there is additional «magnetic» damping ✓
\n
MP4 and MP5 can be expressed in terms of energy transfer from kinetic energy of bob to electrical/thermal energy in bob
Unpolarized light of intensity I0 is incident on the first of two polarizing sheets. Initially the planes of polarization of the sheets are perpendicular.
\nWhich sheet must be rotated and by what angle so that light of intensity can emerge from the second sheet?
\nC
\nWhen a sound wave travels from a region of hot air to a region of cold air, it refracts as shown.
\nWhat changes occur in the frequency and wavelength of the sound as it passes from the hot air to the cold air?
\nB
\nThe diagram below shows four energy levels for the atoms of a gas. The diagram is drawn to scale. The wavelengths of the photons emitted by the energy transitions between levels are shown.
\n
What are the wavelengths of spectral lines, emitted by the gas, in order of decreasing frequency?
\nA.
\nB.
\nC.
\nD.
\nB
\nThe graph shows the variation of current with potential difference for a filament lamp.
\nWhat is the resistance of the filament when the potential difference across it is 6.0 V?
A. 0.5 mΩ
B. 1.5 mΩ
C. 670 Ω
D. 2000 Ω
C
\nWhen a high-energy -particle collides with a beryllium-9 () nucleus, a nucleus of carbon may be produced. What are the products of this reaction?
\nB
\nAn electron is accelerated through a potential difference of 2.5 MV. What is the change in kinetic energy of the electron?
\nA. 0.4μJ
\nB. 0.4 nJ
\nC. 0.4 pJ
\nD. 0.4 fJ
\nC
\nA kaon is made up of two quarks. What is the particle classification of a kaon?
\nA. Exchange boson
\nB. Baryon
\nC. Lepton
\nD. Meson
\nD
\nDuring electron capture, an atomic electron is captured by a proton in the nucleus. The stable nuclide thallium-205 () can be formed when an unstable lead (Pb) nuclide captures an electron.
\nWrite down the equation to represent this decay.
\nThe unstable lead nuclide has a half-life of 15 × 106 years. A sample initially contains 2.0 μmol of the lead nuclide. Calculate the number of thallium nuclei being formed each second 30 × 106 years later.
\n\n
The neutron number N and the proton number Z are not equal for the nuclide . Explain, with reference to the forces acting within the nucleus, the reason for this.
\nThallium-205 () can also form from successive alpha (α) and beta-minus (β−) decays of an unstable nuclide. The decays follow the sequence α β− β− α. The diagram shows the position of on a chart of neutron number against proton number.
\nDraw four arrows to show the sequence of changes to N and Z that occur as the forms from the unstable nuclide.
\n✓
\n✓
\n\n
calculates ✓
\ncalculates nuclei remaining ✓
\nactivity ✓
\n
Accept conversion to seconds at any stage.
Award [3] marks for a bald correct answer.
\nAllow ECF from MP1 and MP2
\nAllow use of decay equation.
\nReference to proton repulsion OR nucleon attraction ✓
\nstrong force is short range OR electrostatic/electromagnetic force is long range ✓
\nmore neutrons «than protons» needed «to hold nucleus together» ✓
\nany α change correct ✓
\nany β change correct ✓
\ndiagram fully correct ✓
\n
Award [2] max for a correct diagram without arrows drawn.
For MP1 accept a (−2, −2 ) line with direction indicated, drawn at any position in the graph.
\nFor MP2 accept a (1, −1) line with direction indicated, drawn at any position in the graph.
\nAward [1] max for a correct diagram with all arrows in the opposite direction.
\nA cell is connected in series with a resistor and supplies a current of 4.0 A for a time of 500 s. During this time, 1.5 kJ of energy is dissipated in the cell and 2.5 kJ of energy is dissipated in the resistor.
\nWhat is the emf of the cell?
\nA. 0.50 V
\nB. 0.75 V
\nC. 1.5 V
\nD. 2.0 V
\nD
\nConsider the Feynman diagram below.
\n
What is the exchange particle X?
\nA. Lepton
\nB. Gluon
\nC. Meson
\nD. Photon
\nD
\nAn electron travelling at speed v perpendicular to a magnetic field of strength B experiences a force F.
\nWhat is the force acting on an alpha particle travelling at 2v parallel to a magnetic field of strength 2B?
\nA. 0
\nB. 2F
\nC. 4F
\nD. 8F
\nA
\nA horizontal disc rotates uniformly at a constant angular velocity about a central axis normal to the plane of the disc.
\nPoint X is a distance 2L from the centre of the disc. Point Y is a distance L from the centre of the disc. Point Y has a linear speed v and a centripetal acceleration a.
\nWhat is the linear speed and centripetal acceleration of point X?
\nB
\nA black-body radiator emits a peak wavelength of and a maximum power of . The peak wavelength emitted by a second black-body radiator with the same surface area is . What is the total power of the second black-body radiator?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhat is the main role of carbon dioxide in the greenhouse effect?
\nA. It absorbs incoming radiation from the Sun.
\nB. It absorbs outgoing radiation from the Earth.
\nC. It reflects incoming radiation from the Sun.
\nD. It reflects outgoing radiation from the Earth.
\nB
\nA block rests on a rough horizontal plane. A force P is applied to the block and the block moves to the right.
\nThere is a coefficient of friction giving rise to a frictional force F between the block and the plane. The force P is doubled. Will and F be unchanged or greater?
\nA
\nA projectile is launched at an angle above the horizontal with a horizontal component of velocity and a vertical component of velocity . Air resistance is negligible. Which graphs show the variation with time of and of ?
\nD
\nWhich graph shows the variation of amplitude with intensity for a wave?
\nB
\nAn object of constant mass is tied to the end of a rope of length l and made to move in a horizontal circle. The speed of the object is increased until the rope breaks at speed v. The length of the rope is then changed. At what other combination of rope length and speed will the rope break?
\nA
\nA nucleus of phosphorus (P) decays to a nucleus of silicon (Si) with the emission of particle X and particle Y.
\n\n
What are X and Y?
\nD
\nA circuit contains a variable resistor of maximum resistance R and a fixed resistor, also of resistance R, connected in series. The emf of the battery is and its internal resistance is negligible.
\nWhat are the initial and final voltmeter readings when the variable resistor is increased from an initial resistance of zero to a final resistance of R?
\nC
\nA sample of a pure radioactive nuclide initially contains atoms. The initial activity of the sample is .
\nA second sample of the same nuclide initially contains atoms.
\nWhat is the activity of the second sample after three half lives?
\nA.
\nB.
\nC.
\nD.
\nB
\nWhat is the definition of the unified atomic mass unit?
\nA. the mass of a neutral atom of carbon-12
\nB. The mass of a neutral atom of hydrogen-1
\nC. the mass of a nucleus of carbon-12
\nD. The mass of a nucleus of hydrogen-1
\nA
\nIn nuclear fission, a nucleus of element X absorbs a neutron (n) to give a nucleus of element Y and a nucleus of element Z.
\nX + n → Y + Z + 2n
\nWhat is and ?
\nA
\nWhat is the energy equivalent to the mass of one proton?
\nA. 9.38 × (3 × 108)2 × 106 J
\nB. 9.38 × (3 × 108)2 × 1.6 × 10–19 J
\nC. J
\nD. 9.38 × 108 × 1.6 × 10–19 J
\nD
\nDuring the nuclear fission of nucleus X into nucleus Y and nucleus Z, energy is released. The binding energies per nucleon of X, Y and Z are , and respectively. What is true about the binding energy per nucleon of X, Y and Z?
\n
A. > and >
B. = and =
\nC. > and >
\nD. = +
\nA
\nA model of an ideal wind turbine with blade length is designed to produce a power when the average wind speed is . A second ideal wind turbine is designed to produce a power when the average wind speed is . What is the blade length for the second wind turbine?
\nA.
\nB.
\nC.
\nD.
\nC
\nThe following are energy sources.
\nI. a battery of rechargeable electric cells
II. crude oil
III. a pumped storage hydroelectric system
Which of these are secondary energy sources?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nOn a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger.
\n
The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with a frequency of 195 Hz. The sounding length of the string is 62 cm.
\nOutline how a standing wave is produced on the string.
\nShow that the speed of the wave on the string is about 240 m s−1.
\nSketch a graph to show how the acceleration of point P varies with its displacement from the rest position.
\n
«travelling» wave moves along the length of the string and reflects «at fixed end» ✓
\nsuperposition/interference of incident and reflected waves ✓
\nthe superposition of the reflections is reinforced only for certain wavelengths ✓
\n✓
\n✓
\nAnswer must be to 3 or more sf or working shown for MP2.
\nstraight line through origin with negative gradient ✓
\nPlanet X and planet Y both emit radiation as black bodies. Planet X has a surface temperature that is less than the surface temperature of planet Y.
\nWhat is the graph of the variation of intensity I with wavelength λ for the radiation emitted by planet Y? The graph for planet X is shown dotted.
\nD
\nA mass–spring system oscillates vertically with a period of at the surface of the Earth. The gravitational field strength at the surface of Mars is . What is the period of the same mass–spring system on the surface of Mars?
\nA.
\nB.
\nC.
\nD.
\nC
\nLight passes through a diffraction grating. Which quantity must be decreased to improve the resolution of the diffraction grating?
\nA. The grating spacing
\nB. The number of grating lines illuminated by the light source
\nC. The number of grating lines per millimetre
\nD. The spectral order being observed
\nA
\nA train is moving in a straight line away from a stationary observer when the train horn emits a sound of frequency . The speed of the train is where is the speed of sound. What is the frequency of the horn as heard by the observer?
\nA.
\nB.
\nC.
\nD.
\nB
\nThe average surface temperature of Mars is approximately 200 K and the average surface temperature of Earth is approximately 300 K. Mars has a radius half that of Earth. Assume that both Mars and Earth act as black bodies.
\nWhat is ?
\nA. 20
B. 5
C. 0.2
D. 0.05
D
\nThe graph shows the variation of the acceleration a of an object with time t.
\nWhat is the change in speed of the object shown by the graph?
\nA. 0.5 m s–1
\nB. 2.0 m s–1
\nC. 36 m s–1
\nD. 72 m s–1
\nC
\nMonochromatic light of wavelength passes through a single-slit of width and produces a diffraction pattern on a screen. Which combination of changes to and will cause the greatest decrease in the width of the central maximum?
\nC
\nA horizontal spring of spring constant k and negligible mass is compressed through a distance y from its equilibrium length. An object of mass m that moves on a frictionless surface is placed at the end of the spring. The spring is released and returns to its equilibrium length.
\nWhat is the speed of the object just after it leaves the spring?
\nA.
\nB.
\nC.
\nD.
\nA
\nAn object of mass released from rest near the surface of a planet has an initial acceleration . What is the gravitational field strength near the surface of the planet?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhich of the following is not a primary energy source?
\nA. Wind turbine
\nB. Jet Engine
\nC. Coal-fired power station
\nD. Nuclear power station
\nB
\nA mass–spring system oscillates horizontally on a frictionless surface. The mass has an acceleration when its displacement from its equilibrium position is .
\nThe variation of with is modelled in two different ways, A and B, by the graphs shown.
\nOutline two reasons why both models predict that the motion is simple harmonic when is small.
\nDetermine the time period of the system when is small.
\nOutline, without calculation, the change to the time period of the system for the model represented by graph B when is large.
\nThe graph shows for model A the variation with of elastic potential energy Ep stored in the spring.
\nDescribe the graph for model B.
\nFor both models:
displacement is ∝ to acceleration/force «because graph is straight and through origin» ✓
displacement and acceleration / force in opposite directions «because gradient is negative»
OR
acceleration/«restoring» force is always directed to equilibrium ✓
attempted use of ✓
\nsuitable read-offs leading to gradient of line = 28 « s-2» ✓
\n«» ✓
\ns ✓
\ntime period increases ✓
\n\n
because average ω «for whole cycle» is smaller
\nOR
\nslope / acceleration / force at large x is smaller
\nOR
\narea under graph B is smaller so average speed is smaller. ✓
\nsame curve OR shape for small amplitudes «to about 0.05 m» ✓
\nfor large amplitudes «outside of 0.05 m» Ep smaller for model B / values are lower than original / spread will be wider ✓ OWTTE
\n\n
Accept answers drawn on graph – e.g.
\nThis item was essentially encouraging candidates to connect concepts about simple harmonic motion to a physical situation described by a graph. The marks were awarded for discussing the physical motion (such as \"the acceleration is in the opposite direction of the displacement\") and not just for describing the graph itself (such as \"the slope of the graph is negative\"). Most candidates were successful in recognizing that the acceleration was proportional to displacement for the first marking point, but many simply described the graph for the second marking point.
\nThis question was well done by many candidates. A common mistake was to select an incorrect gradient, but candidates who showed their work clearly still earned the majority of the marks.
\nMany candidates recognized that the time period would increase for B, and some were able to give a valid reason based on the difference between the motion of B and the motion of A. It should be noted that the prompt specified \"without calculation\", so candidates who simply attempted to calculate the time period of B did not receive marks.
\nCandidates were generally successful in describing one of the two aspects of the graph of B compared to A, but few were able to describe both. It should be noted that this is a two mark question, so candidates should have considered the fact that there are two distinct statements to be made about the graphs. Examiners did accept clearly drawn graphs as well for full marks.
\nThe points X and Y are in a uniform electric field of strength . The distance OX is and the distance OY is .
\n
What is the magnitude of the change in electric potential between X and Y?
\nA.
\nB.
\nC.
\nD.
\nA
\nWhat are the principal energy changes in a photovoltaic cell and in a solar heating panel?
\nA
\nA football player kicks a stationary ball of mass 0.45 kg towards a wall. The initial speed of the ball after the kick is 19 m s−1 and the ball does not rotate. Air resistance is negligible and there is no wind.
\nThe player’s foot is in contact with the ball for 55 ms. Calculate the average force that acts on the ball due to the football player.
\nThe ball leaves the ground at an angle of 22°. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m. Calculate the time taken for the ball to reach the wall.
\nThe top of the wall is 2.4 m above the ground. Deduce whether the ball will hit the wall.
\nIn practice, air resistance affects the ball. Outline the effect that air resistance has on the vertical acceleration of the ball. Take the direction of the acceleration due to gravity to be positive.
\nThe player kicks the ball again. It rolls along the ground without sliding with a horizontal velocity of . The radius of the ball is . Calculate the angular velocity of the ball. State an appropriate SI unit for your answer.
\n✓
\n✓
\nAllow [2] marks for a bald correct answer.
\nAllow ECF for MP2 if 19 sin22 OR 19 cos22 used.
\n✓
\n✓
\nAllow ECF for MP2
\n✓
\n✓
\nball does not hit wall OR 2.5 «m» > 2.4 «m» ✓
\n
Allow ECF from (b)(i) and from MP1
Allow g = 10 m s−2
\nair resistance opposes «direction of» motion
OR
air resistance opposes velocity ✓
on the way up «vertical» acceleration is increased OR greater than g ✓
\non the way down «vertical» acceleration is decreased OR smaller than g ✓
\n
Allow deceleration/acceleration but meaning must be clear
✓
\n
Unit must be seen for mark
Accept Hz
\nAccept
\nA mass is suspended from the ceiling of a train carriage by a string. The string makes an angle θ with the vertical when the train is accelerating along a straight horizontal track.
\nWhat is the acceleration of the train?
\nA. g sin θ
\nB. g cos θ
\nC. g tan θ
\nD.
\nC
\nA satellite orbits planet with a speed at a distance from the centre of planet . Another satellite orbits planet at a speed of at a distance from the centre of planet . The mass of planet is and the mass of planet is . What is the ratio of ?
A. 0.25
B. 0.5
\nC. 2.0
\nD. 4.0
\nB
\nAn object of mass 2kg is thrown vertically downwards with an initial kinetic energy of 100J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
\nA. 2.5m
B. 5.0m
C. 10m
D. 14m
B
\nA parallel-plate capacitor is connected to a cell of constant emf. The capacitor plates are then moved closer together without disconnecting the cell. What are the changes in the capacitance of the capacitor and the energy stored in the capacitor?
\nA
\nTwo identical positive point charges X and Y are placed 0.30 m apart on a horizontal line. O is the point midway between X and Y. The charge on X and the charge on Y is +4.0 µC.
\nA positive charge Z is released from rest 0.010 m from O on the line between X and Y. Z then begins to oscillate about point O.
\nCalculate the electric potential at O.
\nSketch, on the axes, the variation of the electric potential V with distance between X and Y.
\nIdentify the direction of the resultant force acting on Z as it oscillates.
\nDeduce whether the motion of Z is simple harmonic.
\nuse of ✓
\nOR 240 «kV» for one charge calculated ✓
\n480 «kV» for both ✓
\n\n
MP1 can be seen or implied from calculation.
\nAllow ECF from MP2 for MP3.
\nsymmetric curve around 0 with potential always positive, “bowl shape up” and curve not touching the horizontal axis. ✓
\nclear asymptotes at X and Y ✓
\n\n
force is towards O ✓
\nalways ✓
\nALTERNATIVE 1
\nmotion is not SHM ✓
\n«because SHM requires force proportional to r and» this force depends on ✓
\n
ALTERNATIVE 2
motion is not SHM ✓
\nenergy-distance «graph must be parabolic for SHM and this» graph is not parabolic ✓
\nThis question was generally well approached. Two common errors were either starting with the wrong equation (electric potential energy or Coulomb's law) or subtracting the potentials rather than adding them.
\nVery few candidates drew a graph that was awarded two marks. Many had a generally correct shape, but common errors were drawing the graph touching the x-axis at O and drawing a general parabola with no clear asymptotes.
\nMany candidates were able to identify the direction of the force on the particle at position Z, but a common error was to miss that the question was about the direction as the particle was oscillating. Examiners were looking for a clear understanding that the force was always directed toward the equilibrium position, and not just at the moment shown in the diagram.
\nThis was a challenging question for candidates. Most simply assumed that because the charge was oscillating that this meant the motion was simple harmonic. Some did recognize that it was not, and most of those candidates correctly identified that the relationship between force and displacement was an inverse square.
\nA student draws a graph to show the variation with time t of the acceleration a of an object.
\nWhat can the student deduce from this graph only, and what quantity from the graph is used to make this deduction?
\n
B
\nA point source of light of amplitude A0 gives rise to a particular light intensity when viewed at a distance from the source. When the amplitude is increased and the viewing distance is doubled, the light intensity is doubled. What is the new amplitude of the source?
\nA. 2A0
B. 2 A0
\nC. 4A0
\nD. 8A0
\nB
\nThe graph shows the variation of an alternating current with time in a resistor.
\nWhat is the average power dissipated in the resistor?
\nA.
\nB.
\nC.
\nD.
\nB
\nWhich diagram shows the shape of the wavefront as a result of the diffraction of plane waves by an object?
\nA
\nA cyclist accelerates in a straight line. At one instant, when the cyclist is exerting a forward force of 40 N, the air resistance acting on the cyclist is 10 N.
\nWhat is the rate of change of momentum of the cyclist at this instant?
\nA. 10 kg m s–2
\nB. 30 kg m s–2
\nC. 40 kg m s–2
\nD. 50 kg m s–2
\nB
\nA magnet connected to a spring oscillates above a solenoid with a 240 turn coil as shown.
\nThe graph below shows the variation with time of the emf across the solenoid with the period, , of the system shown.
\nThe spring is replaced with one that allows the magnet to oscillate with a higher frequency. Which graph shows the new variation with time of the current in the resistor for this new set-up?
\nA
\nThe diagram below shows part of a downhill ski course which starts at point A, 50 m above level ground. Point B is 20 m above level ground.
\nA skier of mass 65 kg starts from rest at point A and during the ski course some of the gravitational potential energy transferred to kinetic energy.
\nAt the side of the course flexible safety nets are used. Another skier of mass 76 kg falls normally into the safety net with speed 9.6 m s–1.
\nFrom A to B, 24 % of the gravitational potential energy transferred to kinetic energy. Show that the velocity at B is 12 m s–1.
\nSome of the gravitational potential energy transferred into internal energy of the skis, slightly increasing their temperature. Distinguish between internal energy and temperature.
\nThe dot on the following diagram represents the skier as she passes point B.
Draw and label the vertical forces acting on the skier.
The hill at point B has a circular shape with a radius of 20 m. Determine whether the skier will lose contact with the ground at point B.
\nThe skier reaches point C with a speed of 8.2 m s–1. She stops after a distance of 24 m at point D.
\nDetermine the coefficient of dynamic friction between the base of the skis and the snow. Assume that the frictional force is constant and that air resistance can be neglected.
\nCalculate the impulse required from the net to stop the skier and state an appropriate unit for your answer.
\nExplain, with reference to change in momentum, why a flexible safety net is less likely to harm the skier than a rigid barrier.
\n\n
«m s–1»
\n\n
Award GPE lost = 65 × 9.81 × 30 = «19130 J»
\nMust see the 11.9 value for MP2, not simply 12.
\nAllow g = 9.8 ms–2.
\ninternal energy is the total KE «and PE» of the molecules/particles/atoms in an object
\ntemperature is a measure of the average KE of the molecules/particles/atoms
\n\n
Award [1 max] if there is no mention of molecules/particles/atoms.
\narrow vertically downwards from dot labelled weight/W/mg/gravitational force/Fg/Fgravitational AND arrow vertically upwards from dot labelled reaction force/R/normal contact force/N/FN
\nW > R
\n\n
Do not allow gravity.
Do not award MP1 if additional ‘centripetal’ force arrow is added.
Arrows must connect to dot.
Ignore any horizontal arrow labelled friction.
Judge by eye for MP2. Arrows do not have to be correctly labelled or connect to dot for MP2.
ALTERNATIVE 1
recognition that centripetal force is required / seen
= 468 «N»
\nW/640 N (weight) is larger than the centripetal force required, so the skier does not lose contact with the ground
\n\n
ALTERNATIVE 2
\nrecognition that centripetal acceleration is required / seen
\na = 7.2 «ms–2»
\ng is larger than the centripetal acceleration required, so the skier does not lose contact with the ground
\n\n
ALTERNATIVE 3
\nrecognition that to lose contact with the ground centripetal force ≥ weight
\ncalculation that v ≥ 14 «ms–1»
\ncomment that 12 «ms–1» is less than 14 «ms–1» so the skier does not lose contact with the ground
\n\n
ALTERNATIVE 4
\nrecognition that centripetal force is required / seen
\ncalculation that reaction force = 172 «N»
\nreaction force > 0 so the skier does not lose contact with the ground
\n\n
\n
Do not award a mark for the bald statement that the skier does not lose contact with the ground.
\nALTERNATIVE 1
0 = 8.22 + 2 × a × 24 therefore a = «−»1.40 «m s−2»
friction force = ma = 65 × 1.4 = 91 «N»
\ncoefficient of friction = = 0.14
\n\n
ALTERNATIVE 2
KE = mv2 = 0.5 x 65 x 8.22 = 2185 «J»
friction force = KE/distance = 2185/24 = 91 «N»
\ncoefficient of friction = = 0.14
\n\n
Allow ECF from MP1.
\n«76 × 9.6»= 730
Ns OR kg ms–1
safety net extends stopping time
\nF = therefore F is smaller «with safety net»
\nOR
\nforce is proportional to rate of change of momentum therefore F is smaller «with safety net»
\n\n
Accept reverse argument.
\nA capacitor is charged with a constant current . The graph shows the variation of potential difference across the capacitor with time . The gradient of the graph is . What is the capacitance of the capacitor?
\n
A.
\nB.
\nC.
\nD.
\nA
\nA 12V battery has an internal resistance of 2.0Ω. A load of variable resistance is connected across the battery and adjusted to have resistance equal to that of the internal resistance of the battery. Which statement is correct for this circuit?
\nA. The current in the battery is 6A.
B. The potential difference across the load is 12V.
C. The power dissipated in the battery is 18W.
D. The resistance in the circuit is 1.0Ω.
C
\nA travelling wave of period 5.0 ms travels along a stretched string at a speed of 40 m s–1. Two points on the string are 0.050 m apart.
\nWhat is the phase difference between the two points?
\nA. 0
\nB.
\nC.
\nD. 2
\nB
\nWhich of the following lists the particles emitted during radioactive decay in order of increasing ionizing power?
\nA. γ, β, α
B. β, α, γ
C. α, γ, β
D. α, β, γ
A
\nProperties of waves are
\nI. polarization
II. diffraction
III. refraction
Which of these properties apply to sound waves?
\nA. I and II
B. I and III
C. II and III
D. I, II and III
C
\nTwo capacitors C1 and C2 of capacitance 28 µF and 22 µF respectively are connected in a circuit with a two-way switch and a cell of emf 1.5 V with a negligible internal resistance. The capacitors are initially uncharged. The switch is then connected to position A.
\nThe switch is moved to position B.
\nA cell is now connected by a switch to a coil X. A second coil Y of cross-sectional area 6.4 cm2 with 5 turns is looped around coil X and connected to an ideal voltmeter.
\nThe graph shows the variation with t of the magnetic flux density B in coil Y.
\nShow that the charge stored on C1 is about 0.04 mC.
\nCalculate the energy transferred from capacitor C1.
\nExplain why the energy gained by capacitor C2 differs from your answer in (b)(i).
\nThe switch is closed at time t = 0. Explain how the voltmeter reading varies after the switch is closed.
\nDetermine the average emf induced across coil Y in the first 3.0 ms.
\n«» = 0.042 «mC» ✓
\n\n
Award MP for full replacement or correct answer to at least 2 significant figures.
\n«J» ✓
\n\n
total capacitance = 50 «μF»
\nOR
\npd = «» 0.84 «V»
\nOR
\ncharge on C1 after switch moved to B = 0.0235 «mC» ✓
\n\n
«J» ✓
\nenergy lost «μJ» ✓
\nenergy transferred to electromagnetic radiation «to environment»
\nOR
\nenergy is transferred as thermal energy / heat «to circuit components» ✓
\ninitial deflection by voltmeter falling to zero reading ✓
\nemf is induced «only» while the field / flux is changing ✓
\nattempted use of OR ✓
\n✓
\n8.0 «mV» ✓
\nThis was a \"show that\" question, and it was very well done by most candidates.
\nThis question was challenging for many candidates. A large number successfully calculated the initial energy of C1, but then seemed confused about the next steps. Few candidates successfully calculated the energy in C1 after the switch was closed. There was an ECF opportunity for candidates who recognized that the final answer was the difference between these two values.
\nThis question used an \"explain\" command term, so examiners were looking for more than a generic \"energy was lost\". Candidates needed to specify a form of energy that was lost (such as thermal energy) for the mark. A very common incorrect response was simply stating that the difference was due to the capacitors having a different capacitance.
\nThis question was well answered by some candidates who recognized that it was an electromagnetic induction question and understood that eventually the current in coil X would hit a steady state and the voltmeter reading would return to zero. Common issues were candidates thinking that the potential would fluctuate in a manner similar to an alternating current, candidates discussing this more as a transformer, and candidates who missed that there were two separate coils and wrote responses suggesting that a simple circuit had been formed and the voltmeter would read the potential of the cell.
\nThis question was well approached with most candidates recognizing that this was a Faraday's law question. Many made an attempt to use the correct equation, but common errors were choosing incorrect values from the graph and incorrectly converting the given area. Examiners were generous with ECF for candidates who clearly showed work leading to an incorrect result.
\nA particle of energy is incident upon a barrier and has a certain probability of quantum tunnelling through the barrier. Assuming remains constant, which combination of changes in particle mass and barrier length will increase the probability of the particle tunnelling through the barrier?
\nA
\nThe solar constant is the intensity of the Sun’s radiation at
\nA. the surface of the Earth.
B. the mean distance from the Sun of the Earth’s orbit around the Sun.
C. the surface of the Sun.
D. 10km above the surface of the Earth.
B
\nElement X has a nucleon number and a nuclear density . Element Y has a nucleon number of . What is an estimate of the nuclear density of element Y?
\nA.
\nB.
\nC.
\nD.
\nB
\nA vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.
\nThe charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation
\n.
\nDemonstrate that the units of the quantities in this equation are consistent.
\nThe thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg.
\nDetermine the horizontal force that acts on the ball.
\nThe charge on the ball is 1.2 × 10−6 C. Determine σ.
\nThe centre of the ball, still carrying a charge of , is now placed from a point charge Q. The charge on the ball acts as a point charge at the centre of the ball.
\nP is the point on the line joining the charges where the electric field strength is zero.
The distance PQ is .
Calculate the charge on Q. State your answer to an appropriate number of significant figures.
\nidentifies units of as ✓
\nseen and reduced to ✓
\n\n
Accept any analysis (eg dimensional) that yields answer correctly
\nhorizontal force on the ball ✓
\n✓
\n✓
\n
Allow g = 10 N kg−1
Award [3] marks for a bald correct answer.
\nAward [1max] for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.
\n✓
\n✓
\n
Allow ECF from the calculated F in (b)(i)
Award [2] for a bald correct answer.
\n\n
✓
\n✓
\n2sf ✓
\n
Do not award MP2 if charge is negative
Any answer given to 2 sig figs scores MP3
\nX and Y are two spherical black-body radiators that emit the same total power. The absolute temperature of X is half that of Y.
\nWhat is ?
\nA. 4
\nB. 8
\nC. 16
\nD. 32
\nA
\nWater is draining from a vertical tube that was initially full. A vibrating tuning fork is held near the top of the tube. For two positions of the water surface only, the sound is at its maximum loudness.
\nThe distance between the two positions of maximum loudness is x.
\nWhat is the wavelength of the sound emitted by the tuning fork?
\nA.
\nB. x
\nC.
\nD. 2x
\nD
\nElectrons, each with a charge e, move with speed v along a metal wire. The electric current in the wire is I.
\nPlane P is perpendicular to the wire. How many electrons pass through plane P in each second?
\nA.
\nB.
\nC.
\nD.
\nD
\nWhat is true for the Bohr model for the hydrogen atom?
\nA. Angular momentum of electrons is quantized.
\nB. Electrons are described by wave functions.
\nC. Electrons never exist in fixed orbitals.
\nD. Electrons will continuously emit radiation.
\nA
\nA particle is oscillating with simple harmonic motion (shm) of amplitude x0 and maximum kinetic energy Ek. What is the potential energy of the system when the particle is a distance 0.20x0 from its maximum displacement?
\nA. 0.20Ek
\nB. 0.36Ek
\nC. 0.64Ek
\nD. 0.80Ek
\nC
\nA pure sample of nuclide A and a pure sample of nuclide B have the same activity at time t = 0. Nuclide A has a half-life of T, nuclide B has a half-life of 2T.
\nWhat is when t = 4T?
\nA. 4
\nB. 2
\nC.
\nD.
\nD
\nMonochromatic light is incident on a double slit. Both slits have a finite width. The light then forms an interference pattern on a screen some distance away. Which graph shows the variation of intensity with distance from the centre of the pattern?
\nD
\nAn object can lose energy through
\nI. conduction
II. convection
III. radiation
What are the principal means for losing energy for a hot rock resting on the surface of the Moon?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nThe average albedo of glacier ice is 0.25.
\nWhat is ?
\nA. 0.25
\nB. 0.33
\nC. 2.5
\nD. 3.0
\nD
\nAn electron of non-relativistic speed interacts with an atom. All the energy of the electron is transferred to an emitted photon of frequency . An electron of speed now interacts with the same atom and all its energy is transmitted to a second photon. What is the frequency of the second photon?
\nA.
\nB.
\nC.
\nD.
\nD
\nLight of wavelength λ is incident normally on a diffraction grating that has a slit separation of . What is the greatest number of maxima that can be observed using this arrangement?
\nA. 4
B. 6
C. 7
D. 9
C
\nDuring electron capture, an atomic electron is captured by a proton in the nucleus. The stable nuclide thallium-205 () can be formed when an unstable lead (Pb) nuclide captures an electron.
\nWrite down the equation to represent this decay.
\nThe neutron number N and the proton number Z are not equal for the nuclide . Explain, with reference to the forces acting within the nucleus, the reason for this.
\nThallium-205 () can also form from successive alpha (α) and beta-minus (β−) decays of an unstable nuclide. The decays follow the sequence α β− β− α. The diagram shows the position of on a chart of neutron number against proton number.
\nDraw four arrows to show the sequence of changes to N and Z that occur as the forms from the unstable nuclide.
\n✓
\n✓
\n\n
Reference to proton repulsion OR nucleon attraction ✓
\nstrong force is short range OR electrostatic/electromagnetic force is long range ✓
\nmore neutrons «than protons» needed «to hold nucleus together» ✓
\nany α change correct ✓
\nany β change correct ✓
\ndiagram fully correct ✓
\n
Award [2] max for a correct diagram without arrows drawn.
For MP1 accept a (−2, −2 ) line with direction indicated, drawn at any position in the graph.
\nFor MP2 accept a (1, −1) line with direction indicated, drawn at any position in the graph.
\nAward [1] max for a correct diagram
\n\n
A pendulum oscillating near the surface of the Earth swings with a time period T. What is the time period of the same pendulum near the surface of the planet Mercury where the gravitational field strength is 0.4g?
\nA. 0.4T
\nB. 0.6T
\nC. 1.6T
\nD. 2.5T
\nC
\nOutline what is meant by the principle of superposition of waves.
\nRed laser light is incident on a double slit with a slit separation of 0.35 mm.
A double-slit interference pattern is observed on a screen 2.4 m from the slits.
The distance between successive maxima on the screen is 4.7 mm.
Calculate the wavelength of the light. Give your answer to an appropriate number of significant figures.
\nExplain the change to the appearance of the interference pattern when the red-light laser is replaced by one that emits green light.
\nOne of the slits is now covered.
\nDescribe the appearance of the pattern on the screen.
\nwhen 2 waves meet the resultant displacement
\nis the «vector» sum of their individual displacements
\n\n
Displacement should be mentioned at least once in MP 1 or 2.
\nλ =
\n= 6.9 x 10–7 «m»
\nanswer to 2 SF
\n\n
Allow missed powers of 10 for MP1.
\ngreen wavelength smaller than red
\nfringe separation / distance between maxima decreases
\n\n
Allow ECF from MP1.
\nbright central maximum
\nsubsidiary maxima «on either side»
\n\n
the width of the central fringe is twice / larger than the width of the subsidiary/secondary fringes/maxima
\nOR
\nintensity of pattern is decreased
\n\n
Allow marks from a suitably labelled intensity graph for single slit diffraction.
\nA diffraction grating is used to observe light of wavelength 400 nm. The light illuminates 100 slits of the grating. What is the minimum wavelength difference that can be resolved when the second order of diffraction is viewed?
\nA. 1 nm
\nB. 2 nm
\nC. 4 nm
\nD. 8 nm
\nB
\nPotassium-40 decays by two processes.
\nThe first process is that of beta-minus (β−) decay to form a calcium (Ca) nuclide.
\nPotassium-40 decays by a second process to argon-40. This decay accounts for 11 % of the total decay of the potassium-40.
\nRocks can be dated by measuring the quantity of argon-40 gas trapped in them. One rock sample contains 340 µmol of potassium-40 and 12 µmol of argon-40.
\nWrite down the equation for this decay.
\nShow that the initial quantity of potassium-40 in the rock sample was about 450 µmol.
\nThe half-life of potassium-40 is 1.3 × 109 years. Estimate the age of the rock sample.
\nOutline how the decay constant of potassium-40 was determined in the laboratory for a pure sample of the nuclide.
\n✓
\nOR ✓
\n\n
Full equation
\ntotal K-40 decayed = «μmol» ✓
\nso total K-40 originally was 109 + 340 = 449 «μmol»✓
\nALTERNATIVE 1
\nused to give 𝜆 = 5.3 x 10-10 per year ✓
\nOR
\n✓
\n
t = 5.2 x 108 «years» ✓
\n
ALTERNATIVE 2
\n«remaining» ✓
\n✓
\nt = 0.40 x 1.3 x 109 = 5.2 x 108 «years» ✓
\n\n
ALTERNATIVE 3
\n«remaining» ✓
\n✓
\nt = 0.40 x 1.3 x 109 = 5.2 x 108 «years» ✓
\n\n
Allow 5.3 x 108 years for final answer.
\nAllow ECF for MP3 for an incorrect number of half-lives.
\n«use the mass of the sample to» determine number of potassium-40 atoms / nuclei in sample ✓
\n«use a counter to» determine (radio)activity / A of sample ✓
\nuse A = λN «to determine the decay constant / λ» ✓
\nThis question was very well done by candidates. The majority were able to identify the correct nuclide of Calcium and many correctly included an electron/beta particle and a properly written antineutrino.
\nThis was a \"show that\" question that was generally well done by candidates.
\nThis was a more challenging question for candidates. Many were able to calculate the decay constant and recognized that the ratio of initial and final quantities of the potassium-40 was important. A very common error was mixing the two common half-life equations up and using the wrong values in the exponent (using half life instead of the decay constant, or using the decay constant instead of the half life). Examiners were generous with ECF for candidates who clearly showed an incorrect number of half-lives multiplied by the time for one half-life.
\nDescribing methods of determining half-life continues to be a struggle for candidates with very few earning all three marks. Many candidates described a method more appropriate to measuring a short half- life, but even those descriptions fell far short of being acceptable.
\nFor fringes to be observed in a double-slit interference experiment, the slits must emit waves that are coherent.
\nWhat conditions are required for the frequency of the waves and for the phase difference between the waves so that the waves are coherent?
\nB
\nA train moving at speed u relative to the ground, sounds a whistle of constant frequency f as it moves towards a vertical cliff face.
\nThe sound from the whistle reaches the cliff face and is reflected back to the train. The speed of sound in stationary air is c.
\nWhat whistle frequency is observed on the train after the reflection?
\nA.
\nB. (c + u)f
\nC. (c – u)f
\nD.
\nA
\nWhat is the unit of Gε0, where G is the gravitational constant and ε0 is the permittivity of free space?
\nA. C kg–1
B. C2 kg–2
\nC. C kg
\nD. C2 kg2
\nB
\nTwo parallel metal plates are connected to a dc power supply. An electric field forms in the space between the plates as shown.
\nWhat is the shape of the equipotentials surfaces that result from this arrangement?
\nB
\nAn electric field acts in the space between two charged parallel plates. One plate is at zero potential and the other is at potential +V.
\nThe distance x is measured from point P in the direction perpendicular to the plate.
\nWhat is the dependence of the electric field strength E on x and what is the dependence of the electric potential V on x?
\nB
\nTwo renewable energy sources are solar and wind.
\nAn alternative generation method is the use of wind turbines.
\nThe following data are available:
\nLength of turbine blade = 17 m
Density of air = 1.3 kg m–3
Average wind speed = 7.5 m s–1
Describe the difference between photovoltaic cells and solar heating panels.
\nA solar farm is made up of photovoltaic cells of area 25 000 m2. The average solar intensity falling on the farm is 240 W m–2 and the average power output of the farm is 1.6 MW. Calculate the efficiency of the photovoltaic cells.
\nDetermine the minimum number of turbines needed to generate the same power as the solar farm.
\nExplain two reasons why the number of turbines required is likely to be greater than your answer to (c)(i).
\nsolar heating panel converts solar/radiation/photon/light energy into thermal energy AND photovoltaic cell converts solar/radiation/photon/light energy into electrical energy
\n\n
Accept internal energy of water.
\npower received = 240 × 25000 = «6.0 MW»
\nefficiency « = 0.27 / 27%
\narea = × 172 «= 908m2»
\npower = «= 0.249 MW»
\nnumber of turbines «» = 7
\n\n
Only allow integer value for MP3.
\nAward [2 max] for 25 turbines (ECF from incorrect power)
\nAward [2 max] for 26 turbines (ECF from incorrect radius)
\n«efficiency is less than 100% as»
\nnot all KE of air can be converted to KE of blades
\nOR
\nair needs to retain KE to escape
\nthermal energy is lost due to friction in turbine/dynamo/generator
\n\n
Allow velocity of air after turbine is not zero.
\nFour particles, two of charge +Q and two of charge −Q, are positioned on the -axis as shown. A particle P with a positive charge is placed on the -axis. What is the direction of the net electrostatic force on this particle?
\nD
\nA satellite at the surface of the Earth has a weight W and gravitational potential energy Ep. The satellite is then placed in a circular orbit with a radius twice that of the Earth.
\nWhat is the weight of the satellite and the gravitational potential energy of the satellite when placed in orbit?
\nC
\nA satellite of mass 1500 kg is in the Earth’s gravitational field. It moves from a point where the gravitational potential is –30 MJ kg–1 to a point where the gravitational potential is –20 MJ kg–1. What is the direction of movement of the satellite and the change in its gravitational potential energy?
\nA
\nTwo point charges are at rest as shown.
\nAt which position is the electric field strength greatest?
\nB
\nWhich of the following reduces the energy losses in a transformer?
\nA. Using thinner wires for the windings.
\nB. Using a solid core instead of a laminated core.
\nC. Using a core made of steel instead of iron.
\nD. Linking more flux from the primary to the secondary core.
\nD
\nA direct current (dc) of 5A dissipates a power P in a resistor. Which peak value of the alternating current (ac) will dissipate an average power P in the same resistor?
\nA. 5A
\nB.
\nC.
\nD.
\nD
\nThe secondary coil of an alternating current (ac) transformer is connected to two diodes as shown.
\nWhich graph shows the variation with time of the potential difference VXY between X and Y?
\nA
\nA parallel-plate capacitor is connected to a battery. What happens when a sheet of dielectric material is inserted between the plates without disconnecting the battery?
\nA. The capacitance is unchanged.
\nB. The charge stored decreases.
\nC. The energy stored increases.
\nD. The potential difference between the plates decreases.
\nC
\nThree capacitors are arranged as shown.
\nWhat is the total capacitance of the arrangement?
\nA. 1.0F
\nB. 2.5F
\nC. 3.0F
\nD. 4.0F
\nA
\nA heater in an electric shower has a power of 8.5 kW when connected to a 240 V electrical supply. It is connected to the electrical supply by a copper cable.
\nThe following data are available:
\nLength of cable = 10 m
Cross-sectional area of cable = 6.0 mm2
Resistivity of copper = 1.7 × 10–8 Ω m
Calculate the current in the copper cable.
\nCalculate the resistance of the cable.
\nExplain, in terms of electrons, what happens to the resistance of the cable as the temperature of the cable increases.
\nThe heater changes the temperature of the water by 35 K. The specific heat capacity of water is 4200 J kg–1 K–1.
\nDetermine the rate at which water flows through the shower. State an appropriate unit for your answer.
\nI «=» =35«A»
\n\n
\n
R =
\n= 0.028 «Ω»
\n\n
Allow missed powers of 10 for MP1.
\n«as temperature increases» there is greater vibration of the metal atoms/lattice/lattice ions
\nOR
\nincreased collisions of electrons
\n\n
drift velocity decreases «so current decreases»
\n«as V constant so» R increases
\n\n
Award [0] for suggestions that the speed of electrons increases so resistance decreases.
\nrecognition that power = flow rate × cΔT
\nflow rate «»
\n= 0.058 «kg s–1»
\nkg s−1 / g s−1 / l s−1 / ml s−1 / m3 s−1
\n\n
Allow MP4 if a bald flow rate unit is stated. Do not allow imperial units.
\nPair production by a photon occurs in the presence of a nucleus. For this process, which of momentum and energy are conserved?
\nD
\nAn object of weight W is falling vertically at a constant speed in a fluid. What is the magnitude of the drag force acting on the object?
\nA. 0
B.
C. W
D. 2W
C
\nAn electron of mass m has an uncertainty in its position r. What is the uncertainty in the speed of this electron?
\nA.
\nB.
\nC.
\nD.
\nD
\nAn object, initially at rest, is accelerated by a constant force. Which graphs show the variation with time t of the kinetic energy and the variation with time t of the speed of the object?
\nA
\nWhich of the following, observed during a radioactive-decay experiment, provide evidence for the existence of nuclear energy levels?
\nI. The spectrum of alpha particle energies
II. The spectrum of beta particle energies
III. The spectrum of gamma ray energies
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nWhat is the charge on an electron antineutrino and during what process is an electron antineutrino produced?
\nD
\nTwo stationary objects of mass 1kg and 2kg are connected by a thread and suspended from a spring.
\nThe thread is cut. Immediately after the cut, what are the magnitudes of the accelerations of the objects in terms of the acceleration due to gravity g?
\nD
\nWhat are the units of magnetic flux and magnetic field strength?
\nB
\nThe diagram shows a sketch of an ideal step-down transformer.
\nThe number of turns in the primary coil is 1800 and that in the secondary coil is 90.
\nState Faraday’s law of induction.
\nExplain, using Faraday’s law of induction, how the transformer steps down the voltage.
\nThe input voltage is 240 V. Calculate the output voltage.
\nOutline how energy losses are reduced in the core of a practical transformer.
\nStep-up transformers are used in power stations to increase the voltage at which the electricity is transmitted. Explain why this is done.
\nthe size of the induced emf
is proportional/equal to the rate of change of flux linkage
\n
The word ‘induced’ is required here.
Allow correctly defined symbols from a correct equation. ‘Induced’ is required for MP1.
varying voltage/current in primary coil produces a varying magnetic field
\nthis produces a change in flux linkage / change in magnetic field in the secondary coil
\na «varying» emf is induced/produced/generated in the secondary coil
\nvoltage is stepped down as there are more turns on the primary than the secondary
\n\n
Comparison of number of turns is required for MP4.
\noutput voltage
\n= 12 «V»
\nlaminated core reduces eddy currents
\nless thermal energy is transferred to the surroundings
\nfor a certain power to be transmitted, large V means low I
\nless thermal energy loss as P = I2R / joule heating
\nA student of weight 600N climbs a vertical ladder 6.0m tall in a time of 8.0s. What is the power developed by the student against gravity?
\nA. 22W
B. 45W
C. 220W
D. 450W
D
\nA possible decay of a lambda particle () is shown by the Feynman diagram.
\nState the quark structures of a meson and a baryon.
\nExplain which interaction is responsible for this decay.
\nDraw arrow heads on the lines representing and d in the .
\nIdentify the exchange particle in this decay.
\nOutline one benefit of international cooperation in the construction or use of high-energy particle accelerators.
\nMeson: quark-antiquark pair
Baryon: 3 quarks
Alternative 1
\nstrange quark changes «flavour» to an up quark
\nchanges in quarks/strangeness happen only by the weak interaction
\n\n
Alternative 2
\nStrangeness is not conserved in this decay «because the strange quark changes to an up quark»
\nStrangeness is not conserved during the weak interaction
\n\n
Do not allow a bald answer of weak interaction.
\narrows drawn in the direction shown
\n\n
Both needed for [1] mark.
\nW −
\n\n
Do not allow W or W+.
\nit lowers the cost to individual nations, as the costs are shared
\ninternational co-operation leads to international understanding OR historical example of co-operation OR co-operation always allows science to proceed
\nlarge quantities of data are produced that are more than one institution/research group can handle co-operation allows effective analysis
\n\n
Any one.
\nA tennis ball is hit with a racket from a point 1.5 m above the floor. The ceiling is 8.0 m above the floor. The initial velocity of the ball is 15 m s–1 at 50° above the horizontal. Assume that air resistance is negligible.
\nDetermine whether the ball will hit the ceiling.
\nThe tennis ball was stationary before being hit. It has a mass of 5.8×10–2 kg and was in contact with the racket for 23 ms.
\n(i) Calculate the mean force exerted by the racket on the ball.
\n(ii) Explain how Newton’s third law applies when the racket hits the tennis ball.
\ndetermines component correctly / 15 sin 50 seen
Allow method via v = u + at. Allow use of g = 10 m s-2, gives 6.6 m and 8.1 m.
Allow [2 max] for use of 15 cos 50, gives 4.7 m and 6.2 m.
correct reasoning consistent with candidate data
Allow [1 max] (as MP2) if 13 m is obtained due to use of 15 m s−1 rather than 15 sin or 15 cos 50. If no unit given, assume metre.
i
OR 37.8«N»
Do not penalise sf here. Working not required.
ii
force of ball on racket is equal to force of racket on ball or is 38N
Do not accept “same force”.
Allow ECF from force value in bi
ball exerts force in opposite direction to force of racket on ball
Accept “opposite force” for “in opposite direction”.
Do not accept undefined references to “reaction” the direction of the forces must be clear.
A battery is used to charge a capacitor fully through a resistor of resistance R. The energy supplied by the battery is Eb. The energy stored by the capacitor is Ec.
\nWhat is the relationship between Eb and Ec?
\nA. Eb < Ec
\nB. Eb = Ec
\nC. Eb > Ec
\nD. The relationship depends on R.
\nC
\nA ball of mass m strikes a vertical wall with a speed v at an angle of θ to the wall. The ball rebounds at the same speed and angle. What is the change in the magnitude of the momentum of the ball?
\nA. 2 mv sin θ
B. 2 mv cos θ
C. 2 mv
D. zero
B
\nMomentum is a vector quantity so the angle and the direction are both relevant to the answer. Hence C and D can be eliminated. If θ = 900, then the ball is just rolling down the wall and there is no change in momentum. Hence B is correct (cos900 = 0).
\nA capacitor is charged by a constant current of 2.5 μA for 100 s. As a result the potential difference across the capacitor increases by 5.0 V.
\nWhat is the capacitance of the capacitor?
\nA. 20 μF
\nB. 50 μF
\nC. 20 mF
\nD. 50 mF
\nB
\nTwo objects m1 and m2 approach each other along a straight line with speeds v1 and v2 as shown. The objects collide and stick together.
\nWhat is the total change of linear momentum of the objects as a result of the collision?
\nA. m1v1 + m2v2
B. m1v1 – m2v2
C. m2v2 – m1v1
D. zero
D
\nCurling is a game played on a horizontal ice surface. A player pushes a large smooth stone across the ice for several seconds and then releases it. The stone moves until friction brings it to rest. The graph shows the variation of speed of the stone with time.
\nThe total distance travelled by the stone in 17.5 s is 29.8 m.
\nDetermine the maximum speed v of the stone.
\n(i) The stone has a mass of 20 kg. Determine the frictional force on the stone during the last 14.0 s.
\n(ii) Determine the energy dissipated due to friction during the last 14.0 s.
\nevidence that area under graph used
OR
use of mean velocity × time
Award [2] for a bald correct answer.
«» «m s–1»
Award [1] for 1.70 m s−1.
i
«deceleration» = OR 0.243 «m s–2»
Award [2] for a bald correct answer.
Award [1 max] for use of first 3.5 s.
Allow ECF from 2(a).
Ignore slight rounding errors
F = «0.243 × 20 =» 4.87«N»
ii
ALTERNATIVE 1
calculates KE using mv2
Allow ECF from (a).
116 J
Award [2] for a bald correct answer.
ALTERNATIVE 2
calculates distance as 23.9 «m»
Allow ECF from (a).
«4.86 × 23.90» = 116 J
Allow ECF from (a) and (b)(i)
Award [2] for a bald correct answer.
Award [1 max] for use of first 3.5 s
Unit is required for MP2
Energy is supplied at a constant rate to a fixed mass of a material. The material begins as a solid. The graph shows the variation of the temperature of the material with time.
\nThe specific heat capacities of the solid, liquid and gaseous forms of the material are cs cl and cg respectively. What can be deduced about the values of cs cl and cg?
\nA. cs > cg > cl
B. cl > cs > cg
C. cl > cg > cs
D. cg > cs > cl
D
\nTwo observations about the photoelectric effect are
\nObservation 1: For light below the threshold frequency no electrons are emitted from the metal surface.
\nObservation 2: For light above the threshold frequency, the emission of electrons is almost instantaneous.
\nThe graph shows how the maximum kinetic energy Emax of electrons emitted from a surface of barium metal varies with the frequency f of the incident radiation.
\nExplain how each observation provides support for the particle theory but not the wave theory of light.
\nDetermine a value for Planck’s constant.
\nState what is meant by the work function of a metal.
\nCalculate the work function of barium in eV.
\nThe experiment is repeated with a metal surface of cadmium, which has a greater work function. Draw a second line on the graph to represent the results of this experiment.
\nObservation 1:
particle – photon energy is below the work function
OR
E = hf and energy is too small «to emit electrons»
wave – the energy of an em wave is independent of frequency
Observation 2:
particle – a single electron absorbs the energy of a single photon «in an almost instantaneous interaction»
wave – it would take time for the energy to build up to eject the electron
attempt to calculate gradient of graph = «»
\n«Js»
\n\n
Do not allow a bald answer of 6.63 x 10-34 Js or 6.6 x 10-34 Js.
\nALTERNATIVE 1
minimum energy required to remove an electron «from the metal surface»
ALTERNATIVE 2
energy required to remove the least tightly bound electron «from the metal surface»
ALTERNATIVE 1
reading of y intercept from graph in range 3.8 − 4.2 × 10–19 «J»
conversion to eV = 2.4 – 2.6 «eV»
ALTERNATIVE 2
reading of x intercept from graph «5.8 − 6.0 × 1014 Hz» and using hf0 to get 3.8 − 4.2 × 10–19 «J»
conversion to eV = 2.4 – 2.6 «eV»
line parallel to existing line
to the right of the existing line
Define internal energy.
\n0.46 mole of an ideal monatomic gas is trapped in a cylinder. The gas has a volume of 21 m3 and a pressure of 1.4 Pa.
\n(i) State how the internal energy of an ideal gas differs from that of a real gas.
\n(ii) Determine, in kelvin, the temperature of the gas in the cylinder.
\n(iii) The kinetic theory of ideal gases is one example of a scientific model. Identify one reason why scientists find such models useful.
\nmention of atoms/molecules/particles
\nsum/total of kinetic energy and «mutual/intermolecular» potential energy
\nDo not allow “kinetic energy and potential energy” bald.
Do not allow “sum of average ke and pe” unless clearly referring to total ensemble.
i
«intermolecular» potential energy/PE of an ideal gas is zero/negligible
ii
THIS IS FOR USE WITH AN ENGLISH SCRIPT ONLY
use of or
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Award [2] for a bald correct answer in K.
Award [2 max] if correct 7.7 K seen followed by –265°C and mark BOD. However, if only –265°C seen, award [1 max].
7.7 K
Do not penalise use of “°K”
ii
THIS IS FOR USE WITH A SPANISH SCRIPT ONLY
T = 7.7 ×10-6 K
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Uses correct unit conversion for volume
Award [2] for a bald correct answer in K. Finds solution. Allow an ECF from MP2 if unit not converted, ie candidate uses 21 m3 and obtains 7.7 K
Do not penalise use of “°K”
\n
iii
models used to predict/hypothesize
explain
\nsimulate
\nsimplify/approximate
Allow similar responses which have equivalent meanings. Response needs to identify one reason.
A conducting square coil is placed in a region where there is a uniform magnetic field. The magnetic field is directed into the page. There is a clockwise current in the coil.
\nWhat is a correct force that acts on a side of the coil?
\nD
\nA tennis ball is released from rest at a height h above the ground. At each bounce 50 % of its kinetic energy is lost to its surroundings. What is the height reached by the ball after its second bounce?
\nA.
\nB.
\nC.
\nD. zero
\nB
\nAn ideal gas of N molecules is maintained at a constant pressure p. The graph shows how the volume V of the gas varies with absolute temperature T.
\nWhat is the gradient of the graph?
\nA.
\nB.
\nC.
\nD.
\nC
\nA particular K meson has a quark structure s. State the charge on this meson.
\nThe Feynman diagram shows the changes that occur during beta minus (β–) decay.
\nLabel the diagram by inserting the four missing particle symbols.
\nCarbon-14 (C-14) is a radioactive isotope which undergoes beta minus (β–) decay to the stable isotope nitrogen-14 (N-14). Energy is released during this decay. Explain why the mass of a C-14 nucleus and the mass of a N-14 nucleus are slightly different even though they have the same nucleon number.
\ncharge: –1«e» or negative or K−
\nNegative signs required.
\ncorrect symbols for both missing quarks
\nexchange particle and electron labelled W or W– and e or e–
Do not allow W+ or e+ or β+ Allow β or β–
decay products include an electron that has mass
OR
products have energy that has a mass equivalent
OR
mass/mass defect/binding energy converted to mass/energy of decay products
«so»
\nmass C-14 > mass N-14
OR
mass of n > mass of p
OR
mass of d > mass of u
Accept reference to “lighter” and “heavier” in mass.
Do not accept implied comparison, eg “C-14 has greater mass”. Comparison must be explicit as stated in scheme.
The diameter of a silver-108 () nucleus is approximately three times that of the diameter of a nucleus of
\nA.
\nB.
\nC.
\nD.
\nA
\nA heater in an electric shower has a power of 8.5 kW when connected to a 240 V electrical supply. It is connected to the electrical supply by a copper cable.
\nThe following data are available:
\nLength of cable = 10 m
Cross-sectional area of cable = 6.0 mm2
Resistivity of copper = 1.7 × 10–8 Ω m
Calculate the power dissipated in the cable.
\npower = «352 × 0.028» = 34 «W»
\n\n
Allow 35 – 36 W if unrounded figures for R or I are used.
Allow ECF from (a)(i) and (a)(ii).
The pressure of a fixed mass of an ideal gas in a container is decreased at constant temperature. For the molecules of the gas there will be a decrease in
\nA. the mean square speed.
B. the number striking the container walls every second.
C. the force between them.
D. their diameter.
B
\nWhat can be used to calculate the probability of finding an electron in a particular region of space?
\nA.
\nB.
\nC. The magnitude of the wave function
\nD. The magnitude of the (wave function)2
\nD
\nThe initial kinetic energy of a block moving on a horizontal floor is 48 J. A constant frictional force acts on the block bringing it to rest over a distance of 2 m. What is the frictional force on the block?
\nA. 24 N
\nB. 48 N
\nC. 96 N
\nD. 192 N
\nA
\nThe diagram shows the gravitational field lines of planet X.
\nOutline how this diagram shows that the gravitational field strength of planet X decreases with distance from the surface.
\nThe diagram shows part of the surface of planet X. The gravitational potential at the surface of planet X is –3V and the gravitational potential at point Y is –V.
\nSketch on the grid the equipotential surface corresponding to a gravitational potential of –2V.
\nA meteorite, very far from planet X begins to fall to the surface with a negligibly small initial speed. The mass of planet X is 3.1 × 1021 kg and its radius is 1.2 × 106 m. The planet has no atmosphere. Calculate the speed at which the meteorite will hit the surface.
\nAt the instant of impact the meteorite which is made of ice has a temperature of 0 °C. Assume that all the kinetic energy at impact gets transferred into internal energy in the meteorite. Calculate the percentage of the meteorite’s mass that melts. The specific latent heat of fusion of ice is 3.3 × 105 J kg–1.
\nthe field lines/arrows are further apart at greater distances from the surface
\ncircle centred on Planet X
three units from Planet X centre
loss in gravitational potential =
\n«= 1.72 × 105 JKg−1»
\nequate to v2
\nv = 590 «m s−1»
\n\n
Allow ECF from MP1.
\navailable energy to melt one kg 1.72 × 105 «J»
\nfraction that melts is = 0.52 OR 52%
\n\n
Allow ECF from MP1.
\nAllow 53% from use of 590 ms-1.
\nA body undergoes one oscillation of simple harmonic motion (shm). What is correct for the direction of the acceleration of the body and the direction of its velocity?
\nA. Always opposite
B. Opposite for half a period
C. Opposite for a quarter of a period
D. Never opposite
B
\nThe efficiency of an electric motor is 20 %. When lifting a body 500 J of energy are wasted. What is the useful work done by the motor?
\nA. 100 J
\nB. 125 J
\nC. 250 J
\nD. 400 J
\nB
\nA photon of energy E and wavelength λ is scattered from an electron initially at rest.
\nWhat is the energy of the photon and the wavelength of the photon when the electron moves away?
\nD
\nTwo microwave transmitters, X and Y, are placed 12 cm apart and are connected to the same source. A single receiver is placed 54 cm away and moves along a line AB that is parallel to the line joining X and Y.
\nMaxima and minima of intensity are detected at several points along AB.
\n(i) Explain the formation of the intensity minima.
\n(ii) The distance between the central maximum and the first minimum is 7.2 cm. Calculate the wavelength of the microwaves.
\nRadio waves are emitted by a straight conducting rod antenna (aerial). The plane of polarization of these waves is parallel to the transmitting antenna.
\nAn identical antenna is used for reception. Suggest why the receiving antenna needs to be be parallel to the transmitting antenna.
\nThe receiving antenna becomes misaligned by 30° to its original position.
\nThe power of the received signal in this new position is 12 μW.
\n(i) Calculate the power that was received in the original position.
\n(ii) Calculate the minimum time between the wave leaving the transmitting antenna and its reception.
\ni
minima = destructive interference
Allow “crest meets trough”, but not “waves cancel”.
Allow “destructive superposition” but not bald “superposition”.
at minima waves meet 180° or π out of phase
Allow similar argument in terms of effective path difference of .
Allow “antiphase”, allow “completely out of phase”
Do not allow “out of phase” without angle. Do not allow unless qualified to odd integers but accept
ii
or or seen
Award [2] for a bald correct answer.
«» 3.2 «cm»
Award [1 max] for 1.6 «cm»
Award [2 max] to a trigonometric solution in which candidate works out individual path lengths and equates to .
ALTERNATIVE 1
\nthe component of the polarized signal in the direction of the receiving antenna
\nis a maximum «when both are parallel»
\nALTERNATIVE 2:
\nreceiving antenna must be parallel to plane of polarisation
for power/intensity to be maximum
Do not accept “receiving antenna must be parallel to transmitting antenna”
\nALTERNATIVE 3:
\nrefers to Malus’ law or I = I0 cos2θ
\nexplains that I is max when θ = 0
\nALTERNATIVE 4:
\nan electric current is established in the receiving antenna which is proportional to the electric field
\nmaximum current in receiving antenna requires maximum field «and so must be parallel»
\ni
or seen
Award [2] for bald correct answer.
Award [1 max] for MP1 if 9 x 10-6W is the final answer (I and I0 reversed).
Award [1 max] if cos not squared (14 μW).
1.6 × 10-5«W»
Units not required but if absent assume W.
ii
1.9 × 10–4 «s»
A particle oscillates with simple harmonic motion (shm) of period T. Which graph shows the variation with time of the kinetic energy of the particle?
\nD
\nThe equipment shown in the diagram was used by a student to investigate the variation with volume, of the pressure p of air, at constant temperature. The air was trapped in a tube of constant cross-sectional area above a column of oil.
\nThe pump forces oil to move up the tube decreasing the volume of the trapped air.
\nThe student measured the height H of the air column and the corresponding air pressure p. After each reduction in the volume the student waited for some time before measuring the pressure. Outline why this was necessary.
\nThe following graph of p versus was obtained. Error bars were negligibly small.
\nThe equation of the line of best fit is .
\nDetermine the value of b including an appropriate unit.
\nOutline how the results of this experiment are consistent with the ideal gas law at constant temperature.
\nThe cross-sectional area of the tube is 1.3 × 10–3m2 and the temperature of air is 300 K. Estimate the number of moles of air in the tube.
\nThe equation in (b) may be used to predict the pressure of the air at extremely large values of . Suggest why this will be an unreliable estimate of the pressure.
\nin order to keep the temperature constant
\nin order to allow the system to reach thermal equilibrium with the surroundings/OWTTE
\n\n
Accept answers in terms of pressure or volume changes only if clearly related to reaching thermal equilibrium with the surroundings.
\n[1 mark]
\nrecognizes b as gradient
\ncalculates b in range 4.7 × 104 to 5.3 × 104
\nPam
\n\n
Award [2 max] if POT error in b.
Allow any correct SI unit, eg kgs–2.
[3 marks]
\nthus ideal gas law gives
\nso graph should be «a straight line through origin,» as observed
\n[2 marks]
\nOR correct substitution of one point from the graph
\n\n
\n
Answer must be to 1 or 2 SF.
\nAllow ECF from (b).
\n[2 marks]
\nvery large means very small volumes / very high pressures
\nat very small volumes the ideal gas does not apply
OR
at very small volumes some of the assumptions of the kinetic theory of gases do not hold
[2 marks]
\nElectron capture can be represented by the equation
\np + e– → X + Y.
\nWhat are X and Y?
\nD
\nA net force acts on a body. Which characteristic of the body will definitely change?
\nA. Speed
\nB. Momentum
\nC. Kinetic energy
\nD. Direction of motion
\nB
\nA light ray is incident on an air–diamond boundary. The refractive index of diamond is greater than 1. Which diagram shows the correct path of the light ray?
\nA
\nA cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) generator to an electrical load. The copper wires are protected by an insulator.
\nThe cable consists of 32 copper wires each of length 35 km. Each wire has a resistance of 64 Ω. The cable is connected to the ac generator which has an output power of 110 MW when the peak potential difference is 150 kV. The resistivity of copper is 1.7 x 10–8 Ω m.
\noutput power = 110 MW
\n\n
To ensure that the power supply cannot be interrupted, two identical cables are connected in parallel.
\nThe energy output of the ac generator is at a much lower voltage than the 150 kV used for transmission. A step-up transformer is used between the generator and the cables.
\nCalculate the radius of each wire.
\nCalculate the peak current in the cable.
\nDetermine the power dissipated in the cable per unit length.
\nCalculate the root mean square (rms) current in each cable.
\nThe two cables in part (c) are suspended a constant distance apart. Explain how the magnetic forces acting between the cables vary during the course of one cycle of the alternating current (ac).
\nSuggest the advantage of using a step-up transformer in this way.
\nThe use of alternating current (ac) in a transformer gives rise to energy losses. State how eddy current loss is minimized in the transformer.
\narea = «= 9.3 x 10–6 m2»
\nradius = «» 0.00172 m
\nIpeak «» = 730 « A »
\nresistance of cable identified as «» 2 Ω
\nseen in solution
\nplausible answer calculated using «plausible if in range 10 W m–1 to 150 W m–1 when quoted answers in (b)(ii) used» 31 «W m–1»
\n\n
Allow [3] for a solution where the resistance per unit metre is calculated using resistivity and answer to (a) (resistance per unit length of cable = 5.7 x 10–5 m )
\nAward [2 max] if 64 Ω used for resistance (answer x32).
\nAn approach from or VI using 150 kV is incorrect (award [0]), however allow this approach if the pd across the cable has been calculated (pd dropped across cable is 1.47 kV).
\n«» = 260 «A»
\nwires/cable attract whenever current is in same direction
\ncharge flow/current direction in both wires is always same «but reverses every half cycle»
\nforce varies from 0 to maximum
\nforce is a maximum twice in each cycle
\n\n
Award [1 max] if response suggests that there is repulsion between cables at any stage in cycle.
\nhigher voltage gives lower current
\n«energy losses depend on current» hence thermal/heating/power losses reduced
\nlaminated core
\n\n
Do not allow “wires are laminated”.
\nA student is investigating a method to measure the mass of a wooden block by timing the period of its oscillations on a spring.
\nA 0.52 kg mass performs simple harmonic motion with a period of 0.86 s when attached to the spring. A wooden block attached to the same spring oscillates with a period of 0.74 s.
\nWith the block stationary a longitudinal wave is made to travel through the original spring from left to right. The diagram shows the variation with distance x of the displacement y of the coils of the spring at an instant of time.
\nA point on the graph has been labelled that represents a point P on the spring.
\nDescribe the conditions required for an object to perform simple harmonic motion (SHM).
\nCalculate the mass of the wooden block.
\nIn carrying out the experiment the student displaced the block horizontally by 4.8 cm from the equilibrium position. Determine the total energy in the oscillation of the wooden block.
\nA second identical spring is placed in parallel and the experiment in (b) is repeated. Suggest how this change affects the fractional uncertainty in the mass of the block.
\nState the direction of motion of P on the spring.
\nExplain whether P is at the centre of a compression or the centre of a rarefaction.
\nacceleration/restoring force is proportional to displacement
and in the opposite direction/directed towards equilibrium
ALTERNATIVE 1
\n\n
mass = 0.38 / 0.39 «kg»
\n\n
ALTERNATIVE 2
\n«use of T » k = 28 «Nm–1»
\n«use of T » m = 0.38 / 0.39 «kg»
\n\n
Allow ECF from MP1.
\nω = «» = 8.5 «rads–1»
\ntotal energy =
\n= 0.032 «J»
\n\n
Allow ECF from (b) and incorrect ω.
\nAllow answer using k from part (b).
\nspring constant/k/stiffness would increase
T would be smaller
fractional uncertainty in T would be greater, so fractional uncertainty of mass of block would be greater
left
\ncoils to the right of P move right and the coils to the left move left
\nhence P at centre of rarefaction
\n\n
Do not allow a bald statement of rarefaction or answers that don’t include reference to the movement of coils.
\nAllow ECF from MP1 if the movement of the coils imply a compression.
\n(i) Define gravitational field strength.
\n(ii) State the SI unit for gravitational field strength.
\nA planet orbits the Sun in a circular orbit with orbital period T and orbital radius R. The mass of the Sun is M.
\n(i) Show that .
\n(ii) The Earth’s orbit around the Sun is almost circular with radius 1.5×1011 m. Estimate the mass of the Sun.
\n(i) «gravitational» force per unit mass on a «small or test» mass
\n\n
(ii) N kg–1
\nAward mark if N kg-1 is seen, treating any further work as neutral.
Do not accept bald m s–2
i
clear evidence that v in is equated to orbital speed
OR
clear evidence that centripetal force is equated to gravitational force
OR
clear evidence that a in etc is equated to g in with consistent use of symbols
Minimum is a statement that is the orbital speed which is then used in
Minimum is Fc = Fg ignore any signs.
Minimum is g = a.
substitutes and re-arranges to obtain result
Allow any legitimate method not identified here.
Do not allow spurious methods involving equations of shm etc
\n
ii
«T = 365 × 24 × 60 × 60 = 3.15 × 107 s»
2×1030«kg»
Allow use of 3.16 x 107 s for year length (quoted elsewhere in paper).
Condone error in power of ten in MP1.
Award [1 max] if incorrect time used (24 h is sometimes seen, leading to 2.66 x 1035 kg).
Units are not required, but if not given assume kg and mark POT accordingly if power wrong.
Award [2] for a bald correct answer.
No sf penalty here.
A ball of mass 0.2 kg strikes a force sensor and sticks to it. Just before impact the ball is travelling horizontally at a speed of 4.0 m s–1. The graph shows the variation with time t of the force F recorded by the sensor.
\nWhat is Fmax?
\nA. 2 N
\nB. 4 N
\nC. 20 N
\nD. 40 N
\nD
\nThe graph shows how current I varies with potential difference V for a resistor R and a non-ohmic component T.
\n(i) State how the resistance of T varies with the current going through T.
\n(ii) Deduce, without a numerical calculation, whether R or T has the greater resistance at I=0.40 A.
\nComponents R and T are placed in a circuit. Both meters are ideal.
\nSlider Z of the potentiometer is moved from Y to X.
\n(i) State what happens to the magnitude of the current in the ammeter.
\n(ii) Estimate, with an explanation, the voltmeter reading when the ammeter reads 0.20 A.
\ni
\nRT decreases with increasing I
\nOR
\nRT and I are negatively correlated
\nMust see reference to direction of change of current in first alternative.
Do not allow “inverse proportionality”.
May be worth noting any marks on graph relating to 7bii
\n
ii
\nat 0.4 A: VR > VT or VR= 5.6 V and VT = 5.3 V
\nAward [0] for a bald correct answer without deduction or with incorrect reasoning.
\nIgnore any references to graph gradients.
\nso RR >RT because V = IR / V∝ R «and I same for both»
\nBoth elements must be present for MP2 to be awarded.
\ni
\ndecreases
OR
becomes zero at X
\n
ii
\nrealization that V is the same for R and T
OR
identifies that currents are 0.14 A and 0.06 A
Award [0] if pds 2.8 V and 3.7 V or 1.4 V and 2.6V are used in any way. Otherwise award [1 max] for a bald correct answer. Explanation expected.
\n2 V = 2 V OR 2.0 V
\nA spring XY lies on a frictionless table with the end Y free.
\nA horizontal pulse travels along the spring from X to Y. What happens when the pulse reaches Y?
\nA. The pulse will be reflected towards X and inverted.
B. The pulse will be reflected towards X and not be inverted.
C. Y will move and the pulse will disappear.
D. Y will not move and the pulse will disappear.
B
\nThe graph shows the variation with time t of the temperature T of two samples, X and Y. X and Y have the same mass and are initially in the solid phase. Thermal energy is being provided to X and Y at the same constant rate.
\nWhat is the correct comparison of the specific latent heats LX and LY and specific heat capacities in the liquid phase cX and cY of X and Y?
\nD
\nA mass m of ice at a temperature of –5 °C is changed into water at a temperature of 50 °C.
\nSpecific heat capacity of ice = ci
Specific heat capacity of water = cw
Specific latent heat of fusion of ice = L
Which expression gives the energy needed for this change to occur?
\nA. 55 m cw + m L
\nB. 55 m ci + 5 m L
\nC. 5 m ci + 50 m cw + m L
\nD. 5 m ci + 50 m cw + 5 m L
\nC
\nYellow light of photon energy 3.5 x 10–19 J is incident on the surface of a particular photocell.
\nThe photocell is connected to a cell as shown. The photoelectric current is at its maximum value (the saturation current).
\nRadiation with a greater photon energy than that in (b) is now incident on the photocell. The intensity of this radiation is the same as that in (b).
\nCalculate the wavelength of the light.
\nElectrons emitted from the surface of the photocell have almost no kinetic energy. Explain why this does not contradict the law of conservation of energy.
\nRadiation of photon energy 5.2 x 10–19 J is now incident on the photocell. Calculate the maximum velocity of the emitted electrons.
\nDescribe the change in the number of photons per second incident on the surface of the photocell.
\nState and explain the effect on the maximum photoelectric current as a result of increasing the photon energy in this way.
\nwavelength = «» 5.7 x 10–7 «m»
\n\n
If no unit assume m.
\n«potential» energy is required to leave surface
\nDo not allow reference to “binding energy”.
Ignore statements of conservation of energy.
all/most energy given to potential «so none left for kinetic energy»
energy surplus = 1.7 x 10–19 J
\nvmax = «m s–1»
\n\n
Award [1 max] if surplus of 5.2 x 10–19J used (answer: 1.1 x 106 m s–1)
\n\n
«same intensity of radiation so same total energy delivered per square metre per second»
\nlight has higher photon energy so fewer photons incident per second
\n\n
Reason is required
\n1:1 correspondence between photon and electron
\nso fewer electrons per second
\ncurrent smaller
\n\n
Allow ECF from (c)(i)
Allow ECF from MP2 to MP3.
A student stands a distance L from a wall and claps her hands. Immediately on hearing the reflection from the wall she claps her hands again. She continues to do this, so that successive claps and the sound of reflected claps coincide. The frequency at which she claps her hands is f. What is the speed of sound in air?
\nA.
\nB.
\nC. Lf
\nD. 2Lf
\nD
\nA sealed container contains a mixture of oxygen and nitrogen gas.
The ratio is .
The ratio is
\nA. 1.
\nB. .
\nC. .
\nD. dependent on the concentration of each gas.
\nA
\nIn simple harmonic oscillations which two quantities always have opposite directions?
\nA. Kinetic energy and potential energy
\nB. Velocity and acceleration
\nC. Velocity and displacement
\nD. Acceleration and displacement
\nD
\nIn a simple pendulum experiment, a student measures the period T of the pendulum many times and obtains an average value T = (2.540 ± 0.005) s. The length L of the pendulum is measured to be L = (1.60 ± 0.01) m.
\nCalculate, using , the value of the acceleration of free fall, including its uncertainty. State the value of the uncertainty to one significant figure.
\nIn a different experiment a student investigates the dependence of the period T of a simple pendulum on the amplitude of oscillations θ. The graph shows the variation of with θ, where T0 is the period for small amplitude oscillations.
\nThe period may be considered to be independent of the amplitude θ as long as . Determine the maximum value of θ for which the period is independent of the amplitude.
\n\n
«» 0.0997
\nOR
\n1.0%
\nhence g = (9.8 ± 0.1) «ms−2» OR Δg = 0.1 «ms−2»
\n\n
For the first marking point answer must be given to at least 2 dp.
Accept calculations based on
\n
\n
\n
[3 marks]
\n\n
θmax = 22 «º»
\n\n
Accept answer from interval 20 to 24.
\n[2 marks]
\nA –5µC charge and a +10µC charge are a fixed distance apart.
\nWhere can the electric field be zero?
\nA. position I only
B. position II only
C. position III only
D. positions I, II and III
C
\nThe following data are available for a natural gas power station that has a high efficiency.
\n| Rate of consumption of natural gas | \n= 14.6 kg s–1 | \n
| Specific energy of natural gas | \n= 55.5 MJ kg–1 | \n
| Efficiency of electrical power generation | \n= 59.0 % | \n
| Mass of CO2 generated per kg of natural gas | \n= 2.75 kg | \n
| One year | \n= 3.16 × 107 s | \n
Calculate, with a suitable unit, the electrical power output of the power station.
\nCalculate the mass of CO2 generated in a year assuming the power station operates continuously.
\nExplain, using your answer to (b), why countries are being asked to decrease their dependence on fossil fuels.
\nDescribe, in terms of energy transfers, how thermal energy of the burning gas becomes electrical energy.
\n«55.5 × 14.6 × 0.59» = 4.78 × 108 W
\nA unit is required for this mark. Allow use of J s–1.
\nNo sf penalty.
\n«14.6 × 2.75 × 3.16 × 107 =» 1.27 × 109 «kg»
\nIf no unit assume kg
\nCO2 linked to greenhouse gas OR greenhouse effect
\nleading to «enhanced» global warming
OR
climate change
OR
other reasonable climatic effect
Internal energy of steam/particles OR KE of steam/particles
\n«transfers to» KE of turbine
\n«transfers to» KE of generator or dynamo «producing electrical energy»
\nDo not award mark for first and last energies as they are given in the question.
Do not allow “gas” for “steam”
\nDo not accept reference to moving OR turning generator
\nA girl in a stationary boat observes that 10 wave crests pass the boat every minute. What is the period of the water waves?
\nA. min
\nB. min–1
\nC. 10 min
\nD. 10 min–1
\nA
\nAn electrical circuit is shown with loop X and junction Y.
\nWhat is the correct expression of Kirchhoff’s circuit laws for loop X and junction Y?
\nA
\nThe graph shows the variation with distance x of the displacement of the particles of a medium in which a longitudinal wave is travelling from left to right. Displacements to the right of equilibrium positions are positive.
\nWhich point is at the centre of a compression?
\nA. x = 0
\nB. x = 1 m
\nC. x = 2 m
\nD. x = 3 m
\nB
\nA long current-carrying wire is at rest in the reference frame S of the laboratory. A positively charged particle P outside the wire moves with velocity v relative to S. The electrons making up the current in the wire move with the same velocity v relative to S.
\nState what is meant by a reference frame.
\nState and explain whether the force experienced by P is magnetic, electric or both, in reference frame S.
\nState and explain whether the force experienced by P is magnetic, electric or both, in the rest frame of P.
\na set of coordinate axes and clocks used to measure the position «in space/time of an object at a particular time»
OR
a coordinate system to measure x,y,z, and t / OWTTE
[1 mark]
\nmagnetic only
\nthere is a current but no «net» charge «in the wire»
\n[2 marks]
\nelectric only
\nP is stationary so experiences no magnetic force
\nrelativistic contraction will increase the density of protons in the wire
\n[3 marks]
\nA beam of unpolarized light is incident on the first of two parallel polarizers. The transmission axes of the two polarizers are initially parallel.
\nThe first polarizer is now rotated about the direction of the incident beam by an angle smaller than 90°. Which gives the changes, if any, in the intensity and polarization of the transmitted light?
\nA
\nThe frequency of the first harmonic standing wave in a pipe that is open at both ends is 200 Hz. What is the frequency of the first harmonic in a pipe of the same length that is open at one end and closed at the other?
\nA. 50 Hz
\nB. 75 Hz
\nC. 100 Hz
\nD. 400 Hz
\nC
\nA cell of emf 4V and negligible internal resistance is connected to three resistors as shown. Two resistors of resistance 2Ω are connected in parallel and are in series with a resistor of resistance 1Ω.
\nWhat power is dissipated in one of the 2Ω resistors and in the whole circuit?
\nD
\nThe gravitational potential due to the Sun at its surface is –1.9 x 1011 J kg–1. The following data are available.
\n| Mass of Earth | \n= 6.0 x 1024 kg | \n
| Distance from Earth to Sun | \n= 1.5 x 1011 m | \n
| Radius of Sun | \n= 7.0 x 108 m | \n
Outline why the gravitational potential is negative.
\nThe gravitational potential due to the Sun at a distance r from its centre is VS. Show that
\nrVS = constant.
\nCalculate the gravitational potential energy of the Earth in its orbit around the Sun. Give your answer to an appropriate number of significant figures.
\nCalculate the total energy of the Earth in its orbit.
\nAn asteroid strikes the Earth and causes the orbital speed of the Earth to suddenly decrease. Suggest the ways in which the orbit of the Earth will change.
\nOutline, in terms of the force acting on it, why the Earth remains in a circular orbit around the Sun.
\npotential is defined to be zero at infinity
\nso a positive amount of work needs to be supplied for a mass to reach infinity
\nVS = so r x VS «= –GM» = constant because G and M are constants
\nGM = 1.33 x 1020 «J m kg–1»
\nGPE at Earth orbit «= –» = «–» 5.3 x 1033 «J»
\n\n
Award [1 max] unless answer is to 2 sf.
\nIgnore addition of Sun radius to radius of Earth orbit.
\nALTERNATIVE 1
work leading to statement that kinetic energy , AND kinetic energy evaluated to be «+» 2.7 x 1033 «J»
energy «= PE + KE = answer to (b)(ii) + 2.7 x 1033» = «–» 2.7 x 1033 «J»
\n\n
ALTERNATIVE 2
statement that kinetic energy is gravitational potential energy in orbit
so energy «» = «–» 2.7 x 1033 «J»
\n\n
Various approaches possible.
\n«KE will initially decrease so» total energy decreases
OR
«KE will initially decrease so» total energy becomes more negative
Earth moves closer to Sun
\nnew orbit with greater speed «but lower total energy»
\nchanges ellipticity of orbit
\ncentripetal force is required
\nand is provided by gravitational force between Earth and Sun
\n\n
Award [1 max] for statement that there is a “centripetal force of gravity” without further qualification.
\nThe diagram shows two equal and opposite charges that are fixed in place.
\nAt which points is the net electric field directed to the right?
\nA. X and Y only
\nB. Z and Y only
\nC. X and Z only
\nD. X, Y and Z
\nC
\nA wire carrying a current is at right angles to a uniform magnetic field of strength B. A magnetic force F is exerted on the wire. Which force acts when the same wire is placed at right angles to a uniform magnetic field of strength 2B when the current is ?
\nA.
\nB.
\nC. F
D. 2F
\nB
\nA wire has variable cross-sectional area. The cross-sectional area at Y is double that at X.
\nAt X, the current in the wire is I and the electron drift speed is v. What is the current and the electron drift speed at Y?
\nB
\nThe following data are available for a natural gas power station that has a high efficiency.
\n| Rate of consumption of natural gas | \n= 14.6 kg s–1 | \n
| Specific energy of natural gas | \n= 55.5 MJ kg–1 | \n
| Efficiency of electrical power generation | \n= 59.0 % | \n
| Mass of CO2 generated per kg of natural gas | \n= 2.75 kg | \n
| One year | \n= 3.16 × 107 s | \n
Electrical power output is produced by several alternating current (ac) generators which use transformers to deliver energy to the national electricity grid.
\nThe following data are available. Root mean square (rms) values are given.
\n\n\n\nac generator output voltage to a transformer\n= 25 kV\n\n\nac generator output current to a transformer\n= 3.9 kA\n\n\nTransformer output voltage to the grid\n= 330 kV\n\n\nTransformer efficiency\n= 96%\n\n\n\n\n
(i) Calculate the current output by the transformer to the grid. Give your answer to an appropriate number of significant figures.
\n(ii) Electrical energy is often delivered across large distances at 330 kV. Identify the main advantage of using this very high potential difference.
\nIn an alternating current (ac) generator, a square coil ABCD rotates in a magnetic field.
\nThe ends of the coil are connected to slip rings and brushes. The plane of the coil is shown at the instant when it is parallel to the magnetic field. Only one coil is shown for clarity.
\nThe following data are available.
\n\n\n\nDimensions of the coil\n= 8.5 cm×8.5 cm\n\n\nNumber of turns on the coil\n= 80\n\n\nSpeed of edge AB\n= 2.0 ms–1\n\n\nUniform magnetic field strength\n= 0.34 T\n\n\n\n\n
(i) Explain, with reference to the diagram, how the rotation of the generator produces an electromotive force (emf ) between the brushes.
\n(ii) Calculate, for the position in the diagram, the magnitude of the instantaneous emf generated by a single wire between A and B of the coil.
\n(iii) Hence, calculate the total instantaneous peak emf between the brushes.
\ni
Award [2] for a bald correct answer to 2 sf.
Award [1 max] for correct sf if efficiency used in denominator leading to 310 A or if efficiency ignored (e=1) leading to 300 A (from 295 A but 295 would lose both marks).
=280 «A»
Must show two significant figures to gain MP2.
\n
ii
higher V means lower I «for same power»
thermal energy loss depends on I or is ∝I2 or is I2R so thermal energy loss will be less
Accept “heat” or “heat energy” or “Joule heating” for “thermal energy”.
Reference to energy/power dissipation is not enough.
i
\n«long» sides of coil AB/CD cut lines of flux
OR
flux «linkage» in coil is changed
«Faradays law:» induced emf depends on rate of change of flux linked
OR
rate at which lines are cut
“Induced” is required
Allow OWTTE or defined symbols if “induced emf” is given.
Accept “induced” if mentioned at any stage in the context of emf or accept the term “motional emf”.
Award [2 max] if there is no mention of “induced emf”.
emfs acting in sides AB/CD add / act in same direction around coil
\nprocess produces an alternating/sinusoidal emf
\n\n
ii
\nBlv = 0.34×8.5×10–2×2 = 0.058 «V»
\nAccept 0.06V.
\n\n
iii
\n160×(c)(ii) = 9.2 or 9.3 or 9.6 «V»
\nAllow ECF from (c)(ii)
If 80 turns used in cii, give full credit for cii x 2 here.
Curling is a game played on a horizontal ice surface. A player pushes a large smooth stone across the ice for several seconds and then releases it. The stone moves until friction brings it to rest. The graph shows the variation of speed of the stone with time.
\nThe total distance travelled by the stone in 17.5 s is 29.8 m.
\nDetermine the coefficient of dynamic friction between the stone and the ice during the last 14.0 s of the stone’s motion.
\nThe diagram shows the stone during its motion after release.
\nLabel the diagram to show the forces acting on the stone. Your answer should include the name, the direction and point of application of each force.
\nALTERNATIVE 1
\n«deceleration» «»
\nF = 0.243 × m
\n\n
ALTERNATIVE 2
\ndistance travelled after release = 23.85 «m»
KE lost = 5.81m «J»
\n
Award [3] for a bald correct answer.
\nIgnore sign in acceleration.
\nAllow ECF from (a) (note that x candidate answer to (a) ).
\nIgnore any units in answer.
\nCondone omission of m in solution.
\nAllow g = 10 N kg–1 (gives 0.024).
\nnormal force, upwards, ignore point of application
\nForce must be labeled for its mark to be awarded. Blob at poa not required.
Allow OWTTE for normal force. Allow N, R, reaction.
The vertical forces must lie within the middle third of the stone
weight/weight force/force of gravity, downwards, ignore point of application
\nAllow mg, W but not “gravity”.
\nPenalise gross deviations from vertical/horizontal once only
\nfriction/resistive force, to left, at bottom of stone, point of application must be on the interface between ice and stone
\nAllow F, μR. Only allow arrows/lines that lie on the interface. Take the tail of the arrow as the definitive point of application and expect line to be drawn horizontal.
Award [2 max] if any force arrow does not touch the stone
\nDo not award MP3 if a “driving force” is shown acting to the right. This need not be labelled to disqualify the mark. Treat arrows labelled “air resistance” as neutral.
\n\n
N.B: Diagram in MS is drawn with the vertical forces not direction of travel collinear for clarity
\nA circuit contains a cell of electromotive force (emf) 9.0 V and internal resistance 1.0 Ω together with a resistor of resistance 4.0 Ω as shown. The ammeter is ideal. XY is a connecting wire.
\nWhat is the reading of the ammeter?
\nA. 0 A
\nB. 1.8 A
\nC. 9.0 A
\nD. 11 A
\nC
\nAn observer P sitting in a train moving at a speed v measures that his journey takes a time ΔtP. An observer Q at rest with respect to the ground measures that the journey takes a time ΔtQ.
\nAccording to Q there is an instant at which the train is completely within the tunnel.
\nAt that instant two lights at the front and the back of the train are turned on simultaneously according to Q.
\nThe spacetime diagram according to observer Q shows event B (back light turns on) and event F (front light turns on).
\n![\"M17/4/PHYSI/SP3/ENG/TZ1/4d_02\"](\"../assets/uploads/tinymce_asset/asset/3734/Schermafbeelding_2017-09-26_om_14.55.20.png\"/)
State which of the two time intervals is a proper time.
\nCalculate the speed v of the train for the ratio .
\nLater the train is travelling at a speed of 0.60c. Observer P measures the length of the train to be 125 m. The train enters a tunnel of length 100 m according to observer Q.
\nShow that the length of the train according to observer Q is 100 m.
\nDraw the time and space axes for observer P’s reference frame on the spacetime diagram.
\nDeduce, using the spacetime diagram, which light was turned on first according to observer P.
\nApply a Lorentz transformation to show that the time difference between events B and F according to observer P is 2.5 × 10–7 s.
\nDemonstrate that the spacetime interval between events B and F is invariant.
\nA second train is moving at a velocity of –0.70c with respect to the ground.
\nCalculate the speed of the second train relative to observer P.
\n\n
ΔtP / observer sitting in the train
\n[1 mark]
\n«» = 3.3
\nto give v = 0.95c
\n[2 marks]
\n= 1.25
\n«length of train according Q» = 125/1.25
\n«giving 100m»
\n[2 marks]
\naxes drawn with correct gradients of for and 0.6 for
\n\n
Award [1] for one gradient correct and another approximately correct.
\n[1 mark]
\nlines parallel to the axis and passing through B and F
\nintersections on the axis at and shown
\nlight at the front of the train must have been turned on first
\n[3 marks]
\n\n
«2.5 × 10−7»
\n\n
Allow ECF for gamma from (c).
\n[1 mark]
\naccording to P: «−» 10000
\naccording to Q: «−» 10000
\n[2 marks]
\nc
\n= «−» 0.92c
\n[2 marks]
\nAn object at the end of a wooden rod rotates in a vertical circle at a constant angular velocity. What is correct about the tension in the rod?
\nA. It is greatest when the object is at the bottom of the circle.
B. It is greatest when the object is halfway up the circle.
C. It is greatest when the object is at the top of the circle.
D. It is unchanged throughout the motion.
A
\nA positively-charged particle moves parallel to a wire that carries a current upwards.
\nWhat is the direction of the magnetic force on the particle?
\nA. To the left
\nB. To the right
\nC. Into the page
\nD. Out of the page
\nA
\nA glider is an aircraft with no engine. To be launched, a glider is uniformly accelerated from rest by a cable pulled by a motor that exerts a horizontal force on the glider throughout the launch.
\n\n
The glider reaches its launch speed of 27.0 m s–1 after accelerating for 11.0 s. Assume that the glider moves horizontally until it leaves the ground. Calculate the total distance travelled by the glider before it leaves the ground.
\nThe glider and pilot have a total mass of 492 kg. During the acceleration the glider is subject to an average resistive force of 160 N. Determine the average tension in the cable as the glider accelerates.
\nThe cable is pulled by an electric motor. The motor has an overall efficiency of 23 %. Determine the average power input to the motor.
\nThe cable is wound onto a cylinder of diameter 1.2 m. Calculate the angular velocity of the cylinder at the instant when the glider has a speed of 27 m s–1. Include an appropriate unit for your answer.
\nAfter takeoff the cable is released and the unpowered glider moves horizontally at constant speed. The wings of the glider provide a lift force. The diagram shows the lift force acting on the glider and the direction of motion of the glider.
\nDraw the forces acting on the glider to complete the free-body diagram. The dotted lines show the horizontal and vertical directions.
\nExplain, using appropriate laws of motion, how the forces acting on the glider maintain it in level flight.
\nAt a particular instant in the flight the glider is losing 1.00 m of vertical height for every 6.00 m that it goes forward horizontally. At this instant, the horizontal speed of the glider is 12.5 m s–1. Calculate the velocity of the glider. Give your answer to an appropriate number of significant figures.
\ncorrect use of kinematic equation/equations
\n148.5 or 149 or 150 «m»
\n\n
Substitution(s) must be correct.
\na = or 2.45 «m s–2»
\nF – 160 = 492 × 2.45
\n1370 «N»
\n\n
Could be seen in part (a).
Award [0] for solution that uses a = 9.81 m s–2
ALTERNATIVE 1
\n«work done to launch glider» = 1370 x 149 «= 204 kJ»
\n«work done by motor»
\n«power input to motor» or 80.4 or 81 k«W»
\n\n
ALTERNATIVE 2
\nuse of average speed 13.5 m s–1
\n«useful power output» = force x average speed «= 1370 x 13.5»
\npower input = «» 80 or 80.4 or 81 k«W»
\n\n
ALTERNATIVE 3
\nwork required from motor = KE + work done against friction «» = 204 «kJ»
\n«energy input»
\npower input k«W»
\n\n
Award [2 max] for an answer of 160 k«W».
\n«»
\nrad s–1
\n\n
Do not accept Hz.
Award [1 max] if unit is missing.
drag correctly labelled and in correct direction
\nweight correctly labelled and in correct direction AND no other incorrect force shown
\n\n
Award [1 max] if forces do not touch the dot, but are otherwise OK.
\nname Newton's first law
\nvertical/all forces are in equilibrium/balanced/add to zero
OR
vertical component of lift mentioned
as equal to weight
\nany speed and any direction quoted together as the answer
\nquotes their answer(s) to 3 significant figures
\nspeed = 12.7 m s–1 or direction = 9.46º or 0.165 rad «below the horizontal» or gradient of
\nOn Mars, the gravitational field strength is about of that on Earth. The mass of Earth is approximately ten times that of Mars.
\nWhat is ?
\nA. 0.4
\nB. 0.6
\nC. 1.6
\nD. 2.5
\nC
\n0.46 mole of an ideal monatomic gas is trapped in a cylinder. The gas has a volume of 21 m3 and a pressure of 1.4 Pa.
\n(i) State how the internal energy of an ideal gas differs from that of a real gas.
\n(ii) Determine, in kelvin, the temperature of the gas in the cylinder.
\n(iii) The kinetic theory of ideal gases is one example of a scientific model. Identify two reasons why scientists find such models useful.
\ni
«intermolecular» potential energy/PE of an ideal gas is zero/negligible
ii
THIS IS FOR USE WITH AN ENGLISH SCRIPT ONLY
use of or
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Award [2] for a bald correct answer in K.
Award [2 max] if correct 7.7 K seen followed by –265°C and mark BOD. However, if only –265°C seen, award [1 max].
7.7K
Do not penalise use of “°K”
ii
THIS IS FOR USE WITH A SPANISH SCRIPT ONLY
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Uses correct unit conversion for volume
T = 7.7×10-6K
Award [2] for a bald correct answer in K. Finds solution. Allow an ECF from MP2 if unit not converted, ie candidate uses 21m3 and obtains 7.7 K
Do not penalise use of “°K”
\n
iii
«models used to»
predict/hypothesize / lead to further theories
Response needs to identify two different reasons. (N.B. only one in SL).
explain / help with understanding / help to visualize
Do not allow any response that is gas specific. The question is couched in general, nature of science terms and must be answered as such.
simulate
simplify/approximate
Photons of energy 2.3eV are incident on a low-pressure vapour. The energy levels of the atoms in the vapour are shown
\nWhat energy transition will occur when a photon is absorbed by the vapour?
\nA. –3.9eV to –1.6eV
\nB. –1.6eV to 0eV
\nC. –1.6eV to –3.9eV
\nD. 0eV to –1.6eV
\nA
\nA particular K meson has a quark structure s. State the charge, strangeness and baryon number for this meson.
\nThe Feynman diagram shows the changes that occur during beta minus (β–) decay.
\nLabel the diagram by inserting the four missing particle symbols and the direction of the arrows for the decay particles.
\nC-14 decay is used to estimate the age of an old dead tree. The activity of C-14 in the dead tree is determined to have fallen to 21% of its original value. C-14 has a half-life of 5700 years.
\n(i) Explain why the activity of C-14 in the dead tree decreases with time.
\n(ii) Calculate, in years, the age of the dead tree. Give your answer to an appropriate number of significant figures.
\ncharge: –1«e» or negative or K−
\nstrangeness: –1
\nbaryon number: 0
\nNegative signs required.
Award [2] for three correct answers, [1 max] for two correct answer and [0] for one correct answer.
correct symbols for both missing quarks
\nexchange particle and electron labelled W or W– and e or e–
\nDo not allow W+ or e+ or β+. Allow β or β–.
\narrows for both electron and anti-neutrino correct
\nAllow ECF from previous marking point.
\ni
number of C-14 atoms/nuclei are decreasing
OR
decreasing activity proportional to number of C-14 atoms/nuclei
OR
A = A0e–λt so A decreases as t increases
Do not allow “particles”
Must see reference to atoms or nuclei or an equation, just “C-14 is decreasing” is not enough.
ii
0.21 = (0.5)n
OR
n = 2.252 half-lives or t =1 2834 «y»
Early rounding to 2.25 gives 12825 y
13000 y rounded correctly to two significant figures:
Both needed; answer must be in year for MP3.
Allow ECF from MP2.
Award [3] for a bald correct answer.
Two satellites of mass m and 2m orbit a planet at the same orbit radius. If F is the force exerted on the satellite of mass m by the planet and a is the centripetal acceleration of this satellite, what is the force and acceleration of the satellite with mass 2m?
\nA
\nWhen an alpha particle collides with a nucleus of nitrogen-14 , a nucleus X can be produced together with a proton. What is X?
\nA.
\nB.
\nC.
\nD.
\nB
\nAn apparatus is used to investigate the photoelectric effect. A caesium cathode C is illuminated by a variable light source. A variable power supply is connected between C and the collecting anode A. The photoelectric current I is measured using an ammeter.
\nA current is observed on the ammeter when violet light illuminates C. With V held constant the current becomes zero when the violet light is replaced by red light of the same intensity. Explain this observation.
\nThe graph shows the variation of photoelectric current I with potential difference V between C and A when violet light of a particular intensity is used.
\nThe intensity of the light source is increased without changing its wavelength.
\n(i) Draw, on the axes, a graph to show the variation of I with V for the increased intensity.
\n(ii) The wavelength of the violet light is 400 nm. Determine, in eV, the work function of caesium.
\n(iii) V is adjusted to +2.50V. Calculate the maximum kinetic energy of the photoelectrons just before they reach A.
\nreference to photon
OR
energy = hf or =
violet photons have greater energy than red photons
when hf > Φ or photon energy> work function then electrons are ejected
\nfrequency of red light < threshold frequency «so no emission»
OR
energy of red light/photon < work function «so no emission»
i
line with same negative intercept «–1.15V»
otherwise above existing line everywhere and of similar shape with clear plateau
Award this marking point even if intercept is wrong.
\n\n
ii
«» 3.11 «eV»
Intermediate answer is 4.97×10−19 J.
\nAccept approach using f rather than c/λ
«3.10 − 1.15 =» 1.96 «eV»
Award [2] for a bald correct answer in eV.
Award [1 max] if correct answer is given in J (3.12×10−19 J).
\n
iii
\n«KE = qVs =» 1.15 «eV»
\nOR
\n1.84 x 10−19 «J»
\nAllow ECF from MP1 to MP2.
\nadds 2.50 eV = 3.65 eV
\nOR
\n5.84 x 10−19 J
\nMust see units in this question to identify energy unit used.
Award [2] for a bald correct answer that includes units.
Award [1 max] for correct answer without units.
Outline, with reference to energy changes, the operation of a pumped storage hydroelectric system.
\nThe hydroelectric system has four 250 MW generators. The specific energy available from the water is 2.7 kJ kg–1. Determine the maximum time for which the hydroelectric system can maintain full output when a mass of 1.5 x 1010 kg of water passes through the turbines.
\nNot all the stored energy can be retrieved because of energy losses in the system. Explain one such loss.
\nAt the location of the hydroelectric system, an average intensity of 180 W m–2 arrives at the Earth’s surface from the Sun. Solar photovoltaic (PV) cells convert this solar energy with an efficiency of 22 %. The solar cells are to be arranged in a square array. Determine the length of one side of the array that would be required to replace the
hydroelectric system.
PE of water is converted to KE of moving water/turbine to electrical energy «in generator/turbine/dynamo»
\nidea of pumped storage, ie: pump water back during night/when energy cheap to buy/when energy not in demand/when there is a surplus of energy
\ntotal energy = «2.7 x 103 x 1.5 x 1010 =» 4.05 x 1013 «J»
\ntime = «» 11.1h or 4.0 x 104 s
\n\n
For MP2 the unit must be present.
\nfriction/resistive losses in walls of pipe/air resistance/turbulence/turbine and generator bearings
\nthermal energy losses, in electrical resistance of components
\nwater requires kinetic energy to leave system so not all can be transferred
\n\n
Must see “seat of friction” to award the mark.
Do not allow “friction” bald.
area required «= 2.5 x 107 m2»
\nlength of one side k«m»
\nThe gravitational field strength at the surface of Earth is g. Another planet has double the radius of Earth and the same density as Earth. What is the gravitational field strength at the surface of this planet?
\nA.
\nB.
\nC. 2g
\nD. 4g
\nC
\nA boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?
\nA. 5×100 kg m s–1
\nB. 5×101 kg m s–1
\nC. 5×102 kg m s–1
\nD. 5×103 kg m s–1
\nC
\nA student measures the refractive index of water by shining a light ray into a transparent container.
\nIO shows the direction of the normal at the point where the light is incident on the container. IX shows the direction of the light ray when the container is empty. IY shows the direction of the deviated light ray when the container is filled with water.
\nThe angle of incidence θ is varied and the student determines the position of O, X and Y for each angle of incidence.
\nThe table shows the data collected by the student. The uncertainty in each measurement of length is ±0.1 cm.
\n(i) Outline why OY has a greater percentage uncertainty than OX for each pair of data points.
\n(ii) The refractive index of the water is given by when OX is small.
\nCalculate the fractional uncertainty in the value of the refractive index of water for OX = 1.8 cm.
\nA graph of the variation of OY with OX is plotted.
(i) Draw, on the graph, the error bars for OY when OX = 1.8 cm and when OY = 5.8 cm.
\n(ii) Determine, using the graph, the refractive index of the water in the container for values of OX less than 6.0 cm.
\n(iii) The refractive index for a material is also given by where i is the angle of incidence and r is the angle of refraction.
\nOutline why the graph deviates from a straight line for large values of OX.
\ni
OY always smaller than OX AND uncertainties are the same/0.1
« so fraction »
ii
AND
= 0.13 OR 13%
Watch for correct answer even if calculation continues to the absolute uncertainty.
\ni
\ntotal length of bar = 0.2 cm
\nAccept correct error bar in one of the points: OX= 1.8 cm OR OY= 5.8 cm (which is not a measured point but is a point on the interpolated line) OR OX= 5.8 cm.
Ignore error bar of OX.
Allow range from 0.2 to 0.3 cm, by eye.
\n
ii
\nsuitable line drawn extending at least up to 6 cm
OR
gradient calculated using two out of the first three data points
inverse of slope used
\n\n
value between 1.30 and 1.60
\nIf using one value of OX and OY from the graph for any of the first three data points award [2 max].
Award [3] for correct value for each of the three data points and average.
If gradient used, award [1 max].
\n
iii
\n«the equation n=» involves a tan approximation/is true only for small θ «when sinθ = tanθ»
OR
«the equation n=» uses OI instead of the hypotenuse of the ∆IOX or IOY
OWTTE
\n\n
The mass defect for deuterium is 4×10–30 kg. What is the binding energy of deuterium?
\nA. 4×10–7 eV
\nB. 8×10–2 eV
\nC. 2×106 eV
\nD. 2×1012 eV
\nC
\nAs quarks separate from each other within a hadron, the interaction between them becomes larger. What is the nature of this interaction?
\nA. Electrostatic
B. Gravitational
C. Strong nuclear
D. Weak nuclear
C
\nLight of wavelength 400nm is incident on two slits separated by 1000µm. The interference pattern from the slits is observed from a satellite orbiting 0.4Mm above the Earth. The distance between interference maxima as detected at the satellite is
\nA. 0.16Mm.
B. 0.16km.
C. 0.16m.
D. 0.16mm.
B
\nA car moves north at a constant speed of 3m s–1 for 20s and then east at a constant speed of 4m s–1 for 20s. What is the average speed of the car during this motion?
\nA. 7.0m s–1
B. 5.0m s–1
C. 3.5m s–1
D. 2.5m s–1
C
\nAtomic spectra are caused when a certain particle makes transitions between energy levels.
What is this particle?
A. Electron
\nB. Proton
\nC. Neutron
\nD. Alpha particle
\nA
\nThe Sankey diagram represents the energy flow for a coal-fired power station.
\nWhat is the overall efficiency of the power station?
\nA. 0.3
\nB. 0.4
\nC. 0.6
\nD. 0.7
\nA
\nPolice use radar to detect speeding cars. A police officer stands at the side of the road and points a radar device at an approaching car. The device emits microwaves which reflect off the car and return to the device. A change in frequency between the emitted and received microwaves is measured at the radar device.
\nThe frequency change Δf is given by
\n\n
where f is the transmitter frequency, v is the speed of the car and c is the wave speed.
\nThe following data are available.
\n\n\n\nTransmitter frequency f\n= 40 GHz\n\n\nΔf\n= 9.5 kHz\n\n\nMaximum speed allowed\n= 28 m s–1\n\n
(i) Explain the reason for the frequency change.
\n(ii) Suggest why there is a factor of 2 in the frequency-change equation.
\n(iii) Determine whether the speed of the car is below the maximum speed allowed.
\nAirports use radar to track the position of aircraft. The waves are reflected from the aircraft and detected by a large circular receiver. The receiver must be able to resolve the radar images of two aircraft flying close to each other.
\nThe following data are available.
\n\n\n\nDiameter of circular radar receiver\n= 9.3 m\n\n\nWavelength of radar\n= 2.5 cm\n\n\nDistance of two aircraft from the airport\n= 31 km\n\n\n\n\n
Calculate the minimum distance between the two aircraft when their images can just be resolved.
\ni
mention of Doppler effect
OR
«relative» motion between source and observer produces frequency/wavelength change
Accept answers which refer to a double frequency shift.
Award [0] if there is any suggestion that the wave speed is changed in the process.
the reflected waves come from an approaching “source”
OR
the incident waves strike an approaching “observer”
increased frequency received «by the device or by the car»
\n
ii
the car is a moving “observer” and then a moving “source”, so the Doppler effect occurs twice
OR
the reflected radar appears to come from a “virtual image” of the device travelling at 2 v towards the device
\n
iii
ALTERNATIVE 1
For both alternatives, allow ecf to conclusion if v OR Δf are incorrectly calculated.
v = «» 36 «ms–1»
\n«36> 28» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.
ALTERNATIVE 2
reverse argument using speed limit.
«» 7500 «Hz»
\n« 9500> 7500» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.
Award [2] for a bald correct answer.
Award [1 max] for POT error.
100 «m»
Award [1 max] for 83m (omits 1.22).
A radio wave of wavelength is incident on a conductor. The graph shows the variation with wavelength of the maximum distance d travelled inside the conductor.
\nFor = 5.0 x 105 m, calculate the
\nThe graph shows the variation with wavelength of d 2. Error bars are not shown and the line of best-fit has been drawn.
\nA student states that the equation of the line of best-fit is d 2 = a + b. When d 2 and are expressed in terms of fundamental SI units, the student finds that a = 0.040 x 10–4 and b = 1.8 x 10–11.
\nSuggest why it is unlikely that the relation between d and is linear.
\nfractional uncertainty in d.
\npercentage uncertainty in d 2.
\nState the fundamental SI unit of the constant a and of the constant b.
\nDetermine the distance travelled inside the conductor by very high frequency electromagnetic waves.
\nit is not possible to draw a straight line through all the error bars
OR
the line of best-fit is curved/not a straight line
\n
Treat as neutral any reference to the origin.
\nAllow “linear” for “straight line”.
\n[1 mark]
\nd = 0.35 ± 0.01 AND Δd = 0.05 ± 0.01 «cm»
\n«» = 0.14
\nOR
\nor 14% or 0.1
\n\n
Allow final answers in the range of 0.11 to 0.18.
\nAllow [1 max] for 0.03 to 0.04 if = 5 × 106 m is used.
\n[2 marks]
\n28 to 30%
\n\n
Allow ECF from (b)(i), but only accept answer as a %
\n[1 mark]
\na: m2
\nb: m
\n\n
Allow answers in words
\n[2 marks]
\nALTERNATIVE 1 – if graph on page 4 is used
\nd 2 = 0.040 x 10–4 «m2»
\nd = 0.20 x 10–2 «m»
\nALTERNATIVE 2 – if graph on page 2 is used
\nany evidence that d intercept has been determined
\nd = 0.20 ± 0.05 «cm»
\n\n
\n
For MP1 accept answers in range of 0.020 to 0.060 «cm2» if they fail to use given value of “a”.
\nFor MP2 accept answers in range 0.14 to 0.25 «cm» .
\n[2 marks]
\nAn apparatus is used to verify a gas law. The glass jar contains a fixed volume of air. Measurements can be taken using the thermometer and the pressure gauge.
\nThe apparatus is cooled in a freezer and then placed in a water bath so that the temperature of the gas increases slowly. The pressure and temperature of the gas are recorded.
\nThe graph shows the data recorded.
\nIdentify the fundamental SI unit for the gradient of the pressure–temperature graph.
\nThe experiment is repeated using a different gas in the glass jar. The pressure for both experiments is low and both gases can be considered to be ideal.
\n(i) Using the axes provided in (a), draw the expected graph for this second experiment.
\n(ii) Explain the shape and intercept of the graph you drew in (b)(i).
\nkg m–1 s–2 K–1
\ni
\nany straight line that either goes or would go, if extended, through the origin
\n\n
ii
\nfor ideal gas p is proportional to T / P= nRT/V
\ngradient is constant /graph is a straight line
\nline passes through origin / 0,0
\nState what is meant by the event horizon of a black hole.
\nShow that the surface area A of the sphere corresponding to the event horizon is given by
\n.
\nSuggest why the surface area of the event horizon can never decrease.
\nThe diagram shows a box that is falling freely in the gravitational field of a planet.
\nA photon of frequency f is emitted from the floor of the box and is received at the ceiling. State and explain the frequency of the photon measured at the ceiling.
\nthe surface at which the escape speed is the speed for light
OR
the surface from which nothing/not even light can escape to the outside
OR
the surface of a sphere whose radius is the Schwarzschild radius
\n
Accept distance as alternative to surface.
\n[1 mark]
\nuse of and
\n«to get »
\n[1 mark]
\nsince mass and energy can never leave a black hole and
\nOR
\nsome statement that area is increasing with mass
\n«the area cannot decrease»
\n[1 mark]
\nALTERNATIVE 1 — (student/planet frame):
\nphoton energy/frequency decreases with height
OR
there is a gravitational redshift
detector in ceiling is approaching photons so Doppler blue shift
\ntwo effects cancel/frequency unchanged
\nALTERNATIVE 2 – (box frame):
\nby equivalence principle box is an inertial frame
\nso no force on photons
\nso no redshift/frequency unchanged
\n[3 marks]
\nThe half-life of a radioactive element is 5.0 days. A freshly-prepared sample contains 128 g of this element. After how many days will there be 16 g of this element left behind in the sample?
\nA. 5.0 days
\nB. 10 days
\nC. 15 days
\nD. 20 days
\nC
\nA horizontal rigid bar of length 2R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.
\nA point particle of mass moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.
\nThe mass of the rod and the container can be neglected.
\nA torque of 0.010 N m brings the system to rest after a number of revolutions. For this case R = 0.50 m, M = 0.70 kg and v = 2.1 m s–1.
\nWrite down an expression, in terms of M, v and R, for the angular momentum of the system about the vertical axis just before the collision.
\nJust after the collision the system begins to rotate about the vertical axis with angular velocity ω. Show that the angular momentum of the system is equal to .
\nHence, show that .
\nDetermine in terms of M and v the energy lost during the collision.
\nShow that the angular deceleration of the system is 0.043 rads–2.
\nCalculate the number of revolutions made by the system before it comes to rest.
\n\n
[1 mark]
\nevidence of use of:
\n[1 mark]
\nevidence of use of conservation of angular momentum,
\n«rearranging to get »
\n[1 mark]
\ninitial KE =
\nfinal KE =
\nenergy loss =
\n[3 marks]
\n«= » =
\n«to give = 0.04286 rads−2»
\n\n
Working OR answer to at least 3 SF must be shown
\n[1 mark]
\n«from »
\n«» = 12.8 OR 12.9 «rad»
\nnumber of rotations «= » = 2.0 revolutions
\n[3 marks]
\nThe binding energy per nucleon of is 6 MeV. What is the energy required to separate the nucleons of this nucleus?
\nA. 24 MeV
\nB. 42 MeV
\nC. 66 MeV
\nD. 90 MeV
\nC
\nA student investigates how light can be used to measure the speed of a toy train.
\nLight from a laser is incident on a double slit. The light from the slits is detected by a light sensor attached to the train.
\nThe graph shows the variation with time of the output voltage from the light sensor as the train moves parallel to the slits. The output voltage is proportional to the intensity of light incident on the sensor.
\n\n
Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.
\nThe slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.
\nEstimate the speed of the train.
\nIn another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier.
\nThe sound sensor gives a graph of the variation of output voltage with time along the track that is similar in shape to the graph shown in the resource. Explain how this effect arises.
\n«light» superposes/interferes
\npattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»
voltage peaks correspond to interference maxima
\n«» 2.1 x 10–3 «m»
\n\n
If no unit assume m.
Correct answer only.
correct read-off from graph of 25 m s
\nv = «» 8.4 x 10–2 «m s–1»
\n\n
Allow ECF from (b)(i)
\nALTERNATIVE 1
\n«reflection at barrier» leads to two waves travelling in opposite directions
\nmention of formation of standing wave
\nmaximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position
Explain what is meant by the gravitational potential at the surface of a planet.
\nAn unpowered projectile is fired vertically upwards into deep space from the surface of planet Venus. Assume that the gravitational effects of the Sun and the other planets are negligible.
\nThe following data are available.
\n\n\n\nMass of Venus\n= 4.87×1024 kg\n\n\nRadius of Venus\n= 6.05×106 m\n\n\nMass of projectile\n= 3.50×103 kg\n\n\nInitial speed of projectile\n= 1.10×escape speed\n\n\n\n\n
(i) Determine the initial kinetic energy of the projectile.
\n(ii) Describe the subsequent motion of the projectile until it is effectively beyond the gravitational field of Venus.
\nthe «gravitational» work done «by an external agent» per/on unit mass/kg
\nAllow definition in terms of reverse process of moving mass to infinity eg “work done on external agent by…”.
Allow “energy” as equivalent to “work done”
in moving a «small» mass from infinity to the «surface of» planet / to a point
\nN.B.: on SL paper Q5(a)(i) and (ii) is about “gravitational field”.
\ni
escape speed
Care with ECF from MP1.
v = «»
\n or 1.04×104«m s–1»
or «1.1 × 1.04 × 104 m s-1»= 1.14 × 104 «m s–1»
KE = «0.5 × 3500 × (1.1 × 1.04 × 104 m s–1)2 =» 2.27×1011 «J»
\nAward [1 max] for omission of 1.1 – leads to 1.88×1011 m s-1.
Award [2] for a bald correct answer.
\n
ii
Velocity/speed decreases / projectile slows down «at decreasing rate»
«magnitude of» deceleration decreases «at decreasing rate»
Mention of deceleration scores MP1 automatically.
velocity becomes constant/non-zero
OR
deceleration tends to zero
Accept “negative acceleration” for “deceleration”.
\nMust see “velocity” not “speed” for MP3.
\nA student pours a canned carbonated drink into a cylindrical container after shaking the can violently before opening. A large volume of foam is produced that fills the container. The graph shows the variation of foam height with time.
\nDetermine the time taken for the foam to drop to
\n(i) half its initial height.
\n(ii) a quarter of its initial height.
\nThe change in foam height can be modelled using ideas from other areas of physics. Identify one other situation in physics that is modelled in a similar way.
\ni
\n18 «s»
\nAllow answer in the range of 17 «s» to 19 «s».
Ignore wrong unit.
ii
\n36 «s»
\nAllow answer in the range of 35 «s» to 37 «s».
\nradioactive/nuclear decay
OR
capacitor discharge
OR
cooling
Accept any relevant situation, eg: critically damping, approaching terminal velocity
\nA ball is moving in still air, spinning clockwise about a horizontal axis through its centre. The diagram shows streamlines around the ball.
\n![\"M17/4/PHYSI/HP3/ENG/TZ2/10\"](\"../assets/uploads/tinymce_asset/asset/3732/Schermafbeelding_2017-09-26_om_10.55.18.png\"/)
The surface area of the ball is 2.50 x 10–2 m2. The speed of air is 28.4 ms–1 under the ball and 16.6 ms–1 above the ball. The density of air is 1.20 kgm–3.
\nEstimate the magnitude of the force on the ball, ignoring gravity.
\nOn the diagram, draw an arrow to indicate the direction of this force.
\nState one assumption you made in your estimate in (a)(i).
\nΔp = «» 318.6 «Pa»
\nF = «» 1.99 «N»
\n
Allow ECF from MP1.
[2 marks]
\ndownward arrow of any length or position
\nAccept any downward arrow not just vertical.
\n[1 mark]
\nflow is laminar/non-turbulent
OR
Bernoulli’s equation holds
OR
pressure is uniform on each hemisphere
OR
diameter of ball can be ignored /ρgz = constant
[1 mark]
\nThe reaction p+ + n0 → p+ + 0 does not occur because it violates the conservation law of
\nA. electric charge.
\nB. baryon number.
\nC. lepton number.
\nD. strangeness.
\nB
\nThe main role of a moderator in a nuclear fission reactor is to
\nA. slow down neutrons.
\nB. absorb neutrons.
\nC. reflect neutrons back to the reactor.
\nD. accelerate neutrons.
\nA
\nA beam of electrons e– enters a uniform electric field between parallel conducting plates RS. RS are connected to a direct current (dc) power supply. A uniform magnetic field B is directed into the plane of the page and is perpendicular to the direction of motion of the electrons.
\nThe magnetic field is adjusted until the electron beam is undeflected as shown.
\nIdentify, on the diagram, the direction of the electric field between the plates.
\nThe following data are available.
\n\n\n\nSeparation of the plates RS\n= 4.0 cm\n\n\nPotential difference between the plates\n= 2.2 kV\n\n\nVelocity of the electrons\n= 5.0×105 m s–1\n\n
Determine the strength of the magnetic field B.
\nThe velocity of the electrons is now increased. Explain the effect that this will have on the path of the electron beam.
\ndirection indicated downwards, perpendicular to plates
\nArrows must be between plates but allow edge effects if shown. Only one arrow is required.
\n«Vm–1»
\nB = «» 0.11 «T»
\nECF applies from MP1 to MP2 due to math error.
\nAward [2] for a bald correct answer.
\nALTERNATIVE 1
\nmagnetic force increases
OR
magnetic force becomes greater than electric force
\n
electron beam deflects “downwards” / towards S
OR
path of beam is downwards
ALTERNATIVE 2
\nwhen v increases, the B required to maintain horizontal path decreases
«but B is constant» so path of beam is downwards
Do not apply an ecf from (a).
\nAward [1 max] if answer states that magnetic force decreases and therefore path is upwards.
\nIgnore any statement about shape of path
Do not allow “path deviates in direction of magnetic force” without qualification.
\nAn electron X is moving parallel to a current-carrying wire. The positive ions and the wire are fixed in the reference frame of the laboratory. The drift speed of the free electrons in the wire is the same as the speed of the external electron X.
\nDefine frame of reference.
\nIn the reference frame of the laboratory the force on X is magnetic.
\n(i) State the nature of the force acting on X in this reference frame where X is at rest.
\n(ii) Explain how this force arises.
\na coordinate system
OR
a system of clocks and measures providing time and position relative to an observer
OWTTE
\ni
\nelectric
OR
electrostatic
\n
ii
\n«as the positive ions are moving with respect to the charge,» there is a length contraction
\ntherefore the charge density on ions is larger than on electrons
\nso net positive charge on wire attracts X
\nFor candidates who clearly interpret the question to mean that X is now at rest in the Earth frame accept this alternative MS for bii
the magnetic force on a charge exists only if the charge is moving
an electric force on X , if stationary, only exists if it is in an electric field
no electric field exists in the Earth frame due to the wire
and look back at b i, and award mark for there is no electric or magnetic force on X
A room is at a constant temperature of 300 K. A hotplate in the room is at a temperature of 400 K and loses energy by radiation at a rate of P. What is the rate of loss of energy from the hotplate when its temperature is 500 K?
\nA. P
\nB. P
\nC. P
\nD. P
\nD
\nThe circuit shown may be used to measure the internal resistance of a cell.
\n![\"M17/4/PHYSI/SP3/ENG/TZ2/02\"](\"../assets/uploads/tinymce_asset/asset/3736/Schermafbeelding_2017-09-27_om_07.51.02.png\"/)
The ammeter used in the experiment in (b) is an analogue meter. The student takes measurements without checking for a “zero error” on the ammeter.
\nAn ammeter and a voltmeter are connected in the circuit. Label the ammeter with the letter A and the voltmeter with the letter V.
\nIn one experiment a student obtains the following graph showing the variation with current I of the potential difference V across the cell.
\n![\"M17/4/PHYSI/SP3/ENG/TZ2/02b\"](\"../assets/uploads/tinymce_asset/asset/3739/Schermafbeelding_2017-09-27_om_08.05.10.png\"/)
Using the graph, determine the best estimate of the internal resistance of the cell.
\nState what is meant by a zero error.
\nAfter taking measurements the student observes that the ammeter has a positive zero error. Explain what effect, if any, this zero error will have on the calculated value of the internal resistance in (b).
\ncorrect labelling of both instruments
\n\n
[1 mark]
\nV = E – Ir
\nlarge triangle to find gradient and correct read-offs from the line
OR
use of intercept E = 1.5 V and another correct data point
internal resistance = 0.60 Ω
\nFor MP1 – do not award if only is used.
\nFor MP2 points at least 1A apart must be used.
\nFor MP3 accept final answers in the range of 0.55 Ω to 0.65 Ω.
\n[3 marks]
\na non-zero reading when a zero reading is expected/no current is flowing
OR
a calibration error
\n
OWTTE
Do not accept just “systematic error”.
[1 mark]
\nthe error causes «all» measurements to be high/different/incorrect
\neffect on calculations/gradient will cancel out
OR
effect is that value for r is unchanged
Award [1 max] for statement of “no effect” without valid argument.
\nOWTTE
\n[2 marks]
\nA driven system is lightly damped. The graph shows the variation with driving frequency f of the amplitude A of oscillation.
\n![\"M17/4/PHYSI/HP3/ENG/TZ2/11\"](\"../assets/uploads/tinymce_asset/asset/3733/Schermafbeelding_2017-09-26_om_11.02.12.png\"/)
A mass on a spring is forced to oscillate by connecting it to a sine wave vibrator. The graph shows the variation with time t of the resulting displacement y of the mass. The sine wave vibrator has the same frequency as the natural frequency of the spring–mass system.
\nOn the graph, sketch a curve to show the variation with driving frequency of the amplitude when the damping of the system increases.
\nState and explain the displacement of the sine wave vibrator at t = 8.0 s.
\nThe vibrator is switched off and the spring continues to oscillate. The Q factor is 25.
\nCalculate the ratio for the oscillations of the spring–mass system.
\nlower peak
\nidentical behaviour to original curve at extremes
\npeak frequency shifted to the left
\n\n
Award [0] if peak is higher.
\nFor MP2 do not accept curves which cross.
\n[2 marks]
\ndisplacement of vibrator is 0
\nbecause phase difference is or 90º or period
\n\n
Do not penalize sign of phase difference.
\nDo not accept for MP2
\n[2 marks]
\nresonant f = 0.125 « Hz »
\n= 32 «s»
\n\n
Watch for ECF from MP1 to MP2.
\n[2 marks]
\nThe first scientists to identify alpha particles by a direct method were Rutherford and Royds. They knew that radium-226 () decays by alpha emission to form a nuclide known as radon (Rn).
\nWrite down the missing values in the nuclear equation for this decay.
\nRutherford and Royds put some pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.
\nAt the start of the experiment all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder B.
\nThe wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.
\nRutherford and Royds expected 2.7 x 1015 alpha particles to be emitted during the experiment. The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10–5 m3 and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B.
\nRutherford and Royds identified the helium gas in cylinder B by observing its emission spectrum. Outline, with reference to atomic energy levels, how an emission spectrum is formed.
\nThe work was first reported in a peer-reviewed scientific journal. Outline why Rutherford and Royds chose to publish their work in this way.
\n222 AND 4
\n\n
Both needed.
\nalpha particles highly ionizing
OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule
conversion of temperature to 291 K
\np = 4.5 x 10–9 x 8.31 x «»
\nOR
\np = 2.7 x 1015 x 1.38 x 10–23 x «»
\n0.83 or 0.84 «Pa»
\n\n
electron/atom drops from high energy state/level to low state
\nenergy levels are discrete
\nwavelength/frequency of photon is related to energy change or quotes E = hf or E =
\nand is therefore also discrete
\npeer review guarantees the validity of the work
OR
means that readers have confidence in the validity of work
\n
OWTTE
\nIn physics, a paradigm shift denotes the introduction of radically new ideas in order to explain a phenomenon. Which introduces a paradigm shift?
\nA. Multi-loop circuits
\nB. Standing waves
\nC. Total internal reflection
\nD. Atomic spectra
\nD
\nDefine proper length.
\nA charged pion decays spontaneously in a time of 26 ns as measured in the frame of reference in which it is stationary. The pion moves with a velocity of 0.96c relative to the Earth. Calculate the pion’s lifetime as measured by an observer on the Earth.
\nIn the pion reference frame, the Earth moves a distance X before the pion decays. In the Earth reference frame, the pion moves a distance Y before the pion decays. Demonstrate, with calculations, how length contraction applies to this situation.
\nthe length of an object in its rest frame
\n OR
ECF for wrong
93 «ns»
Award [2] for a bald correct answer.
«X is» 7.5 «m» in frame on pion
\n«Y is» 26.8 «m» in frame on Earth
\nidentifies proper length as the Earth measurement
OR
identifies Earth distance according to pion as contracted length
OR
a statement explaining that one of the length is shorter than the other one
A block of weight W is suspended by two strings of equal length. The strings are almost horizontal.
\nWhat is correct about the tension T in one string?
\nA.
\nB.
\nC.
\nD.
\nD
\nThe P–V diagram of the Carnot cycle for a monatomic ideal gas is shown.
\nThe system consists of 0.150 mol of a gas initially at A. The pressure at A is 512 k Pa and the volume is 1.20 × 10–3 m3.
\nAt C the volume is VC and the temperature is TC.
\nState what is meant by an adiabatic process.
\nIdentify the two isothermal processes.
\nDetermine the temperature of the gas at A.
\nThe volume at B is 2.30 × 10–3m3. Determine the pressure at B.
\nShow that
\nThe volume at C is 2.90 × 10–3m3. Calculate the temperature at C.
\nState a reason why a Carnot cycle is of little use for a practical heat engine.
\n«a process in which there is» no thermal energy transferred between the system and the surroundings
\n[1 mark]
\nA to B AND C to D
\n[1 mark]
\n\n
«K»
\n\n
The first mark is for rearranging.
\n[2 marks]
\n\n
\n
\n
The first mark is for rearranging.
\n[2 marks]
\n«B to C adiabatic so» AND PCVC = nRTC «combining to get result»
\n\n
It is essential to see these 2 relations to award the mark.
\n[1 mark]
\n\n
«» = 422 «K»
\n[2 marks]
\nthe isothermal processes would have to be conducted very slowly / OWTTE
\n[1 mark]
\nA block of mass 1.0 kg rests on a trolley of mass 4.0 kg. The coefficient of dynamic friction between the block and the trolley is 0.30.
\nA horizontal force F = 5.0 N acts on the block. The block slides over the trolley. What is the acceleration of the trolley?
\nA. 5.0 m s–2
\nB. 1.0 m s–2
\nC. 0.75 m s–2
\nD. 0.60 m s–2
\nC
\nThe density of muscle is 1075 kg m–3 and the speed of ultrasound in muscle is 1590 m s–1.
\nState a typical frequency used in medical ultrasound imaging.
\nDescribe how an ultrasound transducer produces ultrasound.
\nCalculate the acoustic impedance Z of muscle.
\nUltrasound of intensity 0.012 Wcm–2 is incident on a water–muscle boundary. The acoustic impedance of water is 1.50 x 106 kgm–2s–1.
\nThe fraction of the incident intensity that is reflected is given by
\n\n
where Z1 and Z2 are the acoustic impedances of medium 1 and medium 2.
\nCalculate the intensity of the reflected signal.
\naccept any value between 1 MHz to 20 MHz
\n[1 mark]
\nan alternating electrical signal is applied to a crystal
\ncrystal vibrates emitting sound
\nfrequency of vibration of crystal is the same as the frequency of the ac
\nmention of piezoelectric effect/crystal
\n[3 marks]
\nZmuscle = 1.71 x 106 «kgm–2s–1»
\n[1 mark]
\n«» = 4.3 x 10–3
\nI2 = «0.012 x (4.3 x 10–3) =» 5.1 x 10–5 «Wcm–2»
\n\n
Allow ECF from (c)(i).
\nAllow ECF from MP1 to MP2.
\n[2 marks]
\nA spaceship S leaves the Earth with a speed v = 0.80c. The spacetime diagram for the Earth is shown. A clock on the Earth and a clock on the spaceship are synchronized at the origin of the spacetime diagram.
\nCalculate the angle between the worldline of S and the worldline of the Earth.
\nDraw, on the diagram, the x′-axis for the reference frame of S.
\nAn event Z is shown on the diagram. Label the co-ordinates of this event in the reference frame of S.
\nangle = tan–1 «» = 39 «o» OR 0.67 «rad»
\nadds x′-axis as shown
\nApproximate same angle to v = c as for ct′.
\nIgnore labelling of that axis.
\nadds two lines parallel to ct′ and x′ as shown indicating coordinates
\nThe cable consists of 32 copper wires each of length 35 km. Each wire has a resistance of 64 Ω. The resistivity of copper is 1.7 x 10–8 Ω m.
\nA cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires are protected by an insulator.
\nThe copper wires and insulator are both exposed to an electric field. Discuss, with reference to charge carriers, why there is a significant electric current only in the copper wires.
\nCalculate the radius of each wire.
\nThere is a current of 730 A in the cable. Show that the power loss in 1 m of the cable is about 30 W.
\nWhen the current is switched on in the cable the initial rate of rise of temperature of the cable is 35 mK s–1. The specific heat capacity of copper is 390 J kg–1 K–1. Determine the mass of a length of one metre of the cable.
\nwhen an electric field is applied to any material «using a cell etc» it acts to accelerate any free electrons
\nelectrons are the charge carriers «in copper»
\nAccept “free/valence/delocalised electrons”.
\nmetals/copper have many free electrons whereas insulators have few/no free electrons/charge carriers
\narea = «= 9.3 x 10–6 m2»
\n«resistance of cable = 2Ω»
\npower dissipated in cable = 7302 x 2 «= 1.07 MW»
\npower loss per meter or 30.6 «W m–1»
\n\n
Allow [2] for a solution where the resistance per unit metre is calculated using resistivity and answer to (b)(i) (resistance per unit length of cable =5.7 x 10–5 m)
\n30 = m x 390 x 3.5 x 10–2
\n2.2 k«g»
\n\n
Correct answer only.
\nState one prediction of Maxwell’s theory of electromagnetism that is consistent with special relativity.
\nA current is established in a long straight wire that is at rest in a laboratory.
\nA proton is at rest relative to the laboratory and the wire.
\nObserver X is at rest in the laboratory. Observer Y moves to the right with constant speed relative to the laboratory. Compare and contrast how observer X and observer Y account for any non-gravitational forces on the proton.
\n\n
the speed of light is a universal constant/invariant
OR
c does not depend on velocity of source/observer
electric and magnetic fields/forces unified/frame of reference dependant
\n[1 mark]
\nobserver X will measure zero «magnetic or electric» force
\nobserver Y must measure both electric and magnetic forces
\nwhich must be equal and opposite so that observer Y also measures zero force
\n\n
Allow [2 max] for a comment that both X and Y measure zero resultant force even if no valid explanation is given.
\n[3 marks]
\nA buoy, floating in a vertical tube, generates energy from the movement of water waves on the surface of the sea. When the buoy moves up, a cable turns a generator on the sea bed producing power. When the buoy moves down, the cable is wound in by a mechanism in the generator and no power is produced.
\nThe motion of the buoy can be assumed to be simple harmonic.
\nWater can be used in other ways to generate energy.
\nOutline the conditions necessary for simple harmonic motion (SHM) to occur.
\nA wave of amplitude 4.3 m and wavelength 35 m, moves with a speed of 3.4 m s–1. Calculate the maximum vertical speed of the buoy.
\nSketch a graph to show the variation with time of the generator output power. Label the time axis with a suitable scale.
\nOutline, with reference to energy changes, the operation of a pumped storage hydroelectric system.
\nThe water in a particular pumped storage hydroelectric system falls a vertical distance of 270 m to the turbines. Calculate the speed at which water arrives at the turbines. Assume that there is no energy loss in the system.
\nThe hydroelectric system has four 250 MW generators. Determine the maximum time for which the hydroelectric system can maintain full output when a mass of 1.5 x 1010 kg of water passes through the turbines.
\nNot all the stored energy can be retrieved because of energy losses in the system. Explain two such losses.
\nforce/acceleration proportional to displacement «from equilibrium position»
\nand directed towards equilibrium position/point
OR
and directed in opposite direction to the displacement from equilibrium position/point
\n
Do not award marks for stating the defining equation for SHM.
Award [1 max] for a ω–=2x with a and x defined.
frequency of buoy movement or 0.097 «Hz»
\nOR
\ntime period of buoy or 10.3 «s» or 10 «s»
\nv = « or » or
\n2.6 «m s–1»
\npeaks separated by gaps equal to width of each pulse «shape of peak roughly as shown»
\none cycle taking 10 s shown on graph
\nJudge by eye.
Do not accept cos2 or sin2 graph
At least two peaks needed.
Do not allow square waves or asymmetrical shapes.
Allow ECF from (b)(i) value of period if calculated.
PE of water is converted to KE of moving water/turbine to electrical energy «in generator/turbine/dynamo»
\nidea of pumped storage, ie: pump water back during night/when energy cheap to buy/when energy not in demand/when there is a surplus of energy
\nspecific energy available = «gh =» 9.81 x 270 «= 2650J kg–1»
\nOR
\nmgh mv2
\nOR
\nv2 = 2gh
\nv = 73 «ms–1»
\n\n
Do not allow 72 as round from 72.8
\n\n
total energy = «mgh = 1.5 x 1010 x 9.81 x 270=» 4.0 x 1013 «J»
\nOR
\ntotal energy = « (answer (c)(ii))2 =» 4.0 x 1013 «J»
\ntime = «» 11.1h or 4.0 x 104 s
\n\n
Use of 3.97 x 1013 «J» gives 11 h.
\nFor MP2 the unit must be present.
\nfriction/resistive losses in pipe/fluid resistance/turbulence/turbine or generator «bearings»
OR
sound energy losses from turbine/water in pipe
thermal energy/heat losses in wires/components
water requires kinetic energy to leave system so not all can be transferred
\n
Must see “seat of friction” to award the mark.
\nDo not allow “friction” bald.
\nA stationary nucleus of polonium-210 undergoes alpha decay to form lead-206. The initial speed of the alpha particle is v. What is the speed of the lead-206 nucleus?
\nA. v
\nB. v
\nC. v
\nD. v
\nD
\nAn ideal gas has a volume of 15 ml, a temperature of 20 °C and a pressure of 100 kPa. The volume of the gas is reduced to 5 ml and the temperature is raised to 40 °C. What is the new pressure of the gas?
\nA. 600 kPa
\nB. 320 kPa
\nC. 200 kPa
\nD. 35 kPa
\nB
\nPositive charge is uniformly distributed on a semi-circular plastic rod. What is the direction of the electric field strength at point S?
\nB
\nFour uniform planets have masses and radii as shown. Which planet has the smallest escape speed?
\nC
\nIdentical twins, A and B, are initially on Earth. Twin A remains on Earth while twin B leaves the Earth at a speed of 0.6c for a return journey to a point three light years from Earth.
\nCalculate the time taken for the journey in the reference frame of twin A as measured on Earth.
\nDetermine the time taken for the journey in the reference frame of twin B.
\nDraw, for the reference frame of twin A, a spacetime diagram that represents the worldlines for both twins.
\nSuggest how the twin paradox arises and how it is resolved.
\n«0.6 ct = 6 ly» so t = 10 «years»
\nAccept: If the 6 ly are considered to be measured from B, then the answer is 12.5 years.
\nALTERNATIVE 1
\n102 − 62 = t2 − 02
\nso t is 8 «years»
Accept: If the 6 ly are considered to be measured from B, then the answer is 10 years.
ALTERNATIVE 2
\ngamma is
\n10 × = 8 «years»
\nAllow ECF from a
Allow ECF for incorrect γ in mp1
three world lines as shown
\nAward mark only if axes OR world lines are labelled.
\naccording to both twins, it is the other one who is moving fast therefore clock should run slow
\nAllow explanation in terms of spacetime diagram.
\n«it is not considered a paradox as» twin B is not always in the same inertial frame of reference
\nOR
\ntwin B is actually accelerating «and decelerating»
\nIn nuclear magnetic resonance (NMR) imaging radio frequency electromagnetic radiation is detected by the imaging sensors. Discuss the origin of this radiation.
\n«strong» magnetic field aligns proton «spins»
\nan RF signal is applied to excite protons
OR
change spin up to spin down state
protons de-excite/return to lower energy state
OR
proton relaxation occurs
with emission of RF radiation «that is detected»
\n\n
OWTTE
\nTreat any mention of the following as neutral as they are not strictly relevant to the question:
gradient field, Larmor frequency, precession, resonance, 3-D image
[3 marks]
\nMuons are unstable particles with a proper lifetime of 2.2 μs. Muons are produced 2.0 km above ground and move downwards at a speed of 0.98c relative to the ground. For this speed = 5.0. Discuss, with suitable calculations, how this experiment provides evidence for time dilation.
\nALTERNATIVE 1 — for answers in terms of time
overall idea that more muons are detected at the ground than expected «without time dilation»
«Earth frame transit time = » = 6.8 «μs»
\n«Earth frame dilation of proper half-life = 2.2 μs x 5» = 11 «μs»
OR
«muon’s proper transit time = » = 1.4 «μs»
\n
ALTERNATIVE 2 – for answers in terms of distance
overall idea that more muons are detected at the ground than expected «without time dilation»
«distance muons can travel in a proper lifetime = 2.2 μs x 0.98c» = 650 «m»
\n«Earth frame lifetime distance due to time dilation = 650 m x 5» = 3250 «m»
OR
«muon frame distance travelled = » = 400 «m»
\n
Accept answers from one of the alternatives.
\n[3 marks]
\nThe diagram shows the path of a particle in a region of uniform magnetic field. The field is directed into the plane of the page.
\nThis particle could be
\nA. an alpha particle.
\nB. a beta particle.
\nC. a photon.
\nD. a neutron.
\nA
\nThe diagram shows a bar magnet near an aluminium ring.
\nThe ring is supported so that it is free to move. The ring is initially at rest. In experiment 1 the magnet is moved towards the ring. In experiment 2 the magnet is moved away from the ring. For each experiment what is the initial direction of motion of the ring?
\nB
\nA rocket of proper length 450 m is approaching a space station whose proper length is 9.0 km. The speed of the rocket relative to the space station is 0.80c.
\nX is an observer at rest in the space station.
\n\n
Two lamps at opposite ends of the space station turn on at the same time according to X. Using a Lorentz transformation, determine, according to an observer at rest in the rocket,
\nThe rocket carries a different lamp. Event 1 is the flash of the rocket’s lamp occurring at the origin of both reference frames. Event 2 is the flash of the rocket’s lamp at time ct' = 1.0 m according to the rocket. The coordinates for event 2 for observers in the space station are x and ct.
\n![\"M17/4/PHYSI/SP3/ENG/TZ2/05c\"](\"../assets/uploads/tinymce_asset/asset/3737/Schermafbeelding_2017-09-27_om_07.57.21.png\"/)
Calculate the length of the rocket according to X.
\nA space shuttle is released from the rocket. The shuttle moves with speed 0.20c to the right according to X. Calculate the velocity of the shuttle relative to the rocket.
\nthe time interval between the lamps turning on.
\nwhich lamp turns on first.
\nOn the diagram label the coordinates x and ct.
\nState and explain whether the ct coordinate in (c)(i) is less than, equal to or greater than 1.0 m.
\nCalculate the value of c 2t 2 – x 2.
\nthe gamma factor is or 1.67
\nL = = 270 «m»
\n\n
Allow ECF from MP1 to MP2.
\n[2 marks]
\nu' = «»
OR
0.2c =
u' = «–»0.71c
\n\n
Check signs and values carefully.
\n[2 marks]
\nΔt' = «»
\nΔt' = «–»4.0 x 10–5 «s»
\n\n
Allow ECF for use of wrong from (a)(i).
\n[2 marks]
\nlamp 2 turns on first
\nIgnore any explanation
\n[1 mark]
\nx coordinate as shown
\nct coordinate as shown
\n\n
Labels must be clear and unambiguous.
\nConstruction lines are optional.
\n[2 marks]
\n«in any other frame» ct is greater
\nthe interval ct' = 1.0 «m» is proper time
OR
ct is a dilated time
OR
ct = ct' «= »
\n
MP1 is a statement
\nMP2 is an explanation
\n[2 marks]
\nuse of c 2t 2 – x 2 = c 2t' 2 – x' 2
c 2t 2 – x 2 = 12 – 02 = 1 «m2»
\n
for MP1 equation must be used.
\nAward [2] for correct answer that first finds x (1.33 m) and ct (1.66 m)
\n[2 marks]
\nDerive, using the concept of the cosmological origin of redshift, the relation
\nT
\nbetween the temperature T of the cosmic microwave background (CMB) radiation and the cosmic scale factor R.
\nThe present temperature of the CMB is 2.8 K. This radiation was emitted when the universe was smaller by a factor of 1100. Estimate the temperature of the CMB at the time of its emission.
\nState how the anisotropies in the CMB distribution are interpreted.
\nthe cosmological origin of redshift implies that the wavelength is proportional to the scale factor: R
\ncombining this with Wien’s law
\nOR
\nuse of kT
\n«gives the result»
\n\n
Evidence of correct algebra is needed as relationship T = is given.
\n[2 marks]
\nuse of T
\n= 2.8 x 1100 x 3080 ≈ 3100 «K»
\n[2 marks]
\nCMB anisotropies are related to fluctuations in density which are the cause for the formation of structures/nebulae/stars/galaxies
\n\n
OWTTE
\n[1 mark]
\nA small ball of weight W is attached to a string and moves in a vertical circle of radius R.
\nWhat is the smallest kinetic energy of the ball at position X for the ball to maintain the circular motion with radius R?
\nA.
\nB. W R
\nC. 2 W R
\nD.
\nD
\nA flywheel consists of a solid cylinder, with a small radial axle protruding from its centre.
\nThe following data are available for the flywheel.
\n| Flywheel mass M | \n= 1.22 kg | \n
| Small axle radius r | \n= 60.0 mm | \n
| Flywheel radius R | \n= 240 mm | \n
| Moment of inertia | \n= 0.5 MR2 | \n
An object of mass m is connected to the axle by a light string and allowed to fall vertically from rest, exerting a torque on the flywheel.
The velocity of the falling object is 1.89 m s–1 at 3.98 s. Calculate the average angular acceleration of the flywheel.
\nShow that the torque acting on the flywheel is about 0.3 Nm.
\n(i) Calculate the tension in the string.
\n(ii) Determine the mass m of the falling object.
\nALTERNATIVE 1
\n«rad s–1»
\n« so» «rad s–2»
\nALTERNATIVE 2
«m s–2»
«rad s–2»
\nAward [1 max] for r = 0.24 mm used giving = 1.98 «rad s–2».
\n\n
= 0.278 «Nm»
\nAt least two significant figures required for MP2, as question is a “Show”.
\ni
\n\n
«N»
\nAllow 5 «N» if Γ= 0.3 Νm is used.
\n\n
ii
\n so
m = 0.496 «kg»
Allow ECF
\nThree conducting loops, X, Y and Z, are moving with the same speed from a region of zero magnetic field to a region of uniform non-zero magnetic field.
\nWhich loop(s) has/have the largest induced electromotive force (emf) at the instant when the loops enter the magnetic field?
\nA. Z only
\nB. Y only
\nC. Y and Z only
\nD. X and Y only
\nA
\nThe centre of the Earth is separated from the centre of the Moon by a distance D. Point P lies on a line joining the centre of the Earth and the centre of the Moon, a distance X from the centre of the Earth. The gravitational field strength at P is zero.
\nWhat is the ratio ?
\nA.
\nB.
\nC.
\nD.
\nA
\nThe electrical circuit shown is used to investigate the temperature change in a wire that is wrapped around a mercury-in-glass thermometer.
\nA power supply of emf (electromotive force) 24 V and of negligible internal resistance is connected to a capacitor and to a coil of resistance wire using an arrangement of two switches. Switch S1 is closed and, a few seconds later, opened. Then switch S2 is closed.
\nThe capacitance of the capacitor is 22 mF. Calculate the energy stored in the capacitor when it is fully charged.
\nThe resistance of the wire is 8.0 Ω. Determine the time taken for the capacitor to discharge through the resistance wire. Assume that the capacitor is completely discharged when the potential difference across it has fallen to 0.24 V.
\nThe mass of the resistance wire is 0.61 g and its observed temperature rise is 28 K. Estimate the specific heat capacity of the wire. Include an appropriate unit for your answer.
\nSuggest one other energy loss in the experiment and the effect it will have on the value for the specific heat capacity of the wire.
\n«» = «J»
\n\n
\n
0.81 «s»
\nc =
\nOR
\n\n
370 J kg–1 K–1
\n\n
\n
Allow ECF from 3(a) for energy transferred.
\nCorrect answer only to include correct unit that matches answer power of ten.
\nAllow use of g and kJ in unit but must match numerical answer, eg: 0.37 J kg–1 K–1 receives [1]
\nALTERNATIVE 1
\nsome thermal energy will be transferred to surroundings/along connecting wires/to
thermometer
estimate «of specific heat capacity by student» will be larger «than accepted value»
\n\n
ALTERNATIVE 2
\nnot all energy transferred as capacitor did not fully discharge
\nso estimate «of specific heat capacity by student» will be larger «than accepted value»
\nThe diagram is a partially-completed ray diagram for a compound microscope that consists of two thin converging lenses. The objective lens L1 has a focal length of 3.0 cm. The object is placed 4.0 cm to the left of L1. The final virtual image is formed at the near point of the observer, a distance of 24 cm from the eyepiece lens L2.
\n![\"M17/4/PHYSI/SP3/ENG/TZ1/7a\"](\"../assets/uploads/tinymce_asset/asset/3735/Schermafbeelding_2017-09-26_om_15.46.06.png\"/)
Two converging lenses are used to make an astronomical telescope. The focal length of the objective is 85.0 cm and that of the eyepiece is 2.50 cm. The telescope is used to form a final image of the Moon at infinity.
\nState what is meant by a virtual image.
\nShow that the image of the object formed by L1 is 12 cm to the right of L1.
\nThe distance between the lenses is 18 cm. Determine the focal length of L2.
\nOn the diagram draw rays to locate the focal point of L2. Label this point F.
\nExplain why, for the final image to form at infinity, the distance between the lenses must be 87.5 cm.
\nThe angular diameter of the Moon at the naked eye is 7.8 × 10–3 rad.
\nCalculate the angular diameter of the final image of the Moon.
\nBy reference to chromatic aberration, explain one advantage of a reflecting telescope over a refracting telescope.
\nan image formed by extensions of rays, not rays themselves
OR
an image that cannot be projected on a screen
[1 mark]
\n\n
«v = 12 cm»
\n[1 mark]
\nu = 18 – 12 = 6.0 «cm»
\nv = –24 «cm»
\n«» f = 8.0 «cm»
\n\n
Award [2 max] for answer of 4.8 cm.
Minus sign required for MP2.
[3 marks]
\nline parallel to principal axis from intermediate image meeting eyepiece lens at P
\nline from arrow of final image to P intersecting principal axis at F
\n[2 marks]
\nobject is far away so intermediate image forms at focal plane of objective
\nfor final image at infinity object must also be at focal point of eyepiece
\n«hence 87.5 cm»
\n\n
No mark for simple addition of focal lengths without explanation.
\n[2 marks]
\nangular magnification = = 34
\nangular diameter 3.4 × 7.8 × 10−3 = 0.2652 ≈ 0.27 «rad»
\n[2 marks]
\nchromatic aberration is the dependence of refractive index on wavelength
\nbut mirrors rely on reflection
OR
mirrors do not involve refraction
«so do not suffer chromatic aberration»
\n[2 marks]
\nTwo capacitors of different capacitance are connected in series to a source of emf of negligible internal resistance.
\nWhat is correct about the potential difference across each capacitor and the charge on each capacitor?
\nC
\nIn the nuclear reaction X + Y → Z + W, involving nuclides X, Y, Z and W, energy is released. Which is correct about the masses (M) and the binding energies (BE) of the nuclides?
\nC
\nDescribe what is meant by dark matter.
\nThe distribution of mass in a spherical system is such that the density ρ varies with distance r from the centre as
\nρ =
\nwhere k is a constant.
\nShow that the rotation curve of this system is described by
\nv = constant.
\nCurve A shows the actual rotation curve of a nearby galaxy. Curve B shows the predicted rotation curve based on the visible stars in the galaxy.
\nExplain how curve A provides evidence for dark matter.
\ndark matter is invisible/cannot be seen directly
OR
does not interact with EM force/radiate light/reflect light
interacts with gravitational force
OR
accounts for galactic rotation curves
OR
accounts for some of the “missing” mass/energy of galaxies/the universe
\n
OWTTE
\n[6 marks]
\n«from data booklet formula» substitute to get
\n\n
Substitution of ρ must be seen.
\n[1 mark]
\ncurve A shows that the outer regions of the galaxy are rotating faster than predicted
\nthis suggests that there is more mass in the outer regions that is not visible
OR
more mass in the form of dark matter
\n
OWTTE
\n[2 marks]
\nA fully charged capacitor is connected to a resistor. When the switch is closed the capacitor will discharge through the resistor.
\nWhich graphs correctly show how the charge on the capacitor and the current in the circuit vary with time during the discharging of the capacitor?
\nA
\nWhich of the following leads to a paradigm shift?
\nA. Multi-loop circuits
\nB. Standing waves
\nC. Total internal reflection
\nD. Atomic spectra
\nD
\nThe diagram shows two methods of pedalling a bicycle using a force F.
\nIn method 1 the pedal is always horizontal to the ground. A student claims that method 2 is better because the pedal is always parallel to the crank arm. Explain why method 2 is more effective.
\nin method 1 the perpendicular distance varies from 0 to a maximum value, in method 2 this distance is constant at the maximum value
OR
angle between F and r is 90° in method 2 and less in method 1
OR
Γ = F × perpendicular distance
perpendicular distance/ torque is greater in method 2
\nA mass oscillates with simple harmonic motion (SHM) of amplitude xo. Its total energy is 16 J.
\nWhat is the kinetic energy of the mass when its displacement is ?
\nA. 4 J
\nB. 8 J
\nC. 12 J
\nD. 16 J
\nC
\nA cylindrical space probe of mass 8.00 x 102 kg and diameter 12.0 m is at rest in outer space.
\nRockets at opposite points on the probe are fired so that the probe rotates about its axis. Each rocket produces a force F = 9.60 x 103 N. The moment of inertia of the probe about its axis is 1.44 x 104 kgm2.
\nThe diagram shows a satellite approaching the rotating probe with negligibly small speed. The satellite is not rotating initially, but after linking to the probe they both rotate together.
\nThe moment of inertia of the satellite about its axis is 4.80 x 103 kgm2. The axes of the probe and of the satellite are the same.
\nDeduce the linear acceleration of the centre of mass of the probe.
\nCalculate the resultant torque about the axis of the probe.
\nThe forces act for 2.00 s. Show that the final angular speed of the probe is about 16 rads–1.
\nDetermine the final angular speed of the probe–satellite system.
\nCalculate the loss of rotational kinetic energy due to the linking of the probe with the satellite.
\nzero
\n[1 mark]
\nthe torque of each force is 9.60 x 103 x 6.0 = 5.76 x 104 «Nm»
\nso the net torque is 2 x 5.76 x 104 = 1.15 x 105 «Nm»
\n\n
Allow a one-step solution.
\n[2 marks]
\nthe angular acceleration is given by «= 8.0 s–2»
\nω = «t = 8.0 x 2.00 =» 16 «s–1»
\n[2 marks]
\n1.44 x 104 x 16.0 = (1.44 x 104 + 4.80 x 103) x ω
\nω = 12.0 «s–1»
\n\n
Allow ECF from (b).
\n[2 marks]
\ninitial KE «J»
\nfinal KE «J»
\nloss of KE = 4.6 x 105 «J»
\n\n
Allow ECF from part (c)(i).
\n[3 marks]
\nWhen monochromatic light is incident on a metallic surface, electrons are emitted from the surface. The following changes are considered.
\nI. Increase the intensity of the incident light
II. Increase the frequency of light
III. Decrease the work function of the surface
Which changes will result in electrons of greater energy being emitted from the surface?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nBlue light is incident on two narrow slits. Constructive interference takes place along the lines labelled 1 to 5.
\nThe blue light is now replaced by red light. What additional change is needed so that the lines of constructive interference remain in the same angular positions?
\nA. Make the slits wider
\nB. Make the slits narrower
\nC. Move the slits closer together
\nD. Move the slits further apart
\nD
\nA student investigates how light can be used to measure the speed of a toy train.
\nLight from a laser is incident on a double slit. The light from the slits is detected by a light sensor attached to the train.
\nThe graph shows the variation with time of the output voltage from the light sensor as the train moves parallel to the slits. The output voltage is proportional to the intensity of light incident on the sensor.
\n\n
As the train continues to move, the first diffraction minimum is observed when the light sensor is at a distance of 0.13 m from the centre of the fringe pattern.
\nA student investigates how light can be used to measure the speed of a toy train.
\nLight from a laser is incident on a double slit. The light from the slits is detected by a light sensor attached to the train.
\nThe graph shows the variation with time of the output voltage from the light sensor as the train moves parallel to the slits. The output voltage is proportional to the intensity of light incident on the sensor.
\n\n
Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.
\nThe slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.
\nEstimate the speed of the train.
\nDetermine the width of one of the slits.
\nSuggest the variation in the output voltage from the light sensor that will be observed as the train moves beyond the first diffraction minimum.
\nIn another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier.
\nThe graph shows the variation with time of the output voltage from the sounds sensor.
\nExplain how this effect arises.
\n«light» superposes/interferes
\npattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»
voltage peaks correspond to interference maxima
\n«» 2.1 x 10–3 «m»
\n\n
If no unit assume m.
Correct answer only.
correct read-off from graph of 25 m s
\nv = «» 8.4 x 10–2 «m s–1»
\n\n
Allow ECF from (b)(i)
\nangular width of diffraction minimum = «= 0.026 rad»
\nslit width = «» 2.4 x 10–5 «m»
\n\n
Award [1 max] for solution using 1.22 factor.
\n«beyond the first diffraction minimum» average voltage is smaller
«voltage minimum» spacing is «approximately» same
OR
rate of variation of voltage is unchanged
\n
OWTTE
\n«reflection at barrier» leads to two waves travelling in opposite directions
\nmention of formation of standing wave
\nmaximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position
In the Bohr model for hydrogen an electron in the ground state has orbit radius r and speed v. In the first excited state the electron has orbit radius 4r. What is the speed of the electron in the first excited state?
\nA.
\nB.
\nC.
\nD.
\nA
\nHow many significant figures are there in the number 0.0450?
\nA. 2
\nB. 3
\nC. 4
\nD. 5
\nB
\nA neutron of mass m is confined within a nucleus of diameter d. Ignoring numerical constants, what is an approximate expression for the kinetic energy of the neutron?
\nA.
\nB.
\nC.
\nD.
\nA
\nTwo points illuminated by monochromatic light are separated by a small distance. The light from the two sources passes through a small circular aperture and is detected on a screen far away.
\nTwo points illuminated by monochromatic light are separated by a small distance. The light from the two sources passes through a small circular aperture and is detected on a screen far away.
\nC
\nAn ideal nuclear power plant can be modelled as a heat engine that operates between a hot temperature of 612°C and a cold temperature of 349°C.
\nCalculate the Carnot efficiency of the nuclear power plant.
\nExplain, with a reason, why a real nuclear power plant operating between the stated temperatures cannot reach the efficiency calculated in (a).
\nThe nuclear power plant works at 71.0% of the Carnot efficiency. The power produced is 1.33 GW. Calculate how much waste thermal energy is released per hour.
\nDiscuss the production of waste heat by the power plant with reference to the first law and the second law of thermodynamics.
\ncorrect conversion to K «622 K cold, 885 K hot»
\nor 29.7%
\nAward [1 max] if temperatures are not converted to K, giving result 0.430.
\nthe Carnot efficiency is the maximum possible
\nthe Carnot cycle is theoretical/reversible/impossible/infinitely slow
\nenergy losses to surroundings «friction, electrical losses, heat losses, sound energy»
\nOWTTE
\n0.71 × 0.297 = 0.211
\nAllow solution utilizing wasted power «78.9%».
\n1.33/0.211 × 0.789 = 4.97 «GW»
\n4.97 GW × 3600 = 1.79 × 1013 «J»
\nAward [2 max] if 71% used as the overall efficiency giving an answer of 1.96 × 1012 J.
\nAward [3] for bald correct answer.
Watch for ECF from (a).
\nLaw 1: net thermal energy flow is QIN–QOUT
\nQOUT refers to “waste heat”
\nLaw 1: QIN–QOUT = ∆Q=∆W as ∆U is zero
\nLaw 2: does not forbid QOUT=0
\nLaw 2: no power plant can cover 100% of QIN into work
\nLaw 2: total entropy must increase so some Q must enter surroundings
\nOWTTE
\nAn object is positioned in a gravitational field. The measurement of gravitational force acting on the object has an uncertainty of 3 % and the uncertainty in the mass of the object is 9 %. What is the uncertainty in the gravitational field strength of the field?
\nA. 3 %
\nB. 6 %
\nC. 12 %
\nD. 27 %
\nC
\nA radioactive element has decay constant (expressed in s–1). The number of nuclei of this element at t = 0 is N. What is the expected number of nuclei that will have decayed after 1 s?
\nA.
\nB.
\nC.
\nD.
\nA
\nA train travelling in a straight line emits a sound of constant frequency f. An observer at rest very close to the path of the train detects a sound of continuously decreasing frequency. The train is
\nA. approaching the observer at constant speed.
\nB. approaching the observer at increasing speed.
\nC. moving away from the observer at constant speed.
\nD. moving away from the observer at increasing speed.
\nD
\nA positive charge Q is deposited on the surface of a small sphere. The dotted lines represent equipotentials.
\nA small positive point charge is moved from point P closer to the sphere along three different paths X, Y and Z. The work done along each path is WX, WY and WZ. What is a correct comparison of WX, WY and WZ?
\nA. WZ > WY > WX
\nB. WX > WY = WZ
\nC. WX = WY = WZ
\nD. WZ = WY > WX
\nB
\nCommunication signals are transmitted through optic fibres using infrared radiation.
\nState two advantages of optic fibres over coaxial cables for these transmissions.
\nSuggest why infrared radiation rather than visible light is used in these transmissions.
\nA signal with an input power of 15 mW is transmitted along an optic fibre which has an attenuation per unit length of 0.30 dBkm–1. The power at the receiver is 2.4 mW.
\nCalculate the length of the fibre.
\nState and explain why it is an advantage for the core of an optic fibre to be extremely thin.
\nlonger distance without amplification
\nsignal cannot easily be interfered with
\nless noise
\nno cross talk
\nhigher data transfer rate
\n[2 marks]
\ninfrared radiation suffers lower attenuation
\n[1 mark]
\nloss = «= −7.959 dB»
\nlength = «» 26.53 ≈ 27 «km»
\n[2 marks]
\na thin core means that rays follow essentially the same path / OWTTE
\nand so waveguide (modal) dispersion is minimal / OWTTE
\n[2 marks]
\nThe variation of the displacement of an object with time is shown on a graph. What does the area under the graph represent?
\nA. No physical quantity
\nB. Velocity
\nC. Acceleration
\nD. Impulse
\nA
\nThe graph shows the variation of the gravitational potential V with distance r from the centre of a uniform spherical planet. The radius of the planet is R. The shaded area is S.
\nWhat is the work done by the gravitational force as a point mass m is moved from the surface of the planet to a distance 6R from the centre?
\nA. m (V2 – V1 )
\nB. m (V1 – V2 )
\nC. mS
\nD. S
\nB
\nThe first scientists to identify alpha particles by a direct method were Rutherford and Royds. They knew that radium-226 () decays by alpha emission to form a nuclide known as radon (Rn).
\nAt the start of the experiment, Rutherford and Royds put 6.2 x 10–4 mol of pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.
\nThe experiment lasted for 6 days. The decay constant of radium-226 is 1.4 x 10–11 s–1.
\nAt the start of the experiment, all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder B.
\nWrite down the nuclear equation for this decay.
\nDeduce that the activity of the radium-226 is almost constant during the experiment.
\nShow that about 3 x 1015 alpha particles are emitted by the radium-226 in 6 days.
\nThe wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.
\nThe experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10–5 m3 and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B over the 6 day period. Helium is a monatomic gas.
\n\n
OR
\n\n
\n
\n
These must be seen on the right-hand side of the equation.
\nALTERNATIVE 1
\n6 days is 5.18 x 105 s
\nactivity after 6 days is
\nOR
\nA = 0.9999927 A0 or 0.9999927 N0
\nOR
\nstates that index of e is so small that is ≈ 1
\nOR
\nA – A0 ≈ 10–15 «s–1»
\n\n
ALTERNATIVE 2
shows half-life of the order of 1011 s or 5.0 x 1010 s
converts this to year «1600 y» or days and states half-life much longer than experiment compared to experiment
\n\n
Award [1 max] if calculations/substitutions have numerical slips but would lead to correct deduction.
\neg: failure to convert 6 days to seconds but correct substitution into equation will give MP2.
\nAllow working in days, but for MP1 must see conversion of or half-life to day–1.
\nALTERNATIVE 1
use of A = N0
conversion to number of molecules = nNA = 3.7 x 1020
\nOR
\ninitial activity = 5.2 x 109 «s–1»
\nnumber emitted = (6 x 24 x 3600) x 1.4 x 10–11 x 3.7 x 1020 or 2.7 x 1015 alpha particles
\n\n
ALTERNATIVE 2
use of N = N0
N0 = n x NA = 3.7 x 1020
\nalpha particles emitted «= number of atoms disintegrated = N – N0 =» N0 or 2.7 x 1015 alpha particles
\n\n
Must see correct substitution or answer to 2+ sf for MP3
\nalpha particles highly ionizing
OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule
\n
Do not allow reference to tunnelling.
\nconversion of temperature to 291 K
\np = 4.5 x 10–9 x 8.31 x «»
\nOR
\np = 2.7 x 1015 x 1.3 x 10–23 x «»
0.83 or 0.84 «Pa»
\n
Allow ECF for 2.7 x 1015 from (b)(ii).
\nAn object is projected vertically upwards at time t = 0. Air resistance is negligible. The object passes the same point above its starting position at times 2 s and 8 s.
\nIf g = 10 m s–2, what is the initial speed of the object?
\nA. 50
\nB. 30
\nC. 25
\nD. 4
\nA
\nA heat engine operates on the cycle shown in the pressure–volume diagram. The cycle consists of an isothermal expansion AB, an isovolumetric change BC and an adiabatic compression CA. The volume at B is double the volume at A. The gas is an ideal monatomic gas.
\nAt A the pressure of the gas is 4.00 x 106 Pa, the temperature is 612 K and the volume is 1.50 x 10–4m3. The work done by the gas during the isothermal expansion is 416 J.
\nJustify why the thermal energy supplied during the expansion AB is 416 J.
\nShow that the temperature of the gas at C is 386 K.
\nShow that the thermal energy removed from the gas for the change BC is approximately 330 J.
\nDetermine the efficiency of the heat engine.
\nState and explain at which point in the cycle ABCA the entropy of the gas is the largest.
\nΔU = 0 so Q = ΔU + W = 0 + 416 = 416 «J»
\n\n
Answer given, mark is for the proof.
\n[1 mark]
\nALTERNATIVE 1
\nuse to get
\nhence
\n«TC ≈ 386K»
\nALTERNATIVE 2
\ngiving PC = 1.26 x 106 «Pa»
\ngiving «K»
\n«TC ≈ 386K»
\n\n
Answer of 386K is given. Look carefully for correct working if answers are to 3 SF.
\nThere are other methods:
\nAllow use of PB = 2 x 106 «Pa» and is constant for BC.
\nAllow use of n = 0.118 and TC =
\n[2 marks]
\n\n
\n
«–332 J»
\n\n
Answer of 330 J given in the question.
Look for correct working or more than 2 SF.
[2 marks]
\n\n
e = 0.20
\n\n
Allow .
\nAllow e = 0.21.
\n[2 marks]
\nentropy is largest at B
\nentropy increases from A to B because T = constant but volume increases so more disorder or ΔS = and Q > 0 so ΔS > 0
\nentropy is constant along CA because it is adiabatic, Q = 0 and so ΔS = 0
OR
entropy decreases along BC since energy has been removed, ΔQ < 0 so ΔS < 0
[3 marks]
\nAn object is thrown upwards. The graph shows the variation with time t of the velocity v of the object.
\nWhat is the total displacement at a time of 1.5 s, measured from the point of release?
\nA. 0 m
\nB. 1.25 m
\nC. 2.50 m
\nD. 3.75 m
\nB
\nAn object is released from a stationary hot air balloon at height h above the ground.
\nAn identical object is released at height h above the ground from another balloon that is rising at constant speed. Air resistance is negligible. What does not increase for the object released from the rising balloon?
\nA. The distance through which it falls
\nB. The time taken for it to reach the ground
\nC. The speed with which it reaches the ground
\nD. Its acceleration
\nD
\nA uniform ladder resting in equilibrium on rough ground leans against a smooth wall. Which diagram correctly shows the forces acting on the ladder?
\n![\"M18/4/PHYSI/SPM/ENG/TZ1/04\"](\"../assets/uploads/tinymce_asset/asset/4664/Schermafbeelding_2018-08-10_om_15.49.37.png\"/)
B
\nParticles can be used in scattering experiments to estimate nuclear sizes.
\nElectron diffraction experiments indicate that the nuclear radius of carbon-12 is 2.7 x 10–15 m. The graph shows the variation of nuclear radius with nucleon number. The nuclear radius of the carbon-12 is shown on the graph.
\nThe Feynman diagram shows electron capture.
\nState and explain the nature of the particle labelled X.
\nOutline how these experiments are carried out.
\nOutline why the particles must be accelerated to high energies in scattering experiments.
\nState and explain one example of a scientific analogy.
\nPlot the position of magnesium-24 on the graph.
\nDraw a line on the graph, to show the variation of nuclear radius with nucleon number.
\n«electron» neutrino
\nit has a lepton number of 1 «as lepton number is conserved»
\nit has a charge of zero/is neutral «as charge is conserved»
OR
it has a baryon number of 0 «as baryon number is conserved»
Do not allow antineutrino
\nDo not credit answers referring to energy
\n«high energy particles incident on» thin sample
\ndetect angle/position of deflected particles
\nreference to interference/diffraction/minimum/maximum/numbers of particles
\nAllow “foil” instead of thin
\nλ OR λ
\nso high energy gives small λ
\nto match the small nuclear size
\nAlternative 2
\nE = hf/energy is proportional to frequency
\nfrequency is inversely proportional to wavelength/c = fλ
\nto match the small nuclear size
\nAlternative 3
\nhigher energy means closer approach to nucleus
\nto overcome the repulsive force from the nucleus
\nso greater precision in measurement of the size of the nucleus
\nAccept inversely proportional
\nOnly allow marks awarded from one alternative
\ntwo analogous situations stated
\none element of the analogy equated to an element of physics
\neg: moving away from Earth is like climbing a hill where the contours correspond to the equipotentials
\nAtoms in an ideal gas behave like pool balls
\nThe forces between them only act during collisions
\ncorrectly plotted
\nAllow ECF from (d)(i)
\nsingle smooth curve passing through both points with decreasing gradient
\nthrough origin
\nSpherical converging mirrors are reflecting surfaces which are cut out of a sphere. The diagram shows a mirror, where the dot represents the centre of curvature of the mirror.
\nA ray of light is incident on a converging mirror. On the diagram, draw the reflection of the incident ray shown.
\nThe incident ray shown in the diagram makes a significant angle with the optical axis.
\n(i) State the aberration produced by these kind of rays.
\n(ii) Outline how this aberration is overcome.
\nALTERNATIVE 1
\nfor incident ray, normal drawn which pass through C
\nreflected ray drawn such as i=r
\ni = r by eye
If normal is not visibly constructed using C,do not award MP1.
If no normal is drawn then grazing angles must be equal for MP2.
ALTERNATIVE 2
\ndrawn second ray through C, parallel to incident ray
\nadds focal plane and draws reflected ray so that it meets 2nd ray at focal plane
\nFocal plane position by eye, half-way between C and mirror.
\ni
\nspherical «aberration»
\n\n
ii
\nusing parabolic mirror
OR
reducing the aperture
An object falls from rest from a height h close to the surface of the Moon. The Moon has no atmosphere.
\nWhen the object has fallen to height above the surface, what is
\n?
\nA.
\nB.
\nC.
\nD.
\nA
\nThe diagram shows the forces acting on a block resting on an inclined plane. The angle θ is adjusted until the block is just at the point of sliding. R is the normal reaction, W the weight of the block and F the maximum frictional force.
\nWhat is the maximum coefficient of static friction between the block and the plane?
\nA. sin θ
\nB. cos θ
\nC. tan θ
\nD.
C
\nTheta 1 Orionis is a main sequence star. The following data for Theta 1 Orionis are available.
\n| Luminosity | \nL = 4 × 105 L | \n
| Radius | \nR = 13R | \n
| Apparent brightness | \nb = 4 × 10–11 b | \n
\n
where L, R and b are the luminosity, radius and apparent brightness of the Sun.
\nState what is meant by a main sequence star.
\nShow that the mass of Theta 1 Orionis is about 40 solar masses.
\nThe surface temperature of the Sun is about 6000 K. Estimate the surface temperature of Theta 1 Orionis.
\nDetermine the distance of Theta 1 Orionis in AU.
\nDiscuss how Theta 1 Orionis does not collapse under its own weight.
\nThe Sun and Theta 1 Orionis will eventually leave the main sequence. Compare and contrast the different stages in the evolution of the two stars.
\nstars fusing hydrogen «into helium»
\n[1 mark]
\n\n
«»
\n\n
Accept reverse working.
\n[1 mark]
\n\n
«K»
\n\n
Accept use of substituted values into 4R2T4.
\nAward [2] for a bald correct answer.
\n[2 marks]
\n\n
«AU»
\n\n
Accept use of correct values into .
\n[2 marks]
\nthe gravitation «pressure» is balanced by radiation «pressure»
\nthat is created by the production of energy due to fusion in the core / OWTTE
\n\n
Award [1 max] if pressure and force is inappropriately mixed in the answer.
\nAward [1 max] for unexplained \"hydrostatic equilibrium is reached\".
\n[2 marks]
\nthe Sun will evolve to become a red giant whereas Theta 1 Orionis will become a red super giant
\nthe Sun will explode as a planetary nebula whereas Theta 1 Orionis will explode as a supernova
\nthe Sun will end up as a white dwarf whereas Theta 1 Orionis as a neutron star/black hole
\n[3 marks]
\nA system that consists of a single spring stores a total elastic potential energy Ep when a load is added to the spring. Another identical spring connected in parallel is added to the system. The same load is now applied to the parallel springs.
\nWhat is the total elastic potential energy stored in the changed system?
\nA. Ep
\nB.
\nC.
\nD.
\nB
\nA lamp is located 6.0 m from a screen.
\nSomewhere between the lamp and the screen, a lens is placed so that it produces a real inverted image on the screen. The image produced is 4.0 times larger than the lamp.
\nIdentify the nature of the lens.
\nDetermine the distance between the lamp and the lens.
\nCalculate the focal length of the lens.
\nThe lens is moved to a second position where the image on the screen is again focused. The lamp–screen distance does not change. Compare the characteristics of this new image with the original image.
\nconverging/positive/biconvex/plane convex
\nDo not accept convex.
\n
Award [3] for a bald correct answer.
v + u = 6
Allow [1] if the answer is 4.8 «m».
so lens is 1.2 «m» from object or u = 1.2 «m»
\n«, so , so» f = 0.96 «m» or 1 «m»
\nWatch for ECF from (b)
\nreal AND inverted
\nsmaller OR diminished
\nYellow light from a sodium lamp of wavelength 590 nm is incident at normal incidence on a double slit. The resulting interference pattern is observed on a screen. The intensity of the pattern on the screen is shown.
\nThe double slit is replaced by a diffraction grating that has 600 lines per millimetre. The resulting pattern on the screen is shown.
\nExplain why zero intensity is observed at position A.
\nThe distance from the centre of the pattern to A is 4.1 x 10–2 m. The distance from the screen to the slits is 7.0 m.
\nCalculate the width of each slit.
\nCalculate the separation of the two slits.
\nState and explain the differences between the pattern on the screen due to the grating and the pattern due to the double slit.
\nThe yellow light is made from two very similar wavelengths that produce two lines in the spectrum of sodium. The wavelengths are 588.995 nm and 589.592 nm. These two lines can just be resolved in the second-order spectrum of this diffraction grating. Determine the beam width of the light incident on the diffraction grating.
\nthe diagram shows the combined effect of «single slit» diffraction and «double slit» interference
\nrecognition that there is a minimum of the single slit pattern
\nOR
\na missing maximum of the double slit pattern at A
\nwaves «from the single slit» are in antiphase/cancel/have a path difference of (n + )λ/destructive interference at A
\nθ = OR b = «= »
\n1.0 × 10–4 «m»
\nAward [0] for use of double slit formula (which gives the correct answer so do not award BCA)
\nAllow use of sin or tan for small angles
\nuse of s = with 3 fringes «»
\n3.0 x 10–4 «m»
\nAllow ECF.
\nfringes are further apart because the separation of slits is «much» less
\nintensity does not change «significantly» across the pattern or diffraction envelope is broader because slits are «much» narrower
\nthe fringes are narrower/sharper because the region/area of constructive interference is smaller/there are more slits
\nintensity of peaks has increased because more light can pass through
\nAward [1 max] for stating one or more differences with no explanation
\nAward [2 max] for stating one difference with its explanation
\nAward [MP3] for a second difference with its explanation
\nAllow “peaks” for “fringes”
\nΔλ = 589.592 – 588.995
\nOR
\nΔλ = 0.597 «nm»
\nN = « =» «493»
\nbeam width = « =» 8.2 x 10–4 «m» or 0.82 «mm»
\nChild X throws a ball to child Y. The system consists of the ball, the children and the Earth. What is true for the system when the ball has been caught by Y?
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/06\"](\"../assets/uploads/tinymce_asset/asset/4665/Schermafbeelding_2018-08-10_om_15.54.06.png\"/)
\n
A. The momentum of child Y is equal and opposite to the momentum of child X.
\nB. The speed of rotation of the Earth will have changed.
\nC. The ball has no net momentum while it is in the air.
\nD. The total momentum of the system has not changed.
\nD
\nA moving system undergoes an explosion. What is correct for the momentum of the system and the kinetic energy of the system when they are compared immediately before and after the explosion?
\nA
\nAn increasing force acts on a metal wire and the wire extends from an initial length l0 to a new length l. The graph shows the variation of force with length for the wire. The energy required to extend the wire from l0 to l is E.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/07\"](\"../assets/uploads/tinymce_asset/asset/4666/Schermafbeelding_2018-08-10_om_15.56.50.png\"/)
The wire then contracts to half its original extension.
\nWhat is the work done by the wire as it contracts?
\nA. 0.25E
\nB. 0.50E
\nC. 0.75E
\nD. E
\nC
\nWhat does the constant n represent in the equation of state for an ideal gas pV = nRT?
\nA. The number of atoms in the gas
\nB. The number of moles of the gas
\nC. The number of molecules of the gas
\nD. The number of particles in the gas
\nB
\nThe diagram shows planar wavefronts incident on a converging lens. The focal point of the lens is marked with the letter F.
\n![\"M17/4/PHYSI/SP3/ENG/TZ2/08\"](\"../assets/uploads/tinymce_asset/asset/3738/Schermafbeelding_2017-09-27_om_08.02.15.png\"/)
Wavefront X is incomplete. Point Q and point P lie on the surface of the lens and the principal axis.
\nOn the diagram, sketch the part of wavefront X that is inside the lens.
\nOn the diagram, sketch the wavefront in air that passes through point P. Label this wavefront Y.
\nExplain your sketch in (a)(i).
\nTwo parallel rays are incident on a system consisting of a diverging lens of focal length 4.0 cm and a converging lens of focal length 12 cm.
\nThe rays emerge parallel from the converging lens. Determine the distance between the two lenses.
\nline of correct curvature as shown
\n[1 mark]
\nline of approximately correct curvature as shown
\n\n
Judged by eye.
\nAllow second wavefront Y, to have “passed” P as this is how this question is being interpreted by some.
\nIgnore any waves beyond Y.
\n[1 mark]
\nwave travels slower in glass than in air
OR
RI greater for glass
wavelength less in glass than air
hence wave from Q will cover a shorter distance «than in air» causing the curvature shown
\n\n
OWTTE
\n[2 marks]
\nrealization that the two lenses must have a common focal point
\ndistance is 12 – 4.0 = 8.0 «cm»
\n\n
Accept MP1 from a separate diagram or a sketch on the original diagram.
\nA valid reason from MP1 is expected.
\nAward [1 max] for a bald answer of 12 – 4 = 8 «cm».
\n[2 marks]
\nA 1.0 kW heater supplies energy to a liquid of mass 0.50 kg. The temperature of the liquid changes by 80 K in a time of 200 s. The specific heat capacity of the liquid is 4.0 kJ kg–1 K–1. What is the average power lost by the liquid?
\nA. 0
\nB. 200 W
\nC. 800 W
\nD. 1600 W
\nB
\nThe distances between successive positions of a moving car, measured at equal time intervals, are shown.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/08\"](\"../assets/uploads/tinymce_asset/asset/4667/Schermafbeelding_2018-08-10_om_15.58.31.png\"/)
The car moves with
\nA. acceleration that increases linearly with time.
\nB. acceleration that increases non-linearly with time.
\nC. constant speed.
\nD. constant acceleration.
\nD
\nAn object is moving in a straight line. A force F and a resistive force f act on the object along the straight line.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/09\"](\"../assets/uploads/tinymce_asset/asset/4668/Schermafbeelding_2018-08-10_om_16.20.45.png\"/)
Both forces act for a time t.
\nWhat is the rate of change of momentum with time of the object during time t ?
\nA. F + f
\nB. F – f
\nC. (F + f )t
\nD. (F – f )t
\nB
\nUnder what conditions of pressure and temperature does a real gas approximate to an ideal gas?
\nC
\nA fixed mass of an ideal gas is trapped in a cylinder of constant volume and its temperature is varied. Which graph shows the variation of the pressure of the gas with temperature in degrees Celsius?
\n![\"M18/4/PHYSI/SPM/ENG/TZ1/10\"](\"../assets/uploads/tinymce_asset/asset/4669/Schermafbeelding_2018-08-10_om_16.23.05.png\"/)
A
\nBoth optical refracting telescopes and compound microscopes consist of two converging lenses.
\nCompare the focal lengths needed for the objective lens in an refracting telescope and in a compound microscope.
\nA student has four converging lenses of focal length 5, 20, 150 and 500 mm. Determine the maximum magnification that can be obtained with a refracting telescope using two of the lenses.
\nThere are optical telescopes which have diameters about 10 m. There are radio telescopes with single dishes of diameters at least 10 times greater.
\n(i) Discuss why, for the same number of incident photons per unit area, radio telescopes need to be much larger than optical telescopes.
\n(ii) Outline how is it possible for radio telescopes to achieve diameters of the order of a thousand kilometres.
\nThe diagram shows a schematic view of a compound microscope with the focal points fo of the objective lens and the focal points fe of the eyepiece lens marked on the axis.
\nOn the diagram, identify with an X, a suitable position for the image formed by the objective of the compound microscope.
\nImage 1 shows details on the petals of a flower under visible light. Image 2 shows the same flower under ultraviolet light. The magnification is the same, but the resolution is higher in Image 2.
\nExplain why an ultraviolet microscope can increase the resolution of a compound microscope.
\nfOBJECTIVE for telescope > fOBJECTIVE for microscope
OR
fOBJECTIVE for telescope> fEYEPIECE for telescope but fOBJECTIVE for microscope< fEYEPIECE for microscope
\n
OR
100 times
i
RF photons have smaller energy, so signal requires larger dish
RF waves have greater wavelength, good resolution requires larger dish
Must see both, reason and explanation.
ii
use of an array of dishes/many mutually connected antennas «so the effective diameter equals to the distance between the furthest antennas»
between fe and eyepiece lens, on its left
\nAccept any clear indication of the image (eg: X, arrow, dot).
Accept positions which are slightly off axis.
resolution improves as wavelength decreases AND wavelength of UV is smaller
OR
gives resolution formula AND adds that λ is smaller for UV
The graph shows the variation with time t of the velocity v of an object undergoing simple harmonic motion (SHM). At which velocity does the displacement from the mean position take a maximum positive value?
\nD
\nThe radioactive nuclide beryllium-10 (Be-10) undergoes beta minus (β–) decay to form a stable boron (B) nuclide.
\nThe initial number of nuclei in a pure sample of beryllium-10 is N0. The graph shows how the number of remaining beryllium nuclei in the sample varies with time.
\nAn ice sample is moved to a laboratory for analysis. The temperature of the sample is –20 °C.
\nIdentify the missing information for this decay.
\nOn the graph, sketch how the number of boron nuclei in the sample varies with time.
\nAfter 4.3 × 106 years,
\n\n
Show that the half-life of beryllium-10 is 1.4 × 106 years.
\nBeryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 1011 atoms of beryllium-10. State the number of remaining beryllium-10 nuclei in the sample after 2.8 × 106 years.
\nState what is meant by thermal radiation.
\nDiscuss how the frequency of the radiation emitted by a black body can be used to estimate the temperature of the body.
\nCalculate the peak wavelength in the intensity of the radiation emitted by the ice sample.
\nDerive the units of intensity in terms of fundamental SI units.
\n\n
conservation of mass number AND charge ,
\n\n
Correct identification of both missing values required for [1].
\n[1 mark]
\ncorrect shape ie increasing from 0 to about 0.80 N0
\ncrosses given line at 0.50 N0
\n![\"M18/4/PHYSI/SP2/ENG/TZ1/06.b.i/M\"](\"../assets/uploads/tinymce_asset/asset/4691/Schermafbeelding_2018-08-10_om_19.42.49.png\"/)
[2 marks]
\nALTERNATIVE 1
\nfraction of Be = , 12.5%, or 0.125
\ntherefore 3 half lives have elapsed
\n«≈ 1.4 × 106» «y»
\n\n
ALTERNATIVE 2
\nfraction of Be = , 12.5%, or 0.125
\nleading to λ = 4.836 × 10–7 «y»–1
\n= 1.43 × 106 «y»
\n\n
\n
Must see at least one extra sig fig in final answer.
\n[3 marks]
\n1.9 × 1011
\n[1 mark]
\nemission of (infrared) electromagnetic/infrared energy/waves/radiation.
\n[1 mark]
\nthe (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law
\nso frequency/color depends on temperature
\n[2 marks]
\n\n
= 1.1 × 10–5 «m»
\n\n
Allow ECF from MP1 (incorrect temperature).
\n[2 marks]
\ncorrect units for Intensity (allow W, Nms–1 OR Js–1 in numerator)
\nrearrangement into proper SI units = kgs–3
\n\n
Allow ECF for MP2 if final answer is in fundamental units.
\n[2 marks]
\nWhat is the phase difference, in rad, between the centre of a compression and the centre of a rarefaction for a longitudinal travelling wave?
\nA. 0
\nB.
\nC.
\nD.
\nC
\nWhat are the units of the ratio ?
\nA. no units
\nB. k
\nC. k–1
\nD. k–2
\nC
\nAn isolated hollow metal sphere of radius R carries a positive charge. Which graph shows the variation of potential V with distance x from the centre of the sphere?
\nB
\nA spectral line in the light received from a distant galaxy shows a redshift of z = 0.16.
\nState two characteristics of the cosmic microwave background (CMB) radiation.
\nThe present temperature of the CMB is 2.8 K. Calculate the peak wavelength of the CMB.
\nDescribe how the CMB provides evidence for the Hot Big Bang model of the universe.
\nDetermine the distance to this galaxy using a value for the Hubble constant of H0 = 68 km s–1Mpc–1.
\nEstimate the size of the Universe relative to its present size when the light was emitted by the galaxy in (c).
\nblack body radiation / 3 K
\nhighly isotropic / uniform throughout
OR
filling the universe
\n
Do not accept: CMB provides evidence for the Big Bang model.
\n[2 marks]
\n«» ≈ 1.0 «mm»
\n[1 mark]
\nthe universe is expanding and so the wavelength of the CMB in the past was much smaller
\nindicating a very high temperature at the beginning
\n[2 marks]
\n«» v = 0.16 × 3 × 105 «= 0.48 × 105 kms−1»
\n«» ≈ 710 «Mpc»
\n\n
Award [1 max] for POT error.
\n[2 marks]
\n\n
\n
[2 marks]
\nA sealed cylinder of length l and cross-sectional area A contains N molecules of an ideal gas at kelvin temperature T.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/12\"](\"../assets/uploads/tinymce_asset/asset/4670/Schermafbeelding_2018-08-10_om_16.26.56.png\"/)
What is the force acting on the area of the cylinder marked A due to the gas?
\nA.
\nB.
\nC.
\nD.
\nD
\nA non-uniform electric field, with field lines as shown, exists in a region where there is no gravitational field. X is a point in the electric field. The field lines and X lie in the plane of the paper.
\nOutline what is meant by electric field strength.
\nAn electron is placed at X and released from rest. Draw, on the diagram, the direction of the force acting on the electron due to the field.
\nThe electron is replaced by a proton which is also released from rest at X. Compare, without calculation, the motion of the electron with the motion of the proton after release. You may assume that no frictional forces act on the electron or the proton.
\nforce per unit charge
\nacting on a small/test positive charge
\nhorizontally to the left
\nArrow does not need to touch X
\nproton moves to the right/they move in opposite directions
\nforce on each is initially the same
\nproton accelerates less than electron initially «because mass is greater»
\nfield is stronger on right than left «as lines closer»
\nproton acceleration increases «as it is moving into stronger field»
\nOR
\nelectron acceleration decreases «as it is moving into weaker field»
\nAllow ECF from (b)
\nAccept converse argument for electron
\nTwo wave pulses, each of amplitude A, approach each other. They then superpose before continuing in their original directions. What is the total amplitude during superposition and the amplitudes of the individual pulses after superposition?
\nD
\nTwo converging lenses placed a distance 90 cm apart are used as a simple astronomical refracting telescope at normal adjustment. The angular magnification of this arrangement is 17.
\nDetermine the focal length of each lens.
\nThe telescope is used to form an image of the Moon. The angle subtended by the image of the Moon at the eyepiece is 0.16 rad. The distance to the Moon is 3.8 x 108 m. Estimate the diameter of the Moon.
\nState two advantages of the use of satellite-borne telescopes compared to Earth-based telescopes.
\nstates fo + fe = 90 AND
\nsolves to give fo = 85 AND fe = 5 «cm»
\n\n
Both needed.
\nBoth needed.
\n[2 marks]
\nangle subtended by Moon is «rad»
\n\n
D = 3.6 x 106 «m»
\n\n
Allow ECF from MP1.
\nAllow [2] for an answer of 6.1 x 107 «m» if the factor of 17 is missing in MP1.
\n[3 marks]
\noperation day and night
\noperation at all wavelengths/no atmospheric absorption
\noperation without atmospheric turbulence/light pollution
\n\n
Accept any other sensible advantages.
\n[2 marks]
\nThe plane of a coil is positioned at right angles to a magnetic field of flux density B. The coil has N turns, each of area A. The coil is rotated through 180˚ in time t.
\nWhat is the magnitude of the induced emf?
\nA.
\nB.
\nC.
\nD.
\nD
\nA first-harmonic standing wave is formed on a vertical string of length 3.0 m using a vibration generator. The boundary conditions for this string are that it is fixed at one boundary and free at the other boundary.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/13\"](\"../assets/uploads/tinymce_asset/asset/4671/Schermafbeelding_2018-08-10_om_16.36.00.png\"/)
The generator vibrates at a frequency of 300 Hz.
\nWhat is the speed of the wave on the string?
\nA. 0.90 km s–1
\nB. 1.2 km s–1
\nC. 1.8 km s–1
\nD. 3.6 km s–1
\nD
\nThe ratio for a transformer is 2.5.
\nThe primary coil of the transformer draws a current of 0.25 A from a 200 V alternating current (ac) supply. The current in the secondary coil is 0.5 A. What is the efficiency of the transformer?
\nA. 20 %
\nB. 50 %
\nC. 80 %
\nD. 100 %
\nC
\nThe refractive index for light travelling from medium X to medium Y is . The refractive index for light travelling from medium Y to medium Z is . What is the refractive index for light travelling from medium X to medium Z?
\nA.
\nB.
\nC.
\nD.
\nA
\nOptical fibres can be classified, based on the way the light travels through them, as single-mode or multimode fibres. Multimode fibres can be classified as step-index or graded-index fibres.
\nState the main physical difference between step-index and graded-index fibres.
\nExplain why graded-index fibres help reduce waveguide dispersion.
\nstep-index fibres have constant «core» refracting index, graded index fibres have refracting index that reduces/decreases/gets smaller away from axis
OWTTE but refractive index is variable is not enough for the mark.
Award the mark if these ideas are evident in the answer to 14(b).
«in graded index fibres» rays travelling longer paths travel faster
\nso that rays travelling different paths arrive at same/similar time
\nIgnore statements about different colours/wavelengths.
\nA pipe of fixed length is closed at one end. What is ?
\nA.
\nB.
\nC. 3
\nD. 5
\nC
\nAn alternating current (ac) generator produces a peak emf E0 and periodic time T. What are the peak emf and periodic time when the frequency of rotation is doubled?
\nB
\nTwo travelling waves are moving through a medium. The diagram shows, for a point in the medium, the variation with time t of the displacement d of each of the waves.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/14_01\"](\"../assets/uploads/tinymce_asset/asset/4673/Schermafbeelding_2018-08-10_om_16.38.32.png\"/)
For the instant when t = 2.0 ms, what is the phase difference between the waves and what is the resultant displacement of the waves?
\n![\"M18/4/PHYSI/SPM/ENG/TZ1/14_02\"](\"../assets/uploads/tinymce_asset/asset/4674/Schermafbeelding_2018-08-10_om_16.39.41.png\"/)
D
\nA magnetized needle is oscillating on a string about a vertical axis in a horizontal magneticfield B. The time for 10 oscillations is recorded for different values of B.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/01_01\"](\"../assets/uploads/tinymce_asset/asset/4692/Schermafbeelding_2018-08-11_om_06.15.15.png\"/)
The graph shows the variation with B of the time for 10 oscillations together with the uncertainties in the time measurements. The uncertainty in B is negligible.
\nDraw on the graph the line of best fit for the data.
\nWrite down the time taken for one oscillation when B = 0.005 T with its absolute uncertainty.
\nA student forms a hypothesis that the period of one oscillation P is given by:
\n\n
where K is a constant.
\nDetermine the value of K using the point for which B = 0.005 T.
\nState the uncertainty in K to an appropriate number of significant figures.
\nState the unit of K.
\nThe student plots a graph to show how P2 varies with for the data.
\nSketch the shape of the expected line of best fit on the axes below assuming that the relationship is verified. You do not have to put numbers on the axes.
\nState how the value of K can be obtained from the graph.
\nsmooth line, not kinked, passing through all the error bars.
\n[1 mark]
\n0.84 ± 0.03 «s»
\n\n
Accept any value from the range: 0.81 to 0.87.
\nAccept uncertainty 0.03 OR 0.025.
\n[1 mark]
\n\n
«»
\n\n
«K =(0.059 ± 0.002)»
\nuncertainty given to 1sf
\n\n
Allow ECF [3 max] if 10T is used.
\nAward [3] for BCA.
\n[3 marks]
\n\n
\n
Accept or in words.
\n[1 mark]
\nstraight AND ascending line
\nthrough origin
\n[2 marks]
\n\n
[1 mark]
\nIn the circuit shown, the fixed resistor has a value of 3 Ω and the variable resistor can be varied between 0 Ω and 9 Ω.
\nThe power supply has an emf of 12 V and negligible internal resistance. What is the difference between the maximum and minimum values of voltage V across the 3 Ω resistor?
\nA. 3 V
\nB. 6 V
\nC. 9 V
\nD. 12 V
\nC
\nThe graphs show the variation with time of the intensity of a signal that is being transmitted through an optic fibre. Graph 1 shows the input signal to the fibre and Graph 2 shows the output signal from the fibre. The scales of both graphs are identical.
\nThe diagram shows a ray of light in air that enters the core of an optic fibre.
\nThe ray makes an angle A with the normal at the air–core boundary. The refractive index of the core is 1.52 and that of the cladding is 1.48.
\nDetermine the largest angle A for which the light ray will stay within the core of the fibre.
\nIdentify the features of the output signal that indicate the presence of attenuation and dispersion.
\nThe length of the optic fibre is 5.1 km. The input power of the signal is 320 mW. The output power is 77 mW. Calculate the attenuation per unit length of the fibre in dBkm–1.
\ncalculation of critical angle at core–cladding boundary «» θC = 76.8º
\nrefraction angle at air–core boundary 90º – 76.8º = 13.2º
\n«» A = 20.3º
\n\n
Allow ECF from MP1 to MP2 to MP3.
\n[3 marks]
\nattenuation: output signal has smaller area
\ndispersion: output signal is wider than input signal
\n\n
OWTTE
\nOWTTE
\n[2 marks]
\nattenuation = «» «–» 6.2 «dB»
\n= «–» 1.2 «dBkm–1»
\n\n
Allow intensity ratio to be inverted.
\nAllow ECF from MP1 to MP2.
\n[2 marks]
\nSix identical capacitors, each of value C, are connected as shown.
\nWhat is the total capacitance?
\nA.
\nB.
\nC.
\nD. 6C
\nB
\nThe diagram shows an interference pattern produced by two sources that oscillate on the surface of a liquid.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/15\"](\"../assets/uploads/tinymce_asset/asset/4675/Schermafbeelding_2018-08-10_om_16.41.00.png\"/)
Which of the distances shown in the diagram corresponds to one fringe width of the interference pattern?
\nC
\nAlpha Centauri A and B is a binary star system in the main sequence.
\nState what is meant by a binary star system.
\n(i) Calculate .
\n(ii) The luminosity of the Sun is 3.8 × 1026 W. Calculate the radius of Alpha Centauri A.
\nShow, without calculation, that the radius of Alpha Centauri B is smaller than the radius of Alpha Centauri A.
\nAlpha Centauri A is in equilibrium at constant radius. Explain how this equilibrium is maintained.
\nA standard Hertzsprung–Russell (HR) diagram is shown.
\nUsing the HR diagram, draw the present position of Alpha Centauri A and its expected evolutionary path.
\ntwo stars orbiting about a common centre «of mass/gravity»
Do not accept two stars orbiting each other.
i
stars are roughly at the same distance from Earth
OR
d is constant for binaries
\n
Award [2] for a bald correct answer.
\n\n
ii
= 8.4 × 108 «m»
\nAward [2] for a bald correct answer.
\n«A=» B and A have similar temperatures
\nso areas are in ratio of luminosities
\n«so B radius is less than A»
\nradiation pressure/force outwards
\ngravitational pressure/force inwards
\nforces/pressures balance
\nAlpha Centauri A within allowable region
\nsome indication of star moving right and up then left and down ending in white dwarf region as indicated
\nIn an experiment, data were collected on the variation of specific heat capacity of water with temperature. The graph of the plotted data is shown.
\nThe uncertainty in the values for specific heat capacity is 5%.
\nWater of mass (100 ± 2) g is heated from (75.0 ± 0.5) °C to (85.0 ± 0.5) °C.
\nDraw the line of best-fit for the data.
\nDetermine the gradient of the line at a temperature of 80 °C.
\nState the unit for the quantity represented by the gradient in your answer to (b)(i).
\nCalculate the energy required to raise the temperature of the water from 75 °C to 85 °C.
\nUsing an appropriate error calculation, justify the number of significant figures that should be used for your answer to (c)(i).
\nsingle smooth curve passing through all data points
\n\n
Do not accept straight lines joining the dots
\nCurve must touch some part of every x
\ntangent drawn at 80 °C
\ngradient values separated by minimum of 20 °C
\n9.0 × 10–4 «kJ kg–1 K–2»
\nDo not accept tangent unless “ruler” straight.
\nTangent line must be touching the curve drawn for MP1 to be awarded.
\nAccept values between 7.0 × 10–4 and 10 × 10–4.
\nAccept working in J, giving 0.7 to 1.0
\nkJ kg−1 K−2
\n\n
Accept J instead of kJ
\nAccept °C–2 instead of K−2
\nAccept °C–1 K–1 instead of K−2
\nAccept C for °C
\n«0.1 x 4.198 x 10 =» 4.198 «kJ» or 4198 «J»
\nAccept values between 4.19 and 4.21
\npercentage uncertainty in ΔT = 10%
\n«2% + 5% + 10%» = 17%
\nabsolute uncertainty «0.17 × 4.198 =» 0.7 «kJ» therefore 2 sig figs
\nOR
\nabsolute uncertainty to more than 1 sig fig and consistent final answer
\nAllow fractional uncertainties in MP1 and MP2
\nWatch for ECF from (c)(i)
Watch for ECF from MP1
\nWatch for ECF from MP2
Do not accept an answer without justification
\nA proton is accelerated from rest through a potential difference V to a speed of 0.86c.
\nCalculate the potential difference V.
\nThe proton collides with an antiproton moving with the same speed in the opposite direction. As a result both particles are annihilated and two photons of equal energy are produced.
\nDetermine the momentum of one of the photons.
\nγ = 1.96
\nEk = (γ − 1) m0c2 = 900 «MeV»
\npd ≈ 900 «MV»
Award [2 max] if Energy and Potential difference are not clearly distinguished, eg by the unit.
\n[3 marks]
\nenergy of proton = γmc2 = 1838 «MeV»
\ntotal energy available = energy of proton + energy of antiproton = 1838 + 1838 = 3676 «MeV»
\nmomentum of a one photon = Total energy / 2c = 1838 «MeVc–1»
\n[3 marks]
\nKirchhoff’s laws are applied to the circuit shown.
\nWhat is the equation for the dotted loop?
\nA. 0 = 3I2 + 4I3
\nB. 0 = 4I3 − 3I2
\nC. 6 = 2I1 + 3I2 + 4I3
\nD. 6 = 3I2 + 4I3
\nB
\nA capacitor of capacitance C discharges through a resistor of resistance R. The graph shows the variation with time t of the voltage V across the capacitor.
\nThe capacitor is changed to one of value 2C and the resistor is changed to one of value 2R. Which graph shows the variation with t of V when the new combination is discharged?
\nB
\nA system that is subject to a restoring force oscillates about an equilibrium position.
\nFor the motion to be simple harmonic, the restoring force must be proportional to
\nA. the amplitude of the oscillation.
\nB. the displacement from the equilibrium position.
\nC. the potential energy of the system.
\nD. the period of the oscillation.
\nB
\nWith reference to internal energy conversion and ability to be recharged, what are the characteristics of a primary cell?
\nB
\nA particle is displaced from rest and released at time t = 0. It performs simple harmonic motion (SHM). Which graph shows the variation with time of the kinetic energy Ek of the particle?
\n![\"M18/4/PHYSI/SPM/ENG/TZ1/17\"](\"../assets/uploads/tinymce_asset/asset/4676/Schermafbeelding_2018-08-10_om_16.43.41.png\"/)
D
\nMonochromatic electromagnetic radiation is incident on a metal surface. The kinetic energy of the electrons released from the metal
\nA. is constant because the photons have a constant energy.
\nB. is constant because the metal has a constant work function.
\nC. varies because the electrons are not equally bound to the metal lattice.
\nD. varies because the work function of the metal is different for different electrons.
\nC
\nThe diagram shows two current-carrying wires, P and Q, that both lie in the plane of the paper. The arrows show the conventional current direction in the wires.
\nThe electromagnetic force on Q is in the same plane as that of the wires. What is the direction of the electromagnetic force acting on Q?
\nA
\nThe diagram shows the structure of a typical main sequence star.
\nStar X is likely to evolve into a neutron star.
\nState the most abundant element in the core and the most abundant element in the outer layer.
\nThe Hertzsprung–Russell (HR) diagram shows two main sequence stars X and Y and includes lines of constant radius. R is the radius of the Sun.
\n![\"M17/4/PHYSI/SP3/ENG/TZ2/11b\"](\"../assets/uploads/tinymce_asset/asset/3740/Schermafbeelding_2017-09-27_om_09.35.17.png\"/)
Using the mass–luminosity relation and information from the graph, determine the ratio .
\nOn the HR diagram in (b), draw a line to indicate the evolutionary path of star X.
\nOutline why the neutron star that is left after the supernova stage does not collapse under the action of gravitation.
\nThe radius of a typical neutron star is 20 km and its surface temperature is 106 K. Determine the luminosity of this neutron star.
\nDetermine the region of the electromagnetic spectrum in which the neutron star in (c)(iii) emits most of its energy.
\ncore: helium
\nouter layer: hydrogen
\n\n
Accept no other elements.
\n[2 marks]
\nratio of masses is
\nratio of volumes is
\nso ratio of densities is
\n\n
Allow ECF for MP3 from earlier MPs
\n[3 marks]
\nline to the right of X, possibly undulating, very roughly horizontal
\n\n
Ignore any paths beyond this as the star disappears from diagram.
\n[1 mark]
\ngravitation is balanced by a pressure/force due to neutrons/neutron degeneracy/pauli exclusion principle
\n\n
Do not accept electron degeneracy.
\n[1 mark]
\nL = AT 4 = 5.67 x 10–8 x 4 x (2.0 x 104)2 x (106)4
\nL = 3 x 1026 «W»
OR
L = 2.85 x 1026 «W»
\n
Allow ECF for [1 max] if r 2 used (gives 7 x 1026 «W »)
\nAllow ECF for a POT error in MP1.
\n[2 marks]
\n«m»
\nthis is an X-ray wavelength
\n[2 marks]
\nIn an experiment a source of iron-57 emits gamma rays of energy 14.4 ke V. A detector placed 22.6 m vertically above the source measures the frequency of the gamma rays.
\nCalculate the expected shift in frequency between the emitted and the detected gamma rays.
\nExplain whether the detected frequency would be greater or less than the emitted frequency.
\nf = «» = «3.475 × 1018 Hz»
\nΔf = «» 8550 «Hz»
\n[2 marks]
\n«as the photon moves away from the Earth, » it has to spend energy to overcome the gravitational field
\nsince E = hf, the detected frequency would be lower «than the emitted frequency»
\n[2 marks]
\nThree resistors are connected as shown. What is the value of the total resistance between X and Y?
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/18\"](\"../assets/uploads/tinymce_asset/asset/4677/Schermafbeelding_2018-08-10_om_16.44.55.png\"/)
A. 1.5 Ω
\nB. 1.9 Ω
\nC. 6.0 Ω
\nD. 8.0 Ω
\nA
\nA photon interacts with a nearby nucleus to produce an electron. What is the name of this process?
\nA. Pair annihilation
\nB. Pair production
\nC. Electron diffraction
\nD. Quantum tunnelling
\nB
\nA liquid that contains negative charge carriers is flowing through a square pipe with sides A, B, C and D. A magnetic field acts in the direction shown across the pipe.
\nOn which side of the pipe does negative charge accumulate?
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/19\"](\"../assets/uploads/tinymce_asset/asset/4678/Schermafbeelding_2018-08-10_om_16.46.44.png\"/)
A
\nFive resistors of equal resistance are connected to a cell as shown.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/20\"](\"../assets/uploads/tinymce_asset/asset/4679/Schermafbeelding_2018-08-10_om_16.48.40.png\"/)
What is correct about the power dissipated in the resistors?
\nA. The power dissipated is greatest in resistor X.
\nB. The power dissipated is greatest in resistor Y.
\nC. The power dissipated is greatest in resistor Z.
\nD. The power dissipated is the same in all resistors.
\nC
\nA girl on a sledge is moving down a snow slope at a uniform speed.
\nThe sledge, without the girl on it, now travels up a snow slope that makes an angle of 6.5˚ to the horizontal. At the start of the slope, the speed of the sledge is 4.2 m s–1. The coefficient of dynamic friction of the sledge on the snow is 0.11.
\nDraw the free-body diagram for the sledge at the position shown on the snow slope.
\nAfter leaving the snow slope, the girl on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the girl must be in equilibrium as she moves over the horizontal region.
\nWhen the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is 4.2 m s–1. The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg. Calculate the speed of the sledge immediately before the girl jumps from it.
\nThe girl chooses to jump so that she lands on loosely-packed snow rather than frozen ice. Outline why she chooses to land on the snow.
\nShow that the acceleration of the sledge is about –2 m s–2.
\nCalculate the distance along the slope at which the sledge stops moving. Assume that the coefficient of dynamic friction is constant.
\nThe coefficient of static friction between the sledge and the snow is 0.14. Outline, with a calculation, the subsequent motion of the sledge.
\narrow vertically downwards labelled weight «of sledge and/or girl»/W/mg/gravitational force/Fg/Fgravitational AND arrow perpendicular to the snow slope labelled reaction force/R/normal contact force/N/FN
\nfriction force/F/f acting up slope «perpendicular to reaction force»
\nDo not allow G/g/“gravity”.
\nDo not award MP1 if a “driving force” is included.
\nAllow components of weight if correctly labelled.
\nIgnore point of application or shape of object.
\nIgnore “air resistance”.
\nIgnore any reference to “push of feet on sledge”.
\nDo not award MP2 for forces on sledge on horizontal ground
\nThe arrows should contact the object
\n\n
gravitational force/weight from the Earth «downwards»
\nreaction force from the sledge/snow/ground «upwards»
\nno vertical acceleration/remains in contact with the ground/does not move vertically as there is no resultant vertical force
\nAllow naming of forces as in (a)
\nAllow vertical forces are balanced/equal in magnitude/cancel out
\nmention of conservation of momentum
\nOR
\n5.5 x 4.2 = (55 + 5.5) «v»
\n0.38 «m s–1»
\nAllow p=p′ or other algebraically equivalent statement
\nAward [0] for answers based on energy
\n\n
same change in momentum/impulse
\nthe time taken «to stop» would be greater «with the snow»
\ntherefore F is smaller «with the snow»
\nOR
\nforce is proportional to rate of change of momentum therefore F is smaller «with the snow»
\nAllow reverse argument for ice
\n«friction force down slope» = μmg cos(6.5) = «5.9 N»
\n«component of weight down slope» = mg sin(6.5) «= 6.1 N»
\n«so a = » acceleration = = 2.2 «m s–2»
\nIgnore negative signs
\nAllow use of g = 10 m s–2
\ncorrect use of kinematics equation
\ndistance = 4.4 or 4.0 «m»
\nAlternative 2
\nKE lost=work done against friction + GPE
\ndistance = 4.4 or 4.0 «m»
\nAllow ECF from (e)(i)
\nAllow [1 max] for GPE missing leading to 8.2 «m»
\ncalculates a maximum value for the frictional force = «μR=» 7.5 «N»
\nsledge will not move as the maximum static friction force is greater than the component of weight down the slope
\nAllow correct conclusion from incorrect MP1
\nAllow 7.5 > 6.1 so will not move
\nAn electrical circuit is used during an experiment to measure the current I in a variable resistor of resistance R. The emf of the cell is e and the cell has an internal resistance r.
\nA graph shows the variation of with R.
\nShow that the gradient of the graph is equal to .
\nState the value of the intercept on the R axis.
\n«ε = IR + Ir»
\n\n
identifies equation with y = mx + c
\n«hence m = »
\nNo mark for stating data booklet equation
\nDo not accept working where r is ignored or ε = IR is used
\nOWTTE
\n«–» r
\nAllow answer in words
\nThe first graph shows the variation of apparent brightness of a Cepheid star with time.
\nThe second graph shows the average luminosity with period for Cepheid stars.
\n\n
Determine the distance from Earth to the Cepheid star in parsecs. The luminosity of the Sun is 3.8 × 1026 W. The average apparent brightness of the Cepheid star is 1.1 × 10–9 W m–2.
\nExplain why Cephids are used as standard candles.
\nfrom first graph period=5.7 «days» ±0.3 «days»
\nfrom second graph «»
\nd = «» =250 «pc»
\nAccept answer from interval 240 to 270 pc If unit omitted, assume pc.
Watch for ECF from mp1
\n
Cepheids have a definite/known «average» luminosity
\nwhich is determined from «measurement of» period
OR
determined from period-luminosity graph
Cepheids can be used to estimate the distance of galaxies
\nDo not accept brightness for luminosity.
\nAn air bubble has a radius of 0.25 mm and is travelling upwards at its terminal speed in a liquid of viscosity 1.0 × 10–3 Pa s.
\nThe density of air is 1.2 kg m–3 and the density of the liquid is 1200 kg m–3.
\nExplain the origin of the buoyancy force on the air bubble.
\nWith reference to the ratio of weight to buoyancy force, show that the weight of the air bubble can be neglected in this situation.
\nCalculate the terminal speed.
\nALTERNATIVE 1
\npressure in a liquid increases with depth
\nso pressure at bottom of bubble greater than pressure at top
\nALTERNATIVE 2
\nweight of liquid displaced
\ngreater than weight of bubble
\n[2 marks]
\n\n
since the ratio is very small, the weight can be neglected
\n\n
Award [1 max] if only mass of the bubble is calculated and identified as negligible to mass of liquid displaced.
\n[2 marks]
\nevidence of equating the buoyancy and the viscous force «»
\nvt = «» 0.16 «ms–1»
\n[2 marks]
\nTwo resistors X and Y are made of uniform cylinders of the same material. X and Y are connected in series. X and Y are of equal length and the diameter of Y is twice the diameter of X.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/21\"](\"../assets/uploads/tinymce_asset/asset/4680/Schermafbeelding_2018-08-10_om_16.52.34.png\"/)
The resistance of Y is R.
\nWhat is the resistance of this series combination?
\nA.
\nB.
\nC. 3R
\nD. 5R
\nD
\nA student is running an experiment to determine the acceleration of free-fall g. She drops a small metal ball from a given height and measures the time t taken for it to fall using an electronic timer. She repeats the same experiment several times.
\nSuggest a reason for repeating the experiment in the same conditions.
\nWith the collected data she determines the value of g to be (10.4 ± 0.7) m s–2. Researching scientific literature about the location of her experiment she finds the value of g to be (9.807 ± 0.006) m s–2. State, with a reason, whether her experiment is accurate.
\n«to reduce» random errors
\nto reduce absolute uncertainty
\nto improve precision
\nOWTTE
\nDo not accept just “to find an average” or just “reduce error”
\nIgnore any mention to accuracy
\n[Max 1 Mark]
\nas the literature value is within the range «9.7 − 11.1»
\nhence it is accurate
\nOWTTE
\nMP2 must be correctly justified
\nA mass attached to a string rotates in a gravitational field with a constant period in a vertical plane.
\nHow do the tension in the string and the kinetic energy of the mass compare at P and Q?
\nB
\nAn object of mass m at the end of a string of length r moves in a vertical circle at a constant angular speed ω.
\nWhat is the tension in the string when the object is at the bottom of the circle?
\nA. m(ω2r + g)
\nB. m(ω2r – g)
\nC. mg(ω2r + 1)
\nD. mg(ω2r – 1)
\nA
\nOutline the conclusion from Maxwell’s work on electromagnetism that led to one of the postulates of special relativity.
\nlight is an EM wave
\nspeed of light is independent of the source/observer
\nThe peak wavelength of the cosmic microwave background (CMB) radiation spectrum corresponds to a temperature of 2.76 K.
\nIdentify two other characteristics of the CMB radiation that are predicted from the Hot Big Bang theory.
\nA spectral line in the hydrogen spectrum measured in the laboratory today has a wavelength of 21cm. Since the emission of the CMB radiation, the cosmic scale factor has changed by a factor of 1100. Determine the wavelength of the 21cm spectral line in the CMB radiation when it is observed today.
\nisotropic/appears the same from every viewing angle
\nhomogenous/same throughout the universe
\nblack-body radiation
\n23 100 «cm»
OR
231 «m»
A satellite X of mass m orbits the Earth with a period T. What will be the orbital period of satellite Y of mass 2m occupying the same orbit as X?
\nA.
\nB. T
\nC.
\nD. 2T
\nB
\nNewton’s law of gravitation
\nA. is equivalent to Newton’s second law of motion.
\nB. explains the origin of gravitation.
\nC. is used to make predictions.
\nD. is not valid in a vacuum.
\nC
\nA particular emission line in a distant galaxy shows a redshift z = 0.084.
\nThe Hubble constant is H0 = 68 km s–1 Mpc–1.
\nDescribe what is meant by the Big Bang model of the universe.
\nState two features of the cosmic microwave background (CMB) radiation which are consistent with the Big Bang model.
\nDetermine the distance to the galaxy in Mpc.
\nDescribe how type Ia supernovae could be used to measure the distance to this galaxy.
\ntheory in which all space/time/energy/matter were created at a point/singularity
\nat enormous temperature
\nwith the volume of the universe increasing ever since or the universe expanding
\n\n
OWTTE
\n[2 marks]
\nCMB has a black-body spectrum
\nwavelength stretched by expansion
\nis highly isotropic/homogenous
\nbut has minor anisotropies predicted by BB model
\nT «= 2.7 K» is close to predicted value
\n\n
For MP4 and MP5 idea of “prediction” is needed
\n[2 marks]
\n«kms–1»
\n«Mpc»
\n\n
Allow ECF from MP1 to MP2.
\n[2 marks]
\ntype Ia have a known luminosity/are standard candles
\nmeasure apparent brightness
\ndetermine distance from d =
\n\n
Must refer to type Ia. Do not accept other methods (parallax, Cepheids)
\n[3 marks]
\nTwo rockets, A and B, are moving towards each other on the same path. From the frame of reference of the Earth, an observer measures the speed of A to be 0.6c and the speed of B to be 0.4c. According to the observer on Earth, the distance between A and B is 6.0 x 108 m.
\nDefine frame of reference.
\nCalculate, according to the observer on Earth, the time taken for A and B to meet.
\nIdentify the terms in the formula.
\nu′ =
\n\n
Determine, according to an observer in A, the velocity of B.
\nDetermine, according to an observer in A, the time taken for B to meet A.
\nDeduce, without further calculation, how the time taken for A to meet B, according to an observer in B, compares with the time taken for the same event according to an observer in A.
\na co-ordinate system in which measurements «of distance and time» can be made
\nIgnore any mention to inertial reference frame.
\nclosing speed = c
\n2 «s»
\nu and v are velocities with respect to the same frame of reference/Earth AND u′ the relative velocity
\nAccept 0.4c and 0.6c for u and v
\n\n
«–» 0.81c
\n= 1.25
\nso the time is t = 1.6 «s»
\ngamma is smaller for B
\nso time is greater than for A
\nAn experiment to find the internal resistance of a cell of known emf is to be set. The following equipment is available:
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/02\"](\"../assets/uploads/tinymce_asset/asset/4694/Schermafbeelding_2018-08-11_om_06.19.36.png\"/)
Draw a suitable circuit diagram that would enable the internal resistance to be determined.
\nIt is noticed that the resistor gets warmer. Explain how this would affect the calculated value of the internal resistance.
\nOutline how using a variable resistance could improve the accuracy of the value found for the internal resistance.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/02.a/M\"](\"../assets/uploads/tinymce_asset/asset/4708/Schermafbeelding_2018-08-11_om_07.22.52.png\"/)
ammeter and resistor in series
\n[1 mark]
\nresistance of resistor would increase / be greater than 10 Ω
\nR + r «from ε = I(R + r)» would be overestimated / lower current
\ntherefore calculated r would be larger than real
\n\n
Award MP3 only if at least one previous mark has been awarded.
\n[3 marks]
\nvariable resistor would allow for multiple readings to be made
\ngradient of V-I graph could be found «to give r»
\n\n
Award [1 max] for taking average of multiple.
\n[2 marks]
\nWhich statement about atomic spectra is not true?
\nA. They provide evidence for discrete energy levels in atoms.
\nB. Emission and absorption lines of equal frequency correspond to transitions between the same two energy levels.
\nC. Absorption lines arise when electrons gain energy.
\nD. Emission lines always correspond to the visible part of the electromagnetic spectrum.
\nD
\nWhich Feynman diagram shows beta-plus (β+) decay?
\n![\"M18/4/PHYSI/SPM/ENG/TZ1/24\"](\"../assets/uploads/tinymce_asset/asset/4681/Schermafbeelding_2018-08-10_om_16.58.22.png\"/)
A
\nAn electron and a positron have identical speeds but are travelling in opposite directions. Their collision results in the annihilation of both particles and the production of two photons of identical energy. The initial kinetic energy of the electron is 2.00 MeV.
\nExplain, in terms of a conservation law, why two photons need to be created.
\nDetermine the speed of the incoming electron.
\nCalculate the energy and the momentum for each photon after the collision.
\nas the total initial momentum is zero, it must be zero after the collision
\n2 = (–1)m0c2 = (–1) 0.511
\n= 4.91
\nv = 0.978c
\n\n
«2 + 2 + 2 × 0.511 = 5.02 MeV» so each photon is 2.51«MeV»
«MeVc–1»
The Feynman diagram shows electron capture.
\nDeduce that X must be an electron neutrino.
\nDistinguish between hadrons and leptons.
\nit has a lepton number of 1 «as lepton number is conserved»
\nit has a charge of zero/is neutral «as charge is conserved»
\nOR
\nit has a baryon number of 0 «as baryon number is conserved»
\nDo not credit answers referring to energy
\nhadrons experience strong force
\nOR
\nleptons do not experience the strong force
hadrons made of quarks/not fundamental
OR
\nleptons are not made of quarks/are fundamental
\nhadrons decay «eventually» into protons
\nOR
\nleptons do not decay into protons
\nAccept leptons experience the weak force
\nAllow “interaction” for “force”
\nThe graph below shows the displacement y of an oscillating system as a function of time t.
\nState what is meant by damping.
\nCalculate the Q factor for the system.
\nThe Q factor of the system increases. State and explain the change to the graph.
\nthe loss of energy in an oscillating system
\n[1 mark]
\n\n
\n
Accept calculation based on any two correct values giving answer from interval 10 to 13.
\n[1 mark]
\nthe amplitude decreases at a slower rate
\na higher Q factor would mean that less energy is lost per cycle
\n[2 marks]
\nThe average binding energy per nucleon of the nucleus is 7.5 MeV. What is the total energy required to separate the nucleons of one nucleus of ?
\nA. 53 MeV
\nB. 60 MeV
\nC. 113 MeV
\nD. 173 MeV
\nC
\nWhat gives the total change in nuclear mass and the change in nuclear binding energy as a result of a nuclear fusion reaction?
\nB
\nElectrical resistors can be made by forming a thin film of carbon on a layer of an insulating material.
\nA carbon film resistor is made from a film of width 8.0 mm and of thickness 2.0 μm. The diagram shows the direction of charge flow through the resistor.
\nThe resistance of the carbon film is 82 Ω. The resistivity of carbon is 4.1 x 10–5 Ω m. Calculate the length l of the film.
\nThe film must dissipate a power less than 1500 W from each square metre of its surface to avoid damage. Calculate the maximum allowable current for the resistor.
\nState why knowledge of quantities such as resistivity is useful to scientists.
\nThe current direction is now changed so that charge flows vertically through the film.
\nDeduce, without calculation, the change in the resistance.
\nDraw a circuit diagram to show how you could measure the resistance of the carbon-film resistor using a potential divider arrangement to limit the potential difference across the resistor.
\n«l = »
\n0.032 «m»
\npower = 1500 × 8 × 10–3 × 0.032 «= 0.384»
\n«current ≤ »
\n0.068 «A»
\n\n
Be aware of ECF from (a)(i)
\nAward [1] for 4.3 «A» where candidate has not calculated area
\nquantities such as resistivity depend on the material
\nOR
\nthey allow the selection of the correct material
\nOR
\nthey allow scientists to compare properties of materials
\nas area is larger and length is smaller
\nresistance is «very much» smaller
\nAward [1 max] for answers that involve a calculation
\ncomplete functional circuit with ammeter in series with resistor and voltmeter across it
\npotential divider arrangement correct
\neg:
\nThe Feynman diagram shows a particle interaction involving a W– boson.
\nWhich particles are interacting?
\nA. U and Y
\nB. W– boson and Y
\nC. X and Y
\nD. U and X
\nC
\nTwo pure samples of radioactive nuclides X and Y have the same initial number of atoms. The half-life of X is .
\nAfter a time equal to 4 half-lives of X the ratio is .
\nWhat is the half-life of Y?
\nA.
\nB.
\nC.
\nD.
\nD
\nA train is passing through a tunnel of proper length 80 m. The proper length of the train is 100 m. According to an observer at rest relative to the tunnel, when the front of the train coincides with one end of the tunnel, the rear of the train coincides with the other end of the tunnel.
\nExplain what is meant by proper length.
\nDraw a spacetime diagram for this situation according to an observer at rest relative to the tunnel.
\nCalculate the velocity of the train, according to an observer at rest relative to the tunnel, at which the train fits the tunnel.
\nFor an observer on the train, it is the tunnel that is moving and therefore will appear length contracted. This seems to contradict the observation made by the observer at rest to the tunnel, creating a paradox. Explain how this paradox is resolved. You may refer to your spacetime diagram in (b).
\nthe length of an object in its rest frame
\nOR
\nthe length of an object measured when at rest relative to the observer
\nworld lines for front and back of tunnel parallel to ct axis
\nworld lines for front and back of train
\nwhich are parallel to ct′ axis
\nrealizes that gamma = 1.25
\n0.6c
\nALTERNATIVE 1
\nindicates the two simultaneous events for t frame
\nmarks on the diagram the different times «for both spacetime points» on the ct′ axis «shown as Δt′ on each diagram»
\nALTERNATIVE 2: (no diagram reference)
\nthe two events occur at different points in space
\nstatement that the two events are not simultaneous in the t′ frame
\nThe diagram shows the motion of the electrons in a metal wire carrying an electric current as seen by an observer X at rest with respect to the wire. The distance between adjacent positive charges is d.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/03\"](\"../assets/uploads/tinymce_asset/asset/4696/Schermafbeelding_2018-08-11_om_06.22.52.png\"/)
Observer Y is at rest with respect to the electrons.
\nState whether the field around the wire according to observer X is electric, magnetic or a combination of both.
\nDiscuss the change in d according to observer Y.
\nDeduce whether the overall field around the wire is electric, magnetic or a combination of both according to observer Y.
\nmagnetic field
\n[1 mark]
\n«according to Y» the positive charges are moving «to the right»
\nd decreases
\n\n
For MP1, movement of positive charges must be mentioned explicitly.
\n[2 marks]
\npositive charges are moving, so there is a magnetic field
\nthe density of positive charges is higher than that of negative charges, so there is an electric field
\n\n
The reason must be given for each point to be awarded.
\n[2 marks]
\nThe energy-level diagram for an atom that has four energy states is shown.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/27\"](\"../assets/uploads/tinymce_asset/asset/4682/Schermafbeelding_2018-08-10_om_17.05.37.png\"/)
What is the number of different wavelengths in the emission spectrum of this atom?
\nA. 1
\nB. 3
\nC. 6
\nD. 7
\nC
\nIn the context of nuclear magnetic resonance (NMR) imaging explain the role of
\nOutline why the fracture in a broken bone can be seen in a medical X-ray image.
\nThe diagram shows X-rays incident on tissue and bone.
\nThe thicknesses of bone and tissue are both 0.054 m.
\nThe intensity of X-rays transmitted through bone is Ib and the intensity transmitted through tissue is It.
\nThe following data are available.
\nMass absorption coefficient for bone = mass absorption
coefficient for tissue = 1.2 × 10–2m2kg–1
Density of bone = 1.9 × 103 kgm–3
Density of tissue = 1.1 × 103 kgm–3
Calculate the ratio .
\nthe large uniform magnetic field applied to the patient.
\nthe radio-frequency signal emitted towards the patient.
\nthe non-uniform magnetic field applied to the patient.
\nbone and tissue absorb different amounts of X-rays
OR
bone and tissue have different attenuation coefficients
so boundaries and fractures are delineated in an image
\n[2 marks]
\n\n
\n
\n
[3 marks]
\nto split the energy level of protons in the body
OR
to cause protons in the body to align with the field / precess at Larmor frequency
[1 mark]
\nto force/excite protons that are in the spin up/parallel state
\ninto a transition to the spin down/antiparallel state
\n[2 marks]
\nthe emitted radio frequency signal has a frequency that depends on the magnetic field
\nwith a gradient field different parts of the body have different frequencies and so can be identified
\n[2 marks]
\nWhich of the energy sources are classified as renewable and non-renewable?
\nC
\nA lambda 0 particle at rest decays into a proton p and a pion according to the reaction
\n0 → p + –
\nwhere the rest energy of p = 938 MeV and the rest energy of – = 140 MeV.
\nThe speed of the pion after the decay is 0.579c. For this speed = 1.2265. Calculate the speed of the proton.
\npion momentum is «MeVc–1»
\nuse of momentum conservation to realize that produced particles have equal and opposite momenta
\nso for proton
\nsolving to get v = 0.105c
\n\n
Accept pion momentum calculation using E 2 = p 2c 2 +m 2c 4.
Award [2 max] for a non-relativistic answer of v = 0.0864c
\n[4 marks]
\nThe global positioning system (GPS) uses satellites that orbit the Earth. The satellites transmit information to Earth using accurately known time signals derived from atomic clocks on the satellites. The time signals need to be corrected due to the gravitational redshift that occurs because the satellites are at a height of 20 Mm above the surface of the Earth.
\nThe gravitational field strength at 20 Mm above the surface of the Earth is about 0.6 N kg–1. Estimate the time correction per day needed to the time signals, due to the gravitational redshift.
\nSuggest, whether your answer to (a) underestimates or overestimates the correction required to the time signal.
\nso
\n\n
1.3 × 10–10 × 24 × 3600 = 1.15×10–5 «s» «running fast»
\nAward [3 max] if for g 0.6 OR 9.8 OR average of 0.6 and 9.8 is used.
\nALTERNATIVE 1
\ng is not constant through ∆h so value determined should be larger
\nUse ECF from (a)
Accept under or overestimate for SR argument.
ALTERNATIVE 2
\nthe satellite clock is affected by time dilation due to special relativity/its motion
\nWhat is equivalent to ?
\nA. density of the fuel
\nB.
\nC.
\nD.
\nB
\nThe energy density of a substance can be calculated by multiplying its specific energy with which quantity?
\nA. mass
\nB. volume
\nC.
\nD.
\nC
\nThree energy sources for power stations are
\nI. fossil fuel
\nII. pumped water storage
\nIII. nuclear fuel.
\nWhich energy sources are primary sources?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nThe diagram shows a simple climate model for the Earth.
\n![\"M18/4/PHYSI/SPM/ENG/TZ1/30\"](\"../assets/uploads/tinymce_asset/asset/4683/Schermafbeelding_2018-08-10_om_17.12.57.png\"/)
What does this model predict for the average albedo of the Earth?
\nA. 0.30
\nB. 0.51
\nC. 0.70
\nD. 0.81
\nA
\nMuons are created in the upper atmosphere of the Earth at an altitude of 10 km above the surface. The muons travel vertically down at a speed of 0.995c with respect to the Earth. When measured at rest the average lifetime of the muons is 2.1 μs.
\nCalculate, according to Galilean relativity, the time taken for a muon to travel to the ground.
\nDeduce why only a small fraction of the total number of muons created is expected to be detected at ground level according to Galilean relativity.
\nCalculate, according to the theory of special relativity, the time taken for a muon to reach the ground in the reference frame of the muon.
\nDiscuss how your result in (b)(i) and the outcome of the muon decay experiment support the theory of special relativity.
\n«» 34 «μs»
\n\n
Do not accept 104/c = 33 μs.
\n[1 mark]
\ntime is much longer than 10 times the average life time «so only a small proportion would not decay»
\n[1 mark]
\n\n
«» 3.4 «μs»
\n[2 marks]
\nthe value found in (b)(i) is of similar magnitude to average life time
\nsignificant number of muons are observed on the ground
\n«therefore this supports the special theory»
\n[2 marks]
\nA solid cube of side 0.15 m has an average density of 210 kg m–3.
\n(i) Calculate the weight of the cube.
\n(ii) The cube is placed in gasoline of density 720 kg m–3. Calculate the proportion of the volume of the cube that is above the surface of the gasoline.
\nWater flows through a constricted pipe. Vertical tubes A and B, open to the air, are located along the pipe.
\nDescribe why tube B has a lower water level than tube A.
\ni
Fweight = «ρgVcube = 210×9.81×0.153 =» 6.95«N»
ii
Fbuoyancy = 6.95 = ρgV gives V = 9.8×10−4
=0.29 so 0.71 or 71% of the cube is above the gasoline
\nAward [2] for a bald correct answer.
\n«from continuity equation» v is greater at B
OR
area at B is smaller thus «from continuity equation» velocity at B is greater
increase in speed leads to reduction in pressure «through Bernoulli effect»
\npressure related to height of column
OR
p=gh
An elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The other end of the rope is connected to a block which is initially at position A. The block is released from rest. The mass of the rope is negligible.
\n![\"M18/4/PHYSI/SP2/ENG/TZ1/01\"](\"../assets/uploads/tinymce_asset/asset/4684/Schermafbeelding_2018-08-10_om_17.44.22.png\"/)
The unextended length of the rope is 60.0 m. From position A to position B, the block falls freely.
\nAt position C the speed of the block reaches zero. The time taken for the block to fall between B and C is 0.759 s. The mass of the block is 80.0 kg.
\nFor the rope and block, describe the energy changes that take place
\nAt position B the rope starts to extend. Calculate the speed of the block at position B.
\nDetermine the magnitude of the average resultant force acting on the block between B and C.
\nSketch on the diagram the average resultant force acting on the block between B and C. The arrow on the diagram represents the weight of the block.
\nCalculate the magnitude of the average force exerted by the rope on the block between B and C.
\nbetween A and B.
\nbetween B and C.
\nThe length reached by the rope at C is 77.4 m. Suggest how energy considerations could be used to determine the elastic constant of the rope.
\nuse of conservation of energy
\nOR
\nv2 = u2 + 2as
\n\n
v = «» = 34.3 «ms–1»
\n\n
[2 marks]
\nuse of impulse Fave × Δt = Δp
\nOR
\nuse of F = ma with average acceleration
\nOR
\nF =
\n\n
3620«N»
\n\n
Allow ECF from (a).
\n[2 marks]
\nupwards
\nclearly longer than weight
\n\n
For second marking point allow ECF from (b)(i) providing line is upwards.
\n[2 marks]
\n3620 + 80.0 × 9.81
\n4400 «N»
\n\n
Allow ECF from (b)(i).
\n[2 marks]
\n(loss in) gravitational potential energy (of block) into kinetic energy (of block)
\n\n
Must see names of energy (gravitational potential energy and kinetic energy) – Allow for reasonable variations of terminology (eg energy of motion for KE).
\n[1 mark]
\n(loss in) gravitational potential and kinetic energy of block into elastic potential energy of rope
\n\n
See note for 1(c)(i) for naming convention.
\nMust see either the block or the rope (or both) mentioned in connection with the appropriate energies.
\n[1 mark]
\nk can be determined using EPE = kx2
\ncorrect statement or equation showing
\nGPE at A = EPE at C
\nOR
\n(GPE + KE) at B = EPE at C
\n\n
Candidate must clearly indicate the energy associated with either position A or B for MP2.
\n[2 marks]
\nA black body emits radiation with its greatest intensity at a wavelength of Imax. The surface temperature of the black body doubles without any other change occurring. What is the wavelength at which the greatest intensity of radiation is emitted?
\nA. Imax
\nB.
\nC.
\nD.
\nB
\nA hoop of mass m, radius r and moment of inertia mr2 rests on a rough plane inclined at an angle θ to the horizontal. It is released so that the hoop gains linear and angular acceleration by rolling, without slipping, down the plane.
\nOn the diagram, draw and label the forces acting on the hoop.
\nShow that the linear acceleration a of the hoop is given by the equation shown.
\na =
\nCalculate the acceleration of the hoop when θ = 20°. Assume that the hoop continues to roll without slipping.
\nState the relationship between the force of friction and the angle of the incline.
\nThe angle of the incline is slowly increased from zero. Determine the angle, in terms of the coefficient of friction, at which the hoop will begin to slip.
\nweight, normal reaction and friction in correct direction
\ncorrect points of application for at least two correct forces
\nLabelled on diagram.
\nAllow different wording and symbols
\nIgnore relative lengths
\nALTERNATIVE 1
\nma = mg sin θ – Ff
\nI = Ff x r
\nOR
\nmr = Ff
\n=
\nma = mg sin θ – mr → 2a = g sin θ
\nCan be in any order
\nNo mark for re-writing given answer
Accept answers using the parallel axis theorem (with I = 2mr2) only if clear and explicit mention that the only torque is from the weight
Answer given look for correct working
\nALTERNATIVE 2
\nmgh = Iω2 + mv2
\nsubstituting ω = «giving v = »
\ncorrect use of a kinematic equation
\nuse of trigonometry to relate displacement and height «s = h sin θ»
\nFor alternative 2, MP3 and MP4 can only be awarded if the previous marking points are present
\n1.68 «ms–2»
\nALTERNATIVE 1
\nN = mg cos θ
\nFf ≤ μmg cos θ
\nALTERNATIVE 2
\nFf = ma «from 7(b)»
\nso Ff =
\nFf = μmg cos θ
\n= mg sin θ – μmg cos θ
\nOR
\nmg = μmg cos θ
\nalgebraic manipulation to reach tan θ = 2μ
\nAn elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The other end of the rope is connected to a block which is initially at position A. The block is released from rest. The mass of the rope is negligible.
\n
The unextended length of the rope is 60.0 m. From position A to position B, the block falls freely.
\nIn another test, the block hangs in equilibrium at the end of the same elastic rope. The elastic constant of the rope is 400 Nm–1. The block is pulled 3.50 m vertically below the equilibrium position and is then released from rest.
\nAn elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The other end of the rope is connected to a block which is initially at position A. The block is released from rest. The mass of the rope is negligible.
\n
The unextended length of the rope is 60.0 m. From position A to position B, the block falls freely.
\nAt position C the speed of the block reaches zero. The time taken for the block to fall between B and C is 0.759 s. The mass of the block is 80.0 kg.
\nFor the rope and block, describe the energy changes that take place
\nAt position B the rope starts to extend. Calculate the speed of the block at position B.
\nDetermine the magnitude of the average resultant force acting on the block between B and C.
\nSketch on the diagram the average resultant force acting on the block between B and C. The arrow on the diagram represents the weight of the block.
\nCalculate the magnitude of the average force exerted by the rope on the block between B and C.
\nbetween A and B.
\nbetween B and C.
\nThe length reached by the rope at C is 77.4 m. Suggest how energy considerations could be used to determine the elastic constant of the rope.
\nCalculate the time taken for the block to return to the equilibrium position for the first time.
\nCalculate the speed of the block as it passes the equilibrium position.
\nuse of conservation of energy
\nOR
\nv2 = u2 + 2as
\n\n
v = «» = 34.3 «ms–1»
\n\n
[2 marks]
\nuse of impulse Fave × Δt = Δp
\nOR
\nuse of F = ma with average acceleration
\nOR
\nF =
\n\n
3620«N»
\n\n
Allow ECF from (a).
\n[2 marks]
\nupwards
\nclearly longer than weight
\n\n
For second marking point allow ECF from (b)(i) providing line is upwards.
\n[2 marks]
\n3620 + 80.0 × 9.81
\n4400 «N»
\n\n
Allow ECF from (b)(i).
\n[2 marks]
\n(loss in) gravitational potential energy (of block) into kinetic energy (of block)
\n\n
Must see names of energy (gravitational potential energy and kinetic energy) – Allow for reasonable variations of terminology (eg energy of motion for KE).
\n[1 mark]
\n(loss in) gravitational potential and kinetic energy of block into elastic potential energy of rope
\n\n
See note for 1(c)(i) for naming convention.
\nMust see either the block or the rope (or both) mentioned in connection with the appropriate energies.
\n[1 mark]
\nk can be determined using EPE = kx2
\ncorrect statement or equation showing
\nGPE at A = EPE at C
\nOR
\n(GPE + KE) at B = EPE at C
\n\n
Candidate must clearly indicate the energy associated with either position A or B for MP2.
\n[2 marks]
\nT = 2π = 2.81 «s»
\ntime = = 0.702 «s»
\n\n
Award [0] for kinematic solutions that assume a constant acceleration.
\n[2 marks]
\nALTERNATIVE 1
\nω = = 2.24 «rad s–1»
\nv = 2.24 × 3.50 = 7.84 «ms–1»
\n\n
ALTERNATIVE 2
\nkx2 = mv2 OR 400 × 3.52 = 80v2
\nv = 7.84 «ms–1»
\n\n
Award [0] for kinematic solutions that assume a constant acceleration.
\nAllow ECF for T from (e)(i).
\n[2 marks]
\nThe three statements give possible reasons why an average value should be used for the solar constant.
\nI. The Sun’s output varies during its 11 year cycle.
II. The Earth is in elliptical orbit around the Sun.
III. The plane of the Earth’s spin on its axis is tilted to the plane of its orbit about the Sun.
Which are the correct reasons for using an average value for the solar constant?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nThe diagram shows an analogue meter with a mirror behind the pointer.
\nWhat is the main purpose of the mirror?
\nA. To provide extra light when reading the scale
\nB. To reduce the risk of parallax error when reading the scale
\nC. To enable the pointer to be seen from different angles
\nD. To magnify the image of the pointer
\nB
\nA mass-spring system is forced to vibrate vertically at the resonant frequency of the system. The motion of the system is damped using a liquid.
\nAt time t=0 the vibrator is switched on. At time tB the vibrator is switched off and the system comes to rest. The graph shows the variation of the vertical displacement of the system with time until tB.
\n\n
Explain, with reference to energy in the system, the amplitude of oscillation between
\n(i) t = 0 and tA.
\n(ii) tA and tB.
\nThe system is critically damped. Draw, on the graph, the variation of the displacement with time from tB until the system comes to rest.
\ni
\namplitude is increasing as energy is added
\n\n
ii
\nenergy input = energy lost due to damping
\ncurve from time tB reaching zero displacement
\nin no more than one cycle
\nAward zero if displacement after tB goes to negative values.
\nA large cube is formed from ice. A light ray is incident from a vacuum at an angle of 46˚ to the normal on one surface of the cube. The light ray is parallel to the plane of one of the sides of the cube. The angle of refraction inside the cube is 33˚.
\nEach side of the ice cube is 0.75 m in length. The initial temperature of the ice cube is –20 °C.
\nCalculate the speed of light inside the ice cube.
\nShow that no light emerges from side AB.
\nSketch, on the diagram, the subsequent path of the light ray.
\nDetermine the energy required to melt all of the ice from –20 °C to water at a temperature of 0 °C.
\nSpecific latent heat of fusion of ice = 330 kJ kg–1
Specific heat capacity of ice = 2.1 kJ kg–1 k–1
Density of ice = 920 kg m–3
Outline the difference between the molecular structure of a solid and a liquid.
\n«v = c =»
\n2.3 x 108 «m s–1»
\nlight strikes AB at an angle of 57°
\ncritical angle is «sin–1 =» 50.1°
\n49.2° from unrounded value
\nangle of incidence is greater than critical angle so total internal reflection
\nOR
\nlight strikes AB at an angle of 57°
\ncalculation showing sin of “refracted angle” = 1.1
\nstatement that since 1.1>1 the angle does not exist and the light does not emerge
\n[Max 3 marks]
\ntotal internal reflection shown
\nray emerges at opposite face to incidence
\nJudge angle of incidence=angle of reflection by eye or accept correctly labelled angles
\nWith sensible refraction in correct direction
\nmass = «volume x density» (0.75)3 x 920 «= 388 kg»
\nenergy required to raise temperature = 388 x 2100 x 20 «= 1.63 x 107 J»
\nenergy required to melt = 388 x 330 x 103 «= 1.28 x 108 J»
\n1.4 x 108 «J» OR 1.4 x 105 «kJ»
\nAccept any consistent units
\nAward [3 max] for answer which uses density as 1000 kg–3 (1.5× 108 «J»)
\nin solid state, nearest neighbour molecules cannot exchange places/have fixed positions/are closer to each other/have regular pattern/have stronger forces of attraction
\nin liquid, bonds between molecules can be broken and re-form
\nOWTTE
\nAccept converse argument for liquids
\n[Max 1 Mark]
\nWhat is a correct value for the charge on an electron?
\nA. 1.60 x 10–12 μC
\nB. 1.60 x 10–15 mC
\nC. 1.60 x 10–22 kC
\nD. 1.60 x 10–24 MC
\nC
\nAn observer on Earth watches rocket A travel away from Earth at a speed of 0.80c. The spacetime diagram shows the worldline of rocket A in the frame of reference of the Earth observer who is at rest at x = 0.
\nAnother rocket, B, departs from the same location as A but later than A at ct = 1.2 km according to the Earth observer. Rocket B travels at a constant speed of 0.60c in the opposite direction to A according to the Earth observer.
\nRocket A and rocket B both emit a flash of light that are received simultaneously by the Earth observer. Rocket A emits the flash of light at a time coordinate ct = 1.8 km according to the Earth observer.
\nDraw on the spacetime diagram the worldline of B according to the Earth observer and label it B.
\nDeduce, showing your working on the spacetime diagram, the value of ct according to the Earth observer at which the rocket B emitted its flash of light.
\nExplain whether or not the arrival times of the two flashes in the Earth frame are simultaneous events in the frame of rocket A.
\nCalculate the velocity of rocket B relative to rocket A.
\nstraight line with negative gradient with vertical intercept at ct = 1.2 «km»
\nthrough (–0.6, 2.2) ie gradient = –1.67
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/05.a/M\"](\"../assets/uploads/tinymce_asset/asset/4709/Schermafbeelding_2018-08-11_om_10.02.40.png\"/)
\n
Tolerance: Allow gradient from interval –2.0 to –1.4, (at ct = 2.2, x from interval 0.5 to 0.7).
\nIf line has positive gradient from interval 1.4 to 2.0 and intercepts at ct = 1.2 km then allow [1 max].
\n[2 marks]
\nline for the flash of light from A correctly drawn
\nline for the flash of light of B correctly drawn
\ncorrect reading taken for time of intersection of flash of light and path of B, ct = 2.4 «km»
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/05.b/M\"](\"../assets/uploads/tinymce_asset/asset/4710/Schermafbeelding_2018-08-11_om_10.10.53.png\"/)
\n
Accept values in the range: 2.2 to 2.6.
\n[3 marks]
\nthe two events take place in the same point in space at the same time
\nso all observers will observe the two events to be simultaneous / so zero difference
\n\n
Award the second MP only if the first MP is awarded.
\n[2 marks]
\n\n
= «–»0.95 «c»
\n[2 marks]
\nThe linear attenuation coefficient μ of a material is affected by the energy of the X-ray beam and by the density ρ of the material. The mass absorption coefficient is equal to to take into account the density of the material.
\nThe graph shows the variation of mass absorption coefficient with energy of the X-ray beam for both muscle and bone.
\nShow that the attenuation coefficient for bone of density 1800 kg m–3, for X-rays of 20 keV, is about 7 cm–1.
\nThe density of muscle is 1200 kg m–3. Calculate the ratio of intensities to compare, for a beam of 20 keV, the attenuation produced by 1 cm of bone and 1 cm of muscle.
\nSuggest why more energetic beams of about 150 keV would be unsuitable for imaging a bone–muscle section of a body.
\nreads value on graph at 20 keV as 4 «cm2 g–1»
\n«4 cm2 g–1 × 1800 kg m–3 × = » 7.2 «cm–1»
\nEnsure that the calculation has right POT conversion.
\nAnswer must be to at least two significant figures.
\nALTERNATIVE 1
(finds intensity ratios for muscle and bone separately)
Watch for ECF
for muscle: obtains μ = 0.96 cm−1
Allow answers in the range of 0.90 to 1.02 cm–1.
= e−μx so for muscle 0.38
\nAllow answers in the range of 0.36 to 0.41.
\nAllow answers in dB. Muscle -4dB, Bone -30 or -31dB
\nfor bone: = 7.5 × 10−4 «if μ = 7.2 is used»
\nOR
\n9.1×10−4 «if μ=7 is used»
\nALTERNATIVE 2
for muscle: obtains μ = 0.96 cm−1
Allow answers in the range of 0.90 to 1.02 cm–1.
\n
Frequently the POT will be incorrect for MP1. Allow ECF from MP1 to MP2.
Allow +/- 26 or 27dB
Award [2 max] if μ=960 as they will get = 0.
ratio is about 500 «513»
Allow range 395 to 546
If 7 used, ratio is about 420, if 7.2 is used, ratio is about 510
Allow answer IBONE/IMUSCLE from a range 0.0017 to 0.0026.
similar absorption so poor contrast
\nOutline, with reference to star formation, what is meant by the Jeans criterion.
\nIn the proton–proton cycle, four hydrogen nuclei fuse to produce one nucleus of helium releasing a total of 4.3 × 10–12 J of energy. The Sun will spend 1010 years on the main sequence. It may be assumed that during this time the Sun maintains a constant luminosity of 3.8 × 1026 W.
\n
Show that the total mass of hydrogen that is converted into helium while the Sun is on the main sequence is 2 × 1029 kg.
Massive stars that have left the main sequence have a layered structure with different chemical elements in different layers. Discuss this structure by reference to the nuclear reactions taking place in such stars.
\na star will form out of a cloud of gas
\nwhen the gravitational potential energy of the cloud exceeds the total random kinetic energy of the particles of the cloud
OR
the mass exceeds a critical mass for a particular radius and temperature
[2 marks]
\nnumber of reactions is
\nH mass used is «kg»
\n[2 marks]
\nnuclear fusion reactions produce ever heavier elements depending on the mass of the star / temperature of the core
\nthe elements / nuclear reactions arrange themselves in layers, heaviest at the core lightest in the envelope
\n[2 marks]
\nA sunbather is supported in water by a floating sun bed. Which diagram represents the magnitudes of the forces acting on the sun bed?
\nD
\nA monatomic ideal gas is confined to a cylinder with volume 2.0 x 10–3 m3. The initial pressure of the gas is 100 kPa. The gas undergoes a three-step cycle. First, the gas pressure increases by a factor of five under constant volume. Then, the gas expands adiabatically to its initial pressure. Finally it is compressed at constant pressure to its initial volume.
\nShow that the volume of the gas at the end of the adiabatic expansion is approximately 5.3 x 10–3 m3.
\nUsing the axes, sketch the three-step cycle.
\nThe initial temperature of the gas is 290 K. Calculate the temperature of the gas at the start of the adiabatic expansion.
\nUsing your sketched graph in (b), identify the feature that shows that net work is done by the gas in this three-step cycle.
\n\n
V = 5.3 x 10–3 «m3»
\nLook carefully for correct use of pVγ = constant
\ncorrect vertical and horizontal lines
\ncurve between B and C
\n\n
Allow tolerance ±1 square for A, B and C
\nAllow ECF for MP2
\nPoints do not need to be labelled for marking points to be awarded
\nuse of PV = nRT OR use of = constant
\nT = «5 x 290 =» 1450 «K»
\narea enclosed
\nwork is done by the gas during expansion
\nOR
\nwork is done on the gas during compression
\nthe area under the expansion is greater than the area under the compression
\nA toy car of mass 0.15 kg accelerates from a speed of 10 cm s–1 to a speed of 15 cm s–1. What is the impulse acting on the car?
\nA. 7.5 mN s
\nB. 37.5 mN s
\nC. 0.75 N s
\nD. 3.75 N s
\nA
\nA satellite powered by solar cells directed towards the Sun is in a polar orbit about the Earth.
\nThe satellite is orbiting the Earth at a distance of 6600 km from the centre of the Earth.
\nThe satellite carries an experiment that measures the peak wavelength emitted by different objects. The Sun emits radiation that has a peak wavelength λS of 509 nm. The peak wavelength λE of the radiation emitted by the Earth is 10.1 μm.
\nDetermine the orbital period for the satellite.
\nMass of Earth = 6.0 x 1024 kg
\nDetermine the mean temperature of the Earth.
\nSuggest how the difference between λS and λE helps to account for the greenhouse effect.
\nNot all scientists agree that global warming is caused by the activities of man.
\nOutline how scientists try to ensure agreement on a scientific issue.
\n\n
leading to T2 =
\nT = 5320 «s»
\nAlternative 2
\n«» = or 7800 «ms–1»
\ndistance = 2r = 2 x 6600 x 103 «m» or 4.15 x 107 «m»
\n«T = » = 5300 «s»
\nAccept use of ω instead of v
\nT = «»
\n= 287 «K» or 14 «°C»
\nAward [0] for any use of wavelength from Sun
\nDo not accept 287 °C
\nwavelength of radiation from the Sun is shorter than that emitted from Earth «and is not absorbed by the atmosphere»
\ninfrared radiation emitted from Earth is absorbed by greenhouse gases in the atmosphere
\nthis radiation is re-emitted in all directions «including back to Earth»
\npeer review
\ninternational collaboration
\nfull details of experiments published so that experiments can repeated
\n[Max 1 Mark]
\nThe fraction of the internal energy that is due to molecular vibration varies in the different states of matter. What gives the order from highest fraction to lowest fraction of internal energy due to molecular vibration?
\nA. liquid > gas > solid
\nB. solid > liquid > gas
\nC. solid > gas > liquid
\nD. gas > liquid > solid
\nB
\nState the property of protons used in nuclear magnetic resonance (NMR) imaging.
\nExplain how a gradient field and resonance are produced in NMR to allow for the formation of images at a specific plane.
\n«proton» spin
\nstrong B field applied to align proton spins
OWTTE
cross-field applied to give gradient field
OR
each point in a plane has a unique B
RF field excites spins
\nprotons emit RF at resonant/Larmor frequency dependent on Total B field
\nRF detected shows position in the plane / is used to form 2D images
\nAllow features to be mentioned in any order
\nThe graph shows the variation with position s of the displacement x of a wave undergoing simple harmonic motion (SHM).
\nWhat is the magnitude of the velocity at the displacements X, Y and Z?
\nB
\nRecent evidence from the Planck observatory suggests that the matter density of the universe is ρm = 0.32 ρc, where ρc ≈ 10–26 kgm–3 is the critical density.
\nThe graph shows the variation with time t of the cosmic scale factor R in the flat model of the universe in which dark energy is ignored.
\n![\"M17/4/PHYSI/HP3/ENG/TZ1/17.a\"](\"../assets/uploads/tinymce_asset/asset/3731/Schermafbeelding_2017-09-26_om_10.24.01.png\"/)
On the axes above draw a graph to show the variation of R with time, when dark energy is present.
\nThe density of the observable matter in the universe is only 0.05 ρc. Suggest how the remaining 0.27 ρc is accounted for.
\nThe density of dark energy is ρΛc2 where ρΛ = ρc – ρm. Calculate the amount of dark energy in 1 m3 of space.
\ncurve starting earlier, touching at now and going off to infinity
\n[1 mark]
\nthere is dark matter that does not radiate / cannot be observed
\n\n
Unexplained mention of \"dark matter\" is not sufficient for the mark.
\n[1 mark]
\nρΛ = 0.68ρc = 0.68 × 10−26 «kgm−3»
\nenergy in 1 m3 is therefore 0.68 × 10−26 × 9 × 1016 ≈ 6 × 10−10 «J»
\n[2 marks]
\nA closed box of fixed volume 0.15 m3 contains 3.0 mol of an ideal monatomic gas. The temperature of the gas is 290 K.
\nWhen the gas is supplied with 0.86 kJ of energy, its temperature increases by 23 K. The specific heat capacity of the gas is 3.1 kJ kg–1 K–1.
\nCalculate the pressure of the gas.
\nCalculate, in kg, the mass of the gas.
\nCalculate the average kinetic energy of the particles of the gas.
\nExplain, with reference to the kinetic model of an ideal gas, how an increase in temperature of the gas leads to an increase in pressure.
\n«»
\n48 «kPa»
\n[1 mark]
\nmass = «» 0.012 «kg»
\n\n
Award [1] for a bald correct answer.
\n[1 mark]
\n1.38 × 10–23 × 313 = 6.5 × 10–21 «J»
\n[1 mark]
\nlarger temperature implies larger (average) speed/larger (average) KE of molecules/particles/atoms
\nincreased force/momentum transferred to walls (per collision) / more frequent collisions with walls
\nincreased force leads to increased pressure because P = F/A (as area remains constant)
\n\n
Ignore any mention of PV = nRT.
\n[3 marks]
\nDescribe how some white dwarf stars become type Ia supernovae.
\nHence, explain why a type Ia supernova is used as a standard candle.
\nExplain how the observation of type Ia supernovae led to the hypothesis that dark energy exists.
\nwhite dwarf must have companion «in binary system»
\nwhite dwarf gains material «from companion»
\nwhen dwarf reaches and exceeds the Chandrasekhar limit/1.4 MSUN supernova can occur
\na standard candle represents a «stellar object» with a known luminosity
\nthis supernova occurs at an certain/known/exact mass so luminosity/energy released is also known
\nOWTTE
\nMP1 for indication of known luminosity, MP2 for any relevant supportive argument.
\ndistant supernovae were dimmer/further away than expected
\nhence universe is accelerating
\ndark energy «is a hypothesis to» explain this
\nUnpolarized light of intensity I0 is incident on a polarizing filter. Light from this filter is incident on a second filter, which has its axis of polarization at 30˚ to that of the first filter.
\nThe value of cos 30˚ is . What is the intensity of the light emerging through the second filter?
\nA. I0
\nB. I0
\nC. I0
\nD. I0
\nD
\nA constant force of 50.0 N is applied tangentially to the outer edge of a merry-go-round. The following diagram shows the view from above.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/06\"](\"../assets/uploads/tinymce_asset/asset/4698/Schermafbeelding_2018-08-11_om_06.31.00.png\"/)
The merry-go-round has a moment of inertia of 450 kg m2 about a vertical axis. The merry-go-round has a diameter of 4.00 m.
\nA child of mass 30.0 kg is now placed onto the edge of the merry-go-round. No external torque acts on the system.
\nThe child now moves towards the centre.
\nThe merry-go-round starts from rest and the force is applied for one complete revolution.
\nShow that the angular acceleration of the merry-go-round is 0.2 rad s–2.
\nCalculate, for the merry-go-round after one revolution, the angular speed.
\nCalculate, for the merry-go-round after one revolution, the angular momentum.
\nCalculate the new angular speed of the rotating system.
\nExplain why the angular speed will increase.
\nCalculate the work done by the child in moving from the edge to the centre.
\nΓ «= Fr = 50 × 2» = 100 «Nm»
\nα « » =0.22 «rads–2»
\n\n
Final value to at least 2 sig figs, OR clear working with substitution required for mark.
\n[2 marks]
\n«»
\n«»
\n«rads–1»
\n\n
Accept BCA, values in the range: 1.57 to 1.70.
\n[1 mark]
\n«L = Iω = 450 × 1.66»
\n= 750 «kgm2 rads–1»
\n\n
Accept BCA, values in the range: 710 to 780.
\n[1 mark]
\n«I = 450 + mr2»
\nI «= 450 + 30 × 22» = 570 «kgm2»
\n«L = 570 × ω = 747»
\nω = 1.3 «rads–1»
\n\n
Watch for ECF from (a) and (b).
\nAccept BCA, values in the range: 1.25 to 1.35.
\n[2 marks]
\nmoment of inertia will decrease
\nangular momentum will be constant «as the system is isolated»
\n«so the angular speed will increase»
\n[2 marks]
\nωt = 1.66 from bi AND W = ΔEk
\nW = × 450 × 1.662 – × 570 × 1.312 = 131 «J»
\n\n
ECF from 8bi
\nAccept BCA, value depends on the answers in previous questions.
\n[2 marks]
\nA magnifying glass is constructed from a thin converging lens.
\nA converging lens can also be used to produce an image of a distant object. The base of the object is positioned on the principal axis of the lens at a distance of 10.0 m from the centre of the lens. The lens has a focal length of 2.0 m.
\nThe object is replaced with an L shape that is positioned 0.30 m vertically above the principal axis as shown. A screen is used to form a focused image of part of the L shape. Two points P and Q on the base of the L shape and R on its top, are indicated on the diagram. Point Q is 10.0 m away from the same lens as used in part (b).
\nSketch a ray diagram to show how the magnifying glass produces an upright image.
\n\n
State the maximum possible distance from an object to the lens in order for the lens to produce an upright image.
\nDetermine the position of the image.
\n\n
State three characteristics of the image.
\nOn the diagram, draw two rays to locate the point Q′ on the image that corresponds to point Q on the L shape.
\nCalculate the vertical distance of point Q′ from the principal axis.
\nA screen is positioned to form a focused image of point Q. State the direction, relative to Q, in which the screen needs to be moved to form a focused imaged of point R.
\nThe screen is now correctly positioned to form a focused image of point R. However, the top of the L shape looks distorted. Identify and explain the reason for this distortion.
\nwith object placed between lens and focus
\ntwo rays correctly drawn
\nBackwards extrapolation of refracted rays can be dashes or solid lines
\nDo not penalize extrapolated rays which would meet beyond the edge of page
\nImage need not be shown
\n«just less than» the focal length or f
\n\n
v = 2.5 «m»
\nreal, smaller, inverted
\nAll three required — OWTTE
\ntwo correct rays coming from Q
\nlocating Q′ below the main axis AND beyond f to the right of lens AND at intercept of rays
\nAllow any two of the three conventional rays.
\nOR
\n2.5 or 10 × 0.3 «m»
\n«–» 0.075 «m»
\ntowards Q
\nAccept move to the left
\nspherical aberration
\ntop of the shape «R» is far from axis so no paraxial rays
\nFor MP2 accept rays far from the centre converge at different points
\nThere is a proposal to power a space satellite X as it orbits the Earth. In this model, X is connected by an electronically-conducting cable to another smaller satellite Y.
\nSatellite Y orbits closer to the centre of Earth than satellite X. Outline why
\nThe cable acts as a spring. Satellite Y has a mass m of 3.5 x 102 kg. Under certain circumstances, satellite Y will perform simple harmonic motion (SHM) with a period T of 5.2 s.
\nSatellite X orbits 6600 km from the centre of the Earth.
\nMass of the Earth = 6.0 x 1024 kg
\nShow that the orbital speed of satellite X is about 8 km s–1.
\nthe orbital times for X and Y are different.
\nsatellite Y requires a propulsion system.
\nThe cable between the satellites cuts the magnetic field lines of the Earth at right angles.
\nExplain why satellite X becomes positively charged.
\nSatellite X must release ions into the space between the satellites. Explain why the current in the cable will become zero unless there is a method for transferring charge from X to Y.
\nThe magnetic field strength of the Earth is 31 μT at the orbital radius of the satellites. The cable is 15 km in length. Calculate the emf induced in the cable.
\nEstimate the value of k in the following expression.
\nT =
\nGive an appropriate unit for your answer. Ignore the mass of the cable and any oscillation of satellite X.
\nDescribe the energy changes in the satellite Y-cable system during one cycle of the oscillation.
\n«» =
\n7800 «m s–1»
\nFull substitution required
\nMust see 2+ significant figures.
\nY has smaller orbit/orbital speed is greater so time period is less
\nAllow answer from appropriate equation
\nAllow converse argument for X
\nto stop Y from getting ahead
\nto remain stationary with respect to X
\notherwise will add tension to cable/damage satellite/pull X out of its orbit
\ncable is a conductor and contains electrons
\nelectrons/charges experience a force when moving in a magnetic field
\nuse of a suitable hand rule to show that satellite Y becomes negative «so X becomes positive»
\nAlternative 2
\ncable is a conductor
\nso current will flow by induction flow when it moves through a B field
\nuse of a suitable hand rule to show current to right so «X becomes positive»
\nMarks should be awarded from either one alternative or the other.
\nDo not allow discussion of positive charges moving towards X
\nelectrons would build up at satellite Y/positive charge at X
\npreventing further charge flow
\nby electrostatic repulsion
\nunless a complete circuit exists
\n«ε = Blv =» 31 x 10–6 x 7990 x 15000
\n3600 «V»
\nAllow 3700 «V» from v = 8000 m s–1.
\nuse of k = «»
\n510
\nN m–1 or kg s–2
\nAllow MP1 and MP2 for a bald correct answer
\nAllow 500
\nAllow N/m etc.
\nEp in the cable/system transfers to Ek of Y
\nand back again twice in each cycle
\nExclusive use of gravitational potential energy negates MP1
\nThe diagram shows a second harmonic standing wave on a string fixed at both ends.
\nWhat is the phase difference, in rad, between the particle at X and the particle at Y?
\nA. 0
\nB.
\nC.
\nD.
\n\n
A
\nThe graph shows the observed orbital velocities of stars in a galaxy against their distance from the centre of the galaxy. The core of the galaxy has a radius of 4.0 kpc.
\nCalculate the rotation velocity of stars 4.0 kpc from the centre of the galaxy. The average density of the galaxy is 5.0 × 10–21 kg m–3.
\nExplain why the rotation curves are evidence for the existence of dark matter.
\nv = «»
\nv is about 146000 «m s–1» or 146 «km s–1»
Accept answer in the range of 140000 to 160000 «m s–1».
rotation curves/velocity of stars were expected to decrease outside core of galaxy
\nflat curve suggests existence of matter/mass that cannot be seen – now called dark matter
\nA stone falls from rest to the bottom of a water well of depth d. The time t taken to fall is 2.0 ±0.2 s. The depth of the well is calculated to be 20 m using d = at 2. The uncertainty in a is negligible.
\nWhat is the absolute uncertainty in d?
\nA. ± 0.2 m
\nB. ± 1 m
\nC. ± 2 m
\nD. ± 4 m
\nD
\nTwo wires, X and Y, are made from the same metal. The wires are connected in series. The radius of X is twice that of Y. The carrier drift speed in X is vX and in Y it is vY.
\n
What is the value of the ratio ?
A. 0.25
\nB. 0.50
\nC. 2.00
\nD. 4.00
\nA
\nA beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.
\nThe beam is incident normally on a double slit. The distance between the slits is 0.300 mm. A screen is at a distance D from the slits. The diffraction angle θ is labelled.
\n![\"M18/4/PHYSI/SP2/ENG/TZ1/03.a\"](\"../assets/uploads/tinymce_asset/asset/4685/Schermafbeelding_2018-08-10_om_17.53.34.png\"/)
The air between the slits and the screen is replaced with water. The refractive index of water is 1.33.
\nA series of dark and bright fringes appears on the screen. Explain how a dark fringe is formed.
\nThe wavelength of the beam as observed on Earth is 633.0 nm. The separation between a dark and a bright fringe on the screen is 4.50 mm. Calculate D.
\nCalculate the wavelength of the light in water.
\nState two ways in which the intensity pattern on the screen changes.
\nsuperposition of light from each slit / interference of light from both slits
\nwith path/phase difference of any half-odd multiple of wavelength/any odd multiple of (in words or symbols)
\nproducing destructive interference
\n\n
Ignore any reference to crests and troughs.
\n[3 marks]
\nevidence of solving for D «D = »
\n« × 2» = 4.27 «m»
\n\n
Award [1] max for 2.13 m.
\n[2 marks]
\n= 476 «nm»
\n[1 mark]
\ndistance between peaks decreases
\nintensity decreases
\n[2 marks]
\nWhich is a vector quantity?
\nA. Pressure
\nB. Electric current
\nC. Temperature
\nD. Magnetic field
\nD
\nThe diagram shows the magnetic field surrounding two current-carrying metal wires P and Q. The wires are parallel to each other and at right angles to the plane of the page.
\nWhat is the direction of the electron flow in P and the direction of the electron flow in Q?
\nB
\nSamples of different radioactive nuclides have equal numbers of nuclei. Which graph shows the relationship between the half-life and the activity A for the samples?
\nD
\nA ball is tossed vertically upwards with a speed of 5.0 m s–1. After how many seconds will the ball return to its initial position?
\nA. 0.50 s
\nB. 1.0 s
\nC. 1.5 s
\nD. 2.0 s
\nB
\nA projectile is fired horizontally from the top of a cliff. The projectile hits the ground 4 s later at a distance of 2 km from the base of the cliff. What is the height of the cliff?
\nA. 40 m
\nB. 80 m
\nC. 120 m
\nD. 160 m
\nB
\nA spring loaded with mass m oscillates with simple harmonic motion. The amplitude of the motion is A and the spring has total energy E. What is the total energy of the spring when the mass is increased to 3m and the amplitude is increased to 2A?
\nA. 2E
\nB. 4E
\nC. 12E
\nD. 18E
\nB
\nWhat is the unit of electrical energy in fundamental SI units?
\nA. kg m2 C–1 s
B. kg m s–2
C. kg m2 s–2
D. kg m2 s–1 A
C
\nA cylinder is fitted with a piston. A fixed mass of an ideal gas fills the space above the piston.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/07._01\"](\"../assets/uploads/tinymce_asset/asset/4756/Schermafbeelding_2018-08-12_om_16.52.30.png\"/)
The gas expands isobarically. The following data are available.
\n\n
The gas returns to its original state by an adiabatic compression followed by cooling at constant volume.
\nShow that the final volume of the gas is about 53 m3.
\nCalculate, in J, the work done by the gas during this expansion.
\nDetermine the thermal energy which enters the gas during this expansion.
\nSketch, on the pV diagram, the complete cycle of changes for the gas, labelling the changes clearly. The expansion shown in (a) and (b) is drawn for you.
\nOutline the change in entropy of the gas during the cooling at constant volume.
\nThere are various equivalent versions of the second law of thermodynamics. Outline the benefit gained by having alternative forms of a law.
\nALTERNATIVE 1
\n«Using »
\nV2 =
\nV2 = 52.7 «m3»
\n\n
ALTERNATIVE 2
\n«Using PV = nRT»
\nV =
\nV = 52.6 «m3»
\n\n
[2 marks]
\nW «= PΔV» = 11.2 × 103 × (52.7 – 47.1)
\nW = 62.7 × 103 «J»
\n\n
Accept 66.1 × 103 J if 53 used
\nAccept 61.6 × 103 J if 52.6 used
\n[2 marks]
\nΔU «= nRΔT» = 1.5 × 243 × 8.31 × (19 – (–12)) = 9.39 × 104
\nQ «= ΔU + W» = 9.39 × 104 + 6.27 × 104
\nQ = 1.57 × 105 «J»
\n\n
Accept 1.60 × 105 if 66.1 × 103 J used
\nAccept 1.55 × 105 if 61.6 × 103 J used
\n[3 marks]
\nconcave curve from RHS of present line to point above LHS of present line
\nvertical line from previous curve to the beginning
\n\n
![\"M18/4/PHYSI/SP3/ENG/TZ2/07.d.i/M\"](\"../assets/uploads/tinymce_asset/asset/4769/Schermafbeelding_2018-08-13_om_08.30.16.png\"/)
\n
[2 marks]
\nenergy is removed from the gas and so entropy decreases
\nOR
\ntemperature decreases «at constant volume (less disorder)» so entropy decreases
\n\n
OWTTE
\n[1 mark]
\ndifferent paradigms/ways of thinking/modelling/views
\nallows testing in different ways
\nlaws can be applied different situations
\n\n
OWTTE
\n[1 mark]
\nThe pressure–volume (pV) diagram shows a cycle ABCA of a heat engine. The working substance of the engine is 0.221 mol of ideal monatomic gas.
\nAt A the temperature of the gas is 295 K and the pressure of the gas is 1.10 × 105 Pa. The process from A to B is adiabatic.
\nThe process from B to C is replaced by an isothermal process in which the initial state is the same and the final volume is 5.00 × 10–3 m3.
\nShow that the pressure at B is about 5 × 105 Pa.
\nFor the process BC, calculate, in J, the work done by the gas.
\nFor the process BC, calculate, in J, the change in the internal energy of the gas.
\nFor the process BC, calculate, in J, the thermal energy transferred to the gas.
\nExplain, without any calculation, why the pressure after this change would belower if the process was isothermal.
\nDetermine, without any calculation, whether the net work done by the engine during one full cycle would increase or decrease.
\nOutline why an efficiency calculation is important for an engineer designing a heat engine.
\n«»
\n\n
p2 «= » = 5.066 × 105 «Pa»
\n\n
Volume may be in litres or m3.
\nValue to at least 2 sig figs, OR clear working with substitution required for mark.
\n[2 marks]
\n«W = pΔV»
\n«= 5.07 × 105 × (5 × 10–3 – 2 × 10–3)»
\n= 1.52 × 103 «J»
\n\n
Award [0] if POT mistake.
\n[1 mark]
\nΔU = pΔV = 5.07 × 105 × 3 × 10–3 = 2.28 × 10–3 «J»
\n\n
Accept alternative solution via Tc.
\n[1 mark]
\nQ «= (1.5 + 2.28) × 103 =» 3.80 × 103 «J»
\n\n
Watch for ECF from (b)(i) and (b)(ii).
\n[1 mark]
\nfor isothermal process, PV = constant / ideal gas laws mentioned
\nsince VC > VB, PC must be smaller than PB
\n[2 marks]
\nthe area enclosed in the graph would be smaller
\nso the net work done would decrease
\n\n
Award MP2 only if MP1 is awarded.
\n[2 marks]
\nto reduce energy loss; increase engine performance; improve mpg etc
\n\n
Allow any sensible answer.
\n[1 mark]
\nMonochromatic light is incident on two identical slits to produce an interference pattern on a screen. One slit is then covered so that no light emerges from it. What is the change to the pattern observed on the screen?
\nA. Fewer maxima will be observed.
\nB. The intensity of the central maximum will increase.
\nC. The outer maxima will become narrower.
\nD. The width of the central maximum will decrease.
\nA
\nWhich of the following is a scalar quantity?
\nA. Velocity
B. Momentum
C. Kinetic energy
D. Acceleration
C
\nAn object is released from rest in the gravitational field of the Earth. Air resistance is negligible. How far does the object move during the fourth second of its motion?
\nA. 15 m
B. 25 m
C. 35 m
D. 45 m
\n
C
\nA transparent liquid forms a parallel-sided thin film in air. The diagram shows a ray I incident on the upper air–film boundary at normal incidence (the rays are shown at an angle to the normal for clarity).
\nReflections from the top and bottom surfaces of the film result in three rays J, K and L. Which of the rays has undergone a phase change of rad?
\nA. J only
\nB. J and L only
\nC. J and K only
\nD. J, K and L
\nA
\nA stationary sound source emits waves of wavelength and speed v. The source now moves away from a stationary observer. What are the wavelength and speed of the sound as measured by the observer?
\nA
\nAn ohmic conductor is connected to an ideal ammeter and to a power supply of output voltage V.
\n![\"M18/4/PHYSI/SP2/ENG/TZ1/04\"](\"../assets/uploads/tinymce_asset/asset/4686/Schermafbeelding_2018-08-10_om_17.57.33.png\"/)
The following data are available for the conductor:
\ndensity of free electrons = 8.5 × 1022 cm−3
\nresistivity ρ = 1.7 × 10−8 Ωm
\ndimensions w × h × l = 0.020 cm × 0.020 cm × 10 cm.
\n\n
The ammeter reading is 2.0 A.
\nCalculate the resistance of the conductor.
\nCalculate the drift speed v of the electrons in the conductor in cm s–1. State your answer to an appropriate number of significant figures.
\n1.7 × 10–8 ×
\n0.043 «Ω»
\n[2 marks]
\nv «= » =
\n0.368 «cms–1»
\n0.37 «cms–1»
\n\n
Award [2 max] if answer is not expressed to 2 sf.
\n[3 marks]
\nA cell has an emf of 4.0 V and an internal resistance of 2.0 Ω. The ideal voltmeter reads 3.2 V.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/22\"](\"../assets/uploads/tinymce_asset/asset/4730/Schermafbeelding_2018-08-12_om_10.10.06.png\"/)
What is the resistance of R?
\nA. 0.8 Ω
\nB. 2.0 Ω
\nC. 4.0 Ω
\nD. 8.0 Ω
\nD
\nAn astronomical reflecting telescope is being used to observe the night sky.
\nThe diagram shows an incomplete reflecting telescope.
\nComplete the diagram, with a Newtonian mounting, continuing the two rays to show how they pass through the eyepiece.
\nWhen the Earth-Moon distance is 363 300 km, the Moon is observed using the telescope. The mean radius of the Moon is 1737 km. Determine the focal length of the mirror used in this telescope when the diameter of the Moon’s image formed by the main mirror is 1.20 cm.
\nThe final image of the Moon is observed through the eyepiece. The focal length of the eyepiece is 5.0 cm. Calculate the magnification of the telescope.
\nThe Hubble Space reflecting telescope has a Cassegrain mounting. Outline the main optical difference between a Cassegrain mounting and a Newtonian mounting.
\nplane mirror to the left of principal focus tilted anti-clockwise
\ntwo rays which would go through the principal focus
\ntwo rays cross between mirror and eyepiece AND passing through the eyepiece
\neg:
\n\n
f = 1.25 «m»
\nAllow ECF if factor of 2 omitted answer is 2.5m
\nM = = 25
\nparabolic/convex mirror instead of flat mirror
\neyepiece/image axis same as mirror
\nA charge of −3 C is moved from A to B and then back to A. The electric potential at A is +10 V and the electric potential at B is −20 V. What is the work done in moving the charge from A to B and the total work done?
\nC
\nHydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to the ground state. Photons are emitted and are incident on a photoelectric surface as shown.
\n![\"M18/4/PHYSI/HP2/ENG/TZ1/08\"](\"../assets/uploads/tinymce_asset/asset/4802/Schermafbeelding_2018-08-13_om_12.49.40.png\"/)
The photons cause the emission of electrons from the photoelectric surface. The work function of the photoelectric surface is 5.1 eV.
\nThe electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so that the collecting plate is at –1.2 V.
\nShow that the energy of photons from the UV lamp is about 10 eV.
\nCalculate, in J, the maximum kinetic energy of the emitted electrons.
\nSuggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.
\nThe variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.
\nOn the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.
\nAn electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.
\nE1 = –13.6 «eV» E2 = – = –3.4 «eV»
\nenergy of photon is difference E2 – E1 = 10.2 «≈ 10 eV»
\n\n
Must see at least 10.2 eV.
\n[2 marks]
\n10 – 5.1 = 4.9 «eV»
\n4.9 × 1.6 × 10–19 = 7.8 × 10–19 «J»
\n\n
Allow 5.1 if 10.2 is used to give 8.2×10−19 «J».
\nEPE produced by battery
\nexceeds maximum KE of electrons / electrons don’t have enough KE
\n\n
For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.
\n[2 marks]
\n4.9 «V»
\n\n
Allow 5.1 if 10.2 is used in (b)(i).
\nIgnore sign on answer.
\n[1 mark]
\ntwo equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3
\nlabelled correctly
\n![\"M18/4/PHYSI/HP2/ENG/TZ1/08.c.i/M\"](\"../assets/uploads/tinymce_asset/asset/4807/Schermafbeelding_2018-08-13_om_14.47.13.png\"/)
[2 marks]
\nkinetic energy at collecting plate = 0.9 «eV»
\nspeed = «» = 5.6 × 105 «ms–1»
\n\n
Allow ECF from MP1
\n[2 marks]
\nA mass at the end of a string is swung in a horizontal circle at increasing speed until the string breaks.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/23\"](\"../assets/uploads/tinymce_asset/asset/4731/Schermafbeelding_2018-08-12_om_10.12.50.png\"/)
The subsequent path taken by the mass is a
\nA. line along a radius of the circle.
\nB. horizontal circle.
\nC. curve in a horizontal plane.
\nD. curve in a vertical plane.
\nD
\nA spacecraft moves towards the Earth under the influence of the gravitational field of the Earth.
\nThe three quantities that depend on the distance r of the spacecraft from the centre of the Earth are the
\nI. gravitational potential energy of the spacecraft
II gravitational field strength acting on the spacecraft
III. gravitational force acting on the spacecraft.
Which of the quantities are proportional to ?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nA ray diagram for a converging lens is shown. The object is labelled O and the image is labelled I.
\nUsing the ray diagram,
\ndetermine the focal length of the lens.
\ncalculate the linear magnification.
\nThe diagram shows an incomplete ray diagram which consists of a red ray of light and a blue ray of light which are incident on a converging glass lens. In this glass lens the refractive index for blue light is greater than the refractive index for red light.
\nUsing the diagram, outline the phenomenon of chromatic aberration.
\nconstructs ray parallel to principal axis and then to image position
\nOR
\nu = 8 cm and v = 24 cm and lens formula
\n6 «cm»
\n\n
eg: ![\"M18/4/PHYSI/SP3/ENG/TZ2/08.a.i/M\"](\"../assets/uploads/tinymce_asset/asset/4770/Schermafbeelding_2018-08-13_om_08.38.23.png\"/)
Allow answers in the range of 5.6 to 6.4 cm
\n[2 marks]
\nm = «–»3.0
\n[1 mark]
\ncompletes diagram with blue focal point closer to lens
\n\n
blue light/rays refracted/deviated more
\nOR
\nspeed of blue light is less than speed of red light
\nOR
\ndifferent colors/wavelengths have different focal points/converge at different points
\n\n
First marking point can be explained in words or seen on diagram
\n[2 marks]
\nAn electron moves in circular motion in a uniform magnetic field.
\n![\"M18/4/PHYSI/SP2/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/4687/Schermafbeelding_2018-08-10_om_18.05.11.png\"/)
The velocity of the electron at point P is 6.8 × 105 m s–1 in the direction shown.
\nThe magnitude of the magnetic field is 8.5 T.
\nState the direction of the magnetic field.
\nCalculate, in N, the magnitude of the magnetic force acting on the electron.
\nExplain why the electron moves at constant speed.
\nExplain why the electron moves on a circular path.
\nout of the page plane / ⊙
\n\n
Do not accept just “up” or “outwards”.
\n[1 mark]
\n1.60 × 10–19 × 6.8 × 105 × 8.5 = 9.2 × 10–13 «N»
\n[1 mark]
\nthe magnetic force does not do work on the electron hence does not change the electron’s kinetic energy
\nOR
\nthe magnetic force/acceleration is at right angles to velocity
\n\n
[1 mark]
\nthe velocity of the electron is at right angles to the magnetic field
\n(therefore) there is a centripetal acceleration / force acting on the charge
\n\n
OWTTE
\n[2 marks]
\nA detector, placed close to a radioactive source, detects an activity of 260 Bq. The average background activity at this location is 20 Bq. The radioactive nuclide has a half-life of 9 hours.
\nWhat activity is detected after 36 hours?
\nA. 15 Bq
\nB. 16 Bq
\nC. 20 Bq
\nD. 35 Bq
\nD
\nElement X decays through a series of alpha (α) and beta minus (β–) emissions. Which series of emissions results in an isotope of X?
\nA. 1α and 2β–
\nB. 1α and 4β–
\nC. 2α and 2β–
\nD. 2α and 3β–
\nA
\nTwo of the brightest objects in the night sky are the planet Jupiter and the star Vega.
The light observed from Jupiter has a similar brightness to that received from Vega.
Vega is found in the constellation Lyra. The stellar parallax angle of Vega is about 0.13 arc sec.
\nIdentify the mechanism leading stars to produce the light they emit.
\nOutline why the light detected from Jupiter and Vega have a similar brightness, according to an observer on Earth.
\nOutline what is meant by a constellation.
\nOutline how the stellar parallax angle is measured.
\nShow that the distance to Vega from Earth is about 25 ly.
\n«nuclear» fusion
\nDo not accept “burning’’
\nbrightness depends on luminosity and distance/b =
\nVega is much further away but has a larger luminosity
\nAccept answer in terms of Jupiter for MP2
\na group of stars forming a pattern on the sky AND not necessarily close in distance to each other
\nOWTTE
\nthe star’s position is observed at two times, six months apart, relative to distant stars
\nparallax angle is half the angle of shift
\nAnswers may be given in diagram form, so allow the marking points if clearly drawn
\n= 7.7 «pc»
\nso d = 7.7 x 3.26 = 25.1 «ly»
\nA converging (convex) lens forms an image of an object on a screen.
\nIdentify whether the image is real or virtual.
\nThe lens is 18 cm from the screen and the image is 0.40 times smaller than the object. Calculate the power of the lens, in cm–1.
\nLight passing through this lens is subject to chromatic aberration. Discuss the effect that chromatic aberration has on the image formed on the screen.
\nA system consisting of a converging lens of focal length F1 (lens 1) and a diverging lens (lens 2) are used to obtain the image of an object as shown on the scaled diagram. The focal length of lens 1 (F1) is 30 cm.
\nDetermine, using the ray diagram, the focal length of the diverging lens.
\nimage is real «as projected on a screen»
\n[1 mark]
\n«»
\nu = 45
\n\n
OR
\nf = 13 «cm»
\nP = = «» = 0.078 «cm–1»
\n\n
Accept answer 7.7«D».
\n[3 marks]
\nrefractive index depends on wavelength
\nlight of different wavelengths have different focal points / refract differently
\nthere will be coloured fringes around the image / image will be blurred
\n[3 marks]
\nany 2 correct rays to find image from lens 1
\nray to locate F2
\nfocal length = «–»70 «cm»
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/08.b/M\"](\"../assets/uploads/tinymce_asset/asset/4712/Schermafbeelding_2018-08-12_om_07.38.29.png\"/)
\n
Accept values in the range: 65 cm to 75 cm.
\nAccept correct MP3 from accepted range also if working is incorrect or unclear, award [1].
\n[3 marks]
\nA positive pion decays into a positive muon and a neutrino.
\n\n
The momentum of the muon is measured to be 29.8 MeV c–1 in a laboratory reference frame in which the pion is at rest. The rest mass of the muon is 105.7 MeV c–2 and the mass of the neutrino can be assumed to be zero.
\nFor the laboratory reference frame
\nwrite down the momentum of the neutrino.
\nshow that the energy of the pion is about 140 MeV.
\nState the rest mass of the pion with an appropriate unit.
\n«–»29.8 «MeVc–1»
\n[1 mark]
\nEπ = + pvc OR Eμ = 109.8 «MeV»
\nEπ = « + 29.8 =» 139.6 «MeV»
\n\n
Final value to at least 3 sig figs required for mark.
\n[2 marks]
\n139.6 MeVc–2
\n\n
Units required.
\nAccept 140 MeVc–2.
\n[1 mark]
\nA graph of the variation of average binding energy per nucleon with nucleon number has a maximum. What is indicated by the region around the maximum?
\nA. The position below which radioactive decay cannot occur
\nB. The region in which fission is most likely to occur
\nC. The position where the most stable nuclides are found
\nD. The region in which fusion is most likely to occur
\nC
\nThree of the fundamental forces between particles are
\nI. strong nuclear
\nII. weak nuclear
\nIII. electromagnetic.
\nWhat forces are experienced by an electron?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nThe diagram represents a simple optical astronomical reflecting telescope with the path of some light rays shown.
\nIdentify, with the letter X, the position of the focus of the primary mirror.
\nThis arrangement using the secondary mirror is said to increase the focal length of the primary mirror. State why this is an advantage.
\nDistinguish between this mounting and the Newtonian mounting.
\nA radio telescope also has a primary mirror. Identify one difference in the way radiation from this primary mirror is detected.
\nwhere the extensions of the reflected rays from the primary mirror would meet, with construction lines
\n\n
eg:
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/09.a/M\"](\"../assets/uploads/tinymce_asset/asset/4773/Schermafbeelding_2018-08-13_om_08.46.36.png\"/)
[1 mark]
\ngreater magnification
\n[1 mark]
\nNewtonian mount has
\nplane/not curved «secondary» mirror
\n«secondary» mirror at angle/45° to axis
\neyepiece at side/at 90° to axis
\nmount shown is Cassegrain
\n\n
OWTTE
\nAccept these marking points in diagram form
\n[2 marks]
\nwaves collected above mirror/dish
\nwaves collected at the focus of the mirror/dish
\nwaves detected by radio receiver/antenna
\nwaves converted to electrical signals
\n[1 mark]
\nA wind turbine has a power output p when the wind speed is v. The efficiency of the wind turbine does not change. What is the wind speed at which the power output is ?
\nA.
\nB.
\nC.
\nD.
\nD
\nThree gases in the atmosphere are
\nI. carbon dioxide (CO2)
\nII. dinitrogen monoxide (N2O)
\nIII. oxygen (O2).
\nWhich of these are considered to be greenhouse gases?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nAn observer A is on the surface of planet X. Observer B is in a stationary spaceship above the surface of planet X.
\nObserver A sends a beam of light with a frequency 500 THz towards observer B. When observer B receives the light he observes that the frequency has changed by Δf.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/07_01\"](\"../assets/uploads/tinymce_asset/asset/4808/Schermafbeelding_2018-08-13_om_15.27.24.png\"/)
Observer B then sends a signal with frequency 1500 THz towards observer A.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/07_02\"](\"../assets/uploads/tinymce_asset/asset/4809/Schermafbeelding_2018-08-13_om_15.28.04.png\"/)
Calculate the shift in frequency observed by A in terms of Δf.
\nCalculate the gravitational field strength on the surface of planet X.
\nThe following data is given:
\nΔf = 170 Hz.
\nThe distance between observer A and B is 10 km.
\nObserver A now sends a beam of light initially parallel to the surface of the planet.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/07.c\"](\"../assets/uploads/tinymce_asset/asset/4815/Schermafbeelding_2018-08-13_om_16.50.21.png\"/)
Explain why the path of the light is curved.
\nΔf f
\ntherefore the change is «–»3Δf
\n[2 marks]
\ng = «c2 =» (3 × 108)2
\ng = 3.1 «ms–2»
\n\n
If POT mistake, award [0].
\nAward [2] for BCA.
\n[2 marks]
\nthe mass of the planet warps spacetime around itself
\nthe light will follow the shortest path in spacetime «which is curved»
\n[2 marks]
\nA ray of light travelling in an optic fibre undergoes total internal reflection at point P.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/09\"](\"../assets/uploads/tinymce_asset/asset/4703/Schermafbeelding_2018-08-11_om_06.49.40.png\"/)
The refractive index of the core is 1.56 and that of the cladding 1.34.
\nThe input signal in the fibre has a power of 15.0 mW and the attenuation per unit length is 1.24 dB km–1.
\nCalculate the critical angle at the core−cladding boundary.
\nThe use of optical fibres has led to a revolution in communications across the globe. Outline two advantages of optical fibres over electrical conductors for the purpose of data transfer.
\nDraw on the axes an output signal to illustrate the effect of waveguide dispersion.
\nCalculate the power of the output signal after the signal has travelled a distance of 3.40 km in the fibre.
\nExplain how the use of a graded-index fibre will improve the performance of this fibre optic system.
\n«sin c = »
\nc = 59.2«°»
\n\n
Accept values in the range: 59.0 to 59.5.
\nAccept answer 1.0 rad.
\n[1 mark]
\noptic fibres are not susceptible to earthing problems
\noptic fibres are very thin and so do not require the physical space of electrical cables
\noptic fibres offer greater security as the lines cannot be tapped
\noptic fibres are not affected by external electric/magnetic fields/interference
\noptic fibres have lower attenuation than electrical conductors/require less energy
\nthe bandwidth of an optic fibre is large and so it can carry many communications at once/in a shorter time interval/faster data transfer
\n[2 marks]
\na signal that is wider and lower, not necessarily rectangular, but not a larger area
\n[1 mark]
\nattenuation = –1.24 × 3.4 «= –4.216 dB»
\n–4.216 = 10 log
\nI = 5.68 «mW»
\n\n
Need negative attenuation for MP1, may be shown in MP2.
\nFor mp3 answer must be less than 15 mW (even with ECF) to earn mark.
\nAllow [3] for BCA.
\n[3 marks]
\nrefractive index near the edge of the core is less than at the centre
\nspeed of rays which are reflected from the cladding are greater than the speed of rays which travel along the centre of the core
\nthe time difference for the rays that reflect from the cladding layer compared to those that travel along the centre of the core is less
\nOR
\nthe signal will remain more compact/be less spread out/dispersion is lower
\nbit rate of the system may be greater
\n[3 marks]
\nAn optic fibre of length 185 km has an attenuation of 0.200 dB km–1. The input power to the cable is 400.0 μW. The output power from the cable must not fall below 2.0 μW.
\nAn optic fibre of refractive index 1.4475 is surrounded by air. The critical angle for the core – air boundary interface is 44°. Suggest, with a calculation, why the use of cladding with refractive index 1.4444 improves the performance of the optic fibre.
\nCalculate the maximum attenuation allowed for the signal.
\nAn amplifier can increase the power of the signal by 12 dB. Determine the minimum number of amplifiers required.
\nThe graph shows the variation with wavelength of the refractive index of the glass from which the optic fibre is made.
\n ![\"M18/4/PHYSI/SP3/ENG/TZ2/10.b.ii\"](\"../assets/uploads/tinymce_asset/asset/4774/Schermafbeelding_2018-08-13_om_09.07.29.png\"/)
Two light rays enter the fibre at the same instant along the axes. Ray A has a wavelength of λA and ray B has a wavelength of λB. Discuss the effect that the difference in wavelength has on the rays as they pass along the fibre.
\nIn many places clad optic fibres are replacing copper cables. State one example of how fibre optic technology has impacted society.
\nsin c = or sin c = 0.9978
\ncritical angle = 86.2«°»
\nwith cladding only rays travelling nearly parallel to fibre axis are transmitted
\nOR
\npulse broadening/dispersion will be reduced
\n\n
OWTTE
\n[3 marks]
\nattenuation = «10 log» = 10 log
\nattenuation = «–»23 «dB»
\n\n
Accept 10 log for first marking point
\n[2 marks]
\n185 × 0.200 = 37 loss over length of cable
\n« = 1.17» so two amplifiers are sufficient
\n\n
[2 marks]
\nmention of material dispersion
\nmention that rays become separated in time
\nOR
\nmention that ray A travels slower/arrives later than ray B
\n[2 marks]
\nhigh bandwidth/data transfer rates
\nlow distortion/Low noise/Faithful reproduction
\nhigh security
\nfast «fibre» broadband/internet
\nhigh quality optical audio
\nmedical endoscopy
\n\n
Allow any other verifiable sensible advantage
\n[1 mark]
\nMars and Earth act as black bodies. The and .
\nWhat is the value of ?
\nA.
\nB.
\nC.
\nD.
\nB
\nA negatively charged thundercloud above the Earth’s surface may be modelled by a parallel plate capacitor.
\n![\"M18/4/PHYSI/HP2/ENG/TZ2/08\"](\"../assets/uploads/tinymce_asset/asset/4849/Schermafbeelding_2018-08-14_om_06.28.35.png\"/)
The lower plate of the capacitor is the Earth’s surface and the upper plate is the base of the thundercloud.
\nThe following data are available.
\n\n
Lightning takes place when the capacitor discharges through the air between the thundercloud and the Earth’s surface. The time constant of the system is 32 ms. A lightning strike lasts for 18 ms.
\nShow that the capacitance of this arrangement is C = 6.6 × 10–7 F.
\nCalculate in V, the potential difference between the thundercloud and the Earth’s surface.
\nCalculate in J, the energy stored in the system.
\nShow that about –11 C of charge is delivered to the Earth’s surface.
\nCalculate, in A, the average current during the discharge.
\nState one assumption that needs to be made so that the Earth-thundercloud system may be modelled by a parallel plate capacitor.
\nC = «ε =» 8.8 × 10–12 ×
\n«C = 6.60 × 10–7 F»
\n[1 mark]
\nV = « =»
\nV = 3.8 × 107 «V»
\n\n
Award [2] for a bald correct answer
\n[2 marks]
\nALTERNATIVE 1
\nE = «QV =» × 25 × 3.8 × 107
\nE = 4.7 × 108 «J»
\nALTERNATIVE 2
\nE = «CV2 =» × 6.60 × 10–7 × (3.8 × 107)2
\nE = 4.7 × 108 «J» / 4.8 × 108 «J» if rounded value of V used
\n\n
Award [2] for a bald correct answer
\nAllow ECF from (b)(i)
\n\n
[2 marks]
\nQ = « =» 25 ×
\nQ = 14.2 «C»
\ncharge delivered = Q = 25 – 14.2 = 10.8 «C»
\n«≈ –11 C»
\n\n
Final answer must be given to at least 3 significant figures
\n[3 marks]
\nI «= » ≈ 610 «A»
\n\n
Accept an answer in the range 597 − 611 «A»
\n[1 mark]
\nthe base of the thundercloud must be parallel to the Earth surface
\nOR
\nthe base of the thundercloud must be flat
\nOR
\nthe base of the cloud must be very long «compared with the distance from the surface»
\n[1 mark]
\nSirius is a binary star. It is composed of two stars, Sirius A and Sirius B. Sirius A is a main sequence star.
\nThe Sun’s surface temperature is about 5800 K.
\nThe image shows a Hertzsprung–Russell (HR) diagram.
\nThe mass of Sirius A is twice the mass of the Sun. Using the Hertzsprung–Russell (HR) diagram,
\nState what is meant by a binary star.
\nThe peak spectral line of Sirius B has a measured wavelength of 115 nm. Show that the surface temperature of Sirius B is about 25 000 K.
\nThe mass of Sirius B is about the same mass as the Sun. The luminosity of Sirius B is 2.5 % of the luminosity of the Sun. Show, with a calculation, that Sirius B is not a main sequence star.
\nDetermine the radius of Sirius B in terms of the radius of the Sun.
\nIdentify the star type of Sirius B.
\ndraw the approximate positions of Sirius A, labelled A and Sirius B, labelled B.
\nsketch the expected evolutionary path for Sirius A.
\ntwo stars orbiting a common centre «of mass»
\nDo not accept “stars which orbit each other”
\n« x T = 2.9 x 10–3»
\nT = = 25217 «K»
\nuse of the mass-luminosity relationship or = 1
\nif Sirius B is on the main sequence then = 1 «which it is not»
\nConclusion is given, justification must be stated
\nAllow reverse argument beginning with luminosity
\n= 0.025
\nr Sirius = « =» 0.0085 r Sun
\nwhite dwarf
\nSirius A on the main sequence above and to the left of the Sun AND Sirius B on white dwarf area as shown
\nBoth positions must be labelled
\nAllow the position anywhere within the limits shown.
\narrow goes up and right and then loops to white dwarf area
\nA small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless hemispherical bowl.
\n![\"M18/4/PHYSI/SP2/ENG/TZ2/01.a\"](\"../assets/uploads/tinymce_asset/asset/4732/Schermafbeelding_2018-08-12_om_12.45.38.png\"/)
The normal reaction force N makes an angle θ to the horizontal.
\nState the direction of the resultant force on the ball.
\nOn the diagram, construct an arrow of the correct length to represent the weight of the ball.
\nShow that the magnitude of the net force F on the ball is given by the following equation.
\n\n
The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.
\nOutline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.
\nA second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.
\n ![\"M18/4/PHYSI/SP2/ENG/TZ2/01.d\"](\"../assets/uploads/tinymce_asset/asset/4743/Schermafbeelding_2018-08-12_om_13.41.19.png\"/)
The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.
\ntowards the centre «of the circle» / horizontally to the right
\n\n
Do not accept towards the centre of the bowl
\n[1 mark]
\ndownward vertical arrow of any length
\narrow of correct length
\n\n
Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required
\neg: ![\"M18/4/PHYSI/SP2/ENG/TZ2/01.a.ii\"](\"../assets/uploads/tinymce_asset/asset/4740/Schermafbeelding_2018-08-12_om_13.22.33.png\"/)
[2 marks]
\nALTERNATIVE 1
\nF = N cos θ
\nmg = N sin θ
\ndividing/substituting to get result
\n\n
ALTERNATIVE 2
\nright angle triangle drawn with F, N and W/mg labelled
\nangle correctly labelled and arrows on forces in correct directions
\ncorrect use of trigonometry leading to the required relationship
\n\n
![\"M18/4/PHYSI/SP2/ENG/TZ2/01.a.ii\"](\"../assets/uploads/tinymce_asset/asset/4742/Schermafbeelding_2018-08-12_om_13.28.39.png\"/)
tan θ =
\n[3 marks]
\n= m
\nr = R cos θ
\nv =
\nv = 13.4/13 «ms –1»
\n\n
Award [4] for a bald correct answer
\nAward [3] for an answer of 13.9/14 «ms –1». MP2 omitted
\n[4 marks]
\nthere is no force to balance the weight/N is horizontal
\nso no / it is not possible
\n\n
Must see correct justification to award MP2
\n[2 marks]
\nspeed before collision v = « =» 12.5 «ms–1»
\n«from conservation of momentum» common speed after collision is initial speed «vc = = 6.25 ms–1»
\nh = «» 2.0 «m»
\n\n
Allow 12.5 from incorrect use of kinematics equations
\nAward [3] for a bald correct answer
\nAward [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.
\nAllow ECF from MP1
\nAllow ECF from MP2
\n[3 marks]
\nThe water supply for a hydroelectric plant is a reservoir with a large surface area. An outlet pipe takes the water to a turbine.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/10\"](\"../assets/uploads/tinymce_asset/asset/4810/Schermafbeelding_2018-08-13_om_15.42.28.png\"/)
The following data are available:
\n\n
State the difference in terms of the velocity of the water between laminar and turbulent flow.
\nThe water level is a height H above the turbine. Assume that the flow is laminar in the outlet pipe.
\nShow, using the Bernouilli equation, that the speed of the water as it enters the turbine is given by v = .
\nCalculate the Reynolds number for the water flow.
\nOutline whether it is reasonable to assume that flow is laminar in this situation.
\nin laminar flow, the velocity of the fluid is constant «at any point in the fluid» «whereas it is not constant for turbulent flow»
\n\n
Accept any similarly correct answers.
\n[1 mark]
\nPS = PT «as both are exposed to atmospheric pressure»
\nthen VT = 0 «if the surface area ofthe reservoir is large»
\n« + ρgzS = ρgzT»
\n= g(zT – zS) = gH
\nand so vS =
\n\n
MP1 and MP2 may be implied by the correct substitution showing line 3 in the mark scheme.
\nDo not accept simple use of v = .
\n[3 marks]
\nR = = 2.72 × 107
\n\n
Accept use of radius 0.3 m giving value 1.36 × 107.
\n[1 mark]
\nas R > 1000 it is not reasonable to assume laminar flow
\n[1 mark]
\nThe diagram shows the direction of a sound wave travelling in a metal sheet.
\nThe sound wave in air in (c) enters a pipe that is open at both ends. The diagram shows the displacement, at a particular time T, of the standing wave that is set up in the pipe.
\nA particular air molecule has its equilibrium position at the point labelled M.
\nSound of frequency f = 2500 Hz is emitted from an aircraft that moves with speed v = 280 m s–1 away from a stationary observer. The speed of sound in still air is c = 340 m s–1.
\nParticle P in the metal sheet performs simple harmonic oscillations. When the displacement of P is 3.2 μm the magnitude of its acceleration is 7.9 m s-2. Calculate the magnitude of the acceleration of P when its displacement is 2.3 μm.
\nThe wave is incident at point Q on the metal–air boundary. The wave makes an angle of 54° with the normal at Q. The speed of sound in the metal is 6010 m s–1 and the speed of sound in air is 340 m s–1. Calculate the angle between the normal at Q and the direction of the wave in air.
\nThe frequency of the sound wave in the metal is 250 Hz. Determine the wavelength of the wave in air.
\nOn the diagram, at time T, draw an arrow to indicate the acceleration of this molecule.
\nOn the diagram, at time T, label with the letter C a point in the pipe that is at the centre of a compression.
\nCalculate the frequency heard by the observer.
\nCalculate the wavelength measured by the observer.
\nExpression or statement showing acceleration is proportional to displacement ✔
\nso «» = 5.7«ms–2» ✔
\n✔
\n✔
\n✔
\nhorizontal arrow «at M» pointing left ✔
\nany point labelled C on the vertical line shown below ✔
\neg:
\n✔
\n«Hz» ✔
\n«m» ✔
\nThis was well answered at both levels.
\nMany scored full marks on this question. Common errors were using the calculator in radian mode or getting the equation upside down.
\nThis was very well answered.
\nVery few candidates could interpret this situation and most arrows were shown in a vertical plane.
\nThis was answered well at both levels.
\nThis was answered well with the most common mistake being to swap the speed of sound and the speed of the aircraft.
\nAnswered well with ECF often being awarded to those who answered the previous part incorrectly.
\nThe following data apply to the star Gacrux.
\n\n
A Hertzsprung–Russell (HR) diagram is shown.
\nOn the HR diagram,
\nMain sequence stars are in equilibrium under the action of forces. Outline how this equilibrium is achieved.
\nA main sequence star P, is 1.3 times the mass of the Sun. Calculate the luminosity of P relative to the Sun.
\nThe luminosity of the Sun L is 3.85 × 1026 W. Determine the luminosity of Gacrux relative to the Sun.
\nThe distance to Gacrux can be determined using stellar parallax. Outline why this method is not suitable for all stars.
\ndraw the main sequence.
\nplot the position, using the letter P, of the main sequence star P you calculated in (b).
\nplot the position, using the letter G, of Gacrux.
\nDiscuss, with reference to its change in mass, the evolution of star P from the main sequence until its final stable phase.
\nphoton/fusion/radiation force/pressure balances gravitational force/pressure
\ngives both directions correctly (outwards radiation, inwards gravity)
\n\n
OWTTE
\n[2 marks]
\n«L M35 for main sequence»
\nluminosity of P = 2.5 «luminosity of the Sun»
\n[1 mark]
\nLGacrux = 5.67 × 10–8 × 4π × (58.5 × 109)2 × 36004
\nLGacrux = 4.1 × 10–29 «W»
\n«= » = 1.1 × 103
\n[3 marks]
\nif the star is too far then the parallax angle is too small to be measured
\nOR
\nstellar parallax is limited to closer stars
\n\n
OWTTE
\n[1 mark]
\nline or area roughly inside shape shown – judge by eye
\n\n
Accept straight line or straight area at roughly 45°
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/11.d.i/M\"](\"../assets/uploads/tinymce_asset/asset/4775/Schermafbeelding_2018-08-13_om_09.32.24.png\"/)
[1 mark]
\nP between and on main sequence drawn
\n[1 mark]
\nat , further to right than 5000 K and to the left of 2500 K (see shaded region)
\n\n
![\"M18/4/PHYSI/SP3/ENG/TZ2/11.d.iii/M\"](\"../assets/uploads/tinymce_asset/asset/4776/Schermafbeelding_2018-08-13_om_09.40.07.png\"/)
[1 mark]
\nALTERNATIVE 1
\nMain sequence to red giant
\n\n
planetary nebula with mass reduction/loss
\nOR
\nplanetary nebula with mention of remnant mass
\n\n
white dwarf
\n\n
ALTERNATIVE 2
\nMain sequence to red supergiant region
\n\n
Supernova with mass reduction/loss
\nOR
\nSupernova with mention of remnant mass
\n\n
neutron star
\nOR
\nBlack hole
\n\n
OWTTE for both alternatives
\n[3 marks]
\nRhodium-106 () decays into palladium-106 () by beta minus (β–) decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the β– decay.
\n![\"M18/4/PHYSI/HP2/ENG/TZ2/09.d\"](\"../assets/uploads/tinymce_asset/asset/4850/Schermafbeelding_2018-08-14_om_06.42.36.png\"/)
Bohr modified the Rutherford model by introducing the condition mvr = n. Outline the reason for this modification.
\nShow that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression
\n\n
where k is the Coulomb constant.
\nUsing the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.
\n\n
Calculate the electron’s orbital radius in (c)(ii).
\nExplain what may be deduced about the energy of the electron in the β– decay.
\nSuggest why the β– decay is followed by the emission of a gamma ray photon.
\nCalculate the wavelength of the gamma ray photon in (d)(ii).
\nthe electrons accelerate and so radiate energy
\nthey would therefore spiral into the nucleus/atoms would be unstable
\nelectrons have discrete/only certain energy levels
\nthe only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr = n»
\n[3 marks]
\n\n
OR
\nKE = PE hence mev2 =
\n«solving for v to get answer»
\n\n
Answer given – look for correct working
\n[1 mark]
\ncombining v = with mevr = using correct substitution
\n«eg »
\ncorrect algebraic manipulation to gain the answer
\n\n
Answer given – look for correct working
\nDo not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown
\n[2 marks]
\n« r = »
\nr = 5.3 × 10–11 «m»
\n[1 mark]
\nthe energy released is 3.54 – 0.48 = 3.06 «MeV»
\nthis is shared by the electron and the antineutrino
\nso the electron’s energy varies from 0 to 3.06 «MeV»
\n[3 marks]
\nthe palladium nucleus emits the photon when it decays into the ground state «from the excited state»
\n[1 mark]
\nPhoton energy
\nE = 0.48 × 106 × 1.6 × 10–19 = «7.68 × 10–14 J»
\nλ = « =» 2.6 × 10–12 «m»
\n\n
Award [2] for a bald correct answer
\nAllow ECF from incorrect energy
\n[2 marks]
\nDistinguish between the solar system and a galaxy.
\nDistinguish between a planet and a comet.
\na galaxy is much larger in size than a solar system
\na galaxy contains more than one star system / solar system
\na galaxy is more luminous
\nAny other valid statement.
\n[1 mark]
\na comet is a small icy body whereas a planet is mostly made of rock or gas
\na comet is often accompanied by a tail/coma whereas a planet is not
\ncomets (generally) have larger orbits than planets
\na planet must have cleared other objects out of the way in its orbital neighbourhood
\n[1 mark]
\nThe graph below represents the variation with time t of the horizontal displacement x of a mass attached to a vertical spring.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/11\"](\"../assets/uploads/tinymce_asset/asset/4811/Schermafbeelding_2018-08-13_om_15.49.15.png\"/)
The total mass for the oscillating system is 30 kg. For this system
\nDescribe the motion of the spring-mass system.
\ndetermine the initial energy.
\ncalculate the Q at the start of the motion.
\ndamped oscillation / OWTTE
\n[1 mark]
\nE «= × 30 × π2 × 0.82» = 95 «J»
\n\n
Allow initial amplitude between 0.77 to 0.80, giving range between: 88 to 95 J.
\n[1 mark]
\nΔE = 95 – × 30 × π2 × 0.722 = 18 «J»
\nQ = « 2π =» 33
\n\n
Accept values between 0.70 and 0.73, giving a range of ΔE between 22 and 9, giving Q between 27 and 61.
\nWatch for ECF from (b)(i).
\n[2 marks]
\nThe graph shows the observed spectrum from star X.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/11_01\"](\"../assets/uploads/tinymce_asset/asset/4704/Schermafbeelding_2018-08-11_om_06.55.52.png\"/)
The second graph shows the hydrogen emission spectrum in the visible range.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/11_02\"](\"../assets/uploads/tinymce_asset/asset/4705/Schermafbeelding_2018-08-11_om_06.56.45.png\"/)
The following diagram shows the main sequence.
\n![\"M18/4/PHYSI/SP3/ENG/TZ1/11.b\"](\"../assets/uploads/tinymce_asset/asset/4706/Schermafbeelding_2018-08-11_om_06.58.57.png\"/)
Suggest, using the graphs, why star X is most likely to be a main sequence star.
\nShow that the temperature of star X is approximately 10 000 K.
\nWrite down the luminosity of star X (LX) in terms of the luminosity of the Sun (Ls).
\nDetermine the radius of star X (RX) in terms of the radius of the Sun (Rs).
\nEstimate the mass of star X (MX) in terms of the mass of the Sun (Ms).
\nStar X is likely to evolve into a stable white dwarf star.
\nOutline why the radius of a white dwarf star reaches a stable value.
\nthe wavelengths of the dips correspond to the wavelength in the emission spectrum
\n\n
the absorption lines in the spectrum of star X suggest it contains predominantly hydrogen
\nOR
\nmain sequence stars are rich in hydrogen
\n\n
[2 marks]
\npeak wavelength: 290 ± 10 «nm»
\nT = = «10 000 ± 400 K»
\n\n
Substitution in equation must be seen.
\nAllow ECF from MP1.
\n[2 marks][
\n35 ± 5Ls
\n[1 mark]
\n\n
OR
\n\n
\n
(mark for correct substitution)
\nRX = 2.1Rs
\n\n
Allow ECF from (b)(i).
\nAccept values in the range: 2.0 to 2.3Rs.
\nAllow TS in the range: 5500 K to 6500 K.
\n[3 marks]
\nMX = Ms
\nMX = 2.8Ms
\n\n
Allow ECF from (b)(i).
\nDo not accept MX = (35) for first marking point.
\nAccept values in the range: 2.6 to 2.9Ms.
\n[2 marks]
\nthe star «core» collapses until the «inward and outward» forces / pressures are balanced
\nthe outward force / pressure is due to electron degeneracy pressure «not radiation pressure»
\n[2 marks]
\nA motor of input power 160 W raises a mass of 8.0 kg vertically at a constant speed of 0.50 m s–1.
\nWhat is the efficiency of the system?
\nA. 0.63%
\nB. 25%
\nC. 50%
\nD. 100%
\nB
\nAn ideal monatomic gas is kept in a container of volume 2.1 × 10–4 m3, temperature 310 K and pressure 5.3 × 105 Pa.
\nThe volume of the gas in (a) is increased to 6.8 × 10–4 m3 at constant temperature.
\nState what is meant by an ideal gas.
\nCalculate the number of atoms in the gas.
\nCalculate, in J, the internal energy of the gas.
\nCalculate, in Pa, the new pressure of the gas.
\nExplain, in terms of molecular motion, this change in pressure.
\na gas in which there are no intermolecular forces
\nOR
\na gas that obeys the ideal gas law/all gas laws at all pressures, volumes and temperatures
\nOR
\nmolecules have zero PE/only KE
\n\n
Accept atoms/particles.
\n[1 mark]
\nN = «» 2.6 × 1022
\n[1 mark]
\n«For one atom U = kT» × 1.38 × 10–23 × 310 / 6.4 × 10–21 «J»
\nU = «2.6 × 1022 × × 1.38 × 10–23 × 310» 170 «J»
\n\n
Allow ECF from (a)(ii)
\nAward [2] for a bald correct answer
\nAllow use of U = pV
\n[2 marks]
\np2 = «5.3 × 105 × » 1.6 × 105 «Pa»
\n[1 mark]
\n«volume has increased and» average velocity/KE remains unchanged
\n«so» molecules collide with the walls less frequently/longer time between collisions with the walls
\n«hence» rate of change of momentum at wall has decreased
\n«and so pressure has decreased»
\n\n
The idea of average must be included
\nDecrease in number of collisions is not sufficient for MP2. Time must be included.
\nAccept atoms/particles.
\n[2 marks]
\nA box is accelerated to the right across rough ground by a horizontal force Fa. The force of friction is Ff. The weight of the box is Fg and the normal reaction is Fn. Which is the free-body diagram for this situation?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/04\"](\"../assets/uploads/tinymce_asset/asset/4716/Schermafbeelding_2018-08-12_om_09.07.56.png\"/)
D
\nThe table shows the speed of ultrasound and the acoustic impedance for different media.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/14.d\"](\"../assets/uploads/tinymce_asset/asset/4813/Schermafbeelding_2018-08-13_om_15.56.54.png\"/)
The fraction F of the intensity of an ultrasound wave reflected at the boundary between two media having acoustic impedances Z1 and Z2 is given by F = .
\nOutline how ultrasound is generated for medical imaging.
\nDescribe one advantage and one disadvantage of using high frequencies ultrasound over low frequencies ultra sound for medical imaging.
\nSuggest one reason why doctors use ultrasound rather than X-rays to monitor the development of a fetus.
\nCalculate the density of skin.
\nExplain, with appropriate calculations, why a gel is used between the transducer and the skin.
\ncrystal vibration /piezo-electric effect
\ncaused by an alternating potential difference is applied across a crystal
\n[2 marks]
\nADVANTAGES
\nthe wavelength must be less than the size of the object being imaged to avoid diffraction effects
\nthe frequency must be high to ensure several full wavelengths in the pulse
\nDISADVANTAGES
\nthe depth of the organ being imaged must be considered (no more than 200 wavelengths)
\nattenuation increases at higher frequencies
\n\n
[1] for advantages, [1] for disadvantages.
\n[2 marks]
\nX-rays are an ionizing radiation and so might cause harm to the developing fetus.
\nOR
\nthere are no known harmful effects when using ultrasound
\n\n
Ignore “moving images by ultrasound”.
\n[1 mark]
\nρ = = 1.15 × 103 «kgm–3»
\n[1 mark]
\nF = = 1.0
\nF = = 0.02
\nalmost 100% of the ultrasound will be reflected from the air-skin surface OR almost none is transmitted
\nwhereas only 2% will be reflected from the gel-skin surface and so a much greater proportion is transmitted
\n\n
Need to explain that more is transmitted through gel-skin surface for MP4.
\n[4 marks]
\nThe graph shows the variation with time t of the force F acting on an object of mass 15 000 kg.
\nThe object is at rest at t = 0.
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/4717/Schermafbeelding_2018-08-12_om_09.09.53.png\"/)
What is the speed of the object when t = 30 s?
\nA. 0.18 m s–1
\nB. 6 m s–1
\nC. 12 m s–1
\nD. 180 m s–1
\nB
\nThe collision of two galaxies is being studied. The wavelength of a particular spectral line from the galaxy measured from Earth is 116.04 nm. The spectral line when measured from a source on Earth is 115.00 nm.
\nOutline one reason for the difference in wavelength.
\nDetermine the velocity of the galaxy relative to Earth.
\ngalaxies are moving away
\nOR
\nspace «between galaxies» is expanding
\nDo not accept just red-shift
\n«»
\n0.009c
\nAccept 2.7×106 «m s–1»
\nAward [0] if 116 is used for
\nThree identical light bulbs, X, Y and Z, each of resistance 4.0 Ω are connected to a cell of emf 12 V. The cell has negligible internal resistance.
\nWhen fully charged the space between the plates of the capacitor is filled with a dielectric with double the permittivity of a vacuum.
\nThe switch S is initially open. Calculate the total power dissipated in the circuit.
\nThe switch is now closed. State, without calculation, why the current in the cell will increase.
\nThe switch is now closed. .
\n\n
The cell is used to charge a parallel-plate capacitor in a vacuum. The fully charged capacitor is then connected to an ideal voltmeter.
\nThe capacitance of the capacitor is 6.0 μF and the reading of the voltmeter is 12 V.
\nCalculate the energy stored in the capacitor.
\nCalculate the change in the energy stored in the capacitor.
\nSuggest, in terms of conservation of energy, the cause for the above change.
\ntotal resistance of circuit is 8.0 «Ω» ✔
\n«W» ✔
\n«a resistor is now connected in parallel» reducing the total resistance
OR
current through YZ unchanged and additional current flows through X ✔
\nevidence in calculation or statement that pd across Y/current in Y is the same as before ✔
so ratio is 1 ✔
\n«» ✔
\nALTERNATIVE 1
\ncapacitance doubles and voltage halves ✔
since energy halves ✔
\nso change is «–»2.2×10–4 «J» ✔
\n\n
ALTERNATIVE 2
✔
capacitance doubles and charge unchanged so energy halves ✔
so change is «−»2.2 × 10−4 «J» ✔
\nit is the work done when inserting the dielectric into the capacitor ✔
Most candidates scored both marks. ECF was awarded for those who didn’t calculate the new resistance correctly. Candidates showing clearly that they were attempting to calculate the new total resistance helped examiners to award ECF marks.
\nMost recognised that this decreased the total resistance of the circuit. Answers scoring via the second alternative were rare as the statements were often far too vague.
\nVery few gained any credit for this at both levels. Most performed complicated calculations involving the total circuit and using 12V – they had not realised that the question refers to Y only.
\nMost answered this correctly.
\nBy far the most common answer involved doubling the capacitance without considering the change in p.d. Almost all candidates who did this calculated a change in energy that scored 1 mark.
\nVery few scored on this question.
\nA ball of mass m is thrown with an initial speed of u at an angle θ to the horizontal as shown. Q is the highest point of the motion. Air resistance is negligible.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/4718/Schermafbeelding_2018-08-12_om_09.14.39.png\"/)
What is the momentum of the ball at Q?
\nA. zero
\nB. mu cosθ
\nC. mu
\nD. mu sinθ
\nB
\nThe (Lambda) particle decays spontaneously into a proton and a negatively charged pion of rest mass 140 MeV c–2. After the decay, the particles are moving in the same direction with a proton momentum of 630 MeV c–1 and a pion momentum of 270 MeV c–1.
\nDetermine the rest mass of the particle.
\nDetermine, using your answer to (a), the initial speed of the particle.
\nmomentum = 900
\nEproton = «» 1130 «MeV»
\nEpion = «» 304 «MeV»
\nso rest mass of = «» 1116 «MeV c–2»
\n«E = mc2 so» = « =» 1.28
\nto give 0.64c
\nA loudspeaker emits sound towards the open end of a pipe. The other end is closed. A standing wave is formed in the pipe. The diagram represents the displacement of molecules of air in the pipe at an instant of time.
\nX and Y represent the equilibrium positions of two air molecules in the pipe. The arrow represents the velocity of the molecule at Y.
\nThe loudspeaker in (a) now emits sound towards an air–water boundary. A, B and C are parallel wavefronts emitted by the loudspeaker. The parts of wavefronts A and B in water are not shown. Wavefront C has not yet entered the water.
\nOutline how the standing wave is formed.
\nDraw an arrow on the diagram to represent the direction of motion of the molecule at X.
\nLabel a position N that is a node of the standing wave.
\nThe speed of sound is 340 m s–1 and the length of the pipe is 0.30 m. Calculate, in Hz, the frequency of the sound.
\nThe speed of sound in air is 340 m s–1 and in water it is 1500 m s–1.
\nThe wavefronts make an angle θ with the surface of the water. Determine the maximum angle, θmax, at which the sound can enter water. Give your answer to the correct number of significant figures.
\nDraw lines on the diagram to complete wavefronts A and B in water for θ < θmax.
\nthe incident wave «from the speaker» and the reflected wave «from the closed end»
\nsuperpose/combine/interfere
\n\n
Allow superimpose/add up
\nDo not allow meet/interact
\n[1 mark]
\nHorizontal arrow from X to the right
\n\n
MP2 is dependent on MP1
\nIgnore length of arrow
\n[1 mark]
\nP at a node
\n\n
![\"M18/4/PHYSI/SP2/ENG/TZ2/03.a.iii/M\"](\"../assets/uploads/tinymce_asset/asset/4744/Schermafbeelding_2018-08-12_om_14.17.43.png\"/)
[1 mark]
\nwavelength is λ = « =» 0.40 «m»
\nf = «» 850 «Hz»
\n\n
Award [2] for a bald correct answer
\nAllow ECF from MP1
\n[2 marks]
\n\n
θc = 13«°»
\n\n
Award [2] for a bald correct answer
\nAward [2] for a bald answer of 13.1
\n\n
Answer must be to 2/3 significant figures to award MP2
\nAllow 0.23 radians
\n[2 marks]
\ncorrect orientation
\ngreater separation
\n\n
Do not penalize the lengths of A and B in the water
\nDo not penalize a wavefront for C if it is consistent with A and B
\nMP1 must be awarded for MP2 to be awarded
\n![\"M18/4/PHYSI/SP2/ENG/TZ2/03.b.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4745/Schermafbeelding_2018-08-12_om_14.30.57.png\"/)
[2 marks]
\n\n
Two protons, travelling in opposite directions, collide. Each has a total energy of 3.35 GeV.
\nAs a result of the collision, the protons are annihilated and three particles, a proton, a neutron, and a pion are created. The pion has a rest mass of 140 MeV c–2. The total energy of the emitted proton and neutron from the interaction is 6.20 GeV.
\nCalculate the gamma (γ) factor for one of the protons.
\nDetermine, in terms of MeV c–1, the momentum of the pion.
\nThe diagram shows the paths of the incident protons together with the proton and neutron created in the interaction. On the diagram, draw the path of the pion.
\nγ «= » = 3.37
\n[1 mark]
\nenergy of pion = (3350 × 2) – 6200 = 500 «MeV»
\n5002 = p2c2 + 1402
\np = 480 «MeV c–1»
\n\n
[3 marks]
\npath of pion constructed in direction around 4–5 o’clock by eye
\n\n
![\"M18/4/PHYSI/HP3/ENG/TZ2/06.b.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4864/Schermafbeelding_2018-08-14_om_09.50.44.png\"/)
[1 mark]
\nA boy runs along a straight horizontal track. The graph shows how his speed v varies with time t.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/07_01\"](\"../assets/uploads/tinymce_asset/asset/4719/Schermafbeelding_2018-08-12_om_09.15.54.png\"/)
After 15 s the boy has run 50 m. What is his instantaneous speed and his average speed when t = 15 s?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/07_02\"](\"../assets/uploads/tinymce_asset/asset/4720/Schermafbeelding_2018-08-12_om_09.16.57.png\"/)
C
\nData from distant galaxies are shown on the graph.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/12\"](\"../assets/uploads/tinymce_asset/asset/4761/Schermafbeelding_2018-08-12_om_17.19.55.png\"/)
Estimate, using the data, the age of the universe. Give your answer in seconds.
\nIdentify the assumption that you made in your answer to (a).
\nOn the graph, one galaxy is labelled A. Determine the size of the universe, relative to its present size, when light from the galaxy labelled A was emitted.
\nuse of gradient or any coordinate pair to find H0 «= » or «= »
\nconvert Mpc to m and km to m «for example »
\nage of universe «= » = 3.8 × 1017 «s»
\n\n
\n
Allow final answers between
\n3.7 × 1017 and 3.9 × 1017 «s» or 4 × 1017 «s»
\n[3 marks]
\nnon-accelerated/uniform rate of expansion
\nOR
\nH0 constant over time
\n\n
OWTTE
\n[1 mark]
\nz « = » = = 0.15
\n= «z + 1» = 1.15
\n\n
= « =» 0.87
\nOR
\n87% of the present size
\n\n
[3 marks]
\nThe Schwarzschild radius of a black hole is 6.0 x 105 m. A rocket is 7.0 x 108 m from the black hole and has a clock. The proper time interval between the ticks of the clock on the rocket is 1.0 s. These ticks are transmitted to a distant observer in a region free of gravitational fields.
\nOutline why the clock near the black hole runs slowly compared to a clock close to the distant observer.
\nCalculate the number of ticks detected in 10 ks by the distant observer.
\nthis is gravitational time dilation
\nOR
\nblack hole gives rise to a «strong» gravitational field
\nclocks in stronger field run more slowly
\nOR
\nthe clock «signal» is subject to gravitational red-shift
\nthe clock is subject to gravitational red shift
\nOR
\nthe clock has lost gravitational potential energy in moving close to the black hole
\n[Max 2 Marks]
\nALTERNATIVE 1 (10 ks is in observer frame):
\nΔt' =
\n9995.7 so 9995 «ticks»
\nAllow 9996
\nAllow ECF if 10 is used instead of 10000
\nALTERNATIVE 2 (10 ks is in rocket frame):
\nΔt =
\n10004 «ticks»
\nAllow ECF if 10 is used instead of 10000
\nA weight W is tied to a trolley of mass M by a light string passing over a frictionless pulley. The trolley has an acceleration a on a frictionless table. The acceleration due to gravity is g.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/08\"](\"../assets/uploads/tinymce_asset/asset/4721/Schermafbeelding_2018-08-12_om_09.18.54.png\"/)
What is W ?
\nA.
\nB.
\nC.
\nD.
\nA
\nThe Hubble constant is accepted to be 70 km s–1 Mpc–1. This value of the Hubble constant gives an age for the universe of 14.0 billion years.
\nThe accepted value of the Hubble constant has changed over the past decades.
\nThe redshift of a galaxy is measured to be z = 0.19.
\nExplain how international collaboration has helped to refine this value.
\nEstimate, in Mpc, the distance between the galaxy and the Earth.
\nDetermine, in years, the approximate age of the universe at the instant when the detected light from the distant galaxy was emitted.
\nexperiments and collecting data are extremely costly
\ndata from many projects around the world can be collated
\n\n
OWTTE
\n[1 mark]
\nv = «zc = 0.19 × 3 × 108 =» 5.7 × 107 «ms–1»
\nd = «» = 810Mpc OR 8.1× 108 pc
\n\n
Correct units must be present for MP2 to be awarded.
\nAward [2] for BCA.
\n[2 marks]
\nALTERNATIVE 1
\n= 1 + z = 1.19
\nso (assuming constant expansion rate) = 1.19
\nt = = 11.7By = 12«By (billion years)»
\n\n
ALTERNATIVE 2
\nlight has travelled a distance: (810 × 106 × 3.26 =) 2.6 × 109ly
\nso light was emitted: 2.6 billion years ago
\nso the universe was 11.4 billion years old
\n\n
MP1 can be awarded if MP2 clearly seen.
\nAccept 2.5 × 1025 m for mp1.
\nMP1 can be awarded if MP2 clearly seen.
\n[3 marks]
\nType Ia supernovae typically have a peak luminosity of around 5 × 105 Ls, where Ls is the luminosity of the Sun (3.8 × 1026 W). A type Ia supernova is observed with an apparent peak brightness of 1.6 × 10–6 W m–2.
\nDescribe the formation of a type Ia supernova.
\nShow that the distance to the supernova is approximately 3.1 × 1018 m.
\nState one assumption made in your calculation.
\na white dwarf accretes mass «from a binary partner»
\nwhen the mass becomes more than the Chandrasekhar limit (1.4Ms) «then asupernova explosion takes place»
\n[2 marks]
\nd =
\nd = 3.07 × 1018 «m»
\n\n
At least 3 sig fig required for MP2.
\n[2 marks]
\ntype Ia supernova can be used as standard candles
\nthere is no dust absorbing light between Earth and supernova
\ntheir supernova is a typical type Ia
\n[1 mark]
\nTwo balls X and Y with the same diameter are fired horizontally with the same initial momentum from the same height above the ground. The mass of X is greater than the mass of Y. Air resistance is negligible.
\nWhat is correct about the horizontal distances travelled by X and Y and the times taken by X and Y to reach the ground?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/09\"](\"../assets/uploads/tinymce_asset/asset/4722/Schermafbeelding_2018-08-12_om_09.22.08.png\"/)
C
\nThe diagram shows a simplified model of a Galilean thermometer. The thermometer consists of a sealed glass cylinder that contains ethanol, together with glass spheres. The spheres are filled with different volumes of coloured water. The mass of the glass can be neglected as well as any expansion of the glass through the temperature range experienced. Spheres have tags to identify the temperature. The mass of the tags can be neglected in all calculations.
\nEach sphere has a radius of 3.0 cm and the spheres, due to the different volumes of water in them, are of varying densities. As the temperature of the ethanol changes the individual spheres rise or fall, depending on their densities, compared with that of the ethanol.
\nThe graph shows the variation with temperature of the density of ethanol.
\nUsing the graph, determine the buoyancy force acting on a sphere when the ethanol is at a temperature of 25 °C.
\nWhen the ethanol is at a temperature of 25 °C, the 25 °C sphere is just at equilibrium. This sphere contains water of density 1080 kg m–3. Calculate the percentage of the sphere volume filled by water.
\nThe room temperature slightly increases from 25 °C, causing the buoyancy force to decrease. For this change in temperature, the ethanol density decreases from 785.20 kg m–3 to 785.16 kg m–3. The average viscosity of ethanol over the temperature range covered by the thermometer is 0.0011 Pa s. Estimate the steady velocity at which the 25 °C sphere falls.
\ndensity = 785 «kgm−3»
\n« =» 0.87 «N»
\nAccept answer in the range 784 to 786
\n\n
\n
OR
\n\n
OR
\n\n
0.727 or 73%
\nAllow ECF from (a)(i)
\nuse of drag force to obtain r3 x 0.04 x g = 6 x x 0.0011 x r x v
\nv = 0.071 «ms–1»
\nIt is believed that a non-rotating supermassive black hole is likely to exist near the centre of our galaxy. This black hole has a mass equivalent to 3.6 million times that of the Sun.
\nOutline what is meant by the event horizon of a black hole.
\nCalculate the distance of the event horizon of the black hole from its centre.
\nMass of Sun = 2 × 1030 kg
\nStar S-2 is in an elliptical orbit around a black hole. The distance of S-2 from the centre of the black hole varies between a few light-hours and several light-days. A periodic event on S-2 occurs every 5.0 s.
\n![\"M18/4/PHYSI/HP3/ENG/TZ2/07.b\"](\"../assets/uploads/tinymce_asset/asset/4865/Schermafbeelding_2018-08-14_om_10.00.36.png\"/)
Discuss how the time for the periodic event as measured by an observer on the Earth changes with the orbital position of S-2.
\nboundary inside which events cannot be communicated to an outside observer
\nOR
\ndistance/surface at which escape velocity = c
\n\n
OWTTE
\n[1 mark]
\nmass of black hole = 7.2 × 1036 «kg»
\n« =» 1 × 1010 «m»
\n\n
[2 marks]
\nwherever S-2 is in orbit, time observed is longer than 5.0 s
\nwhen closest to the star S-2 periodic time dilated more than when at greatest distance
\nJustification using formula or time is more dilated in stronger gravitational fields
\n\n
[2 marks]
\nWhich is a unit of force?
\nA. J m
\nB. J m–1
\nC. J m s–1
\nD. J m–1 s
\nB
\nA parachutist of total mass 70 kg is falling vertically through the air at a constant speed of 8 m s–1.
\nWhat is the total upward force acting on the parachutist?
\nA. 0 N
\nB. 70 N
\nC. 560 N
\nD. 700 N
\nD
\nWhat is the best estimate for the diameter of a helium nucleus?
\nA. 10–21 m
\nB. 10–18 m
\nC. 10–15 m
\nD. 10–10 m
\nC
\nA galaxy can be modelled as a sphere of radius R0. The distance of a star from the centre of the galaxy is r.
\n![\"M18/4/PHYSI/HP3/ENG/TZ1/19\"](\"../assets/uploads/tinymce_asset/asset/4814/Schermafbeelding_2018-08-13_om_16.21.46.png\"/)
For this model the graph is a simplified representation of the variation with r of the mass of visible matter enclosed inside r.
\nThe mass of visible matter in the galaxy is M.
\nShow that for stars where r > R0 the velocity of orbit is v = .
\nDraw on the axes the observed variation with r of the orbital speed v of stars in a galaxy.
\nExplain, using the equation in (a) and the graphs, why the presence of visible matter alone cannot account for the velocity of stars when r > R0.
\nand correct rearranging
\n[1 mark]
\nlinear / rising until R0
\nthen «almost» constant
\n[2 marks]
\nfor v to stay constant for r greater than R0, M has to be proportional to r
\n\n
but this contradicts the information from the M-r graph
\nOR
\nif M is constant for r greater than R0, then we would expect v
\n\n
but this contradicts the information from the v-r graph
\n\n
[2 marks]
\nThe graph shows how the temperature of a liquid varies with time when energy is supplied to the liquid at a constant rate P. The gradient of the graph is K and the liquid has a specific heat capacity c.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/11\"](\"../assets/uploads/tinymce_asset/asset/4723/Schermafbeelding_2018-08-12_om_09.24.20.png\"/)
What is the mass of the liquid?
\nA.
\nB.
\nC.
\nD.
\nA
\nA stopper of mass 8 g leaves the opening of a container that contains pressurized gas.The stopper accelerates from rest for a time of 16 ms and leaves the container at a speed of 20 m s–1.
\nWhat is the order of magnitude of the force acting on the stopper?
\nA. 10–3 N
\nB. 100 N
\nC. 101 N
\nD. 103 N
\nC
\nThe velocities vX and vY of two boats, X and Y, are shown.
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/02_01\"](\"../assets/uploads/tinymce_asset/asset/4714/Schermafbeelding_2018-08-12_om_09.03.42.png\"/)
Which arrow represents the direction of the vector vX – vY?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/02_02\"](\"../assets/uploads/tinymce_asset/asset/4715/Schermafbeelding_2018-08-12_om_09.04.57.png\"/)
D
\nThe diagram shows a potential divider circuit used to measure the emf E of a cell X. Both cells have negligible internal resistance.
\n![\"M18/4/PHYSI/SP2/ENG/TZ2/04\"](\"../assets/uploads/tinymce_asset/asset/4736/Schermafbeelding_2018-08-12_om_13.01.10.png\"/)
AB is a wire of uniform cross-section and length 1.0 m. The resistance of wire AB is 80 Ω. When the length of AC is 0.35 m the current in cell X is zero.
\nState what is meant by the emf of a cell.
\nShow that the resistance of the wire AC is 28 Ω.
\nDetermine E.
\nthe work done per unit charge
\nin moving charge from one terminal of a cell to the other / all the way round the circuit
\n\n
Award [1] for “energy per unit charge provided by the cell”/“power per unit current”
\nAward [1] for “potential difference across the terminals of the cell when no current is flowing”
\nDo not accept “potential difference across terminals of cell”
\n[2 marks]
\nthe resistance is proportional to length / see 0.35 AND 1«.00»
\nso it equals 0.35 × 80
\n«= 28 Ω»
\n[2 marks]
\ncurrent leaving 12 V cell is = 0.15 «A»
\nOR
\nE = × 28
\nE = «0.15 × 28 =» 4.2 «V»
\n\n
Award [2] for a bald correct answer
\nAllow a 1sf answer of 4 if it comes from a calculation.
\nDo not allow a bald answer of 4 «V»
\nAllow ECF from incorrect current
\n[2 marks]
\nA container that contains a fixed mass of an ideal gas is at rest on a truck. The truck now moves away horizontally at a constant velocity. What is the change, if any, in the internal energy of the gas and the change, if any, in the temperature of the gas when the truck has been travelling for some time?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/12\"](\"../assets/uploads/tinymce_asset/asset/4724/Schermafbeelding_2018-08-12_om_09.27.13.png\"/)
A
\nA farmer is driving a vehicle across an uneven field in which there are undulations every 3.0 m.
\nThe farmer’s seat is mounted on a spring. The system, consisting of the mass of the farmer and the spring, has a natural frequency of vibration of 1.9 Hz.
\nExplain why it would be uncomfortable for the farmer to drive the vehicle at a speed of 5.6 m s–1.
\nOutline what change would be required to the value of Q for the mass–spring system in order for the drive to be more comfortable.
\nALTERNATIVE 1
\nthe time between undulations is = 0.536 «s»
\nf = = 1.87 «Hz»
\n«frequencies match» resonance occurs so amplitude of vibration becomes greater
\nMust see mention of “resonance” for MP3
\nALTERNATIVE 2
\nf =
\nf = 1.87 «Hz»
\n«frequencies match» resonance occurs so amplitude of vibration becomes greater
\nMust see mention of “resonance” for MP3
\n\n
\n
«to increase damping» reduce Q
\nA sealed container contains water at 5 °C and ice at 0 °C. This system is thermally isolated from its surroundings. What happens to the total internal energy of the system?
\nA. It remains the same.
\nB. It decreases.
\nC. It increases until the ice melts and then remains the same.
\nD. It increases.
\nA
\nThe ball is now displaced through a small distance x from the bottom of the bowl and is then released from rest.
\n![\"M18/4/PHYSI/HP2/ENG/TZ2/01.d\"](\"../assets/uploads/tinymce_asset/asset/4848/Schermafbeelding_2018-08-14_om_06.19.20.png\"/)
The magnitude of the force on the ball towards the equilibrium position is given by
\n\n
where R is the radius of the bowl.
\nA small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless hemispherical bowl.
\n
The normal reaction force N makes an angle θ to the horizontal.
\nState the direction of the resultant force on the ball.
\nOn the diagram, construct an arrow of the correct length to represent the weight of the ball.
\nShow that the magnitude of the net force F on the ball is given by the following equation.
\n\n
The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.
\nOutline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.
\nOutline why the ball will perform simple harmonic oscillations about the equilibrium position.
\nShow that the period of oscillation of the ball is about 6 s.
\nThe amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time t of the velocity v of the ball during one period.
\nA second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.
\n
The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.
\ntowards the centre «of the circle» / horizontally to the right
\n\n
Do not accept towards the centre of the bowl
\n[1 mark]
\ndownward vertical arrow of any length
\narrow of correct length
\n\n
Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required
\neg:
[2 marks]
\nALTERNATIVE 1
\nF = N cos θ
\nmg = N sin θ
\ndividing/substituting to get result
\n\n
ALTERNATIVE 2
\nright angle triangle drawn with F, N and W/mg labelled
\nangle correctly labelled and arrows on forces in correct directions
\ncorrect use of trigonometry leading to the required relationship
\n\n
tan θ =
\n[3 marks]
\n= m
\nr = R cos θ
\nv =
\nv = 13.4/13 «ms –1»
\n\n
Award [4] for a bald correct answer
\nAward [3] for an answer of 13.9/14 «ms –1». MP2 omitted
\n[4 marks]
\nthere is no force to balance the weight/N is horizontal
\nso no / it is not possible
\n\n
Must see correct justification to award MP2
\n[2 marks]
\nthe «restoring» force/acceleration is proportional to displacement
\n\n
Direction is not required
\n[1 mark]
\nω = «» = «= 1.107 s–1»
\nT = « = =» 5.7 «s»
\n\n
Allow use of or g = 9.8 or 10
\nAward [0] for a substitution into T = 2π
\n[2 marks]
\nsine graph
\ncorrect amplitude «0.13 m s–1»
\ncorrect period and only 1 period shown
\n\n
Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.
\nAmplitude should be correct to ± square for MP2
\neg: v /m s–1 ![\"M18/4/PHYSI/HP2/ENG/TZ2/01.d.iii\"](\"../assets/uploads/tinymce_asset/asset/4852/Schermafbeelding_2018-08-14_om_06.59.06.png\"/)
[3 marks]
\nspeed before collision v = « =» 12.5 «ms–1»
\n«from conservation of momentum» common speed after collision is initial speed «vc = = 6.25 ms–1»
\nh = «» 2.0 «m»
\n\n
Allow 12.5 from incorrect use of kinematics equations
\nAward [3] for a bald correct answer
\nAward [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.
\nAllow ECF from MP1
\nAllow ECF from MP2
\n[3 marks]
\nTwo tubes, A and B, are inserted into a fluid flowing through a horizontal pipe of diameter 0.50 m. The openings X and Y of the tubes are at the exact centre of the pipe. The liquid rises to a height of 0.10 m in tube A and 0.32 m in tube B. The density of the fluid = 1.0 × 103 kg m–3.
\n![\"M18/4/PHYSI/HP3/ENG/TZ2/10\"](\"../assets/uploads/tinymce_asset/asset/4862/Schermafbeelding_2018-08-14_om_09.34.50.png\"/)
The viscosity of water is 8.9 × 10–4 Pa s.
\nShow that the velocity of the fluid at X is about 2 ms–1, assuming that the flow is laminar.
\nEstimate the Reynolds number for the fluid in your answer to (a).
\nOutline whether your answer to (a) is valid.
\n\n
vX =
\nvx = 2.08 «ms–1»
\n[3 marks]
\nR = «» 5.9 × 105
\n[1 mark]
\n(R > 1000) flow is not laminar, so assumption is invalid
\n\n
OWTTE
\n[1 mark]
\nA proton moves along a circular path in a region of a uniform magnetic field. The magnetic field is directed into the plane of the page.
\nThe speed of the proton is 2.16 × 106 m s-1 and the magnetic field strength is 0.042 T.
\nLabel with arrows on the diagram the magnetic force F on the proton.
\nLabel with arrows on the diagram the velocity vector v of the proton.
\nFor this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.
\nFor this proton, calculate, in s, the time for one full revolution.
\nF towards centre ✔
v tangent to circle and in the direction shown in the diagram ✔
\n
« ✔
\nR = 0.538«m» ✔
\nR = 0.54«m» ✔
\n ✔
«s» ✔
\nExaminers were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.
\nExaminers were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.
\nThis was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.
\nThis was well answered with many candidates scoring ECF from the previous part.
\nTwo sound waves from a point source on the ground travel through the ground to a detector. The speed of one wave is 7.5 km s–1, the speed of the other wave is 5.0 km s–1. The waves arrive at the detector 15 s apart. What is the distance from the point source to the detector?
\nA. 38 km
\nB. 45 km
\nC. 113 km
\nD. 225 km
\nD
\nA ball starts from rest and moves horizontally. Six positions of the ball are shown at time intervals of 1.0 ms. The horizontal distance between X, the initial position, and Y, the final position, is 0.050 m.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/4818/Schermafbeelding_2018-08-13_om_18.24.32.png\"/)
What is the average acceleration of the ball between X and Y?
\nA. 2000 m s–2
\nB. 4000 m s–2
\nC. 5000 m s–2
\nD. 8000 m s–2
\nB
\nA ball of mass m collides with a vertical wall with an initial horizontal speed u and rebounds with a horizontal speed v. The graph shows the variation of the speed of the ball with time.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/07\"](\"../assets/uploads/tinymce_asset/asset/4819/Schermafbeelding_2018-08-13_om_18.26.37.png\"/)
What is the magnitude of the mean net force on the ball during the collision?
\nA.
\nB.
\nC.
\nD.
\nD
\nQ and R are two rigid containers of volume 3V and V respectively containing molecules of the same ideal gas initially at the same temperature. The gas pressures in Q and R are p and 3p respectively. The containers are connected through a valve of negligible volume that is initially closed.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/09\"](\"../assets/uploads/tinymce_asset/asset/4820/Schermafbeelding_2018-08-13_om_18.30.51.png\"/)
The valve is opened in such a way that the temperature of the gases does not change. What is the change of pressure in Q?
\nA. +p
\nB.
\nC.
\nD. –p
\nB
\nSome optic fibres consist of a core surrounded by cladding as shown in the diagram.
\nCalculate the maximum angle β for light to travel through the fibre.
\nRefractive index of core = 1.50
Refractive index of cladding = 1.48
Outline how the combination of core and cladding reduces the overall dispersion in the optic fibres.
\nrealization that 𝜃 min is the critical angle
\n𝜃 = «sin–1 =» 80.6 «°»
\nAccept 1.4 rad
\nβ = «90 – 80.6 =» 9.4 «°»
\nAccept 0.16 rad
\nbecause the critical angle is nearly 90°
\nthen only rays that are «almost» parallel to the fibre pass down it
\nso pulse broadening is reduced
\nOWTTE
\nAn ideal monatomic gas is kept in a container of volume 2.1 × 10–4 m3, temperature 310 K and pressure 5.3 × 105 Pa.
\nThe volume of the gas in (a) is increased to 6.8 × 10–4 m3 at constant temperature.
\nState what is meant by an ideal gas.
\nCalculate the number of atoms in the gas.
\nCalculate, in J, the internal energy of the gas.
\nCalculate, in Pa, the new pressure of the gas.
\nExplain, in terms of molecular motion, this change in pressure.
\na gas in which there are no intermolecular forces
\nOR
\na gas that obeys the ideal gas law/all gas laws at all pressures, volumes and temperatures
\nOR
\nmolecules have zero PE/only KE
\n\n
Accept atoms/particles.
\n[1 mark]
\nN = «» 2.6 × 1022
\n[1 mark]
\n«For one atom U = kT» × 1.38 × 10–23 × 310 / 6.4 × 10–21 «J»
\nU = «2.6 × 1022 × × 1.38 × 10–23 × 310» 170 «J»
\n\n
Allow ECF from (a)(ii)
\nAward [2] for a bald correct answer
\nAllow use of U = pV
\n[2 marks]
\np2 = «5.3 × 105 × » 1.6 × 105 «Pa»
\n[1 mark]
\n«volume has increased and» average velocity/KE remains unchanged
\n«so» molecules collide with the walls less frequently/longer time between collisions with the walls
\n«hence» rate of change of momentum at wall has decreased
\n«and so pressure has decreased»
\n\n
The idea of average must be included
\nDecrease in number of collisions is not sufficient for MP2. Time must be included.
\nAccept atoms/particles.
\n[2 marks]
\nWhat is true about the acceleration of a particle that is oscillating with simple harmonic motion (SHM)?
\nA. It is in the opposite direction to its velocity
\nB. It is decreasing when the potential energy is increasing
\nC. It is proportional to the frequency of the oscillation
\nD. It is at a minimum when the velocity is at a maximum
\nD
\nA ray of light passes from the air into a long glass plate of refractive index n at an angle θ to the edge of the plate.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/13_01\"](\"../assets/uploads/tinymce_asset/asset/4777/Schermafbeelding_2018-08-13_om_10.40.22.png\"/)
The ray is incident on the internal surface of the glass plate and the refracted ray travels along the external surface of the plate.
\nWhat change to n and what change to θ will cause the ray to travel entirely within the plate after incidence?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/13_02\"](\"../assets/uploads/tinymce_asset/asset/4778/Schermafbeelding_2018-08-13_om_10.41.08.png\"/)
A
\nA string stretched between two fixed points sounds its second harmonic at frequency f.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/13\"](\"../assets/uploads/tinymce_asset/asset/4821/Schermafbeelding_2018-08-13_om_18.33.18.png\"/)
Which expression, where n is an integer, gives the frequencies of harmonics that have a node at the centre of the string?
\nA.
\nB. nf
\nC. 2nf
\nD. (2n + 1)f
\nB
\nWhat are the changes in the speed and in the wavelength of monochromatic light when the light passes from water to air?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/16\"](\"../assets/uploads/tinymce_asset/asset/4725/Schermafbeelding_2018-08-12_om_09.30.27.png\"/)
A
\nA cell of emf 6.0 V and negligible internal resistance is connected to three resistors as shown.
\nThe resistors have resistance of 3.0 Ω and 6.0 Ω as shown.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/16\"](\"../assets/uploads/tinymce_asset/asset/4823/Schermafbeelding_2018-08-13_om_18.36.47.png\"/)
What is the current in resistor X?
\nA. 0.40 A
\nB. 0.50 A
\nC. 1.0 A
\nD. 2.0 A
\nC
\nThe natural frequency of a driven oscillating system is 6 kHz. The frequency of the driver for the system is varied from zero to 20 kHz.
\nDraw a graph to show the variation of amplitude of oscillation of the system with frequency.
\nThe Q factor for the system is reduced significantly. Describe how the graph you drew in (a) changes.
\ngeneral shape as shown
\npeak at 6 kHz
\ngraph does not touch the f axis
\n\n
![\"M18/4/PHYSI/HP3/ENG/TZ2/11.a/M\"](\"../assets/uploads/tinymce_asset/asset/4867/Schermafbeelding_2018-08-14_om_10.21.57.png\"/)
[3 marks]
\npeak broadens
\nreduced maximum amplitude / graph shifted down
\nresonant frequency decreases / graph shifted to the left
\n[2 marks]
\nAn ion of charge +Q moves vertically upwards through a small distance s in a uniform vertical electric field. The electric field has a strength E and its direction is shown in the diagram.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/15\"](\"../assets/uploads/tinymce_asset/asset/4779/Schermafbeelding_2018-08-13_om_10.43.27.png\"/)
What is the electric potential difference between the initial and final position of the ion?
\nA.
\nB. EQs
\nC. Es
\nD.
\nC
\nIn a pumped storage hydroelectric system, water is stored in a dam of depth 34 m.
\n![\"M18/4/PHYSI/SP2/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/4737/Schermafbeelding_2018-08-12_om_13.07.03.png\"/)
The water leaving the upper lake descends a vertical distance of 110 m and turns the turbine of a generator before exiting into the lower lake.
\nWater flows out of the upper lake at a rate of 1.2 × 105 m3 per minute. The density of water is 1.0 × 103 kg m–3.
\nEstimate the specific energy of water in this storage system, giving an appropriate unit for your answer.
\nShow that the average rate at which the gravitational potential energy of the water decreases is 2.5 GW.
\nThe storage system produces 1.8 GW of electrical power. Determine the overall efficiency of the storage system.
\nAfter the upper lake is emptied it must be refilled with water from the lower lake and this requires energy. Suggest how the operators of this storage system can still make a profit.
\nAverage height = 127 «m»
\nSpecific energy «= = 9.81 × 127» = 1.2 × 103 J kg–1
\n\n
Unit is essential
\nAllow g = 10 gives 1.3 × 103 J kg–1
\nAllow ECF from 110 m
\n(1.1 × 103 J kg–1) or 144 m
\n(1.4 × 103 J kg–1)
\n[2 marks]
\nmass per second leaving dam is × 103 = «2.0 × 106 kg s–1»
\nrate of decrease of GPE is = 2.0 × 106 × 9.81 × 127
\n= 2.49 × 109 «W» /2.49 «GW»
\n\n
Do not award ECF for the use of 110 m or 144 m
\nAllow 2.4 GW if rounded value used from (a)(i) or 2.6 GW if g = 10 is used
\n[3 marks]
\nefficiency is « =» 0.72 / 72%
\n[1 mark]
\nwater is pumped back up at times when the demand for/price of electricity is low
\n[1 mark]
\nAn object of mass m moves in a horizontal circle of radius r with a constant speed v. What is the rate at which work is done by the centripetal force?
\nA.
\nB.
\nC.
\nD. zero
\nD
\nIdentify the conservation law violated in the proposed reaction.
\np+ + p+ → p+ + n0 + μ+
\nA. Strangeness
\nB. Lepton number
\nC. Charge
\nD. Baryon number
\nB
\nWhen an electric cell of negligible internal resistance is connected to a resistor of resistance 4R, the power dissipated in the resistor is P.
\nWhat is the power dissipated in a resistor of resistance value R when it is connected to the same cell?
\nA.
\nB. P
\nC. 4P
\nD. 16P
\nC
\nA sound wave has a wavelength of 0.20 m. What is the phase difference between two points along the wave which are 0.85 m apart?
\nA. zero
\nB. 45°
\nC. 90°
\nD. 180°
\nC
\nAn X-ray beam of intensity I0 is incident on lead. After travelling a distance x through the lead the intensity of the beam is reduced to I.
\nThe graph shows the variation of ln with x.
\n\n
Show that the attenuation coefficient of lead is 60 cm–1.
\nA technician operates an X-ray machine that takes 100 images each day. Estimate the width of the lead screen that is required so that the total exposure of the technician in 250 working days is equal to the exposure that the technician would receive from one X-ray exposure without the lead screen.
\nevidence of finding the gradient
\nμ = «– gradient =» 59.9 «cm–1»
\nI =
\n«ln 25000 = μx» x = 0.17 «cm» or 1.7 «mm»
\nWhat is correct about the Higgs Boson?
\nA. It was predicted before it was observed.
\nB. It was difficult to detect because it is charged.
\nC. It is not part of the Standard Model.
\nD. It was difficult to detect because it has no mass.
\nA
\nThe Sankey diagram shows the energy input from fuel that is eventually converted to useful domestic energy in the form of light in a filament lamp.
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/22\"](\"../assets/uploads/tinymce_asset/asset/4825/Schermafbeelding_2018-08-13_om_18.42.25.png\"/)
What is true for this Sankey diagram?
\nA. The overall efficiency of the process is 10%.
\nB. Generation and transmission losses account for 55% of the energy input.
\nC. Useful energy accounts for half of the transmission losses.
\nD. The energy loss in the power station equals the energy that leaves it.
\nA
\nDeuterium, , undergoes fusion according to the following reaction.
\n\n
\n
The following data are available for binding energies per nucleon.
\n\n
\n
\n
Particle Y is produced in the collision of a proton with a K- in the following reaction.
\nThe quark content of some of the particles involved are
\nIdentify particle X.
\nDetermine, in MeV, the energy released.
\nSuggest why, for the fusion reaction above to take place, the temperature of deuterium must be very high.
\nIdentify, for particle Y, the charge.
\nIdentify, for particle Y, the strangeness.
\nproton / / p ✔
\n«3 x 2.78 − 2 × 2 × 1.12»
See 3 × 2.78/8.34 OR 2 × 2 × 1.12/4.48✔
3.86 «MeV» ✔
\n\n
the deuterium nuclei are positively charged/repel ✔
high KE/energy is required to overcome «Coulomb/electrostatic» repulsion /potential barrier
\nOR
high KE/energy is required to bring the nuclei within range of the strong nuclear force ✔
high temperatures are required to give high KEs/energies ✔
\n\n
−1 / -e ✔
\n\n
−3 ✔
\nAt HL this was well answered with the most common wrong answer being ‘neutron’. At SL however, this was surprisingly wrongly answered by many. Suggestions given included most smallish particles, alpha, positron, beta, antineutrino and even helium.
\nThe majority of candidates missed the fact that the figures given were the binding energies per nucleon. Many complicated calculations were also seen, particularly at SL, that involved E = mc2.
\nThe most common mark to be awarded here was the one for linking high temperature to high KE. A large number of candidates talked about having to overcome the strong nuclear force before fusion could happen.
\nAt SL many answers of just ‘negative’ were seen.
\nThis was poorly answered at both levels with the most common answer being zero.
\nWhat part of a nuclear power station is principally responsible for increasing the chance that a neutron will cause fission?
\nA. Moderator
\nB. Control rod
\nC. Pressure vessel
\nD. Heat exchanger
\nA
\nA pair of slits in a double slit experiment are illuminated with monochromatic light of wavelength 480 nm. The slits are separated by 1.0 mm. What is the separation of the fringes when observed at a distance of 2.0 m from the slits?
\nA. 2.4 × 10–4 mm
\nB. 9.6 × 10–4 mm
\nC. 2.4 × 10–1 mm
\nD. 9.6 × 10–1 mm
\nD
\nA nuclear reactor contains atoms that are used for moderation and atoms that are used for control.
\nWhat are the ideal properties of the moderator atoms and the control atoms in terms of neutron absorption?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/23\"](\"../assets/uploads/tinymce_asset/asset/4780/Schermafbeelding_2018-08-13_om_10.48.15.png\"/)
B
\nA simple pendulum bob oscillates as shown.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/24\"](\"../assets/uploads/tinymce_asset/asset/4826/Schermafbeelding_2018-08-13_om_18.44.27.png\"/)
At which position is the resultant force on the pendulum bob zero?
\nA. At position A
\nB. At position B
\nC. At position C
\nD. Resultant force is never zero during the oscillation
\nD
\nThe Sun is a second generation star. Outline, with reference to the Jeans criterion (MJ), how the Sun is likely to have been formed.
\nSuggest how fluctuations in the cosmic microwave background (CMB) radiation are linked to the observation that galaxies collide.
\nShow that the critical density of the universe is
\n\n
where H is the Hubble parameter and G is the gravitational constant.
\ninterstellar gas/dust «from earlier supernova»
\ngravitational attraction between particles
\nif the mass is greater than the Jean’s mass/Mj the interstellar gas coalesces
\nas gas collapses temperature increases leading to nuclear fusion
\nMP3 can be expressed in terms of potential and kinetic energy
\nfluctuations in CMB due to differences in temperature/mass/density
\nduring the inflationary period/epoch/early universe
\nleading to the formation of galaxies/stars/structures
\ngravitational interaction between galaxies can lead to collision
\n[Max 3 Marks]
\nALTERNATIVE 1
\nkinetic energy of galaxy mv2 = mH2r2 «uses Hubble’s law»
\npotential energy = = Gr3 «introduces density»
\nKE=PE to get expression for critical
\n\n
ALTERNATIVE 2
\nescape velocity of distant galaxy v =
\nwhere H0r =
\nsubstitutes M = r3 to get result
\nThe diagram represents a simple optical astronomical reflecting telescope with the path of some light rays shown.
\nIt is proposed to build an array of radio telescopes such that the maximum distance between them is 3800 km. The array will operate at a wavelength of 2.1 cm.
\nComment on whether it is possible to build an optical telescope operating at 580 nm that is to have the same resolution as the array.
\n«use of to get» resolution of 6.7 × 10–9 «rad»
\n= 87 «m»
\nsome reference to difficulty in making optical mirrors/lenses of this size
\n\n
Allow = 105 «m»
\n[3 marks]
\nA beam of monochromatic light is incident on a single slit and a diffraction pattern forms on the screen.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/25\"](\"../assets/uploads/tinymce_asset/asset/4827/Schermafbeelding_2018-08-13_om_18.46.01.png\"/)
What change will increase θs?
\nA. Increase the width of the slit
\nB. Decrease the width of the slit
\nC. Increase the distance between the slit and the screen
\nD. Decrease the distance between the slit and the screen
\nB
\nThe dashed line on the graph shows the variation with wavelength of the intensity of solar radiation before passing through the Earth’s atmosphere.
\nThe solid line on the graph shows the variation with wavelength of the intensity of solar radiation after it has passed through the Earth’s atmosphere.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/24\"](\"../assets/uploads/tinymce_asset/asset/4781/Schermafbeelding_2018-08-13_om_10.49.37.png\"/)
Which feature of the graph helps explain the greenhouse effect?
\nA. Infrared radiation is absorbed at specific wavelengths.
\nB. There is little absorption at infrared wavelengths.
\nC. There is substantial absorption at visible wavelengths.
\nD. There is little absorption at UV wavelengths.
\nA
\nA cell with negligible internal resistance is connected as shown. The ammeter and the voltmeter are both ideal.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/19_01\"](\"../assets/uploads/tinymce_asset/asset/4726/Schermafbeelding_2018-08-12_om_09.33.11.png\"/)
What changes occur in the ammeter reading and in the voltmeter reading when the resistance of the variable resistor is increased?
\n![\"M18/4/PHYSI/SPM/ENG/TZ2/19_02\"](\"../assets/uploads/tinymce_asset/asset/4727/Schermafbeelding_2018-08-12_om_09.34.17.png\"/)
C
\nThe attenuation values for fat and muscle at different X-ray energies are shown.
\n![\"M18/4/PHYSI/HP3/ENG/TZ2/15.b\"](\"../assets/uploads/tinymce_asset/asset/4863/Schermafbeelding_2018-08-14_om_09.40.21.png\"/)
Outline the formation of a B scan in medical ultrasound imaging.
\nState what is meant by half-value thickness in X-ray imaging.
\nA monochromatic X-ray beam of energy 20 keV and intensity I0 penetrates 5.00 cm of fat and then 4.00 cm of muscle.
\n ![\"M18/4/PHYSI/HP3/ENG/TZ2/X15.b.ii\"](\"../assets/uploads/tinymce_asset/asset/4868/Schermafbeelding_2018-08-14_om_10.35.20.png\"/)
Calculate, in terms of I0, the final beam intensity that emerges from the muscle.
\nCompare the use of high and low energy X-rays for medical imaging.
\nmany/array of transducers send ultrasound through body/object
\nB scan made from many A scans in different directions
\nthe reflection from organ boundaries gives rise to position
\nthe amplitude/size gives brightness to the B scan
\n2D/3D image formed «by computer»
\n[3 marks]
\nthe thickness of tissue that reduces the intensity «of the X-rays» by a half
\nOR
\nwhere is the half value thickness and μ is attenuation coefficient
\n\n
Symbols must be defined for mark to be awarded
\n[1 mark]
\nafter fat layer, Ifat = I0e–0.4499 × 5.00
\nafter muscle layer, I = Ifate–0.8490 × 4.00
\nI = 0.003533 I0 or 0.35%
\n\n
[3 marks]
\n«high energies factors:»
\nless attenuation/more penetration
\nmore damage to the body
\n\n
«so» stronger signal leaves the body
\nOR
\n«so» used in «most» medical imaging techniques
\n\n
«low energy factors:»
\nmust be used with enhancement techniques
\ngreater attenuation/less penetration
\n\n
«so» more damage to the body «on surface layers»
\nOR
\n«so» unwanted in «most» medical imaging techniques
\n\n
[3 marks]
\nRhodium-106 () decays into palladium-106 () by beta minus (β–) decay.
\nThe binding energy per nucleon of rhodium is 8.521 MeV and that of palladium is 8.550 MeV.
\nβ– decay is described by the following incomplete Feynman diagram.
\nRutherford constructed a model of the atom based on the results of the alpha particle scattering experiment. Describe this model.
\nState what is meant by the binding energy of a nucleus.
\nShow that the energy released in the β– decay of rhodium is about 3 MeV.
\nDraw a labelled arrow to complete the Feynman diagram.
\nIdentify particle V.
\n«most of» the mass of the atom is confined within a very small volume/nucleus
\n«all» the positive charge is confined within a very small volume/nucleus
\nelectrons orbit the nucleus «in circular orbits»
\n[2 marks]
\nthe energy needed to separate the nucleons of a nucleus
\nOR
\nenergy released when a nucleus is formed from its nucleons
\n\n
Allow neutrons AND protons for nucleons
\nDon’t allow constituent parts
\n[1 mark]
\nQ = 106 × 8.550 − 106 × 8.521 = 3.07 «MeV»
\n«Q ≈ 3 Me V»
\n[1 mark]
\nline with arrow as shown labelled anti-neutrino/
\n\n
Correct direction of the “arrow” is essential
\nThe line drawn must be “upwards” from the vertex in the time direction i.e. above the horizontal
\n![\"M18/4/PHYSI/SP2/ENG/TZ2/06.c.i/M\"](\"../assets/uploads/tinymce_asset/asset/4746/Schermafbeelding_2018-08-12_om_15.34.15.png\"/)
[1 mark]
\nV = W–
\n[1 mark]
\nA loudspeaker emits sound towards the open end of a pipe. The other end is closed. A standing wave is formed in the pipe. The diagram represents the displacement of molecules of air in the pipe at an instant of time.
\nX and Y represent the equilibrium positions of two air molecules in the pipe. The arrow represents the velocity of the molecule at Y.
\nThe loudspeaker in (a) now emits sound towards an air–water boundary. A, B and C are parallel wavefronts emitted by the loudspeaker. The parts of wavefronts A and B in water are not shown. Wavefront C has not yet entered the water.
\nOutline how the standing wave is formed.
\nDraw an arrow on the diagram to represent the direction of motion of the molecule at X.
\nLabel a position N that is a node of the standing wave.
\nThe speed of sound is 340 m s–1 and the length of the pipe is 0.30 m. Calculate, in Hz, the frequency of the sound.
\nThe speed of sound in air is 340 m s–1 and in water it is 1500 m s–1.
\nThe wavefronts make an angle θ with the surface of the water. Determine the maximum angle, θmax, at which the sound can enter water. Give your answer to the correct number of significant figures.
\nDraw lines on the diagram to complete wavefronts A and B in water for θ < θmax.
\nthe incident wave «from the speaker» and the reflected wave «from the closed end»
\nsuperpose/combine/interfere
\n\n
Allow superimpose/add up
\nDo not allow meet/interact
\n[1 mark]
\nHorizontal arrow from X to the right
\n\n
MP2 is dependent on MP1
\nIgnore length of arrow
\n[1 mark]
\nP at a node
\n\n
[1 mark]
\nwavelength is λ = « =» 0.40 «m»
\nf = «» 850 «Hz»
\n\n
Award [2] for a bald correct answer
\nAllow ECF from MP1
\n[2 marks]
\n\n
θc = 13«°»
\n\n
Award [2] for a bald correct answer
\nAward [2] for a bald answer of 13.1
\n\n
Answer must be to 2/3 significant figures to award MP2
\nAllow 0.23 radians
\n[2 marks]
\ncorrect orientation
\ngreater separation
\n\n
Do not penalize the lengths of A and B in the water
\nDo not penalize a wavefront for C if it is consistent with A and B
\nMP1 must be awarded for MP2 to be awarded
\n
[2 marks]
\n\n
A beam of monochromatic light is incident on a diffraction grating of N lines per unit length. The angle between the first orders is θ1.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/26\"](\"../assets/uploads/tinymce_asset/asset/4828/Schermafbeelding_2018-08-13_om_18.48.12.png\"/)
What is the wavelength of the light?
\nA.
\nB. N sin θ1
\nC. N sin
\nD.
\nD
\nA mass at the end of a vertical spring and a simple pendulum perform oscillations on Earth that are simple harmonic with time period T. Both the pendulum and the mass-spring system are taken to the Moon. The acceleration of free fall on the Moon is smaller than that on Earth. What is correct about the time periods of the pendulum and the mass-spring system on the Moon?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/26\"](\"../assets/uploads/tinymce_asset/asset/4782/Schermafbeelding_2018-08-13_om_10.51.38.png\"/)
B
\nAn electron enters the region between two charged parallel plates initially moving parallel to the plates.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/20\"](\"../assets/uploads/tinymce_asset/asset/4728/Schermafbeelding_2018-08-12_om_10.07.01.png\"/)
The electromagnetic force acting on the electron
\nA. causes the electron to decrease its horizontal speed.
\nB. causes the electron to increase its horizontal speed.
\nC. is parallel to the field lines and in the opposite direction to them.
\nD. is perpendicular to the field direction.
\nC
\nMonochromatic light of wavelength λ in air is incident normally on a thin film of refractive index n. The film is surrounded by air. The intensity of the reflected light is a minimum. What is a possible thickness of the film?
\nA.
\nB.
\nC.
\nD.
\nC
\nA train is approaching an observer with constant speed
\n\n
where c is the speed of sound in still air. The train emits sound of wavelength λ. What is the observed speed of the sound and observed wavelength as the train approaches?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/27\"](\"../assets/uploads/tinymce_asset/asset/4829/Schermafbeelding_2018-08-13_om_18.51.53.png\"/)
A
\nA beam of electrons moves between the poles of a magnet.
\n ![\"M18/4/PHYSI/SPM/ENG/TZ2/21\"](\"../assets/uploads/tinymce_asset/asset/4729/Schermafbeelding_2018-08-12_om_10.08.39.png\"/)
What is the direction in which the electrons will be deflected?
\nA. Downwards
\nB. Towards the N pole of the magnet
\nC. Towards the S pole of the magnet
\nD. Upwards
\nD
\nA student measures the radius r of a sphere with an absolute uncertainty Δr. What is the fractional uncertainty in the volume of the sphere?
\nA.
\nB.
\nC.
\nD.
\nB
\nA river flows north. A boat crosses the river so that it only moves in the direction east of its starting point.
\nWhat is the direction in which the boat must be steered?
\n ![\"M18/4/PHYSI/SPM/ENG/TZ1/02\"](\"../assets/uploads/tinymce_asset/asset/4663/Schermafbeelding_2018-08-10_om_15.46.12.png\"/)
C
\nMonochromatic light is incident on 4 rectangular, parallel slits. The first principal maximum is observed at an angle θ to the direction of the incident light. The number of slits is increased to 8 each having the same width and spacing as the first 4.
\nThree statements about the first principal maximum with 8 slits are
\nI. the angle at which it is observed is greater than θ
\nII. its intensity increases
\nIII. its width decreases.
\nWhich statements are correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nThe spacetime diagram is in the reference frame of an observer O on Earth. Observer O and spaceship A are at the origin of the spacetime diagram when time . The worldline for spaceship A is shown.
\n\n
Event E is the emission of a flash of light. Observer O sees light from the flash when years and calculates that event E is away, in the positive direction.
\nCalculate in terms of the velocity of spaceship A relative to observer O.
\nDraw the axis for the reference frame of spaceship A.
\nPlot the event E on the spacetime diagram and label it E.
\nDetermine the time, according to spaceship A, when light from event E was observed on spaceship A.
\n✓
\nAccept if unit given.
\nline through origin and through one small square at this coordinate ✓
\nAnswers shown for 5(a)(ii) and (b)(i) and (b)(ii).
\nX value of E at «» ✓
Y value of E at «» ✓
\nlight cone from E «crosses at so» intersection on «on scale» ✓
\n✓
\nso « after leaving Earth» ✓
\n\n
MP1 accept use of linear equations to find
\nAllow ECF from (b)(i) and (a)
\nVery well answered.
\nMost answers successfully drew the correct x' axis.
\nEvent E is the event when the light is emitted (4,5). Some candidates missed that and placed it at (4,9).
\nTo determine these coordinates candidates were expected to construct a light path from (0,9) which intercepts at x=4 and t=5. This is a very common mistake which needs careful explanation by teachers.
\nThose candidates who placed E at the correct position were then able to calculate the time appropriately.
\nIn an experiment to measure the specific latent heat of vaporization of water Lv, a student uses an electric heater to boil water. A mass m of water vaporizes during time t. Lv may be calculated using the relation
\n\n
where V is the voltage applied to the heater and I the current through it.
\nOutline why, during the experiment, V and I should be kept constant.
\nOutline whether the value of Lv calculated in this experiment is expected to be larger or smaller than the actual value.
\nA student suggests that to get a more accurate value of Lv the experiment should be performed twice using different heating rates. With voltage and current V1, I1 the mass of water that vaporized in time t is m1. With voltage and current V2, I2 the mass of water that vaporized in time t is m2. The student now uses the expression
\n\n
\n
\n
to calculate Lv. Suggest, by reference to heat losses, why this is an improvement.
\nto provide a constant heating rate / power
\nOR
\nto have m proportional to t ✔
\ndue to heat losses «VIt is larger than heat into liquid» ✔
\nLv calculated will be larger ✔
\nheat losses will be similar / the same for both experiments
\nOR
\nheat loss presents systematic error ✔
\n\n
taking the difference cancels/eliminates the effect of these losses
\nOR
\nuse a graph to eliminate the effect ✔
\nA moon of mass M orbits a planet of mass 100M. The radius of the planet is R and the distance between the centres of the planet and moon is 22R.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/28\"](\"../assets/uploads/tinymce_asset/asset/4830/Schermafbeelding_2018-08-13_om_18.53.09.png\"/)
What is the distance from the centre of the planet at which the total gravitational potential has a maximum value?
\nA. 2R
\nB. 11R
\nC. 20R
\nD. 2R and 20R
\nC
\nTwo lines X and Y in the emission spectrum of hydrogen gas are measured by an observer stationary with respect to the gas sample.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/29_01\"](\"../assets/uploads/tinymce_asset/asset/4783/Schermafbeelding_2018-08-13_om_10.56.15.png\"/)
The emission spectrum is then measured by an observer moving away from the gas sample.
\nWhat are the correct shifts X* and Y* for spectral lines X and Y?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/29_02\"](\"../assets/uploads/tinymce_asset/asset/4784/Schermafbeelding_2018-08-13_om_10.56.59.png\"/)
C
\nThe average temperature of ocean surface water is 289 K. Oceans behave as black bodies.
\nThe intensity in (b) returned to the oceans is 330 W m-2. The intensity of the solar radiation incident on the oceans is 170 W m-2.
\nShow that the intensity radiated by the oceans is about 400 W m-2.
\nExplain why some of this radiation is returned to the oceans from the atmosphere.
\nCalculate the additional intensity that must be lost by the oceans so that the water temperature remains constant.
\nSuggest a mechanism by which the additional intensity can be lost.
\n5.67 × 10−8 × 2894
\nOR
= 396«W m−2» ✔
«≈ 400 W m−2»
\n«most of the radiation emitted by the oceans is in the» infrared ✔
«this radiation is» absorbed by greenhouse gases/named greenhouse gas in the atmosphere ✔
«the gases» reradiate/re-emit ✔
partly back towards oceans/in all directions/awareness that radiation in other directions is also present ✔
\nwater loses 396 − 330/66 «W m −2» ✔
extra intensity that must be lost is «170 − 66» = 104 ≈ 100 «W m−2» ✔
OR
absorbed by water 330 + 170/500 «W m−2»✔
extra intensity that must be lost is «500 − 396» = 104 ≈ 100 «W m−2» ✔
\nconduction to the air above
OR
«mainly» evaporation
OR
melting ice at the poles
OR
reflection of sunlight off the surface of the ocean ✔
\nDo not accept convection or radiation.
\nThis was well answered with candidates scoring the mark for either a correct substitution or an answer given to at least one more sf than the show that value. Some candidates used 298 rather than 289.
\nFor many this was a well-rehearsed answer which succinctly scored full marks. For others too many vague terms were used. There was much talk about energy being trapped or reflected and the ozone layer was often included. The word ‘albedo’ was often written down with no indication of what it means and ‘the albedo effect also featured.
\nThis was well-answered, a very straightforward 2 marks.
\nMany candidates didn’t understand this question and thought that the answer needed to be some form of human activity that would reduce global temperature rise.
\nFour identical, positive, point charges of magnitude Q are placed at the vertices of a square of side 2d. What is the electric potential produced at the centre of the square by the four charges?
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/30\"](\"../assets/uploads/tinymce_asset/asset/4785/Schermafbeelding_2018-08-13_om_10.58.10.png\"/)
A. 0
\nB.
\nC.
\nD.
\nD
\nThe diagram shows the electric field and the electric equipotential surfaces between two charged parallel plates. The potential difference between the plates is 200 V.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/29_01\"](\"../assets/uploads/tinymce_asset/asset/4832/Schermafbeelding_2018-08-13_om_18.54.51.png\"/)
What is the work done, in nJ, by the electric field in moving a negative charge of magnitude 1 nC from the position shown to X and to Y?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/29_02\"](\"../assets/uploads/tinymce_asset/asset/4833/Schermafbeelding_2018-08-13_om_18.56.05.png\"/)
A
\nA positive point charge is placed above a metal plate at zero electric potential. Which diagram shows the pattern of electric field lines between the charge and the plate?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/30\"](\"../assets/uploads/tinymce_asset/asset/4834/Schermafbeelding_2018-08-13_om_18.57.24.png\"/)
C
\nThe diagram shows 5 gravitational equipotential lines. The gravitational potential on each line is indicated. A point mass m is placed on the middle line and is then released. Values given in MJ kg–1.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/31_01\"](\"../assets/uploads/tinymce_asset/asset/4787/Schermafbeelding_2018-08-13_om_11.00.49.png\"/)
Which is correct about the direction of motion and the acceleration of the point mass?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/31_02\"](\"../assets/uploads/tinymce_asset/asset/4788/Schermafbeelding_2018-08-13_om_11.01.52.png\"/)
D
\nTo determine the acceleration due to gravity, a small metal sphere is dropped from rest and the time it takes to fall through a known distance and open a trapdoor is measured.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/01\"](\"../assets/uploads/tinymce_asset/asset/4747/Schermafbeelding_2018-08-12_om_15.43.42.png\"/)
The following data are available.
\n\n
Determine the distance fallen, in m, by the centre of mass of the sphere including an estimate of the absolute uncertainty in your answer.
\nUsing the following equation
\n\n
calculate, for these data, the acceleration due to gravity including an estimate of the absolute uncertainty in your answer.
\ndistance fallen = 654 – 12 = 642 «mm»
\nabsolute uncertainty = 2 + 0.1 «mm» ≈ 2 × 10–3 «m» or = 2.1 × 10–3 «m» or 2.0 × 10–3 «m»
\n\n
Accept answers in mm or m
\n[2 marks]
\n«a = » = 9.744 «ms–2»
\nfractional uncertainty in distance = AND fractional uncertainty in time =
\ntotal fractional uncertainty = «= 0.00311 + 2 × 0.00551»
\ntotal absolute uncertainty = 0.1 or 0.14 AND same number of decimal places in value and uncertainty, ie: 9.7 ± 0.1 or 9.74 ± 0.14
\n\n
Accept working in % for MP2 and MP3
\nFinal uncertainty must be the absolute uncertainty
\n[4 marks]
\nThe diagram shows a potential divider circuit used to measure the emf E of a cell X. Both cells have negligible internal resistance.
\n
AB is a wire of uniform cross-section and length 1.0 m. The resistance of wire AB is 80 Ω. When the length of AC is 0.35 m the current in cell X is zero.
\nState what is meant by the emf of a cell.
\nShow that the resistance of the wire AC is 28 Ω.
\nDetermine E.
\nCell X is replaced by a second cell of identical emf E but with internal resistance 2.0 Ω. Comment on the length of AC for which the current in the second cell is zero.
\nthe work done per unit charge
\nin moving charge from one terminal of a cell to the other / all the way round the circuit
\n\n
Award [1] for “energy per unit charge provided by the cell”/“power per unit current”
\nAward [1] for “potential difference across the terminals of the cell when no current is flowing”
\nDo not accept “potential difference across terminals of cell”
\n[2 marks]
\nthe resistance is proportional to length / see 0.35 AND 1«.00»
\nso it equals 0.35 × 80
\n«= 28 Ω»
\n[2 marks]
\ncurrent leaving 12 V cell is = 0.15 «A»
\nOR
\nE = × 28
\nE = «0.15 × 28 =» 4.2 «V»
\n\n
Award [2] for a bald correct answer
\nAllow a 1sf answer of 4 if it comes from a calculation.
\nDo not allow a bald answer of 4 «V»
\nAllow ECF from incorrect current
\n[2 marks]
\nsince the current in the cell is still zero there is no potential drop across the internal resistance
\nand so the length would be the same
\n\n
OWTTE
\n[2 marks]
\nOutline, with reference to the Jeans criterion, why a cold dense gas cloud is more likely to form new stars than a hot diffuse gas cloud.
\nExplain how neutron capture can produce elements with an atomic number greater than iron.
\n«For a star to form»: magnitude of PE of gas cloud > KE of gas cloud
\nOR
\nMass of cloud > Jean's mass
\nOR
\nJean’s criterion is the critical mass
\n\n
hence a hot diffuse cloud could have KE which is too large/PE too small
\nOR
\nhence a cold dense cloud will have low KE/high PE
\nOR
\na cold dense cloud is more likely to exceed Jeans mass
\nOR
\na hot diffuse cloud is less likely to exceed the Jeans mass
\n\n
Accept Ep + Ek < 0
\n[2 marks]
\nNeutron capture creates heavier isotopes / heavier nuclei / more unstable nucleus
\nβ– decay of heavy elements/iron increases atomic number «by 1»
\n\n
OWTTE
\n[2 marks]
\nThe diagram shows the axes for two inertial reference frames. Frame S represents the ground and frame S′ is a box that moves to the right relative to S with speed v.
\nWhen the origins of the two frames coincide all clocks show zero. At that instant a beam of light of speed c is emitted from the left wall of the box towards the right wall. The box has proper length L. Consider the event E = light arrives at the right wall of the box.
\n
Using Galilean relativity,
State what is meant by a reference frame.
\nexplain why the time coordinate of E in frame S is .
\nhence show that the space coordinate of E in frame S is .
\na set of rulers and clocks / set of coordinates to record the position and time of events ✔
\nALTERNATIVE 1:
\nthe time in frame S′ is ✔
\nbut time is absolute in Galilean relativity so is the same in S ✔
\n\n
ALTERNATIVE 2:
\nIn frame S, light rays travel at c + v ✔
\nso ✔
\n\n
In Alternative 1, they must refer to S'
\nx = x' + vt and x' = L ✔
\n«substitution to get answer»
\nA satellite orbiting a planet moves from orbit X to orbit Y.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/31_01\"](\"../assets/uploads/tinymce_asset/asset/4837/Schermafbeelding_2018-08-13_om_18.58.58.png\"/)
What is the change in the kinetic energy and the change in the gravitational potential energy as a result?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/31_02\"](\"../assets/uploads/tinymce_asset/asset/4838/Schermafbeelding_2018-08-13_om_19.00.39.png\"/)
C
\nAn electron of mass me orbits an alpha particle of mass mα in a circular orbit of radius r. Which expression gives the speed of the electron?
\nA.
\nB.
\nC.
\nD.
\nA
\nTwo identical circular coils are placed one below the other so that their planes are both horizontal. The top coil is connected to a cell and a switch.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/33_01\"](\"../assets/uploads/tinymce_asset/asset/4789/Schermafbeelding_2018-08-13_om_11.06.13.png\"/)
The switch is closed and then opened. What is the force between the coils when the switch is closing and when the switch is opening?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/33_02\"](\"../assets/uploads/tinymce_asset/asset/4790/Schermafbeelding_2018-08-13_om_11.06.53.png\"/)
C
\nThe mass of the Earth is ME and the mass of the Moon is MM. Their respective radii are RE and RM.
\nWhich is the ratio ?
\nA.
\nB.
\nC.
\nD.
\nC
\nA student carries out an experiment to determine the variation of intensity of the light with distance from a point light source. The light source is at the centre of a transparent spherical cover of radius C. The student measures the distance x from the surface of the cover to a sensor that measures the intensity I of the light.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/02\"](\"../assets/uploads/tinymce_asset/asset/4749/Schermafbeelding_2018-08-12_om_15.49.35.png\"/)
The light source emits radiation with a constant power P and all of this radiation is transmitted through the cover. The relationship between I and x is given by
\n\n
The student obtains a set of data and uses this to plot a graph of the variation of with x.
\nThis relationship can also be written as follows.
\n\n
Show that .
\nEstimate C.
\nDetermine P, to the correct number of significant figures including its unit.
\nExplain the disadvantage that a graph of I versus has for the analysis in (b)(i) and (b)(ii).
\ncombines the two equations to obtain result
\n\n
«for example = K2(C + x)2 = (C + x)2»
\nOR
\nreverse engineered solution – substitute K = into = K2(C + x)2 to get I =
\n\n
There are many ways to answer the question, look for a combination of two equations to obtain the third one
\n[1 mark]
\nextrapolating line to cross x-axis / use of x-intercept
\nOR
\nUse C =
\nOR
\nuse of gradient and one point, correctly substituted in one of the formulae
\n\n
accept answers between 3.0 and 4.5 «cm»
\n\n
Award [1 max] for negative answers
\n[2 marks]
\nALTERNATIVE 1
\nEvidence of finding gradient using two points on the line at least 10 cm apart
\nGradient found in range: 115–135 or 1.15–1.35
\nUsing P = to get value between 6.9 × 10–4 and 9.5 × 10–4 «W» and POT correct
\nCorrect unit, W and answer to 1, 2 or 3 significant figures
\n\n
ALTERNATIVE 2
\nFinds from use of one point (x and y) on the line with x > 6 cm and C from(b)(i)to use in I = or = Kx + KC
\nCorrect re-arrangementto get P between 6.9 × 10–4 and 9.5 × 10–4 «W» and POT correct
\nCorrect unit, W and answer to 1, 2 or 3 significant figures
\n\n
Award [3 max] for an answer between 6.9 W and 9.5 W (POT penalized in 3rd marking point)
\nAlternative 2 is worth [3 max]
\n[4 marks]
\nthis graph will be a curve / not be a straight line
\n\n
more difficult to determine value of K
\nOR
\nmore difficult to determine value of C
\nOR
\nsuitable mathematical argument
\n\n
OWTTE
\n[2 marks]
\nExplain the evidence that indicates the location of dark matter in galaxies.
\nOutline why a hypothesis of dark energy has been developed.
\n«rotational» velocity of stars are expected to decrease as distance from centre of galaxy increases
\nthe observed velocity of outer stars is constant/greater than predicted
\nimplying large mass on the edge «which is dark matter»
\n\n
OWTTE
\n1st and 2nd marking points can be awarded from an annotated sketch with similar shape as the one below
\n![\"M18/4/PHYSI/HP3/ENG/TZ2/19.a/M\"](\"../assets/uploads/tinymce_asset/asset/4869/Schermafbeelding_2018-08-14_om_10.48.40.png\"/)
[3 marks]
\ndata from type 1a supernovae shows universe expanding at an accelerated rate
\n\n
gravity was expected to slow down the expansion of the universe
\nOR
\nthis did not fit the hypotheses at that time
\n\n
dark energy counteracts/opposes gravity
\nOR
\ndark energy causes the acceleration
\n\n
OWTTE
\n[3 marks]
\nA bar rotates horizontally about its centre, reaching a maximum angular velocity in six complete rotations from rest. The bar has a constant angular acceleration of . The moment of inertia of the bar about the axis of rotation is .
\nShow that the final angular velocity of the bar is about .
\nDraw the variation with time of the angular displacement of the bar during the acceleration.
\nCalculate the torque acting on the bar while it is accelerating.
\nThe torque is removed. The bar comes to rest in complete rotations with constant angular deceleration. Determine the time taken for the bar to come to rest.
\n✓
\n✓
\n\n
Other methods are possible.
Answer 3 given so look for correct working
At least 2 sig figs for MP2.
concave up from origin ✓
\n✓
\nOR ✓
\n✓
\n\n
Other methods are possible.
\nAllow if used
\nAllow if used
\nAward [2] marks for a bald correct answer
\nThe graph shows the variation with time t of the current I in the primary coil of an ideal transformer.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/34_01\"](\"../assets/uploads/tinymce_asset/asset/4791/Schermafbeelding_2018-08-13_om_11.08.10.png\"/)
The number of turns in the primary coil is 100 and the number of turns in the secondary coil is 200. Which graph shows the variation with time of the current in the secondary coil?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/34_02\"](\"../assets/uploads/tinymce_asset/asset/4792/Schermafbeelding_2018-08-13_om_11.08.54.png\"/)
D
\nThe current I flowing in loop A in a clockwise direction is increasing so as to induce a current both in loops B and C. All three loops are on the same plane.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/33_01\"](\"../assets/uploads/tinymce_asset/asset/4840/Schermafbeelding_2018-08-13_om_19.06.33.png\"/)
What is the direction of the induced currents in loop B and loop C?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/33_02\"](\"../assets/uploads/tinymce_asset/asset/4841/Schermafbeelding_2018-08-13_om_19.06.57.png\"/)
C
\nMonochromatic coherent light is incident on two parallel slits of negligible width a distance d apart. A screen is placed a distance D from the slits. Point M is directly opposite the midpoint of the slits.
\nInitially the lower slit is covered and the intensity of light at M due to the upper slit alone is 22 W m-2. The lower slit is now uncovered.
\nThe width of each slit is increased to 0.030 mm. D, d and λ remain the same.
\nDeduce, in W m-2, the intensity at M.
\nP is the first maximum of intensity on one side of M. The following data are available.
\nd = 0.12 mm
\nD = 1.5 m
\nDistance MP = 7.0 mm
\nCalculate, in nm, the wavelength λ of the light.
\nSuggest why, after this change, the intensity at P will be less than that at M.
\nShow that, due to single slit diffraction, the intensity at a point on the screen a distance of 28 mm from M is zero.
\nthere is constructive interference at M
OR
the amplitude doubles at M ✔
intensity is «proportional to» amplitude2 ✔
88 «W m−2» ✔
\n« ✔
\n» ✔
\n\n
«the interference pattern will be modulated by»
single slit diffraction ✔
«envelope and so it will be less»
\nALTERNATIVE 1
the angular position of this point is «rad» ✔
\nwhich coincides with the first minimum of the diffraction envelope
\n«rad» ✔
\n«so intensity will be zero»
\n\n
ALTERNATIVE 2
the first minimum of the diffraction envelope is at «rad» ✔
\ndistance on screen is «mm» ✔
\n«so intensity will be zero»
\n\n
This was generally well answered by those who attempted it but was the question that was most left blank. The most common mistake was the expected one of simply doubling the intensity.
\nThis was very well answered. As the question asks for the answer to be given in nm a bald answer of 560 was acceptable. Candidates could also gain credit for an answer of e.g. 5.6 x 10-7 m provided that the m was included.
\nMany recognised the significance of the single slit diffraction envelope.
\nCredit was often gained here for a calculation of an angle for alternative 2 in the markscheme but often the final substitution 1.50 was omitted to score the second mark. Both marks could be gained if the calculation was done in one step. Incorrect answers often included complicated calculations in an attempt to calculate an integer value.
\nThe diagram shows a diode bridge rectification circuit and a load resistor.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/35_01\"](\"../assets/uploads/tinymce_asset/asset/4793/Schermafbeelding_2018-08-13_om_11.10.14.png\"/)
The input is a sinusoidal signal. Which of the following circuits will produce the most smoothed output signal?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/35_02\"](\"../assets/uploads/tinymce_asset/asset/4794/Schermafbeelding_2018-08-13_om_11.11.06.png\"/)
C
\nA rocket of proper length 120 m moves to the right with speed 0.82c relative to the ground.
\nA probe is released from the back of the rocket at speed 0.40c relative to the rocket.
\nCalculate the speed of the probe relative to the ground.
\nDetermine the time it takes the probe to reach the front of the rocket according to an observer at rest in the rocket.
\nDetermine the time it takes the probe to reach the front of the rocket according to an observer at rest on the ground.
\n✔
\n0.92c ✔
\n✔
\n«s» ✔
\n«» 1.747 ✔
\n\n
Δt = «» =
\nOR
\nΔt = ✔
\n\n
2.3 × 10−6 «s» ✔
\nA beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.
\nThe beam is incident normally on a double slit. The distance between the slits is 0.300 mm. A screen is at a distance D from the slits. The diffraction angle θ is labelled.
\n
The graph of variation of intensity with diffraction angle for this experiment is shown.
\n![\"M18/4/PHYSI/HP2/ENG/TZ1/03.b\"](\"../assets/uploads/tinymce_asset/asset/4801/Schermafbeelding_2018-08-13_om_12.36.49.png\"/)
A beam of coherent monochromatic light from a distant galaxy is used in an optics experiment on Earth.
\nThe beam is incident normally on a double slit. The distance between the slits is 0.300 mm. A screen is at a distance D from the slits. The diffraction angle θ is labelled.
\n
A series of dark and bright fringes appears on the screen. Explain how a dark fringe is formed.
\nOutline why the beam has to be coherent in order for the fringes to be visible.
\nThe wavelength of the beam as observed on Earth is 633.0 nm. The separation between a dark and a bright fringe on the screen is 4.50 mm. Calculate D.
\nCalculate the angular separation between the central peak and the missing peak in the double-slit interference intensity pattern. State your answer to an appropriate number of significant figures.
\nDeduce, in mm, the width of one slit.
\nThe wavelength of the light in the beam when emitted by the galaxy was 621.4 nm.
\nExplain, without further calculation, what can be deduced about the relative motion of the galaxy and the Earth.
\nsuperposition of light from each slit / interference of light from both slits
\nwith path/phase difference of any half-odd multiple of wavelength/any odd multiple of (in words or symbols)
\nproducing destructive interference
\n\n
Ignore any reference to crests and troughs.
\n[3 marks]
\nlight waves (from slits) must have constant phase difference / no phase difference / be in phase
\n\n
OWTTE
\n[1 mark]
\nevidence of solving for D «D » ✔
\n«» = 4.27 «m» ✔
\n\n
Award [1] max for 2.13 m.
\nsin θ =
\nsin θ = 0.0084401…
\nfinal answer to three sig figs (eg 0.00844 or 8.44 × 10–3)
\n\n
Allow ECF from (a)(iii).
\nAward [1] for 0.121 rad (can award MP3 in addition for proper sig fig)
\nAccept calculation in degrees leading to 0.481 degrees.
\nAward MP3 for any answer expressed to 3sf.
\n[3 marks]
\nuse of diffraction formula «b = »
\nOR
\n\n
«=» 7.5«00» × 10–2 «mm»
\n\n
Allow ECF from (b)(i).
\n[2 marks]
\nwavelength increases (so frequency decreases) / light is redshifted
\ngalaxy is moving away from Earth
\n\n
Allow ECF for MP2 (ie wavelength decreases so moving towards).
\n[2 marks]
\nA parallel plate capacitor is connected to a cell of negligible internal resistance.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/36_01\"](\"../assets/uploads/tinymce_asset/asset/4795/Schermafbeelding_2018-08-13_om_11.12.32.png\"/)
The energy stored in the capacitor is 4 J and the electric field in between the plates is 100 N C–1. The distance between the plates of the capacitor is doubled. What are the energy stored and the electric field strength?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/36_02\"](\"../assets/uploads/tinymce_asset/asset/4796/Schermafbeelding_2018-08-13_om_11.13.23.png\"/)
A
\nA rectangular flat coil moves at constant speed through a uniform magnetic field. The direction of the field is into the plane of the paper.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/34_01\"](\"../assets/uploads/tinymce_asset/asset/4842/Schermafbeelding_2018-08-13_om_19.08.23.png\"/)
Which graph shows the variation with time t, of the induced emf ε in the coil as it moves from P to Q?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/34_02\"](\"../assets/uploads/tinymce_asset/asset/4843/Schermafbeelding_2018-08-13_om_19.09.09.png\"/)
A
\nAn ohmic conductor is connected to an ideal ammeter and to a power supply of output voltage V.
\n
The following data are available for the conductor:
\ndensity of free electrons = 8.5 × 1022 cm−3
\nresistivity ρ = 1.7 × 10−8 Ωm
\ndimensions w × h × l = 0.020 cm × 0.020 cm × 10 cm.
\n\n
The ammeter reading is 2.0 A.
\nThe electric field E inside the sample can be approximated as the uniform electric field between two parallel plates.
\nAn ohmic conductor is connected to an ideal ammeter and to a power supply of output voltage V.
\n
The following data are available for the conductor:
\ndensity of free electrons = 8.5 × 1022 cm−3
\nresistivity ρ = 1.7 × 10−8 Ωm
\ndimensions w × h × l = 0.020 cm × 0.020 cm × 10 cm.
\n\n
The ammeter reading is 2.0 A.
\nCalculate the resistance of the conductor.
\nCalculate the drift speed v of the electrons in the conductor in cm s–1.
\nDetermine the electric field strength E.
\nShow that .
\n1.7 × 10–8 ×
\n0.043 «Ω»
\n[2 marks]
\nv «= » =
\n0.37 «cms–1»
\n\n
[2 marks]
\nV = RI = 0.086 «V»
\n«» 0.86 «V m–1»
\n\n
Allow ECF from 4(a).
\nAllow ECF from MP1.
\n[2 marks]
\nALTERNATIVE 1
\nclear use of Ohm’s Law (V = IR)
\nclear use of R =
\ncombining with I = nAve and V = EL to reach result.
\n\n
ALTERNATIVE 2
\nattempts to substitute values into equation.
\ncorrectly calculates LHS as 4.3 × 109.
\ncorrectly calculates RHS as 4.3 × 109.
\n\n
For ALTERNATIVE 1 look for:
\nV = IR
\nR =
\nV = EL
\nI = nAve
\nV = I
\nEL = I
\nE = I
\nE = nAve = nveρ
\n\n
[3 marks]
\nThe graph shows the power dissipated in a resistor of 100 Ω when connected to an alternating current (ac) power supply of root mean square voltage (Vrms) 60 V.
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/35_01\"](\"../assets/uploads/tinymce_asset/asset/4844/Schermafbeelding_2018-08-13_om_19.10.44.png\"/)
What are the frequency of the ac power supply and the average power dissipated in the resistor?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/35_02\"](\"../assets/uploads/tinymce_asset/asset/4845/Schermafbeelding_2018-08-13_om_19.11.38.png\"/)
A
\nTwo radioactive nuclides, X and Y, have half-lives of 50 s and 100 s respectively. At time t = 0 samples of X and Y contain the same number of nuclei.
\nWhat is when t = 200 s?
\nA. 4
\nB. 2
\nC.
\nD.
\nD
\nWhat is the unit of power expressed in fundamental SI units?
\n\n
A. kg m s–2
\nB. kg m2 s–2
\nC. kg m s–3
\nD. kg m2 s–3
\nD
\nRocket A and rocket B are travelling in opposite directions from the Earth along the same straight line.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/03\"](\"../assets/uploads/tinymce_asset/asset/4751/Schermafbeelding_2018-08-12_om_16.33.49.png\"/)
In the reference frame of the Earth, the speed of rocket A is 0.75c and the speed of rocket B is 0.50c.
\nCalculate, for the reference frame of rocket A, the speed of rocket B according to the Galilean transformation.
\nCalculate, for the reference frame of rocket A, the speed of rocket B according to the Lorentz transformation.
\nOutline, with reference to special relativity, which of your calculations in (a) is more likely to be valid.
\n1.25c
\n[1 mark]
\nALTERNATIVE 1
\n\n
0.91c
\nALTERNATIVE 2
\n\n
–0.91c
\n\n
[2 marks]
\nnothing can travel faster than the speed of light (therefore (a)(ii) is the valid answer)
\n\n
OWTTE
\n[1 mark]
\nThree capacitors, each one with a capacitance C, are connected such that their combined capacitance is 1.5C. How are they connected?
\n![\"M18/4/PHYSI/HPM/ENG/TZ2/36\"](\"../assets/uploads/tinymce_asset/asset/4846/Schermafbeelding_2018-08-13_om_19.12.40.png\"/)
C
\nThe first diagram shows a person standing on a turntable which can rotate freely. The person is stationary and holding a bicycle wheel. The wheel rotates anticlockwise when seen from above.
\n© International Baccalaureate Organization 2020.
\nThe wheel is flipped, as shown in the second diagram, so that it rotates clockwise when seen from above.
\n© International Baccalaureate Organization 2020.
\nExplain the direction in which the person-turntable system starts to rotate.
\nExplain the changes to the rotational kinetic energy in the person-turntable system.
\n«person rotates» anticlockwise ✓
\nthe person gains angular momentum «in the opposite direction to the new wheel motion» ✓
\nso that the total angular momentum is conserved ✓
\n\n
OWTTE
\nAward [1 max] for a bald statement of conservation of angular momentum
\nthe rotational kinetic energy has increased ✓
\nenergy is provided by the person doing work «flipping the wheel» ✓
\n\n
OWTTE
\nThe spacetime diagram shows the axes of an inertial reference frame S and the axes of a second inertial reference frame S′ that moves relative to S with speed 0.745c. When clocks in both frames show zero the origins of the two frames coincide.
\nEvent E has coordinates x = 1 m and ct = 0 in frame S. Show that in frame S′ the space coordinate and time coordinate of event E are
\nA rod at rest in frame S has proper length 1.0 m. At t = 0 the left-hand end of the rod is at x = 0 and the right-hand end is at x = 1.0 m.
\nx′ = 1.5 m.
\nct′ = –1.1 m.
\nLabel, on the diagram, the space coordinate of event E in the S′ frame. Label this event with the letter P.
\nLabel, on the diagram, the event that has coordinates x′ = 1.0 m and ct′ = 0. Label this event with the letter Q.
\nUsing the spacetime diagram, outline without calculation, why observers in frame S′ measure the length of the
rod to be less than 1.0 m.
Using the spacetime diagram, estimate, in m, the length of this rod in the S′ frame.
\n«» 1.499 ✔
\nx′ = «» 1.499 × (1.0 − 0) ✔
\n«x′ = 1.5 m»
\nt′ = « =» «= »
\n«ct′ = –1.1 m»
\nOR
\nusing spacetime interval 0 − 12 = (ct′)2 − 1.52 ⇒ «ct′ = –1.1» ✔
\nline through event E parallel to ct′ axis meeting x' axis and labelled P ✔
\n\n
point on x' axis about of the way to P labelled Q ✔
\n\n
ends of rod must be recorded at the same time in frame S′ ✔
\nany vertical line from E crossing x’, no label required ✔
\nright-hand end of rod intersects at R «whose co-ordinate is less than 1.0 m» ✔
\n\n
0.7 m ✔
\nA photoelectric cell is connected in series with a battery of emf 2 V. Photons of energy 6 eV are incident on the cathode of the photoelectric cell. The work function of the surface of the cathode is 3 eV.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ2/37\"](\"../assets/uploads/tinymce_asset/asset/4847/Schermafbeelding_2018-08-13_om_19.13.41.png\"/)
What is the maximum kinetic energy of the photoelectrons that reach the anode?
\nA. 1 eV
\nB. 3 eV
\nC. 5 eV
\nD. 8 eV
\nA
\nThe length of the side of a cube is 2.0 cm ± 4 %. The mass of the cube is 24.0 g ± 8 %. What is the percentage uncertainty of the density of the cube?
\n\n
A. ± 2 %
\nB. ± 8 %
\nC. ± 12 %
\nD. ± 20 %
\nD
\nWhich of the following is evidence for the wave nature of the electron?
\nA. Continuous energy spectrum in β– decay
\nB. Electron diffraction from crystals
\nC. Existence of atomic energy levels
\nD. Existence of nuclear energy levels
\nB
\nAccording to the Bohr model for hydrogen, visible light is emitted when electrons make transitions from excited states down to the state with n = 2. The dotted line in the following diagram represents the transition from n = 3 to n = 2 in the spectrum of hydrogen.
\n ![\"M18/4/PHYSI/HPM/ENG/TZ1/38_01\"](\"../assets/uploads/tinymce_asset/asset/4797/Schermafbeelding_2018-08-13_om_11.17.45.png\"/)
Which of the following diagrams could represent the visible light emission spectrum of hydrogen?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/38_02\"](\"../assets/uploads/tinymce_asset/asset/4798/Schermafbeelding_2018-08-13_om_11.18.20.png\"/)
B
\nThe radioactive nuclide beryllium-10 (Be-10) undergoes beta minus (β–) decay to form a stable boron (B) nuclide.
\nThe initial number of nuclei in a pure sample of beryllium-10 is N0. The graph shows how the number of remaining beryllium nuclei in the sample varies with time.
\nAn ice sample is moved to a laboratory for analysis. The temperature of the sample is –20 °C.
\nIdentify the missing information for this decay.
\nOn the graph, sketch how the number of boron nuclei in the sample varies with time.
\nAfter 4.3 × 106 years,
\n\n
Show that the half-life of beryllium-10 is 1.4 × 106 years.
\nBeryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 1011 atoms of beryllium-10. The present activity of the sample is 8.0 × 10−3 Bq.
\nDetermine, in years, the age of the sample.
\nState what is meant by thermal radiation.
\nDiscuss how the frequency of the radiation emitted by a black body can be used to estimate the temperature of the body.
\nCalculate the peak wavelength in the intensity of the radiation emitted by the ice sample.
\nThe temperature in the laboratory is higher than the temperature of the ice sample. Describe one other energy transfer that occurs between the ice sample and the laboratory.
\n\n
antineutrino AND charge AND mass number of electron ,
\nconservation of mass number AND charge ,
\n\n
Do not accept V.
\nAccept without subscript e.
\n[2 marks]
\ncorrect shape ie increasing from 0 to about 0.80 N0
\ncrosses given line at 0.50 N0
\n
[2 marks]
\nALTERNATIVE 1
\nfraction of Be = , 12.5%, or 0.125
\ntherefore 3 half lives have elapsed
\n«≈ 1.4 × 106» «y»
\n\n
ALTERNATIVE 2
\nfraction of Be = , 12.5%, or 0.125
\nleading to λ = 4.836 × 10–7 «y»–1
\n= 1.43 × 106 «y»
\n\n
\n
Must see at least one extra sig fig in final answer.
\n[3 marks]
\nλ «= » = 4.95 × 10–7 «y–1»
\nrearranging of A = λN0e–λt to give –λt = ln «= –0.400»
\nt = «y»
\n\n
\n
Allow ECF from MP1
\n[3 marks]
\nemission of (infrared) electromagnetic/infrared energy/waves/radiation.
\n[1 mark]
\nthe (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law
\nso frequency/color depends on temperature
\n[2 marks]
\n\n
= 1.1 × 10–5 «m»
\n\n
Allow ECF from MP1 (incorrect temperature).
\n[2 marks]
\nfrom the laboratory to the sample
\nconduction – contact between ice and lab surface.
\nOR
\nconvection – movement of air currents
\n\n
Must clearly see direction of energy transfer for MP1.
\nMust see more than just words “conduction” or “convection” for MP2.
\n[2 marks]
\nA planet of mass m is in a circular orbit around a star. The gravitational potential due to the star at the position of the planet is V.
\nShow that the total energy of the planet is given by the equation shown.
\n\n
Suppose the star could contract to half its original radius without any loss of mass. Discuss the effect, if any, this has on the total energy of the planet.
\nThe diagram shows some of the electric field lines for two fixed, charged particles X and Y.
\nThe magnitude of the charge on X is and that on Y is . The distance between X and Y is 0.600 m. The distance between P and Y is 0.820 m.
\nAt P the electric field is zero. Determine, to one significant figure, the ratio .
\n✔
\ncomparison with ✔
\n«to give answer»
\n\n
ALTERNATIVE 1
«at the position of the planet» the potential depends only on the mass of the star /does not depend on the radius of the star ✔
the potential will not change and so the energy will not change ✔
ALTERNATIVE 2
r / distance between the centres of the objects / orbital radius remains unchanged ✔
since , energy will not change ✔
\n\n
✔
\n✔
\n\n
This was generally well answered but with candidates sometimes getting in to trouble over negative signs but otherwise producing well-presented answers.
\nA large number of candidates thought that the total energy of the planet would change, mostly double.
\nThe majority of candidates had an idea of the basic technique here but it was surprisingly common to see the squared missing from the expression for field strengths.
\nA truck has an initial speed of 20 m s–1. It decelerates at 4.0 m s–2. What is the distance taken by the truck to stop?
\n\n
A. 2.5 m
\nB. 5.0 m
\nC. 50 m
\nD. 100 m
\nC
\nA particle of fixed energy is close to a potential barrier.
\nWhich changes to the width of the barrier and to the height of the barrier will always make the tunnelling probability greater?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/39\"](\"../assets/uploads/tinymce_asset/asset/4799/Schermafbeelding_2018-08-13_om_11.19.21.png\"/)
D
\nAn electron of initial energy E tunnels through a potential barrier. What is the energy of the electron after tunnelling?
\nA. greater than E
\nB. E
\nC. less than E
\nD. zero
\nB
\nWhen a spaceship passes the Earth, an observer on the Earth and an observer on the spaceship both start clocks. The initial time on both clocks is 12 midnight. The spaceship is travelling at a constant velocity with γ = 1.25. A space station is stationary relative to the Earth and carries clocks that also read Earth time.
\nSome of the radio signal is reflected at the surface of the Earth and this reflected signal is later detected at the spaceship. The detection of this signal is event B. The spacetime diagram is shown for the Earth, showing the space station and the spaceship. Both axes are drawn to the same scale.
\nCalculate the velocity of the spaceship relative to the Earth.
\nThe spaceship passes the space station 90 minutes later as measured by the spaceship clock. Determine, for the reference frame of the Earth, the distance between the Earth and the space station.
\nAs the spaceship passes the space station, the space station sends a radio signal back to the Earth. The reception of this signal at the Earth is event A. Determine the time on the Earth clock when event A occurs.
\nConstruct event A and event B on the spacetime diagram.
\nEstimate, using the spacetime diagram, the time at which event B occurs for the spaceship.
\n0.60c
\nOR
\n1.8 × 108 «m s–1»
\n\n
[1 mark]
\nALTERNATIVE 1
\ntime interval in the Earth frame = 90 × γ = 112.5 minutes
\n«in Earth frame it takes 112.5 minutes for ship to reach station»
\nso distance = 112.5 × 60 × 0.60c
\n1.2 × 10m12 «m»
\n\n
ALTERNATIVE 2
\nDistance travelled according in the spaceship frame = 90 × 60 × 0.6c
\n= 9.72 × 1011 «m»
\nDistance in the Earth frame «= 9.72 × 1011 × 1.25» = 1.2 × 1012 «m»
\n\n
[3 marks]
\nsignal will take «112.5 × 0.60 =» 67.5 «minutes» to reach Earth «as it travels at c»
\nOR
\nsignal will take « =» 4000 «s»
\n\n
total time «= 67.5 + 112.5» = 180 minutes or 3.00 h or 3:00am
\n\n
[2 marks]
\nline from event E to A, upward and to left with A on ct axis (approx correct)
\nline from event A to B, upward and to right with B on ct' axis (approx correct)
\nboth lines drawn with ruler at 45 (judge by eye)
\n\n
eg:
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/04.d.i/M\"](\"../assets/uploads/tinymce_asset/asset/4765/Schermafbeelding_2018-08-13_om_06.36.34.png\"/)
[3 marks]
\nALTERNATIVE 1
\n«In spaceship frame»
\nFinds the ratio (or by similar triangles on x or ct axes), value is approximately 4
\nhence time elapsed ≈ 4 × 90 mins ≈ 6h «so clock time is ≈ 6:00»
\n\n
Alternative 1:
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/04.d.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4764/Schermafbeelding_2018-08-13_om_06.33.50.png\"/)
Allow similar triangles using x-axis or ct-axis, such as from diagrams below
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/04.d.ii_02/M\"](\"../assets/uploads/tinymce_asset/asset/4766/Schermafbeelding_2018-08-13_om_06.52.27.png\"/)
\n
ALTERNATIVE 2
\n«In Earth frame»
\nFinds the ratio
\n, value is approximately 2.5
\nhence time elapsed ≈ ≈ 6h
\n«so clocktime is ≈ 6:00»
\n\n
\n
ALTERNATIVE 2:
\n ![\"M18/4/PHYSI/SP3/ENG/TZ2/04.d.ii_03/M\"](\"../assets/uploads/tinymce_asset/asset/4767/Schermafbeelding_2018-08-13_om_06.53.38.png\"/)
\n
[2 marks]
\nA projectile is fired at an angle to the horizontal. Air resistance is negligible. The path of the projectile is shown.
\nWhich gives the magnitude of the horizontal component and the magnitude of the vertical component of the velocity of the projectile between O and P?
\n\n
A
\nAlpha particles with energy E are directed at nuclei with atomic number Z. Small deviations from the predictions of the Rutherford scattering model are observed.
\nWhich change in E and which change in Z is most likely to result in greater deviations from the Rutherford scattering model?
\n![\"M18/4/PHYSI/HPM/ENG/TZ1/40\"](\"../assets/uploads/tinymce_asset/asset/4800/Schermafbeelding_2018-08-13_om_11.20.27.png\"/)
B
\nTwo samples X and Y of different radioactive isotopes have the same initial activity. Sample X has twice the number of atoms as sample Y. The half-life of X is T. What is the half-life of Y?
\nA. 2T
\nB. T
\nC.
\nD.
\nC
\nA runner starts from rest and accelerates at a constant rate throughout a race. Which graph shows the variation of speed v of the runner with distance travelled s?
\n\n
A
\nA solid sphere of radius and mass is released from rest and rolls down a slope, without slipping. The vertical height of the slope is . The moment of inertia of this sphere about an axis through its centre is .
\nShow that the linear velocity of the sphere as it leaves the slope is .
\nconservation of rotational and linear energy
\nOR
✓
\n
using AND ✓
\nwith correct manipulation to find the requested relationship ✓
\nThe diagram represents an ideal, monatomic gas that first undergoes a compression, then an increase in pressure.
\nAn adiabatic process then increases the volume of the gas to .
\nCalculate the work done during the compression.
\nCalculate the work done during the increase in pressure.
\nCalculate the pressure following this process.
\nOutline how an approximate adiabatic change can be achieved.
\n«–» «» ✓
\n\n
«» ✓
\n\n
OWTTE
\nuse of is constant «» ✓
\nOR «» ✓
\n\n
Award [2] marks for a bald correct answer
\nadiabatic means no transfer of heat in or out of the system ✓
\nshould be fast ✓
\n«can be slow if» the system is insulated ✓
\n\n
OWTTE
\nTwo blocks X and Y rest on a frictionless horizontal surface as shown. A horizontal force is now applied to the larger block and the two blocks move together with the same speed and acceleration.
\nWhich free-body diagram shows the frictional forces between the two blocks?
\n\n
A
\nA small magnet is dropped from rest above a stationary horizontal conducting ring. The south (S) pole of the magnet is upwards.
\nWhile the magnet is moving towards the ring, state why the magnetic flux in the ring is increasing.
\nWhile the magnet is moving towards the ring, sketch, using an arrow on Diagram 2, the direction of the induced current in the ring.
\nWhile the magnet is moving towards the ring, deduce the direction of the magnetic force on the magnet.
\nthe magnetic field at the position of the ring is increasing «because the magnet gets closer to the ring» ✔
\nthe current must be counterclockwise «in diagram 2» ✔
eg:
\nsince the induced magnetic field is upwards
OR
by Lenz law the change «of magnetic field/flux» must be opposed
OR
by conservation of energy the movement of the magnet must be opposed ✔
therefore the force is repulsive/upwards ✔
\nThis was well-answered.
\nAnswers here were reasonably evenly split between clockwise and anti-clockwise, with the odd few arrows pointing left or right.
\nThe majority of candidates recognised that the magnetic force would be upwards and the most common way of explaining this was via Lenz’s law. Students needed to get across that the force is opposing a change or a motion.
\nThe mass at the end of a pendulum is made to move in a horizontal circle of radius r at constant speed. The magnitude of the net force on the mass is F.
\nWhat is the direction of F and the work done by F during half a revolution?
\n\n
A
\nA closed box of fixed volume 0.15 m3 contains 3.0 mol of an ideal monatomic gas. The temperature of the gas is 290 K.
\nWhen the gas is supplied with 0.86 kJ of energy, its temperature increases by 23 K. The specific heat capacity of the gas is 3.1 kJ kg–1 K–1.
\nA closed box of fixed volume 0.15 m3 contains 3.0 mol of an ideal monatomic gas. The temperature of the gas is 290 K.
\nDetermine, in kJ, the total kinetic energy of the particles of the gas.
\nExplain, with reference to the kinetic model of an ideal gas, how an increase in temperature of the gas leads to an increase in pressure.
\nALTERNATIVE 1
\naverage kinetic energy = 1.38 × 10–23 × 313 = 6.5 × 10–21 «J»
\nnumber of particles = 3.0 × 6.02 × 1023 = 1.8 × 1024
\ntotal kinetic energy = 1.8 × 1024 × 6.5 × 10–21 = 12 «kJ»
\n\n
ALTERNATIVE 2
\nideal gas so U = KE
\nKE = 8.31 × 131 × 3
\ntotal kinetic energy = 12 «kJ»
\n\n
[3 marks]
\nlarger temperature implies larger (average) speed/larger (average) KE of molecules/particles/atoms
\nincreased force/momentum transferred to walls (per collision) / more frequent collisions with walls
\nincreased force leads to increased pressure because P = F/A (as area remains constant)
\n\n
Ignore any mention of PV = nRT.
\n[3 marks]
\nA uniform rod of weight 36.0 N and length 5.00 m rests horizontally. The rod is pivoted at its left-hand end and is supported at a distance of 4.00 m from the frictionless pivot.
\nThe support is suddenly removed and the rod begins to rotate clockwise about the pivot point. The moment of inertia of the rod about the pivot point is 30.6 kg m2.
\nCalculate the force the support exerts on the rod.
\nCalculate, in rad s–2, the initial angular acceleration of the rod.
\nAfter time t the rod makes an angle θ with the horizontal. Outline why the equation cannot be used to find the time it takes θ to become (that is for the rod to become vertical for the first time).
\nAt the instant the rod becomes vertical show that the angular speed is ω = 2.43 rad s–1.
\nAt the instant the rod becomes vertical calculate the angular momentum of the rod.
\ntaking torques about the pivot R × 4.00 = 36.0 × 2.5 ✔
\nR = 22.5 «N» ✔
\n36.0 × 2.50 = 30.6 × ✔
\n= 2.94 «rad s–2» ✔
\n\n
the equation can be applied only when the angular acceleration is constant ✔
\nany reasonable argument that explains torque is not constant, giving non constant acceleration ✔
\n\n
«from conservation of energy» Change in GPE = Change in rotational KE ✔
\n✔
\n✔
\n«ω = 2.4254 rad s–1»
\nL = 30.6 × 2.43 = 74.4 «J s» ✔
\nMonochromatic light from two identical lamps arrives on a screen.
\n ![\"M18/4/PHYSI/HP2/ENG/TZ2/05.a_01\"](\"../assets/uploads/tinymce_asset/asset/4853/Schermafbeelding_2018-08-14_om_07.02.59.png\"/)
The intensity of light on the screen from each lamp separately is I0.
\nOn the axes, sketch a graph to show the variation with distance x on the screen of the intensity I of light on the screen.
\nMonochromatic light from a single source is incident on two thin, parallel slits.
\n![\"M18/4/PHYSI/HP2/ENG/TZ2/05.b_01\"](\"../assets/uploads/tinymce_asset/asset/4856/Schermafbeelding_2018-08-14_om_07.10.48.png\"/)
The following data are available.
\n\n
The intensity I of light at the screen from each slit separately is I0. Sketch, on the axes, a graph to show the variation with distance x on the screen of the intensity of light on the screen for this arrangement.
\nThe slit separation is increased. Outline one change observed on the screen.
\nhorizontal straight line through I = 2
\n\n
![\"M18/4/PHYSI/HP2/ENG/TZ2/05.a\"](\"../assets/uploads/tinymce_asset/asset/4855/Schermafbeelding_2018-08-14_om_07.07.00.png\"/)
Accept a curve that falls from I = 2 as distance increases from centre but not if it falls to zero.
\n[1 mark]
\n«standard two slit pattern»
\ngeneral shape with a maximum at x = 0
\nmaxima at 4I0
\nmaxima separated by « =» 2.0 cm
\n\n
Accept single slit modulated pattern provided central maximum is at 4. ie height of peaks decrease as they go away from central maximum. Peaks must be of the same width
\n![\"M18/4/PHYSI/HP2/ENG/TZ2/05.b/M\"](\"../assets/uploads/tinymce_asset/asset/4858/Schermafbeelding_2018-08-14_om_07.15.48.png\"/)
[3 marks]
\nfringe width/separation decreases
\nOR
\nmore maxima seen
\n[1 mark]
\nA compressed spring is used to launch an object along a horizontal frictionless surface. When the spring is compressed through a distance and released, the object leaves the spring at speed . What is the distance through which the spring must be compressed for the object to leave the spring at ?
\n\n
A.
\nB.
\nC.
\nD.
\nB
\nObserver A detects the creation (event 1) and decay (event 2) of a nuclear particle. After creation, the particle moves at a constant speed relative to A. As measured by A, the distance between the events is 15 m and the time between the events is 9.0 × 10–8 s.
\nObserver B moves with the particle.
\nFor event 1 and event 2,
\nExplain what is meant by the statement that the spacetime interval is an invariant quantity.
\ncalculate the spacetime interval.
\ndetermine the time between them according to observer B.
\nOutline why the observed times are different for A and B.
\nquantity that is the same/constant in all inertial frames
\n[1 mark]
\nspacetime interval = 272 – 152 = 504 «m2»
\n[1 mark]
\nALTERNATIVE 1
\nEvidence of x′ = 0
\nt′ «» = 7.5 × 10–8 «s»
\nALTERNATIVE 2
\nγ = 1.2
\nt′ «= » = 7.5 × 10–8 «s»
\n\n
[2 marks]
\nobserver B measures the proper time and this is the shortest time measured
\nOR
\ntime dilation occurs «for B's journey» according to A
\nOR
\nobserver B is stationary relative to the particle, observer A is not
\n[1 mark]
\nThe diagram represents a diverging mirror being used to view an object.
\nConstruct a single ray showing one path of light between the eye, the mirror and the object, to view the object.
\nThe image observed is virtual. Outline the meaning of virtual image.
\nattempt to connect object and eye with ray showing equal angles of reflection such that reflection occurs within 1 hatch mark of position shown ✓
\nconstruction showing normal at point of reflection ✓
\n\n
Allow rays that are drawn freehand without a ruler - use judgement
\nlight rays do not pass through the image
OR
do not form an image on a screen
OR
appear to have come from a point
OR
formed by extension of rays ✓
\n
OWTTE.
\nA ball of mass m collides with a wall and bounces back in a straight line. The ball loses 75 % of the initial energy during the collision. The speed before the collision is v.
\nWhat is the magnitude of the impulse on the ball by the wall?
\n\n
A.
\nB.
\nC.
\nD.
\nD
\nIn an experiment to determine the radius of a carbon-12 nucleus, a beam of neutrons is scattered by a thin film of carbon-12. The graph shows the variation of intensity of the scattered neutrons with scattering angle. The de Broglie wavelength of the neutrons is 1.6 × 10-15 m.
\nA pure sample of copper-64 has a mass of 28 mg. The decay constant of copper-64 is 5.5 × 10-2 hour–1.
\nSuggest why de Broglie’s hypothesis is not consistent with Bohr’s conclusion that the electron’s orbit in the hydrogen atom has a well defined radius.
\nEstimate, using the graph, the radius of a carbon-12 nucleus.
\nThe ratio is approximately A.
Comment on this observation by reference to the strong nuclear force.
\nEstimate, in Bq, the initial activity of the sample.
\nCalculate, in hours, the time at which the activity of the sample has decreased to one-third of the initial activity.
\n«de Broglie’s hypothesis states that the» electron is represented by a wave ✔
therefore it cannot be localized/it is spread out/it does not have a definite position ✔
\nAward MP1 for any mention of wavelike property of an electron.
\n«» «m» ✔
«m» ✔
\nthis implies that the nucleons are very tightly packed/that there is very little space in between the nucleons ✔
because the nuclear force is stronger than the electrostatic force ✔
\nnumber of nuclei is ✔
«» «Bq» ✔
\n ✔
t = 20«hr» ✔
\nA 700 W electric heater is used to heat 1 kg of water without energy losses. The specific heat capacity of water is 4.2 kJ kg–1 K–1. What is the time taken to heat the water from 25 °C to 95 °C?
\n\n
A. 7 s
\nB. 30 s
\nC. 7 minutes
\nD. 420 minutes
\nC
\nA planet has radius R. At a distance h above the surface of the planet the gravitational field strength is g and the gravitational potential is V.
\nState what is meant by gravitational field strength.
\nShow that V = –g(R + h).
\nDraw a graph, on the axes, to show the variation of the gravitational potential V of the planet with height h above the surface of the planet.
\nA planet has a radius of 3.1 × 106 m. At a point P a distance 2.4 × 107 m above the surface of the planet the gravitational field strength is 2.2 N kg–1. Calculate the gravitational potential at point P, include an appropriate unit for your answer.
\nThe diagram shows the path of an asteroid as it moves past the planet.
\n ![\"M18/4/PHYSI/HP2/ENG/TZ2/06.c\"](\"../assets/uploads/tinymce_asset/asset/4861/Schermafbeelding_2018-08-14_om_07.33.17.png\"/)
When the asteroid was far away from the planet it had negligible speed. Estimate the speed of the asteroid at point P as defined in (b).
\nThe mass of the asteroid is 6.2 × 1012 kg. Calculate the gravitational force experienced by the planet when the asteroid is at point P.
\nthe «gravitational» force per unit mass exerted on a point/small/test mass
\n[1 mark]
\nat height h potential is V = –
\nfield is g =
\n«dividing gives answer»
\n\n
Do not allow an answer that starts with g = – and then cancels the deltas and substitutes R + h
\n[2 marks]
\ncorrect shape and sign
\nnon-zero negative vertical intercept
\n\n
![\"M18/4/PHYSI/HP2/ENG/TZ2/06.a.iii/M\"](\"../assets/uploads/tinymce_asset/asset/4860/Schermafbeelding_2018-08-14_om_07.26.11.png\"/)
[2 marks]
\nV = «–2.2 × (3.1 × 106 + 2.4 × 107) =» «–» 6.0 × 107 J kg–1
\n\n
Unit is essential
\nAllow eg MJ kg–1 if power of 10 is correct
\nAllow other correct SI units eg m2s–2, N m kg–1
\n[1 mark]
\ntotal energy at P = 0 / KE gained = GPE lost
\n«mv2 + mV = 0 ⇒» v =
\nv = « =» 1.1 × 104 «ms–1»
\n\n
\n
Award [3] for a bald correct answer
\nIgnore negative sign errors in the workings
\nAllow ECF from 6(b)
\n[3 marks]
\nALTERNATIVE 1
\nforce on asteroid is «6.2 × 1012 × 2.2 =» 1.4 × 1013 «N»
\n«by Newton’s third law» this is also the force on the planet
\nALTERNATIVE 2
\nmass of planet = 2.4 x 1025 «kg» «from V = –»
\nforce on planet «» = 1.4 × 1013 «N»
\n\n
MP2 must be explicit
\n[2 marks]
\nA student wants to determine the angular speed ω of a rotating object. The period T is 0.50 s ±5 %. The angular speed ω is
\n\n
What is the percentage uncertainty of ω?
\nA. 0.2 %
\nB. 2.5 %
\nC. 5 %
\nD. 10 %
\nC
\nThis question was well answered by candidates.
\nThe pV diagram of a heat engine using an ideal gas consists of an isothermal expansion A → B, an isobaric compression B → C and an adiabatic compression C → A.
\nThe following data are available:
\nTemperature at A = 385 K
\nPressure at A = 2.80 × 106 Pa
\nVolume at A = 1.00 × 10–4 m3
\nVolume at B = 2.80 × 10–4 m3
\nVolume at C = 1.85 × 10–4 m3
\nShow that at C the pressure is 1.00 × 106 Pa.
\nShow that at C the temperature is 254 K.
\nShow that the thermal energy transferred from the gas during the change B → C is 238 J.
\nThe work done by the gas from A → B is 288 J. Calculate the efficiency of the cycle.
\nState, without calculation, during which change (A → B, B → C or C → A) the entropy of the gas decreases.
\nALTERNATIVE 1:
\n✔
\n= «= 1.00 × 106 Pa» ✔
\n\n
ALTERNATIVE 2:
\n✔
\n«= 1.00 × 106 Pa» ✔
\nALTERNATIVE 1:
\nSince TB = TA then Tc = ✔
\n= «= 254.4 K» ✔
\n\n
ALTERNATIVE 2:
\n«K»✔
\n«= 254.4 K» ✔
\nwork done = «pΔV = 1.00 × 106 × (1.85 × 10−4 − 2.80 × 10−4 =» −95 «J» ✔
\nchange in internal energy = «pΔV = − × 95 =» −142.5 «J» ✔
\nQ = −95 − 142.5 ✔
\n«−238 J»
\n\n
Allow positive values.
\nnet work is 288 −238 = 50 «J» ✔
\nefficiency = « =» 0.17 ✔
\n\n
along B→C ✔
\n\n
A container is filled with a mixture of helium and oxygen at the same temperature. The molar mass of helium is 4 g mol–1 and that of oxygen is 32 g mol–1.
\nWhat is the ratio ?
\n\n
A.
\nB.
\nC.
\nD. 8
\nC
\nA beam of monochromatic light from infinity is incident on a converging lens A. The diagram shows three wavefronts of the light and the focal point of the lens.
\nDraw on the diagram the three wavefronts after they have passed through the lens.
\nLens A has a focal length of . An object is placed to the left of A. Show by calculation that a screen should be placed about from A to display a focused image.
\nThe screen is removed and the image is used as the object for a second diverging lens B, to form a final image. Lens B has a focal length of and the final real image is from the lens. Calculate the distance between lens A and lens B.
\nCalculate the total magnification of the object by the lens combination.
\nwavefront separation identical and equal to separation before the lens ✓
wavefronts converging, approximately centered on ✓
\n\n
By eye.
\nDotted construction lines are not required, allow wavefronts to extend beyond or be inside the dotted lines here.
\nAllow [1 max] if only two wavefronts drawn.
\n✓
\n✓
\n\n
✓
\n✓
\ndistance necessary = ✓
\n\n
Allow [2 max] for ECF for no negative in MP1. Gives and distance of
\nAllow ECF from (b) in MP3.EG use of .
\n« for A or for B»
\nOR ✓
\ntotal magnification ✓
\n\n
Allow [2] marks for a bald correct answer
Allow ECF from (b) and (c).
Eg if in (c) then and total
\nA student models the relationship between the pressure p of a gas and its temperature T as p = + T.
\nThe units of p are pascal and the units of T are kelvin. What are the fundamental SI units of and ?
\nA
\nContainer X contains 1.0 mol of an ideal gas. Container Y contains 2.0 mol of the ideal gas. Y has four times the volume of X. The pressure in X is twice that in Y.
\nWhat is ?
\n\n
A.
\nB.
\nC. 1
\nD. 2
\nC
\nA student investigates the electromotive force (emf) ε and internal resistance r of a cell.
\nThe current I and the terminal potential difference V are measured.
For this circuit V = ε - Ir .
The table shows the data collected by the student. The uncertainties for each measurement
are shown.
The graph shows the data plotted.
\nThe student has plotted error bars for the potential difference. Outline why no error bars are shown for the current.
\nDetermine, using the graph, the emf of the cell including the uncertainty for this value. Give your answer to the correct number of significant figures.
\nOutline, without calculation, how the internal resistance can be determined from this graph.
\nΔI is too small to be shown/seen
OR
Error bar of negligible size compared to error bar in V ✔
\nevidence that ε can be determined from the y-intercept of the line of best-fit or lines of min and max gradient ✔
states ε=1.59 OR 1.60 OR 1.61V«» ✔
states uncertainty in ε is 0.02 V«» OR 0.03«V» ✔
\ndetermine the gradient «of the line of best-fit» ✔
r is the negative of this gradient ✔
\nAlmost all candidates realised that the uncertainty in I was too small to be shown. A common mistake was to mention that since I is the independent variable the uncertainty is negligible.
\nThe number of candidates who realised that the V intercept was EMF was disappointing. Large numbers of candidates tried to calculate ε using points on the graph, often ending up with unrealistic values. Another common mistake was not giving values of ε and Δε to the correct number of digits - 2 decimal places on this occasion. Very few candidates drew maximum and minimum gradient lines as a way of determining Δε.
\nA wheel of mass 0.25 kg consists of a cylinder mounted on a central shaft. The shaft has a radius of 1.2 cm and the cylinder has a radius of 4.0 cm. The shaft rests on two rails with the cylinder able to spin freely between the rails.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/4753/Schermafbeelding_2018-08-12_om_16.44.54.png\"/)
The stationary wheel is released from rest and rolls down a slope with the shaft rolling on the rails without slipping from point A to point B.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/06.a\"](\"../assets/uploads/tinymce_asset/asset/4754/Schermafbeelding_2018-08-12_om_16.47.09.png\"/)
The wheel leaves the rails at point B and travels along the flat track to point C. For a short time the wheel slips and a frictional force F exists on the edge of the wheel as shown.
\n![\"M18/4/PHYSI/SP3/ENG/TZ2/06.b\"](\"../assets/uploads/tinymce_asset/asset/4755/Schermafbeelding_2018-08-12_om_16.49.45.png\"/)
The moment of inertia of the wheel is 1.3 × 10–4 kg m2. Outline what is meant by the moment of inertia.
\nIn moving from point A to point B, the centre of mass of the wheel falls through a vertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 m s–1 after its displacement.
\nDetermine the angular velocity of the wheel at B.
\nDescribe the effect of F on the linear speed of the wheel.
\nDescribe the effect of F on the angular speed of the wheel.
\nan object’s resistance to change in rotational motion
\nOR
\nequivalent of mass in rotational equations
\n\n
OWTTE
\n[1 mark]
\nΔKE + Δrotational KE = ΔGPE
\nOR
\nmv2 + I = mgh
\n\n
× 0.250 × v2 + × 1.3 × 10–4 × = 0.250 × 9.81 × 0.36
\nv = 1.2 «m s–1»
\n\n
[3 marks]
\nω «= » = 100 «rad s–1»
\n[1 mark]
force in direction of motion
\nso linear speed increases
\n[2 marks]
\nforce gives rise to anticlockwise/opposing torque on
\nwheel ✓ so angular speed decreases ✓
\n\n
OWTTE
\n[2 marks]
\nA particle moving in a circle completes 5 revolutions in 3 s. What is the frequency?
\n\n
A. Hz
\nB. Hz
\nC. Hz
\nD. Hz
\nB
\nA sky diver is falling at terminal speed when she opens her parachute. What are the direction of her velocity vector and the direction of her acceleration vector before she reaches the new terminal speed?
\nC
\nA longitudinal wave moves through a medium. Relative to the direction of energy transfer through the medium, what are the displacement of the medium and the direction of propagation of the wave?
\n\n
B
\nThe graphs show the variation of the displacement y of a medium with distance and with time t for a travelling wave.
\nWhat is the speed of the wave?
\n\n
A. 0.6 m s–1
\nB. 0.8 m s–1
\nC. 600 m s–1
\nD. 800 m s–1
\nD
\nA stone is thrown downwards from the edge of a cliff with a speed of 5.0 m s–1. It hits the ground 2.0 s later. What is the height of the cliff?
\nA. 20 m
\nB. 30 m
\nC. 40 m
\nD. 50 m
\nB
\nThis question was well answered by the majority of candidates and had a high discrimination index.
\nOutline the meaning of normal adjustment for a compound microscope.
\nSketch a ray diagram to find the position of the images for both lenses in the compound microscope at normal adjustment. The object is at O and the focal lengths of the objective and eyepiece lenses are shown.
\n\n
the final image lies at the near point «often assumed to be » ✓
\nany 2 correct rays from O for objective lens ✓
\n
forming an intermediate image at approximate position shown
OR
use of image from objective lens as object for eyepiece lens ✓
any 2 correct rays for eyepiece lens from intermediate image ✓
ray extension to form a final image ✓
\n\n
Allow ECF for MP2, MP3 & MP4 for badly drawn rays.
\nMP4 allow final image to be off the page
\nIn a double-slit experiment, a source of monochromatic red light is incident on slits S1 and S2 separated by a distance . A screen is located at distance from the slits. A pattern with fringe spacing is observed on the screen.
\nThree changes are possible for this arrangement
\nI. increasing
\nII. increasing
\nIII. using green monochromatic light instead of red.
\nWhich changes will cause a decrease in fringe spacing ?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II, and III
\nC
\nA capacitor consists of two parallel square plates separated by a vacuum. The plates are 2.5 cm × 2.5 cm squares. The capacitance of the capacitor is 4.3 pF.
\nCalculate the distance between the plates.
\nThe capacitor is connected to a 16 V cell as shown.
\n ![\"M18/4/PHYSI/HP2/ENG/TZ1/07.b\"](\"../assets/uploads/tinymce_asset/asset/4805/Schermafbeelding_2018-08-13_om_13.59.46.png\"/)
Calculate the magnitude and the sign of the charge on plate A when the capacitor is fully charged.
\nThe capacitor is fully charged and the space between the plates is then filled with a dielectric of permittivity ε = 3.0ε0.
\nExplain whether the magnitude of the charge on plate A increases, decreases or stays constant.
\nIn a different circuit, a transformer is connected to an alternating current (ac) supply.
\n![\"M18/4/PHYSI/HP2/ENG/TZ1/07.d\"](\"../assets/uploads/tinymce_asset/asset/4806/Schermafbeelding_2018-08-13_om_14.19.37.png\"/)
The transformer has 100 turns in the primary coil and 1200 turns in the secondary coil. The peak value of the voltage of the ac supply is 220 V. Determine the root mean square (rms) value of the output voltage.
\nDescribe the use of transformers in electrical power distribution.
\nd = « =» 1.3 × 10–3 «m»
\n\n
[1 mark]
\n6.9 × 10–11 «C»
\nnegative charge/sign
\n[2 marks]
\ncharge increases
\nbecause capacitance increases AND pd remains the same.
\n[2 marks]
\nALTERNATIVE 1
\nεs = × 220
\n= 2640 «V»
\nVrms = = 1870 «V»
\n\n
ALTERNATIVE 2
\n(Primary) Vrms = = 156 «V»
\n(Secondary) Vrms =
\nVrms = 1870 «V»
\n\n
Allow ECF from MP1 and MP2.
\nAward [2] max for 12.96 V (reversing Np and Ns).
\n[3 marks]
\nstep-up transformers increase voltage/step-down transformers decrease voltage
\n(step-up transformers increase voltage) from plants to transmission lines / (step-down transformers decrease voltage) from transmission lines to final utilizers
\nthis decreases current (in transmission lines)
\nto minimize energy/power losses in transmission
\n[3 marks]
\nTwo strings of lengths L1 and L2 are fixed at both ends. The wavespeed is the same for both strings. They both vibrate at the same frequency. L1 vibrates at its first harmonic. L2 vibrates at its third harmonic.
\nWhat is ?
\n\n
A.
\nB. 1
\nC. 2
\nD. 3
\nD
\nA ball is thrown upwards at an angle to the horizontal. Air resistance is negligible. Which statement about the motion of the ball is correct?
\nA. The acceleration of the ball changes during its flight.
\nB. The velocity of the ball changes during its flight.
\nC. The acceleration of the ball is zero at the highest point.
\nD. The velocity of the ball is zero at the highest point.
\nB
\nCandidate responses were divided between responses B (correct), D, and to a lesser extent, C. Many candidates appeared to focus on vertical velocity only or confused vertical velocity and acceleration values. This question had the highest discrimination index, suggesting that it would be a useful question for class discussion.
\nA single pulse of light enters an optic fibre which contains small impurities that scatter the light. Explain the effect of these impurities on the pulse.
\nmention of attenuation ✓
\nmention of dispersion or pulse broadening ✓
\ngives explanation for at least one of above ✓
\nTwo copper wires X and Y are connected in series. The diameter of Y is double that of X. The drift speed in X is v. What is the drift speed in Y?
\n\n
A.
\nB.
\nC. 2v
\nD. 4v
\nA
\nL is a point source of light. The intensity of the light at a distance 2 from L is I. What is the intensity at a distance 3 from L?
\n\n
A. I
\nB. I
\nC. I
\nD. I
\nA
\nX and Y are two coherent sources of waves. The phase difference between X and Y is zero. The intensity at P due to X and Y separately is I. The wavelength of each wave is 0.20 m.
\nWhat is the resultant intensity at P?
\n\n
A. 0
\nB. I
\nC. 2I
\nD. 4I
\nD
\nAn image of a comet is shown.
\nComet P/Halley as taken March 8, 1986 by W. Liller, Easter Island, part of the International Halley Watch (IHW) Large Scale Phenomena Network.
\nThe astronomical unit () and light year () are convenient measures of distance in astrophysics. Define each unit.
\n:
\n:
\nComets develop a tail as they approach the Sun. Identify one other characteristic of comets.
\nIdentify one object visible in the image that is outside our Solar System.
\n: «average» distance from the Earth to the Sun ✓
: distance light travels in one year ✓
\nmade of ice «and dust» ✓
\n«highly» eccentric/elliptical orbit around the Sun ✓
\nformed in the Oort Cloud ✓
\nstar / named star / stellar cluster/ galaxy/ constellation ✓
\n\n
Answer may be indicated on the photograph.
\nAn object of mass m is sliding down a ramp at constant speed. During the motion it travels a distance along the ramp and falls through a vertical distance h. The coefficient of dynamic friction between the ramp and the object is μ. What is the total energy transferred into thermal energy when the object travels distance ?
\n\n
A. mgh
\nB. mgx
\nC. μmgh
\nD. μmgx
\nA
\nThe refractive index of glass decreases with increasing wavelength. The diagram shows rays of light incident on a converging lens made of glass. The light is a mixture of red and blue light.
\nOn the diagram, draw lines to show the rays after they have refracted through the lens. Label the refracted red rays with the letter R and the refracted blue rays with the letter B.
\nSuggest how the refracted rays in (a) are modified when the converging lens is replaced by a diverging lens.
\nHence state how the defect of the converging lens in (a) may be corrected.
\neach incident ray shown splitting into two ✔
\neach pair symmetrically intersecting each other on principal axis ✔
\nfor red, intersection further to the right ✔
\nFor MP3, at least one of the rays must be labelled.
\nrays diverge after passing through lens
\nOR
\nthe extension of the rays will intersect the principal axis on the side of incident rays/as if they were coming from the focal point/points in the left side/OWTTE ✔
\nby placing a diverging lens next to the converging lens
\nOR
\nmake an achromatic doublet ✔
\nLight is incident at the boundary between air and diamond. The speed of light in diamond is less than the speed of light in air. The angle of incidence i of the light is greater than the critical angle. Which diagram is correct for this situation?
\nA
\nTwo blocks of masses m and 2m are travelling directly towards each other. Both are moving at the same constant speed v. The blocks collide and stick together.
\nWhat is the total momentum of the system before and after the collision?
\nA
\nResponse A was the most common (correct) response, however response D was a significant distractor for candidates who took momentum to be a scalar quantity.
\nA wire of length L is used in an electric heater. When the potential difference across the wire is 200 V, the power dissipated in the wire is 1000 W. The same potential difference is applied across a second similar wire of length 2L. What is the power dissipated in the second wire?
\n\n
A. 250 W
\nB. 500 W
\nC. 2000 W
\nD. 4000 W
\nB
\nA combination of four identical resistors each of resistance R are connected to a source of emf ε of negligible internal resistance. What is the current in the resistor X?
\n\n
A.
\nB.
\nC.
\nD.
\nC
\nTwo parallel wires P and Q are perpendicular to the page and carry equal currents. Point S is the same distance from both wires. The arrow shows the magnetic field at S due to P and Q.
\nWhat are the correct directions for the current at P and the current at Q?
\n\n
A
\nTwo parallel wires are perpendicular to the page. The wires carry equal currents in opposite directions. Point S is at the same distance from both wires. What is the direction of the magnetic field at point S?
\nA
\nShow that the apparent brightness , where is the distance of the object from Earth, is the surface temperature of the object and is the surface area of the object.
\nTwo of the brightest objects in the night sky seen from Earth are the planet Venus and the star Sirius. Explain why the equation is applicable to Sirius but not to Venus.
\nsubstitution of into giving
\n\n
Removal of constants and is optional
\nequation applies to Sirius/stars that are luminous/emit light «from fusion» ✓
\nbut Venus reflects the Sun’s light/does not emit light «from fusion» ✓
\n\n
OWTTE
\nThe diagram shows an alternating current generator with a rectangular coil rotating at a constant frequency in a uniform magnetic field.
\nThe graph shows how the generator output voltage varies with time .
\nElectrical power produced by the generator is delivered to a consumer some distance away.
\nExplain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.
\nThe average power output of the generator is . Calculate the root mean square (rms) value of the generator output current.
\nThe voltage output from the generator is stepped up before transmission to the consumer. Estimate the factor by which voltage has to be stepped up in order to reduce power loss in the transmission line by a factor of .
\nThe frequency of the generator is doubled with no other changes being made. Draw, on the axes, the variation with time of the voltage output of the generator.
\nthere is a magnetic flux «linkage» in the coil / coil cuts magnetic field ✓
\nthis flux «linkage» changes as the angle varies/coil rotates ✓
\n«Faraday’s law» connects induced emf with rate of change of flux «linkage» with time ✓
\n
Do not award MP2 or 3 for answers that don’t discuss flux.
✓
\n\n
✓
\n«power loss proportional to hence the step-up factor is ✓
\npeak emf doubles ✓
\nhalves ✓
\n
Must show at least 1 cycle.
This question was well answered with the majority discussing changes in flux rather than wires cutting field lines, which was good to see.
\nGenerally well answered.
\nThis was well answered by many, but some candidates left the answer as a surd. The most common guess here involved the use of root 2.
\nWell answered, with the majority of candidates scoring at least 1 mark.
\nThe diagram shows two light rays that form an intermediate image by the objective lens of an optical compound microscope. These rays are incident on the eyepiece lens. The focal points of the two lenses are marked.
\nThe object O is placed at a distance of 24.0 mm from the objective lens and the final image is formed at a distance 240 mm from the eyepiece lens. The focal length of the objective lens is 20.0 mm and that of the eyepiece lens is 60.0 mm. The near point of the observer is at a distance of 240 mm from the eyepiece lens.
\nDraw rays on the diagram to show the formation of the final image.
\nCalculate the distance between the lenses.
\nDetermine the magnification of the microscope.
\nproper construction lines ✔
\nimage at intersection of proper construction lines ✔
\n\n
distance of intermediate image from objective is
\nie: v = 120 «mm» ✔
\ndistance of intermediate image from eyepiece is
\nie: u = 48 «mm» ✔
\nlens separation 168 «mm» ✔
\nALTERNATIVE 1:
\neyepiece: m = = 5 ✔
\nAND
\nobjective m = = −5 ✔
\nTotal m = −5 × 5 = −25 ✔
\n\n
ALTERNATIVE 2:
\nm = ✔
\nm = −25 ✔
\n\n
Accept positive or negative values throughout.
\nIn the Rutherford-Geiger-Marsden scattering experiment it was observed that a small percentage of alpha particles are deflected through large angles.
\nThree features of the atom are
\nI. the nucleus is positively charged
\nII. the nucleus contains neutrons
\nIII. the nucleus is much smaller than the atom.
\nWhich features can be inferred from the observation?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nA particle of mass m and charge of magnitude q enters a region of uniform magnetic field B that is directed into the page. The particle follows a circular path of radius R. What are the sign of the charge of the particle and the speed of the particle?
\nB
\nAn experiment is conducted to determine how the fundamental frequency f of a vibrating wire varies with the tension T in the wire.
The data are shown in the graph, the uncertainty in the tension is not shown.
\nIt is proposed that the frequency of oscillation is given by f2 = kT where k is a constant.
\nDraw the line of best fit for the data.
\nDetermine the fundamental SI unit for k.
\nWrite down a pair of quantities that, when plotted, enable the relationship f2 = kT to be verified.
\nDescribe the key features of the graph in (b)(ii) if it is to support this relationship.
\nAny curve that passes through ALL the error bars ✔
kg–1 m–1 ✔
f2 AND T
OR
f AND
OR
log f AND log T
OR
ln f AND ln T ✔
graph would be a straight line/constant gradient/linear ✔
passing through the origin ✔
Most candidates correctly drew curves which passed through all the error bars, some tried to draw straight lines. Quite a few did not draw any line, leaving the question unanswered. Candidates need to make sure to check that they read the question paper carefully.
\nDetermining the fundamental units of K (kg-1 m-1 ) was difficult for most candidates.
\nThese questions were not well understood, but a few candidates were able to state that a plot of f2 versus T would give a straight line through the origin.
\nThese questions were not well understood, but a few candidates were able to state that a plot of f2 versus T would give a straight line through the origin.
\nThe graph shows the variation with time of the resultant net force acting on an object. The object has a mass of 1kg and is initially at rest.
\nWhat is the velocity of the object at a time of 200 ms?
\nA. 8 m s–1
\nB. 16 m s–1
\nC. 8 km s–1
\nD. 16 km s–1
\nA
\nMany candidates (incorrectly) selected response B, perhaps neglecting the changing value of force over time.
\nThe following decay is observed.
\nμ− → e− + vμ + X
\nWhat is particle X?
\n\n
A. γ
\nB. e
\nC. Z0
\nD. ve
\nB
\nAn insulated tube is filled with a large number n of lead spheres, each of mass m. The tube is inverted s times so that the spheres completely fall through an average distance L each time. The temperature of the spheres is measured before and after the inversions and the resultant change in temperature is ΔT.
\nWhat is the specific heat capacity of lead?
\n
\n
A.
\nB.
\nC.
\nD.
\n\n
B
\nA block is on the surface of a horizontal rotating disk. The block is at rest relative to the disk. The disk is rotating at constant angular velocity.
\nWhat is the correct arrow to represent the direction of the frictional force acting on the block at the instant shown?
\nC
\nCandidate responses were largely divided between responses C and D, suggesting some confusion around the direction of frictional force in a rotating object (vs. linear motion).
\nThe light from a distant galaxy shows that .
\nCalculate the ratio .
\nOutline how Hubble’s law is related to .
\n« »
\nOR OR ✓
\n«Hubble’s » measure of v/recessional speed uses redshift which is
OR
redshift () of galaxies is proportional to distance «from earth»
OR
combines AND into one expression, e.g. . ✓
OWTTE
\nTwo isolated point particles of mass 4M and 9M are separated by a distance 1 m. A point particle of mass M is placed a distance from the particle of mass 9M. The net gravitational force on M is zero.
\nWhat is ?
\n\n
A. m
\nB. m
\nC. m
\nD. m
\nC
\nAn object undergoing simple harmonic motion (SHM) has a period T and total energy E. The amplitude of oscillations is halved. What are the new period and total energy of the system?
\n\n
C
\nBoiling water is heated in a 2 kW electric kettle. The initial mass of water is 0.4 kg. Assume the specific latent heat of vaporization of water is 2 MJ kg–1.
\nWhat is the time taken for all the water to vaporize?
\nA. 250 s
\nB. 400 s
\nC. 2500 s
\nD. 4000 s
\nB
\nThis question was well answered by candidates.
\nThe graph shows the variation with time of the activity of a pure sample of a radioactive nuclide.
\nWhat percentage of the nuclide remains after 200 s?
\n\n
A. 3.1 %
\nB. 6.3 %
\nC. 13 %
\nD. 25 %
\nB
\nEnergy is transferred to water in a flask at a rate P. The water reaches boiling point and then P is increased. What are the changes to the temperature of the water and to the rate of vaporization of the water after the change?
\n\n
D
\nThe graph shows the variation with diffraction angle of the intensity of light when monochromatic light is incident on four slits.
\nThe number of slits is increased keeping the width and the separation of the slits unchanged.
\nThree possible changes to the pattern are
\nI. the separation of the primary maxima increases
\nII. the intensity of the primary maxima increases
\nIII. the width of the primary maxima decreases.
\nWhich of the possible changes are correct?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nA distinctive feature of the constellation Orion is the Trapezium, an open cluster of stars within Orion.
\nMintaka is one of the stars in Orion.
\nDistinguish between a constellation and an open cluster.
\nThe parallax angle of Mintaka measured from Earth is 3.64 × 10–3 arc-second. Calculate, in parsec, the approximate distance of Mintaka from Earth.
\nState why there is a maximum distance that astronomers can measure using stellar parallax.
\nThe Great Nebula is located in Orion. Describe, using the Jeans criterion, the necessary condition for a nebula to form a star.
\nIn cluster, stars are gravitationally bound OR constellation not ✔
\nIn cluster, stars are the same/similar age OR in constellation not ✔
\nStars in cluster are close in space/the same distance OR in constellation not ✔
\nCluster stars appear closer in night sky than constellation ✔
\nClusters originate from same gas cloud OR constellation does not ✔
\nd = 275 «pc» ✔
\nbecause of the difficulty of measuring very small angles ✔
\nmass of gas cloud > Jeans mass ✔
\n«magnitude of» gravitational potential energy > Ek of particles ✔
\ncloud collapses/coalesces «to form a protostar» ✔
\nA gas storage tank of fixed volume V contains N molecules of an ideal gas at temperature T. The pressure at kelvin temperature T is 20 MPa. molecules are removed and the temperature changed to 2T. What is the new pressure of the gas?
\nA. 10 MPa
\nB. 15 MPa
\nC. 30 MPa
\nD. 40 MPa
\nC
\nThe de Broglie wavelength of a particle accelerated close to the speed of light is approximately
\n\nwhere is the energy of the particle.
A beam of electrons of energy is produced in an accelerator.
The electron beam is used to study the nuclear radius of carbon-12. The beam is directed from the left at a thin sample of carbon-12. A detector is placed at an angle relative to the direction of the incident beam.
\nThe graph shows the variation of the intensity of electrons with . There is a minimum of intensity for .
\nShow that the wavelength of an electron in the beam is about .
\nDiscuss how the results of the experiment provide evidence for matter waves.
\nThe accepted value of the diameter of the carbon-12 nucleus is . Estimate the angle at which the minimum of the intensity is formed.
\nOutline why electrons with energy of approximately would be unsuitable for the investigation of nuclear radii.
\nExperiments with many nuclides suggest that the radius of a nucleus is proportional to , where is the number of nucleons in the nucleus. Show that the density of a nucleus remains approximately the same for all nuclei.
\nOR ✓
\n
Answer to at least s.f. (i.e. 3.0)
«the shape of the graph suggests that» electrons undergo diffraction «with carbon nuclei» ✓
\nonly waves diffract ✓
\n✓
\nOR ✓
\nthe de Broglie wavelength of electrons is «much» longer than the size of a nucleus ✓
\n
hence electrons would not undergo diffraction
OR
no diffraction pattern would be observed ✓
volume of a nucleus proportional to AND mass proportional to ✓
\nthe ratio independent of «hence density the same for all nuclei» ✓
\n\n
Both needed for MP1
\nAn easy calculation with only one energy conversion to consider and a 'show' answer to help.
\nThis question was challenging for candidates many of whom seemed to have little idea of the experiment. Many answers discussed deflection, with the idea that forces between the electron and the nucleus causing it to deflect at a particular angle. This was often combined with the word interference to suggest evidence of matter waves. A number of answers described a demonstration the candidates remembered seeing so answers talked about fuzzy green rings.
\nThis was answered reasonably well with only the odd omission of the sine in the equation.
\nCandidates generally scored poorly on this question. There was confusion between this experiment and another diffraction one, so often the new wavelength was compared to the spacing between atoms. Also, in line with answers to b(i) there were suggestions that the electrons did not have sufficient energy to reach the nucleus or would be deflected by too great an angle to be seen.
\nThis question proved challenging and it wasn't common to find answers that scored both marks. Of those that had the right approach some missed out on both marks by describing A as the mass of the nucleus rather than proportional to the mass of the nucleus.
\nA beam of monochromatic light is incident normally on a diffraction grating. The grating spacing is d. The angles between the different orders are shown on the diagram.
\nWhat is the expression for the wavelength of light used?
\n\n
A.
\nB.
\nC. d sin α
\nD. d sin β
\nB
\nThe graph shows the variation of the number of neutrons N with the atomic number Z for stable nuclei. The same scale is used in the N and Z axes.
\nWhich information can be inferred from the graph?
\nI. For stable nuclei with high Z, N is larger than Z.
\nII. For stable nuclei with small Z, N = Z.
\nIII. All stable nuclei have more neutrons than protons.
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nThe data for the star Eta Aquilae A are given in the table.
\nis the luminosity of the Sun and is the mass of the Sun.
\nShow by calculation that Eta Aquilae A is not on the main sequence.
\nEstimate, in , the distance to Eta Aquilae A using the parallax angle in the table.
\nEstimate, in , the distance to Eta Aquilae A using the luminosity in the table, given that .
\nSuggest why your answers to (b)(i) and (b)(ii) are different.
\nEta Aquilae A is a Cepheid variable. Explain why the brightness of Eta Aquilae A varies.
\n✓
\nthe luminosity of Eta (2630) is very different «so it is not main sequence» ✓
\n\n
Allow calculation of to give so not main sequence
\nOWTTE
✓
\nUse of ✓
\n✓
\n✓
\n\n
Award [3] marks for a bald correct answer between and
\n\n
\n
parallax angle in milliarc seconds/very small/at the limits of measurement ✓
\nuncertainties/error in measuring οr or ✓
\nvalues same order of magnitude, so not significantly different ✓
\n\n
Accept answers where MP1 and MP2 both refer to parallax angle
\nOWTTE
\nreference to change in size ✓
reference to change in temperature ✓
reference to periodicity of the process ✓
reference to transparency / opaqueness ✓
\n
A particle performs simple harmonic motion (shm). What is the phase difference between the displacement and the acceleration of the particle?
\nA. 0
\nB.
\nC.
\nD.
\n\n
C
\nAn optic fibre consists of a glass core of refractive index 1.52 surrounded by cladding of refractive index n. The critical angle at the glass–cladding boundary is 84°.
\nThe diagram shows the longest and shortest paths that a ray can follow inside the fibre.
\nFor the longest path the rays are incident at the core–cladding boundary at an angle just slightly greater than the critical angle. The optic fibre has a length of 12 km.
\n\n
Calculate n.
\nThe refractive indices of the glass and cladding are only slightly different. Suggest why this is desirable.
\nShow that the longest path is 66 m longer than the shortest path.
\nDetermine the time delay between the arrival of signals created by the extra distance in (b)(i).
\nSuggest whether this fibre could be used to transmit information at a frequency of 100 MHz.
\n«» n1 = 1.52 × sin 84.0° ✔
\n\n
n1 = 1.51 ✔
\nto have a critical angle close to 90° ✔
\nso only rays parallel to the axis are transmitted ✔
\nto reduce waveguide/modal dispersion ✔
\nlong path is ✔
\n= 12066 «m» ✔
\n«so 66 m longer»
\nspeed of light in core is «m s−1» ✔
\ntime delay is «s» ✔
\nno, period of signal is 1 × 10−8 «s» which is smaller than the time delay/OWTTE ✔
\n\n
An ambulance siren emits a sound of frequency 1200 Hz. The speed of sound in air is 330 m s–1. The ambulance moves towards a stationary observer at a constant speed of 40 m s–1. What is the frequency heard by the observer?
\n\n
A. Hz
\nB. Hz
\nC. Hz
\nD. Hz
\nD
\nA student uses a Young’s double-slit apparatus to determine the wavelength of light emitted by a monochromatic source. A portion of the interference pattern is observed on a screen.
\nThe distance D from the double slits to the screen is measured using a ruler with a smallest scale division of 1 mm.
The fringe separation s is measured with uncertainty ± 0.1 mm.
The slit separation d has negligible uncertainty.
The wavelength is calculated using the relationship .
\nWhen d = 0.200 mm, s = 0.9 mm and D = 280 mm, determine the percentage uncertainty in the wavelength.
\nExplain how the student could use this apparatus to obtain a more reliable value for λ.
\nEvidence of used ✔
«add fractional/% uncertainties»
obtains 11 % (or 0.11) OR 10 % (or 0.1) ✔
\nALTERNATIVE 1:
measure the combined width for several fringes
OR
repeat measurements ✓
take the average
OR
so the «percentage» uncertainties are reduced ✓
ALTERNATIVE 2:
increase D «hence s»
OR
Decrease d ✓
so the «percentage» uncertainties are reduced ✓
\nDo not accept answers which suggest using different apparatus.
\nA very easy question about percentage uncertainty which most candidates got completely correct. Many candidates gave the uncertainty to 4 significant figures or more. The process used to obtain the final answer was often difficult to follow.
\nThe most common correct answer was the readings should be repeated and an average taken. Another common answer was that D could be increased to reduce uncertainties in s. The best candidates knew that it was good practice to measure many fringe spacings and find the mean value. Quite a few candidates incorrectly stated that different apparatus should be used to give more precise results.
\nCopper () decays to nickel (). What are the particles emitted and the particle that mediates the interaction?
\n\n
D
\nTwo point charges Q1 and Q2 are one metre apart. The graph shows the variation of electric potential V with distance from Q1.
\nWhat is ?
\n\n
A.
\nB.
\nC. 4
\nD. 16
\nD
\nWhich graph shows the variation with time t of the kinetic energy (KE) of an object undergoing simple harmonic motion (shm) of period T?
\nD
\nThe surface temperature of the star Epsilon Indi is 4600 K.
\nDetermine the peak wavelength of the radiation emitted by Epsilon Indi.
\nUsing the axis, draw the variation with wavelength of the intensity of the radiation emitted by Epsilon Indi.
\nThe following data are available for the Sun.
\nSurface temperature = 5800 K
\nLuminosity =
\nMass =
\nRadius =
\nEpsilon Indi has a radius of 0.73 . Show that the luminosity of Epsilon Indi is 0.2 .
\nEpsilon Indi is a main sequence star. Show that the mass of Epsilon Indi is 0.64 .
\n\n
The Sun will spend about nine billion years on the main sequence. Calculate how long Epsilon Indi will spend on the main sequence.
\n\n
Describe the stages in the evolution of Epsilon Indi from the point when it leaves the main sequence until its final stable state.
\nλ = « =» 630 «nm» ✔
\nblack body curve shape ✔
\npeaked at a value from range 600 to 660 nm ✔
\n✔
\nL = 0.211 ✔
\nM = « =» 0.640 ✔
\n«» 3.0 ✔
\nT ≈ 27 billion years ✔
\nred giant ✔
\nplanetary nebula ✔
\nwhite dwarf ✔
\n\n
do NOT accept supernova, red supergiant, neutron star or black hole as stages
\nWhat are the changes in speed, frequency and wavelength of light as it travels from a material of low refractive index to a material of high refractive index?
\nD
\nThis question was well answered by the majority of candidates.
\nMonochromatic light of wavelength λ is normally incident on a diffraction grating. The diagram shows adjacent slits of the diffraction grating labelled V, W and X. Light waves are diffracted through an angle θ to form a second-order diffraction maximum. Points Z and Y are labelled.
\n
State the effect on the graph of the variation of sin θ with n of:
\nState the phase difference between the waves at V and Y.
\nState, in terms of λ, the path length between points X and Z.
\nThe separation of adjacent slits is d. Show that for the second-order diffraction maximum .
\nMonochromatic light of wavelength 633 nm is normally incident on a diffraction grating. The diffraction maxima incident on a screen are detected and their angle θ to the central beam is determined. The graph shows the variation of sinθ with the order n of the maximum. The central order corresponds to n = 0.
\nDetermine a mean value for the number of slits per millimetre of the grating.
\nusing a light source with a smaller wavelength.
\nincreasing the distance between the diffraction grating and the screen.
\n0 OR 2π OR 360° ✓
\n\n
4λ ✓
\n✓
\n
Do not award ECF from(a)(ii).
identifies gradient with OR use of ✓
\ngradient = 0.08 OR correct replacement in equation with coordinates of a point ✓
\n✓
\n✓
\n
Allow ECF from MP3
gradient smaller ✓
\nno change ✓
\nWhich of these waves cannot be polarized?
\nA. microwaves
\nB. ultrasound
\nC. ultraviolet
\nD. X rays
\nB
\nThis question was very well answered by candidates, with a high difficulty index.
\nA string fixed at both ends vibrates in the first harmonic with frequency 400 Hz. The speed of sound in the string is 480 m s–1. What is the length of the string?
\nA. 0.42 m
\nB. 0.60 m
\nC. 0.84 m
\nD. 1.2 m
\nB
\nResponse D was the most common option selected, perhaps by students equating the wavelength of the sound with the length of the string, or incorrectly taking the first harmonic to be the fundamental frequency.
\nThe following interaction is proposed between a proton and a pion.
\np+ + – → K– + +
\nThe quark content of the – is ūd and the quark content of the K– is ūs.
\nThree conservation rules are considered
\nI. baryon number
\nII. charge
\nIII. strangeness.
\nWhich conservation rules are violated in this interaction?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nThe gravitational potential at point P due to Earth is V.
\nWhat is the definition of the gravitational potential at P?
\n\n
A. Work done per unit mass to move a point mass from infinity to P
\nB. Work done per unit mass to move a point mass from P to infinity
\nC. Work done to move a point mass from infinity to P
\nD. Work done to move a point mass from P to infinity
\nA
\nIn science, models are extensively used to study real life situations.
\nA person X on the beach wants to reach a person Y in the sea in the shortest possible time. The speed of person X on land is different from the speed of person X in the water. Which physical phenomenon will best model the path with the least time?
\n\n
A. Conservation of momentum
\nB. Diffraction
\nC. Flow of charge in a conductor
\nD. Refraction
\nD
\nA deuterium nucleus (rest mass ) is accelerated by a potential difference of .
\nDefine rest mass.
\nCalculate the total energy of the deuterium particle in .
\nIn relativistic reactions the mass of the products may be less than the mass of the reactants. Suggest what happens to the missing mass.
\ninvariant mass
OR
mass of object when not in motion/in object’s rest frame ✓
«rest energy =» ✓
✓
\n\n
Final answer accept if unit given
\nAward [2] marks for a bald correct answer.
\nis converted to energy ✓
\nas kinetic energy of the products ✓
\nThe definition of rest mass proved to be known by most candidates.
\nMost were able to score full marks. A few candidates messed up by trying to convert to J, although some were successful with this additional difficulty.
\nAnswers scored at least one or usually both marks.
\nThe escape speed for the Earth is esc. Planet X has half the density of the Earth and twice the radius. What is the escape speed for planet X?
\n\n
A.
\nB.
\nC. esc
\nD. esc
\nD
\nTwo loudspeakers, A and B, are driven in phase and with the same amplitude at a frequency of . Point P is located from A and from B. The speed of sound is .
\nIn another experiment, loudspeaker A is stationary and emits sound with a frequency of . The microphone is moving directly away from the loudspeaker with a constant speed . The frequency of sound recorded by the microphone is .
\nDeduce that a minimum intensity of sound is heard at P.
\nA microphone moves along the line from P to Q. PQ is normal to the line midway between the loudspeakers.
\nThe intensity of sound is detected by the microphone. Predict the variation of detected intensity as the microphone moves from P to Q.
\nWhen both loudspeakers are operating, the intensity of sound recorded at Q is . Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to .
\nEstimate .
\nExplain why the frequency recorded by the microphone is lower than the frequency emitted by the loudspeaker.
\nCalculate .
\nwavelength ✓
path difference ✓
OR «half-wavelengths» ✓
waves meet in antiphase «at P»
OR
destructive interference/superposition «at P» ✓
\n
Allow approach where path length is calculated in terms of number of wavelengths; along path A () and
path B () for MP2, hence path difference wavelengths for MP3
«equally spaced» maxima and minima ✓
\na maximum at Q ✓
\nfour «additional» maxima «between P and Q» ✓
\nthe amplitude of sound at Q is halved ✓
«intensity is proportional to amplitude squared hence» ✓
speed of sound relative to the microphone is less ✓
\n
wavelength unchanged «so frequency is lower»
OR
fewer waves recorded in unit time/per second «so frequency is lower» ✓
✓
\n\n
✓
\nThis was answered very well, with those not scoring full marks able to, at least, calculate the wavelength.
\nMost candidates were able to score at least one mark by referring to a maximum at Q.
\nMost candidates earned 2 marks or nothing. A common answer was that intensity was 1/2 the original.
\nHL only. The majority of candidates answered this by describing the Doppler Effect for a moving source. Others reworded the question without adding any explanation. Correct explanations were rare.
\nHL only. This was answered well with the majority of candidates able to identify the correct formula and the correct values to substitute.
\nTwo charges, +Q and −Q, are placed as shown.
\nWhat is the magnitude of the electric field strength, in descending order, at points X, Y and Z.
\n\n
A. YXZ
\nB. ZXY
\nC. ZYX
\nD. YZX
\nC
\nThis question was well answered by candidates, with a high difficulty index.
\nThe graph shows how current varies with potential difference across a component X.
\nComponent X and a cell of negligible internal resistance are placed in a circuit.
\nA variable resistor R is connected in series with component X. The ammeter reads .
\nComponent X and the cell are now placed in a potential divider circuit.
\n\n
Outline why component X is considered non-ohmic.
\nDetermine the resistance of the variable resistor.
\nCalculate the power dissipated in the circuit.
\nState the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.
\nSlider S of the potential divider is positioned so that the ammeter reads . Explain, without further calculation, any difference in the power transferred by the potential divider arrangement over the arrangement in (b).
\ncurrent is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓
ALTERNATIVE 1
\nvoltage across X ✓
\nvoltage across R ✓
\nresistance of variable resistor ✓
\n\n
ALTERNATIVE 2
\noverall resistance ✓
\nresistance of X ✓
\nresistance of variable resistor ✓
\npower ✓
\nfrom to ✓
\nALTERNATIVE 1
\ncurrent from the cell is greater «than » ✓
\nbecause some of the current must flow through section SQ of the potentiometer ✓
\noverall power greater «than in part (b)» ✓
\n\n
ALTERNATIVE 2
\ntotal/overall resistance decreases ✓
\nbecause SQ and X are in parallel ✓
\noverall power greater «than in part (b)» ✓
\n
Allow the reverse argument.
Most answers that didn't score simply referred to the shape of the graph without any explanation as to what this meant to the relationship between the variables.
\nThis question produced a mixture of answers from the 2 alternatives given in the markscheme. As a minimum, many candidates were able to score a mark for the overall resistance of the circuit.
\nA straightforward calculation question that most candidates answered correctly.
\nSurprisingly a significant number of candidates had difficulty with this. Answers of 20 mA and 4 V were often seen.
\nHL only. This question challenged candidate's ability to describe clearly the changes in an electrical circuit. It revealed many misconceptions about the nature of electrical current and potential difference, of those who did have a grasp of what was going on the explanations often missed the second point in each of the markscheme alternatives as detail was missed about where the current was flowing or what was in parallel with what.
\nIn the Pound–Rebka–Snider experiment, a source of gamma rays was placed vertically above a gamma ray detector, in a tower on Earth.
\nCalculate the fractional change in frequency of the gamma rays at the detector.
\nExplain the cause of the frequency shift for the gamma rays in your answer in (a) in the Earth’s gravitational field.
\nExplain the cause of the frequency shift for the gamma rays in your answer in (a) if the tower and detector were accelerating towards the gamma rays in free space.
\n✓
\ngained by photons so increases ✓
\n, so frequency increases ✓
\ngamma rays travel at ✓
\ndetector accelerates towards source so «by Doppler effect» reduced so frequency increases ✓
\n\n
Award [1 max] for reference to principle of equivalence without further explanation.
\nThe calculation was easily done by most candidates.
\nA simple conceptual explanation in terms of the energy changes was not present.
\nSome candidates did score marks although it was not as simple as anticipated for a classical question.
\nWhat is the function of control rods in a nuclear power plant?
\n\n
A. To slow neutrons down
\nB. To regulate fuel supply
\nC. To exchange thermal energy
\nD. To regulate the reaction rate
\nD
\nA ring of area S is in a uniform magnetic field X. Initially the magnetic field is perpendicular to the plane of the ring. The ring is rotated by 180° about the axis in time T.
\nWhat is the average induced emf in the ring?
\n\n
A. 0
\nB.
\nC.
D.
D
\nA distinctive feature of the constellation Orion is the Trapezium, an open cluster of stars within Orion.
\nMintaka is one of the stars in Orion.
\nDistinguish between a constellation and an open cluster.
\nThe parallax angle of Mintaka measured from Earth is 3.64 × 10–3 arc-second. Calculate, in parsec, the approximate distance of Mintaka from Earth.
\nState why there is a maximum distance that astronomers can measure using stellar parallax.
\nIn cluster, stars are gravitationally bound OR constellation not ✔
\nIn cluster, stars are the same/similar age OR in constellation not ✔
\nStars in cluster are close in space/the same distance OR in constellation not ✔
\nCluster stars appear closer in night sky than constellation ✔
\nClusters originate from same gas cloud OR constellation does not ✔
\nd = 275 «pc» ✔
\nbecause of the difficulty of measuring very small angles ✔
\nTwo cells each of emf 9.0 V and internal resistance 3.0 Ω are connected in series. A 12.0 Ω resistor is connected in series to the cells. What is the current in the resistor?
\nA. 0.50 A
\nB. 0.75 A
\nC. 1.0 A
\nD. 1.5 A
\nC
\nThis question was well answered by candidates and had a higher discrimination index.
\nA photovoltaic panel of area S has an efficiency of 20 %. A second photovoltaic panel has an efficiency of 15 %. What is the area of the second panel so that both panels produce the same power under the same conditions?
\n\n
A.
\nB.
\nC.
\nD.
\nD
\nThe graph shows the variation of the peak output power P with time of an alternating current (ac) generator.
\nWhich graph shows the variation of the peak output power with time when the frequency of rotation is decreased?
\n\n
C
\nThe speed of a spaceship is measured to be 0.50c by an observer at rest in the Earth’s reference frame.
\nDefine an inertial reference frame.
\nAs the spaceship passes the Earth it emits a flash of light that travels in the same direction as the spaceship with speed c as measured by an observer on the spaceship. Calculate, according to the Galilean transformation, the speed of the light in the Earth’s reference frame.
\nUse your answer to (a)(ii) to describe the paradigm shift that Einstein’s theory of special relativity produced.
\na coordinate system which is not accelerating/has constant velocity/Newtons 1st law applies ✔
\nOWTTE
\nBoth “inertial” and “reference frame” need to be defined
\n1.5c ✔
\nc is the same in all frames
OR
c is maximum velocity possible ✔
velocity addition frame dependent ✔
length/time/mass/fields relative measurements ✔
Newtonian/Galilean mechanics valid only at low speed ✔
\nIn defining an inertial frame of reference far too many candidates started with the words ‘ a frame of reference that...... ’ instead of ‘a coordinate system that.....’
\nAlmost no incorrect answers were seen.
\nMost candidates correctly stated that in special relativity the velocity of light, c, is the maximum possible velocity or is invariant. Only a few added that Galilean relativity only applies at speeds much less than the speed of light.
\nLight of intensity I0 is incident on a snow-covered area of Earth. In a model of this situation, the albedo of the cloud is 0.30 and the albedo for the snow surface is 0.80. What is the intensity of the light at P due to the incident ray I0?
\n\n
A. 0.14 I0
\nB. 0.24 I0
\nC. 0.50 I0
\nD. 0.55 I0
\nB
\nA current of 1.0 × 10–3 A flows in the primary coil of a step-up transformer. The number of turns in the primary coil is Np and the number of turns in the secondary coil is Ns. One coil has 1000 times more turns than the other coil.
\nWhat is and what is the current in the secondary coil for this transformer?
\n\n
A
\nThe graph shows the variation with distance from the Earth of the recessional velocities of distant galaxies.
\nOutline how Hubble measured the recessional velocities of galaxies.
\nUse the graph to determine the age of the universe in s.
\nmeasured redshift «z» of star ✔
\nuse of Doppler formula OR z∼v/c OR v = to find v ✔
\n\n
OWTTE
\nuse of gradient or any point on the line to obtain any expression for either or ✔
\ncorrect conversion of d to m and v to m/s ✔
\n= 4.6 × 1017 «s» ✔
\nFour identical capacitors of capacitance X are connected as shown in the diagram.
\nWhat is the effective capacitance between P and Q?
\n\n
A.
\nB. X
\nC.
\nD. 4X
\nC
\nIn an experiment to demonstrate the photoelectric effect, monochromatic electromagnetic radiation from source A is incident on the surfaces of metal P and metal Q. Observations of the emission of electrons from P and Q are made.
\nThe experiment is then repeated with two other sources of electromagnetic radiation: B and C. The table gives the results for the experiment and the wavelengths of the radiation sources.
\nOutline the cause of the electron emission for radiation A.
\nOutline why electrons are never emitted for radiation C.
\nOutline why radiation B gives different results.
\nExplain why there is no effect on the table of results when the intensity of source B is doubled.
\nPhotons with energy 1.1 × 10−18 J are incident on a third metal surface. The maximum energy of electrons emitted from the surface of the metal is 5.1 × 10−19 J.
\nCalculate, in eV, the work function of the metal.
\nphoton transfers «all» energy to electron ✓
\nphoton energy is less than both work functions
OR
photon energy is insufficient «to remove an electron» ✓
Answer must be in terms of photon energy.
Identifies P work function lower than Q work function ✓
\nchanging/doubling intensity «changes/doubles number of photons arriving but» does not change energy of photon ✓
\n✓
\nwork function ✓
\n
Award [2] marks for a bald correct answer.
The surface temperature of the star Epsilon Indi is 4600 K.
\nDetermine the peak wavelength of the radiation emitted by Epsilon Indi.
\nUsing the axis, draw the variation with wavelength of the intensity of the radiation emitted by Epsilon Indi.
\nThe following data are available for the Sun.
\nSurface temperature = 5800 K
\nLuminosity =
\nMass =
\nRadius =
\nEpsilon Indi has a radius of 0.73 . Show that the luminosity of Epsilon Indi is 0.2 .
\nEpsilon Indi is a main sequence star. Show that the mass of Epsilon Indi is 0.64 .
\n\n
Describe how the chemical composition of a star may be determined.
\n\n
Describe the stages in the evolution of Epsilon Indi from the point when it leaves the main sequence until its final stable state.
\nλ = « =» 630 «nm» ✔
\nblack body curve shape ✔
\npeaked at a value from range 600 to 660 nm ✔
\n✔
\nL = 0.211 ✔
\nM = « =» 0.640 ✔
\nObtain «line» spectrum of star ✔
\nCompare to «laboratory» spectra of elements ✔
\nred giant ✔
\nplanetary nebula ✔
\nwhite dwarf ✔
\nOne possible fission reaction of uranium-235 (U-235) is
\n\nMass of one atom of U-235
Binding energy per nucleon for U-235
Binding energy per nucleon for Xe-140
Binding energy per nucleon for Sr-94
A nuclear power station uses U-235 as fuel. Assume that every fission reaction of U-235 gives rise to of energy.
\nA sample of waste produced by the reactor contains of strontium-94 (Sr-94). Sr-94 is radioactive and undergoes beta-minus () decay into a daughter nuclide X. The reaction for this decay is
\n.
\n\n
The graph shows the variation with time of the mass of Sr-94 remaining in the sample.
\nState what is meant by binding energy of a nucleus.
\nOutline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units.
\nShow that the energy released in the reaction is about .
\nEstimate, in , the specific energy of U-235.
\nThe power station has a useful power output of and an efficiency of . Determine the mass of U-235 that undergoes fission in one day.
\nThe specific energy of fossil fuel is typically . Suggest, with reference to your answer to (b)(i), one advantage of U-235 compared with fossil fuels in a power station.
\nWrite down the proton number of nuclide X.
\nState the half-life of Sr-94.
\nCalculate the mass of Sr-94 remaining in the sample after minutes.
\nenergy required to «completely» separate the nucleons
OR
energy released when a nucleus is formed from its constituent nucleons ✓
Allow protons AND neutrons.
the values «in SI units» would be very small ✓
\nOR ✓
\nsee AND ✓
\n✓
\nenergy produced in one day ✓
\nmass ✓
\n«specific energy of uranium is much greater than that of coal, hence» more energy can be produced from the same mass of fuel / per
OR
less fuel can be used to create the same amount of energy ✓
✓
\n
Do not allow unless the proton number is indicated.
✓
\nALTERNATIVE 1
\n✓
\nmass remaining ✓
\n\n
ALTERNATIVE 2
\ndecay constant ✓
\nmass remaining ✓
\nGenerally, well answered but candidates did miss the mark by discussing the constituents of a nucleus rather than the nucleons, or protons and neutrons. There seemed to be fewer comments than usual about 'the energy required to bind the nucleus together'.
\nWell answered with some candidates describing the values as too large or small.
\nWell answered.
\nThis caused problems for some with mass often correctly calculated but energy causing more difficulty with the eV conversion either being inaccurate or omitted. Candidates were allowed error carried forward for the second mark as long as they were dividing an energy by a mass.
\nMost candidates had the right idea, but common problems included forgetting the efficiency or not converting to days.
\nHL only. This was well answered.
\nMost candidates answered this correctly.
\nMost candidates answered this correctly.
\nThis was answered well with most candidates (even at HL) going down the number of half-lives route rather than the exponential calculation route.
\nCharge flows through a liquid. The charge flow is made up of positive and negative ions. In one second 0.10 C of negative ions flow in one direction and 0.10 C of positive ions flow in the opposite direction.
\nWhat is the magnitude of the electric current flowing through the liquid?
\nA. 0 A
\nB. 0.05 A
\nC. 0.10 A
\nD. 0.20 A
\nD
\nWhen green light is incident on a clean zinc plate no photoelectrons are emitted. What change may cause the emission of photoelectrons?
\n\n
A. Using a metal plate with larger work function
\nB. Changing the angle of incidence of the green light on the zinc plate
\nC. Using shorter wavelength radiation
\nD. Increasing the intensity of the green light
\nC
\nWhich is the correct Feynman diagram for pair annihilation and pair production?
\n\n
D
\nA projectile is fired at an angle to the horizontal. The path of the projectile is shown.
\nWhich gives the magnitude of the horizontal component and the magnitude of the vertical component of the velocity of the projectile between O and P?
\n\n
A
\nThe graph shows the variation of the natural log of activity, ln (activity), against time for a radioactive nuclide.
\nWhat is the decay constant, in days–1, of the radioactive nuclide?
\n\n
A.
\nB.
\nC. 3
\nD. 6
\nD
\nA beam of negative ions flows in the plane of the page through the magnetic field due to two bar magnets.
\nWhat is the direction in which the negative ions will be deflected?
\nA. Out of the page
\nB. Into the page X
\nC. Up the page ↑
\nD. Down the page ↓
\nA
\nThis question was well answered by candidates.
\nA radioactive nuclide is known to have a very long half-life.
\nThree quantities known for a pure sample of the nuclide are
\nI. the activity of the nuclide
\nII. the number of nuclide atoms
\nIII. the mass number of the nuclide.
\nWhat quantities are required to determine the half-life of the nuclide?
\n\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nA mass m attached to a string of length R moves in a vertical circle with a constant speed. The tension in the string at the top of the circle is T. What is the kinetic energy of the mass at the top of the circle?
\n\n
A.
\nB.
\nC.
\nD.
\nA
\nThe table gives data for Jupiter and three of its moons, including the radius r of each object.
\nA spacecraft is to be sent from to infinity.
\nCalculate, for the surface of , the gravitational field strength gIo due to the mass of . State an appropriate unit for your answer.
\nShow that the is about 80.
\nOutline, using (b)(i), why it is not correct to use the equation to calculate the speed required for the spacecraft to reach infinity from the surface of .
\nAn engineer needs to move a space probe of mass 3600 kg from Ganymede to Callisto. Calculate the energy required to move the probe from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the mass of the moons in your calculation.
\n✓
\nN kg−1 OR m s−2 ✓
\nAND seen ✓
\n✓
\n
For MP1, potentials can be seen individually or as a ratio.
«this is the escape speed for alone but» gravitational potential / field of Jupiter must be taken into account ✓
\n
OWTTE
✓
\n« multiplies by 3600 kg to get » 1.9 × 1011 «J» ✓
\n
Award [2] marks if factor of ½ used, taking into account orbital kinetic energies, leading to a final answer of 9.4 x 1010 «J».
Allow ECF from MP1
\nAward [2] marks for a bald correct answer.
\nThree forces act at a point. In which diagram is the point in equilibrium?
\nB
\nA solid substance has just reached its melting point. Thermal energy is supplied to the substance at a constant rate. Which graph shows the variation of the temperature T of the substance with energy E supplied?
\nC
\nA motorcyclist is cornering on a curved race track.
\nWhich combination of changes of banking angle θ and coefficient of friction μ between the tyres and road allows the motorcyclist to travel around the corner at greater speed?
\nA
\nIon-thrust engines can power spacecraft. In this type of engine, ions are created in a chamber and expelled from the spacecraft. The spacecraft is in outer space when the propulsion system is turned on. The spacecraft starts from rest.
\nThe mass of ions ejected each second is 6.6 × 10–6 kg and the speed of each ion is 5.2 × 104 m s–1. The initial total mass of the spacecraft and its fuel is 740 kg. Assume that the ions travel away from the spacecraft parallel to its direction of motion.
\nAn initial mass of 60 kg of fuel is in the spacecraft for a journey to a planet. Half of the fuel will be required to slow down the spacecraft before arrival at the destination planet.
\nIn practice, the ions leave the spacecraft at a range of angles as shown.
\nOn arrival at the planet, the spacecraft goes into orbit as it comes into the gravitational field of the planet.
\nDetermine the initial acceleration of the spacecraft.
\nEstimate the maximum speed of the spacecraft.
\nOutline why scientists sometimes use estimates in making calculations.
\nOutline why the ions are likely to spread out.
\nExplain what effect, if any, this spreading of the ions has on the acceleration of the spacecraft.
\nOutline what is meant by the gravitational field strength at a point.
\nNewton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting a spherical planet of uniform density.
\nchange in momentum each second = 6.6 × 10−6 × 5.2 × 104 «= 3.4 × 10−1 kg m s−1» ✔
\nacceleration = « =» 4.6 × 10−4 «m s−2» ✔
\nALTERNATIVE 1:
\n(considering the acceleration of the spacecraft)
\ntime for acceleration = = «4.6 × 106» «s» ✔
\nmax speed = «answer to (a) × 4.6 × 106 =» 2.1 × 103 «m s−1» ✔
\n\n
ALTERNATIVE 2:
\n(considering the conservation of momentum)
\n(momentum of 30 kg of fuel ions = change of momentum of spacecraft)
\n30 × 5.2 × 104 = 710 × max speed ✔
\nmax speed = 2.2 × 103 «m s−1» ✔
\nproblem may be too complicated for exact treatment ✔
\nto make equations/calculations simpler ✔
\nwhen precision of the calculations is not important ✔
\nsome quantities in the problem may not be known exactly ✔
\nions have same (sign of) charge ✔
\nions repel each other ✔
\nthe forces between the ions do not affect the force on the spacecraft. ✔
\nthere is no effect on the acceleration of the spacecraft. ✔
\nforce per unit mass ✔
\nacting on a small/test/point mass «placed at the point in the field» ✔
\nsatellite has a much smaller mass/diameter/size than the planet «so approximates to a point mass» ✔
\nOutline what is meant by dark energy.
\nState two candidates for dark matter.
\nenergy filling all space ✔
\nresulting in a repulsive force/force opposing gravity ✔
\naccounts for the accelerating universe ✔
\nmakes up about 70% of «the energy» of universe ✔
\nblack hole ✔
\nbrown dwarf ✔
\nmassive compact halo object /MACHO✔
\nneutrinos ✔
\nweakly interacting massive particle /WIMP ✔
\nThe refractive index of a material is the ratio of the speed of light in a vacuum , to the speed of light in the material or .
The speed of light in a vacuum is 2.99792 × 108 m s-1. The following data are available for the refractive indices of the fibre core for two wavelengths of light:
\nOutline the differences between step-index and graded-index optic fibres.
\nDetermine the difference between the speed of light corresponding to these two wavelengths in the core glass.
\nAn input signal to the fibre consists of wavelengths that range from 1299 nm to 1301 nm. The diagram shows the variation of intensity with time of the input signal.
\nSketch, on the axes, the variation of signal intensity with time after the signal has travelled a long distance along the fibre.
\nExplain the shape of the signal you sketched in (b)(ii).
\nA signal consists of a series of pulses. Outline how the length of the fibre optic cable limits the time between transmission of pulses in a practical system.
\nrefractive index of step index fibre is constant ✔
refractive index of graded index fibre decreases with distance from axis/centre ✔
graded index fibres have less dispersion ✔
step index fibre: path of rays is in a zig-zag manner ✔
graded index fibre: path of rays is in curved path ✔
For MP2 do not accept vague statements such as “index increases/varies with distance from centre”.
\n«ms–1» AND
\n«ms–1»
\nOR
\n✔
\nOR «ms–1»✔
\n\n
pulse wider ✔
pulse area smaller ✔
\nFor MP2 do not accept lower amplitude unless pulse area is also smaller.
\nreference to dispersion
OR
reference to time/speed/path difference ✔
reference to power loss/energy loss/scattering/attenuation ✔
\nlonger cables give wider pulses ✔
which overlap/interfere if T too small/f too high ✔
\nOWTTE
\nThe differences between step index fibres and graded index fibres seem well-known.
\nThe calculation of the difference in the speed of light for two different wavelengths was well answered. Candidates often rounded answers to a small number of significant figures when finding the individual speeds.
\nMost candidates correctly drew a wider pulse with smaller area.
\nCorrect answers mentioning dispersion and attenuation were common but few candidates were able to relate those phenomena to the shape of the pulse drawn.
\nMost candidates did not mention the fact that if the time between pulses was too small then the pulses would overlap for longer fibres.
\nSatellite X orbits a planet with orbital radius R. Satellite Y orbits the same planet with orbital radius 2R. Satellites X and Y have the same mass.
\nWhat is the ratio ?
\nA.
\nB.
\nC. 2
\nD. 4
\nD
\nThe graph shows the variation with distance from the Earth of the recessional velocities of distant galaxies.
\nOutline how Hubble measured the recessional velocities of galaxies.
\nUsing the graph, determine in s, the age of the universe.
\nmeasured redshift «z» of star ✔
\nuse of Doppler formula OR z∼v/c OR v = to find v ✔
\nuse of gradient or any point on the line to obtain any expression for either or ✔
\ncorrect conversion of d to m and v to m/s ✔
\n= 4.6 × 1017 «s» ✔
\nA sphere is dropped into a container of oil.
The following data are available.
Density of oil
Viscosity of oil
Volume of sphere
Mass of sphere
The sphere is now suspended from a spring so that the sphere is below the surface of the oil.
\nState two properties of an ideal fluid.
\nDetermine the terminal velocity of the sphere.
\nDetermine the force exerted by the spring on the sphere when the sphere is at rest.
\nThe sphere oscillates vertically within the oil at the natural frequency of the sphere-spring system. The energy is reduced in each cycle by . Calculate the factor for this system.
\nOutline the effect on of changing the oil to one with greater viscosity.
\nincompressible ✓
\nnon-viscous ✓
\nlaminar/streamlined flow ✓
\nradius of sphere ✓
\n\n
weight of sphere
\nOR
\n✓
\n\n
✓
\n\n
Accept use of leading to
\nAllow implicit calculation of radius for MP1
\nDo not allow ECF for MP3 if buoyant force omitted.
\nOR
\n✓
\n\n
✓
\n\n
Accept use of leading to
\n✓
\ndrag force increases OR damping increases OR more energy lost per cycle ✓
\nwill decrease ✓
\nThe properties of fluids proved to be a very well-studied topic.
\nOnly those candidates who forgot to include the buoyant force missed marks here.
\nContinuing from b, most candidates scored full marks.
\nThe calculation needed to obtain the Q-factor proved to be known by many.
\nVery well answered.
\nTwo protons are moving to the right with the same speed v with respect to an observer at rest in the laboratory frame.
\nOutline why there is an attractive magnetic force on each proton in the laboratory frame.
\nExplain why there is no magnetic force on each proton in its own rest frame.
\nExplain why there must be a resultant repulsive force on the protons in all reference frames.
\nmoving charges give rise to magnetic fields
OR
magnetic attraction between parallel currents ✔
\nprotons at rest produce no magnetic field
OR
mention of F = Bev where B and/or v =0 ✔
\nthere is a repulsive electric/electrostatic force «in both frames» ✔
\nthe attractive magnetic force «in the lab frame» is smaller than the repulsive electric force ✔
\nin all frames the net force is repulsive as all must agree that protons move apart
\nOR
\nmention of the first postulate of relativity ✔
\nCandidates usually realised that the magnetic field was due to the motion of the protons and that in the proton rest frame there could be no magnetic field. The answers were too often poorly worded and the candidates appeared to reword the question without providing a clear explanation.
\nCandidates usually realised that the magnetic field was due to the motion of the protons and that in the proton rest frame there could be no magnetic field. The answers were too often poorly worded and the candidates appeared to reword the question without providing a clear explanation.
\nA few candidates mentioned that there was an electrostatic repulsive force between the protons in both frames. However very few realised that there had to be an overall repulsive force in both frames because of the relativity postulate.
\nTwo sets of data, shown below with circles and squares, are obtained in two experiments. The size of the error bars is the same for all points.
\nWhat is correct about the absolute uncertainty and the fractional uncertainty of the y intercept of the two lines of best fit?
\nD
\nShow that the temperature of the universe is inversely proportional to the cosmic scale factor.
\nThe present temperature of the cosmic microwave background (CMB) radiation is 3 K. Estimate the size of the universe relative to the present size of the universe when the temperature of the CMB was 300 K.
\n«wavelength of light/CBR» λ ∝ R ✔
\nreference to Wien’s law showing that λ ∝ ✔
\ncombine to get result ✔
\n\n
OWTTE
\n\n
✔
\n\n
An electron with total energy 1.50 MeV collides with a positron at rest. As a result two photons are produced. One photon moves in the same direction as the electron and the other in the opposite direction.
\nThe momenta of the photons produced have magnitudes p1 and p2. A student writes the following correct equations.
\np1 – p2 = 1.41 MeV c–1
\np1 + p2 = 2.01 MeV c–1
\nShow that the momentum of the electron is 1.41 MeV c–1.
\nExplain the origin of each equation.
\nCalculate, in MeV c–1, p1 and p2.
\n«= 1.410 MeV» ✔
\nfirst equation is due to momentum conservation ✔
\nsecond equation is due to total energy conservation ✔
\nadding 2p1 = 3.42 MeV c–1 ⇒ p1 = 1.17 MeV c–1 ✔
\np2 = 0.30 MeV c–1 ✔
\nA large stone is dropped from a tall building. What is correct about the speed of the stone after 1 s?
\nA. It is decreasing at increasing rate.
\nB. It is decreasing at decreasing rate.
\nC. It is increasing at increasing rate.
\nD. It is increasing at decreasing rate.
\nD
\nWhich property of a nuclide does not change as a result of beta decay?
\nA. Nucleon number
\nB. Neutron number
\nC. Proton number
\nD. Charge
\nA
\nResponse A was the most common (correct) response from a minority of candidates (38 %). Incorrect responses were evenly divided among the remaining options.
\nThe graph shows how the position of an object varies with time in the interval from 0 to 3 s.
\nAt which point does the instantaneous speed of the object equal its average speed over the interval from 0 to 3 s?
\nC
\nAn ultrasound A-scan is performed on a patient.
\nThe graph shows a received signal incident upon a transducer to produce an A-scan. The density of the soft tissue being examined is approximately 1090 kg m-3.
\nState one advantage and one disadvantage of using ultrasound imaging in medicine compared to using x-ray imaging.
\nAdvantage:
\n\n
Disadvantage:
\nSuggest why ultrasound gel is necessary during an ultrasound examination.
\nUltrasound of intensity 50 mW m-2 is incident on a muscle. The reflected intensity is 10 mW m-2. Calculate the relative intensity level between the reflected and transmitted signals.
\nThe acoustic impedance of soft tissue is 1.65 × 106 kg m-2 s-1. Show that the speed of sound in the soft tissue is approximately 1500 m s–1.
\nEstimate, using data from the graph, the depth of the organ represented by the dashed line.
\nIn the ultrasound scan the frequency is chosen so that the distance between the transducer and the organ is at least 200 ultrasound wavelengths. Estimate, based on your response to (b)(ii), the minimum ultrasound frequency that is used.
\nA physician has a range of frequencies available for ultrasound. Comment on the use of higher frequency sound waves in an ultrasound imaging study.
\nAdvantage of ultrasound compared to X-rays:
no exposure to radiation
OR
relatively harmless
OR
can be performed in a doctor’s office
OR
can be used to measure blood flow rate
OR
Video image possible <<eg heart, foetus>> ✔
Accept any reasonable advantage.
\nDisadvantage:
limited resolution
OR
difficulty imaging lungs or gastrointestinal system
OR
difficulty imaging any body part with a gas in it ✔
Accept any reasonable disadvantage.
Do not allow answers that contradict each other.
\n\n
gel has similar Z to skin
OR
gel prevents acoustic mismatch ✔
without gel much ultrasound is reflected at skin
OR
gel increases ultrasound transmission ✔
\nOWTTE
\n«ms–1» ✔
«≈1500ms–1»
\nAnswer 1500 is given, check working or look for at least 3 significant figures.
\n4.5 × 10−2 «m»✔
\n«m» ✔
\n«Hz» ✔
\n\n
«compared to lower frequencies, higher frequencies»
have better resolution ✔
have greater attenuation ✔
used for superficial structures/organs ✔
have greater heating effect ✔
\nOWTTE
Award [0] for contradictory comments or for any incorrect comment
\nThe question was well answered by almost all candidates.
\nMost candidates mentioned that the gel improves the transmission of ultrasound. On quite a few occasions candidates seemed to confuse acoustic impedance and refractive index.
\nThe question was generally well answered with a few candidates simply taking the ratio of intensities instead of 10x log ratio (Intensity level)
\nAlmost all candidates managed to obtain the result given.
\nMany candidates did not seem to know how to start answering the question. The factor of two was often omitted when finding the depth of the organ in the A scan.
\nFew candidates managed to understand how to approach the problem and to obtain the correct answer. ECF from bii was frequently need.
\nMost candidates mentioned that the resolution would be better at higher frequencies.
\nTwo players are playing table tennis. Player A hits the ball at a height of 0.24 m above the edge of the table, measured from the top of the table to the bottom of the ball. The initial speed of the ball is 12.0 m s−1 horizontally. Assume that air resistance is negligible.
\nThe ball bounces and then reaches a peak height of 0.18 m above the table with a horizontal speed of 10.5 m s−1. The mass of the ball is 2.7 g.
\nShow that the time taken for the ball to reach the surface of the table is about 0.2 s.
\nSketch, on the axes, a graph showing the variation with time of the vertical component of velocity vv of the ball until it reaches the table surface. Take g to be +10 m s−2.
\nThe net is stretched across the middle of the table. The table has a length of 2.74 m and the net has a height of 15.0 cm.
\nShow that the ball will go over the net.
\nDetermine the kinetic energy of the ball immediately after the bounce.
\nPlayer B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic.
\nCalculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.
\nt = «=» 0.22 «s»
OR
t = ✓
\nAnswer to 2 or more significant figures or formula with variables replaced by correct values.
\nincreasing straight line from zero up to 0.2 s in x-axis ✓
\nwith gradient = 10 ✓
\nALTERNATIVE 1
\n«0.114 s» ✓
\nm ✓
\nso (0.24 − 0.065) = 0.175 > 0.15 OR 0.065 < (0.24 − 0.15) «so it goes over the net» ✓
\n\n
ALTERNATIVE 2
\n«0.24 − 0.15 = 0.09 = so» t = 0.134 s ✓
\n0.134 × 12 = 1.6 m ✓
\n1.6 > 1.37 «so ball passed the net already» ✓
\n\n
Allow use of g = 9.8.
\nALTERNATIVE 1
\nKE = mv2 + mgh = 0.0027 ×10.52 + 0.0027 × 9.8 × 0.18 ✓
\n0.15 «J» ✓
\n\n
ALTERNATIVE 2
\nUse of vx = 10.5 AND vy = 1.88 to get v = «» = 10.67 «m s−1» ✓
\nKE = × 0.0027 × 10.672 = 0.15 «J» ✓
\n«m s−1» ✓
\n\nOR
\n5.67 «N» ✓
\nany answer to 2 significant figures «N» ✓
\nA student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of 64 m s–1. The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms.
\nThe student strikes the tennis ball at point P. The tennis ball is initially directed at an angle of 7.00° to the horizontal.
\nThe following data are available.
Height of P = 2.80 m
Distance of student from net = 11.9 m
Height of net = 0.910 m
Initial speed of tennis ball = 64 m s-1
Calculate the average force exerted by the racquet on the ball.
\nCalculate the average power delivered to the ball during the impact.
\nCalculate the time it takes the tennis ball to reach the net.
\nShow that the tennis ball passes over the net.
\nDetermine the speed of the tennis ball as it strikes the ground.
\nA student models the bounce of the tennis ball to predict the angle θ at which the ball leaves a surface of clay and a surface of grass.
\nThe model assumes
\n• during contact with the surface the ball slides.
• the sliding time is the same for both surfaces.
• the sliding frictional force is greater for clay than grass.
• the normal reaction force is the same for both surfaces.
Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.
\n✔
\n= 148 «N»≈150«N» ✔
\n\n
\n
ALTERNATIVE 1
\n✔
\n«» ✔
\n\n
ALTERNATIVE 2
\n✔
\n«» ✔
\n\n
horizontal component of velocity is «» ✔
«»«» ✔
\n\n
ALTERNATIVE 1
\nuy=64sin7/7.80«ms–1» ✔
\ndecrease in height = 7.80 × 0.187 + × 9.81 × 0.1872 / 1.63«m» ✔
\nfinal height = «2.80 – 1.63» = 1.1/1.2«m» ✔
«higher than net so goes over»
\n
ALTERNATIVE 2
vertical distance to fall to net «=2.80 – 0.91» = 1.89«m» ✔
\ntime to fall this distance found using «»
\n= 0.21«s» ✔
\n0.21«s» > 0.187«s» ✔
«reaches the net before it has fallen far enough so goes over»
\nALTERNATIVE 1
\nInitial KE + PE = final KE /
\n✔
\n«» ✔
ALTERNATIVE 2
«» = 10.8«» ✔
\n«»
\n«» ✔
\n\n
so horizontal velocity component at lift off for clay is smaller ✔
normal force is the same so vertical component of velocity is the same ✔
so bounce angle on clay is greater ✔
At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.
\nThis was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.
\nThis question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.
\nThere were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.
\nA common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.
\nThis proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.
\nAs the command term in this question is ‘predict’ a bald answer of clay was acceptable for one mark. This was a testing question that candidates found demanding but there were some very well-reasoned answers. The most common incorrect answer involved suggesting that the greater frictional force on the clay court left the ball with less kinetic energy and so a smaller angle. At SL many gained the answer that the angle on clay would be greater with the argument that frictional force is greater and so the distance the ball slides is less.
\nA probe launched from a spacecraft moves towards the event horizon of a black hole.
\nState what is meant by the event horizon of a black hole.
\nThe mass of the black hole is 4.0 × 1036 kg. Calculate the Schwarzschild radius of the black hole.
\nThe probe is stationary above the event horizon of the black hole in (a). The probe sends a radio pulse every 1.0 seconds (as measured by clocks on the probe). The spacecraft receives the pulses every 2.0 seconds (as measured by clocks on the spacecraft). Determine the distance of the probe from the centre of the black hole.
\nthe distance from the black hole at which the escape speed is the speed of light ✔
\nRs = «» 5.9 × 109 «m» ✔
\n✔
\nrearranged to give r
\nOR
\nr = 1.33 × 5.9 × 109 «m» ✔
\nr = 7.9 × 109 «m» ✔
\n\n
The rest mass of the helium isotope is m.
\nWhich expression gives the binding energy per nucleon for ?
\nA.
\nB.
\nC.
\nD.
\n\n
B
\nA car takes 20 minutes to climb a hill at constant speed. The mass of the car is 1200 kg and the car gains gravitational potential energy at a rate of 6.0 kW. Take the acceleration of gravity to be 10 m s−2. What is the height of the hill?
\nA. 0.6 m
\nB. 10 m
\nC. 600 m
\nD. 6000 m
C
\nA lighting system consists of two long metal rods with a potential difference maintained between them. Identical lamps can be connected between the rods as required.
\nThe following data are available for the lamps when at their working temperature.
\n\n
Lamp specifications 24 V, 5.0 W
\nPower supply emf 24 V
\nPower supply maximum current 8.0 A
\nLength of each rod 12.5 m
\nResistivity of rod metal 7.2 × 10–7 Ω m
\nEach rod is to have a resistance no greater than 0.10 Ω. Calculate, in m, the minimum radius of each rod. Give your answer to an appropriate number of significant figures.
\nCalculate the maximum number of lamps that can be connected between the rods. Neglect the resistance of the rods.
\nOne advantage of this system is that if one lamp fails then the other lamps in the circuit remain lit. Outline one other electrical advantage of this system compared to one in which the lamps are connected in series.
\nALTERNATIVE 1:
\nOR ✔
\nr = 5.352 × 10−3 ✔
\n5.4 × 10−3 «m» ✔
\n\n
ALTERNATIVE 2:
\n✔
\nr = 5.352 × 10−3 ✔
\n5.4 × 10−3 «m» ✔
\ncurrent in lamp = «= 0.21» «A»
\nOR
\nn = 24 × ✔
\n\n
so «38.4 and therefore» 38 lamps ✔
\n\n
when adding more lamps in parallel the brightness stays the same ✔
\nwhen adding more lamps in parallel the pd across each remains the same/at the operating value/24 V ✔
\nwhen adding more lamps in parallel the current through each remains the same ✔
\nlamps can be controlled independently ✔
\nthe pd across each bulb is larger in parallel ✔
\nthe current in each bulb is greater in parallel ✔
\nlamps will be brighter in parallel than in series ✔
\nIn parallel the pd across the lamps will be the operating value/24 V ✔
\n\n
Accept converse arguments for adding lamps in series:
\nwhen adding more lamps in series the brightness decreases
\nwhen adding more lamps in series the pd decreases
\nwhen adding more lamps in series the current decreases
\nlamps can’t be controlled independently
\nthe pd across each bulb is smaller in series
\nthe current in each bulb is smaller in series
\n\n
in series the pd across the lamps will less than the operating value/24 V
\nDo not accept statements that only compare the overall resistance of the combination of bulbs.
\nSilicon-30 can be formed from phosphorus-30 by a process of beta-plus decay.
\nWrite down the nuclear equation that represents this reaction.
\nSketch the Feynman diagram that represents this reaction. The diagram has been started for you.
\nEnergy is transferred to a hadron in an attempt to separate its quarks. Describe the implications of quark confinement for this situation.
\nThe Standard Model was accepted by many scientists before the observation of the Higgs boson was made.
\nOutline why it is important to continue research into a topic once a scientific model has been accepted by the scientific community.
\n✔
\n✔
\n\n
Subscript on neutrino not necessary to award MP2.
\nAllow the use of β for e.
\nDo not allow an anti-neutrino for MP2.
\ncorrect change of either u to d ✔
\nW+ shown ✔
\ncorrect arrow directions for positron and electron neutrino ✔
\nAllow ECF from MP2 in ai for MP3.
\nquarks cannot be directly observed as free particles/must remain bound to other quarks/quarks cannot be isolated ✔
\nbecause energy given to nucleon creates other particles rather than freeing quarks/OWTTE ✔
\nmodels need testing/new information may change models/new technology may bring new information/Models can be revised/OWTTE ✔
\nLook for responses that suggest changes/improvements to models.
\nDon’t accept answers specifically about the Standard Model.
\nDon’t accept answers about simply proving the model correct.
\nFew candidates were awarded full marks for a variety of reasons for the Feynman diagram they drew, and many left this question blank. It should be noted that on the exam the time axis can either be vertical or horizontal, so candidates should be familiar with both methods of drawing Feynman diagrams. Candidates should be able to draw Feynman diagrams from scratch either way. The examiners were looking for the basics of drawing a diagram (proper change in quark structure, proper exchange particle, and proper arrow directions for the positron and neutrino).
\nFew candidates recognized that quarks cannot exist in isolation, and fewer still could discuss the effect of adding energy to attempt to separate quarks. Some recognized that the added energy would ultimately be converted into mass, but few clearly specified that this would form new particles (such as mesons) rather than just new quarks.
\nThis was a “nature of science” question. The examiners were looking for the idea that models can be improved on and revised by new data rather than just proven right or wrong.
\nWhich of the following atomic energy level transitions corresponds to photons of the shortest wavelength?
\nC
\nThe most common (incorrect) response was A, where students apparently assumed energy difference was proportional to the wavelength of the emitted photon.
\nA mass is released from the top of a smooth ramp of height . After leaving the ramp, the mass slides on a rough horizontal surface.
\nThe mass comes to rest in a distance d. What is the coefficient of dynamic friction between the mass and the horizontal surface?
\n\n\n\n\nD
\nA train of proper length 85 m moves with speed 0.60c relative to a stationary observer on a platform.
\nDefine proper length.
\nIn the reference frame of the train a ball travels with speed 0.50c from the back to the front of the train, as the train passes the platform. Calculate the time taken for the ball to reach the front of the train in the reference frame of the train.
\nIn the reference frame of the train a ball travels with speed 0.50c from the back to the front of the train, as the train passes the platform. Calculate the time taken for the ball to reach the front of the train in the reference frame of the platform.
\nthe length measured «in a reference frame» where the object is at rest ✔
ALTERNATIVE 1:
ALTERNATIVE 2:
\nv of ball is 0.846c for platform ✔
length of train is 68m for platform ✔
\nALTERNATIVE 3:
\nProper length is quite well understood. A common mistake is to mention that it is the length measured by a reference frame at rest.
\nBecause there were three frames of reference in this question many candidates struggled to find the simple value for the time of the ball’s travel down the train in the train’s frame of reference.
\nAlmost no candidates could use a Lorentz transformation to find the time of the ball’s travel in the frame of reference of the platform. Most just applied some form of t=γt’. Elapsed time and instantaneous time in different frames were easily confused. Candidates rarely mention which reference frame is used when making calculations, however this is crucial in relativity.
\nMasses X and Y rest on a smooth horizontal surface and are connected by a massless spring. The mass of X is 3.0 kg and the mass of Y is 6.0 kg. The masses are pushed toward each other until the elastic potential energy stored in the spring is 1.0 J.
\nThe masses are released. What is the maximum speed reached by mass Y?
\nA. 0.11 m s−1
\nB. 0.33 m s−1
\nC. 0.45 m s−1
\nD. 0.66 m s−1
\nB
\nThe photograph shows an X-ray image of a hand.
\n© International Baccalaureate Organization 2020.
\n
Explain how attenuation causes the contrast between soft tissue and bone in the image.
\nX-ray images of other parts of the body require the contrast to be enhanced. State one technique used in X-ray medical imaging to enhance contrast.
\nbone «denser so» absorb rays «and appear white in the negative» ✓
\nlarger attenuation for bone ✓
\nmuscles have less attenuation, so rays pass through «and appear darker» ✓
\n\n
Accept the reversed argument
\ncollimation ✓
\nfluorescent screens «each side of photographic plate» ✓
\nbarium/magnesium meal ✓
\nCandidates successfully answered in terms of absorption and managed to score two marks but not all of them.
\nAlmost all candidates were familiar with different techniques to enhance contrast. A barium meal was the most popular one.
\nA force acts on an object of mass 40 kg. The graph shows how the acceleration a of the object varies with its displacement d.
\nWhat is the work done by the force on the object?
\nA. 50 J
\nB. 2000 J
\nC. 2400 J
\nD. 3200 J
\nB
\nA beaker containing 1 kg of water at room temperature is heated on a 400 W hot plate. The specific heat capacity of water is 4200 J kg–1 K–1.
\nThe temperature of the water increases until it reaches a constant value. It is then removed from the hot plate.
\nWhat will be the initial rate of change of temperature?
\nA. 10 K s–1
\nB. 1 K s–1
\nC. 0.1 K s–1
\nD. 0.01 K s–1
\nC
\nA horizontal pipe is inserted into the cylindrical tube so that its centre is at a depth of 5.0 m from the surface of the water. The diameter D of the pipe is half that of the tube.
\nWhen the pipe is opened, water exits the pipe with speed u and the surface of the water in the tube moves downwards with speed v.
\nAn ice cube floats in water that is contained in a tube.
\nThe ice cube melts.
\nSuggest what happens to the level of the water in the tube.
\nOutline why u = 4v.
\nThe density of water is 1000 kg m–3. Calculate u.
\nice displaces its own weight of water / OWTTE
\nOR
\nmelted ice volume equals original volume displaced / OWTTE ✔
\n\n
no change will take place ✔
\ncontinuity equation says v × A1 = u × A2 ✔
\n«and» A1 = 4A2 ✔
\n«giving result»
\nBernoulli:
\n«» gives ✔
\nu = 10.2 «m s–1» ✔
\n\n
Accept solving directly via conservation of energy.
\nWhich aspect of thermal physics is best explained by the molecular kinetic model?
\nA. The equation of state of ideal gases
\nB. The difference between Celsius and Kelvin temperature
\nC. The value of the Avogadro constant
\nD. The existence of gaseous isotopes
\nA
\nWhen 40 kJ of energy is transferred to a quantity of a liquid substance, its temperature increases by 20 K. When 600 kJ of energy is transferred to the same quantity of the liquid at its boiling temperature, it vaporizes completely at constant temperature. What is
\n\nfor this substance?
\nA. 15 K−1
\nB. 15 K
\nC. 300 K−1
\nD. 300 K
\nD
\nA chicken’s egg of mass 58 g is dropped onto grass from a height of 1.1 m. The egg comes to rest in a time of 55 ms. Assume that air resistance is negligible and that the egg does not bounce or break.
\nDetermine the magnitude of the average decelerating force that the ground exerts on the egg.
\nExplain why the egg is likely to break when dropped onto concrete from the same height.
\nALTERNATIVE 1:
\ninitial momentum = mv = «= 0.27 kg m s−1»
\nOR
\nmv = «= 0.27 kg m s−1» ✔
\nforce = « =» ✔
\n4.9 «N» ✔
\nF − mg = 4.9 so F= 5.5 «N» ✔
\n\n
ALTERNATIVE 2:
\n«Ek = mv2 = 0.63 J» v = 4.7 m s−1 ✔
\nacceleration = « =» = «85 m s−2» ✔
\n4.9 «N» ✔
\nF − mg = 4.9 so F= 5.5 «N» ✔
\nALTERNATIVE 1:
\nconcrete reduces the stopping time/distance ✔
\nimpulse/change in momentum same so force greater
\nOR
\nwork done same so force greater ✔
\n\n
ALTERNATIVE 2:
\nconcrete reduces the stopping time ✔
\ndeceleration is greater so force is greater ✔
\n\n
Allow reverse argument for grass.
\nTwo ideal gases X and Y are at the same temperature. The mass of a particle of gas X is larger than the mass of a particle of gas Y. Which is correct about the average kinetic energy and the average speed of the particles in gases X and Y?
\nB
\nMost power stations rely on a turbine and a generator to produce electrical energy. Which power station works on a different principle?
\nA. Nuclear
\nB. Solar
\nC. Fossil fuel
\nD. Wind
\nB
\nA planet is in a circular orbit around a star. The speed of the planet is constant. The following data are given:
\nMass of planet kg
Mass of star kg
Distance from the star to the planet R m.
A spacecraft is to be launched from the surface of the planet to escape from the star system. The radius of the planet is 9.1 × 103 km.
\nExplain why a centripetal force is needed for the planet to be in a circular orbit.
\nCalculate the value of the centripetal force.
\nShow that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 109 J kg−1.
\nEstimate the escape speed of the spacecraft from the planet–star system.
\n«circular motion» involves a changing velocity ✓
\n«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓
\nthere must be a force/acceleration towards centre/star ✓
\nwithout a centripetal force the planet will move in a straight line ✓
\n«N» ✓
\nVplanet = «−»«−» 5.9 × 107 «J kg−1» ✓
\nVstar = «−»«−» 4.9 × 109 «J kg−1» ✓
\nVplanet + Vstar = «−» 4.9 «09» × 109 «J kg−1» ✓
\n
Must see substitutions and not just equations.
use of vesc = ✓
\nv = 9.91 × 104 «m s−1» ✓
\n\n
\n
An object performs simple harmonic motion (shm). The graph shows how the velocity v of the object varies with time t.
\nThe displacement of the object is x and its acceleration is a. What is the variation of x with t and the variation of a with t?
\nA
\nExplain the cause of the radio-frequency emissions from a patient’s body during nuclear magnetic resonance (NMR) imaging.
\nOutline how a gradient field allows NMR to be used in medical resonance imaging.
\nIdentify one advantage of NMR over ultrasound in medical situations.
\nuse of strong magnetic field ✓
\nprotons are aligned ✓
\nradio wave at «nuclear» resonant frequency flips «some of» them into higher energy state ✓
\nproton de-excites emitting energy at known «radio» wavelength/frequency/Larmor frequency ✓
\n«which can be located and detected»
\nmention of gradient field «added to the NMR uniform magnetic field» ✓
\nreference to «the total field that determines» the output «Larmor» frequency from the de-excitation ✓
\ndifferent positions «in the body» give rise to different frequencies ✓
\n«and this can be mapped»
\nNMR higher resolution ✓
\nNMR less attenuation ✓
\n\n
Accept the reverse argument
\nCandidates proved to be familiar with the use of strong magnetic fields to produce proton alignment and its consequence on excitation de-excitation of protons by emission of radio frequencies (Larmor frequency).
\nMost candidates continued successfully here by referring to a gradient field eventually leading to a mapping of the position of the protons from (a).
\nHigher resolution was the most popular answer in a high scoring question overall.
\nA pipe is open at both ends. A first-harmonic standing wave is set up in the pipe. The diagram shows the variation of displacement of air molecules in the pipe with distance along the pipe at time t = 0. The frequency of the first harmonic is f.
\nA transmitter of electromagnetic waves is next to a long straight vertical wall that acts as a plane mirror to the waves. An observer on a boat detects the waves both directly and as an image from the other side of the wall. The diagram shows one ray from the transmitter reflected at the wall and the position of the image.
\nAn air molecule is situated at point X in the pipe at t = 0. Describe the motion of this air molecule during one complete cycle of the standing wave beginning from t = 0.
\nThe speed of sound c for longitudinal waves in air is given by
\n\n
where ρ is the density of the air and K is a constant.
\nA student measures f to be 120 Hz when the length of the pipe is 1.4 m. The density of the air in the pipe is 1.3 kg m–3. Determine, in kg m–1 s–2, the value of K for air.
\nDemonstrate, using a second ray, that the image appears to come from the position indicated.
\nOutline why the observer detects a series of increases and decreases in the intensity of the received signal as the boat moves along the line XY.
\n«air molecule» moves to the right and then back to the left ✔
\nreturns to X/original position ✔
\nwavelength = 2 × 1.4 = «2.8 m» ✔
\nc = «f λ =» 120 × 2.8 «= 340 m s−1» ✔
\nK = «ρc2 = 1.3 × 3402 =» 1.5 × 105 ✔
\nconstruction showing formation of image ✔
\nAnother straight line/ray from image through the wall with line/ray from intersection at wall back to transmitter. Reflected ray must intersect boat.
\ninterference pattern is observed
\nOR
\ninterference/superposition mentioned ✔
\n
maximum when two waves occur in phase/path difference is nλ
OR
\nminimum when two waves occur 180° out of phase/path difference is (n + ½)λ ✔
\nA container of volume 3.2 × 10-6 m3 is filled with helium gas at a pressure of 5.1 × 105 Pa and temperature 320 K. Assume that this sample of helium gas behaves as an ideal gas.
\n\n
\n
A helium atom has a volume of 4.9 × 10-31 m3.
\nThe mass of a helium atom is 6.6 × 10-27 kg. Estimate the average speed of the helium atoms in the container.
\nShow that the number of helium atoms in the container is 4 × 1020.
\nCalculate the ratio .
\nDiscuss, by reference to the kinetic model of an ideal gas and the answer to (c)(i), whether the assumption that helium behaves as an ideal gas is justified.
\n✔
\nv = 1.4 × 103«ms–1» ✔
\n\n
\n
OR
\n✔
\n\n
✔
\n«» ✔
\n«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas
\nOR
«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas
OR
«For an ideal gas» particles are assumed to be point objects ✔
\ncalculation/ratio/result in (c)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔
\nAt HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.
\nAgain at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by NA to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by NA.
\nThis was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.
\nAt HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.
\nThe following data are available for the Cepheid variable δ-Cephei.
Peak luminosity = 7.70 × 1029 W
Distance from Earth = 273 pc
Peak wavelength of light = 4.29 × 10–7 m
\nOutline the processes that produce the change of luminosity with time of Cepheid variables.
\nExplain how Cepheid variables are used to determine distances.
\nDetermine the peak apparent brightness of δ-Cephei as observed from Earth.
\nCalculate the peak surface temperature of δ-Cephei.
\nAstronomers claim to know the properties of distant stars. Outline how astronomers can be certain that their measurement methods yield correct information.
\nCepheid variables expand and contract
OR
Radius increases and decreases
OR
Surface area increases and decreases ✔
Surface temperature decreases then increases✔
Surface becomes transparent then opaque ✔
\nOWTTE
Do not reward ‘change in luminosity/brightness’ as this is given in the question.
Accept changes in reverse order
\nthe «peak» luminosity/actual brightness depends on the period
OR
More luminous Cepheid variables have greater period✔
measurements of apparent brightness allow distance determination
OR
Mention of ✔
OWTTE
\n«m» ✔
\n«Wm–2» ✔
\n\n
«»
\n=6800«K» ✔
\nData subject to peer review/checks by others ✔
Compare light from stars with Earth based light sources ✔
measurements are corroborated by different instruments/methods from different teams ✔
\nOWTTE
The expansion and contraction of Cepheid stars was commonly mentioned. Changes in surface temperature and opacity were less commonly mentioned. A common misconception seems to be that the variation of luminosity is due to a change of the rate of fusion. A few candidates left this question unanswered.
\nMany candidates knew that if the luminosity of the Cepheid is known then the absolute brightness can be used to determine distance. But far fewer candidates could link luminosity with the period of the Cepheid star. Many seemed to think that the luminosity of all Cepheids is the same.
\nCalculating the brightness of a star from its luminosity was an easy question for most candidates. But quite a few did not convert parsecs into metres especially at SL.
\nThis simple calculation using Wien’s law was very well answered.
\nMany candidates correctly stated that astronomers can use peer review or different methods in checking that the information obtained from stars is correct.
\nA sound wave has a frequency of 1.0 kHz and a wavelength of 0.33 m. What is the distance travelled by the wave in 2.0 ms and the nature of the wave?
\nC
\nA girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is 16 V and it can deliver a charge of 43 kC when discharging completely from a full charge.
\nThe maximum speed of the girl on a horizontal road is 7.0 m s–1 with energy from the battery alone. The maximum distance that the girl can travel under these conditions is 20 km.
\nThe bicycle and the girl have a total mass of 66 kg. The girl rides up a slope that is at an angle of 3.0° to the horizontal.
\nThe bicycle has a meter that displays the current and the terminal potential difference (pd) for the battery when the motor is running. The diagram shows the meter readings at one instant. The emf of the cell is 16 V.
\nThe battery is made from an arrangement of 10 identical cells as shown.
\nShow that the time taken for the battery to discharge is about 3 × 103 s.
\nDeduce that the average power output of the battery is about 240 W.
\nFriction and air resistance act on the bicycle and the girl when they move. Assume that all the energy is transferred from the battery to the electric motor. Determine the total average resistive force that acts on the bicycle and the girl.
\nCalculate the component of weight for the bicycle and girl acting down the slope.
\nThe battery continues to give an output power of 240 W. Assume that the resistive forces are the same as in (a)(iii).
\nCalculate the maximum speed of the bicycle and the girl up the slope.
\nOn another journey up the slope, the girl carries an additional mass. Explain whether carrying this mass will change the maximum distance that the bicycle can travel along the slope.
\nDetermine the internal resistance of the battery.
\nCalculate the emf of one cell.
\nCalculate the internal resistance of one cell.
\ntime taken «= 2860 s» = 2900«s» ✔
\nMust see at least two s.f.
\nuse of E = qV OR energy = 4.3 × 103 × 16 «= 6.88 × 105 J» ✔
\npower = 241 «W» ✔
\nAccept 229 W − 241 W depending on the exact value of t used from ai.
\nMust see at least three s.f.
\nuse of power = force × speed OR force × distance = power × time ✔
\n«34N» ✔
\nAward [2] for a bald correct answer.
\nAccept 34 N – 36 N.
\n66 g sin(3°) = 34 «N» ✔
\ntotal force 34 + 34 = 68 «N» ✔
3.5 «ms-1»✔
If you suspect that the incorrect reference in this question caused confusion for a particular candidate, please refer the response to the PE.
\nLook for ECF from aiii and bi.
\nAccept 3.4 − 3.5 «ms-1».
\nAward [0] for solutions involving use of KE.
\nAward [0] for v = 7 ms-1.
\nAward [2] for a bald correct answer.
\n«maximum» distance will decrease OWTTE ✔
\nbecause opposing/resistive force has increased
OR
because more energy is transferred to GPE
OR
because velocity has decreased
OR
increased mass means more work required «to move up the hill» ✔
V dropped across battery OR Rcircuit = 1.85 Ω ✔
\nso internal resistance = = 0.62«Ω» ✔
\nFor MP1 allow use of internal resistance equations that leads to 16V − 12V (=4V).
\nAward [2] for a bald correct answer.
\n= 3.2 «V» ✔
\nALTERNATIVE 1:
\n2.5r = 0.62 ✔
\nr = 0.25 «Ω» ✔
\nALTERNATIVE 2:
\n= 0.124 «Ω» ✔
\nr = 2(0.124)= 0.248 «Ω» ✔
\nAllow ECF from (d) and/or e(i).
\nThis question was generally well answered. Candidates should be reminded on questions where a given value is being calculated that they should include an unrounded answer. This whole question set was a blend of electricity and mechanics concepts, and it was clear that some candidates struggled with applying the correct concepts in the various sub-questions.
\nMany candidates struggled with this question. They either simply calculated the weight, used the cosine rather than the sine function, or failed to multiply by the acceleration due to gravity. Candidates need to be able to apply free-body diagram skills in a variety of “real world” situations.
\nThis question was well answered in general, with the vast majority of candidates specifying that the maximum distance would decrease. This is an “explain” command term, so the examiners were looking for a detailed reason why the distance would decrease for the second marking point. Unfortunately, some candidates simply wrote that because the mass increased so did the weight without making it clear why this would change the maximum distance.
\nMonochromatic light of wavelength λ is incident on a double slit. The resulting interference pattern is observed on a screen a distance y from the slits. The distance between consecutive fringes in the pattern is 55 mm when the slit separation is a.
\nλ, y and a are all doubled. What is the new distance between consecutive fringes?
\nA. 55 mm
\nB. 110 mm
\nC. 220 mm
\nD. 440 mm
\nB
\nA mass is attached to a vertical spring. The other end of the spring is attached to the driver of an oscillator.
\nThe mass is performing very lightly damped harmonic oscillations. The frequency of the driver is higher than the natural frequency of the system. At one instant the driver is moving downwards.
\nState and explain the direction of motion of the mass at this instant.
\nThe oscillator is switched off. The system has a Q factor of 22. The initial amplitude is 10 cm. Determine the amplitude after one complete period of oscillation.
\nbecause the mass and the driver are out of phase «by » ✔
\nso upwards ✔
\n\n
Justification needed for MP2
\nALTERNATIVE 1:
\n«» ⇒
\n« =» A1 = 8.5 «cm»
\n\n
ALTERNATIVE 2:
\ndriver amplitude is constant ✔
\nso mass amplitude is unchanged at 10 cm ✔
\nThe diagram shows the position of the principal lines in the visible spectrum of atomic hydrogen and some of the corresponding energy levels of the hydrogen atom.
\nDetermine the energy of a photon of blue light (435nm) emitted in the hydrogen spectrum.
\nIdentify, with an arrow labelled B on the diagram, the transition in the hydrogen spectrum that gives rise to the photon with the energy in (a).
\nExplain your answer to (b).
\nidentifies λ = 435 nm ✔
\nE = « =» ✔
\n4.6 ×10−19 «J» ✔
\n–0.605 OR –0.870 OR –1.36 to –5.44 AND arrow pointing downwards ✔
\nArrow MUST match calculation in (a)(i)
\nAllow ECF from (a)(i)
\nDifference in energy levels is equal to the energy of the photon ✔
\nDownward arrow as energy is lost by hydrogen/energy is given out in the photon/the electron falls from a higher energy level to a lower one ✔
\nAllow ECF from (a)(i)
\nA mass of 1.0 kg of water is brought to its boiling point of 100 °C using an electric heater of power 1.6 kW.
\nA mass of 0.86 kg of water remains after it has boiled for 200 s.
\nThe electric heater has two identical resistors connected in parallel.
\nThe circuit transfers 1.6 kW when switch A only is closed. The external voltage is 220 V.
\nThe molar mass of water is 18 g mol−1. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.
\nState one assumption of the kinetic model of an ideal gas.
\nEstimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.
\nExplain why the temperature of water remains at 100 °C during this time.
\nThe heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.
\nDetermine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.
\nSpecific heat capacity of pasta = 1.8 kJ kg−1 K−1
Specific heat capacity of water = 4.2 kJ kg−1 K−1
Show that each resistor has a resistance of about 30 Ω.
\nCalculate the power transferred by the heater when both switches are closed.
\nEk = « » = «J» ✓
\nv = «» = 720 «m s−1» ✓
\nparticles can be considered points «without dimensions» ✓
\nno intermolecular forces/no forces between particles «except during collisions»✓
\nthe volume of a particle is negligible compared to volume of gas ✓
\ncollisions between particles are elastic ✓
\ntime between particle collisions are greater than time of collision ✓
\nno intermolecular PE/no PE between particles ✓
\n\n
Accept reference to atoms/molecules for “particle”
\n«mL = P t» so «» = 2.3 x 106 «J kg-1» ✓
\nJ kg−1 ✓
\n«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓
\nAccept reference to atoms/molecules for “particle”
\nuse of mcΔT ✓
\n0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓
\nTeq = 85.69«°C» ≅ 86«°C» ✓
\nAccept Teq in Kelvin (359 K).
\n«Ω» ✓
\nMust see either the substituted values OR a value for R to at least three s.f.
\n\n
use of parallel resistors addition so Req = 15 «Ω» ✓
\nP = 3200 «W» ✓
\nThe light from a distant galaxy shows that .
\nCalculate the ratio .
\nOutline how Hubble’s law is related to .
\nHubble originally linked galactic redshift to a Doppler effect arising from galactic recession. Hubble’s law is now regarded as being due to cosmological redshift, not the Doppler effect. Explain the observed galactic redshift in cosmological terms.
\n«»
\nOR OR ✓
\n\n
«Hubble’s » measure of v/recessional speed uses redshift which is
OR
redshift () of galaxies is proportional to distance «from earth»
OR
combines AND into one expression, e.g. ✓
OWTTE
\nreference to «redshift due to» expansion of the universe, «not recessional speed» ✓
\nexpansion of universe stretches spacetime / increases distance between objects ✓
\n«so» wavelength stretches / increases leading to observed redshift ✓
\nMany candidates got the ratio upside down and ended up with R/R0 as 1.11.
\n
This was not accepted as it would have required an identification of the variables. Perhaps candidates need to look more carefully at which R is which here. R is the current value of the scale factor in the data book, so R0/R = 0.9 was required.
To show the link between z and Hubble's law many rearranged formulae to obtain zc = Hd or similar. Others stated that Hubble used redshift z to determine that v was proportional to distance. Either approach was allowed.
\nThe galactic redshift was successfully explained by many in terms of the stretching of spacetime.
\nA metal rod of length 45 cm is clamped at its mid point. The speed of sound in the metal rod is 1500 m s−1 and the speed of sound in air is 300 m s−1. The metal rod vibrates at its first harmonic. What is the wavelength in air of the sound wave produced by the metal rod?
\nA. 4.5 cm
\nB. 9.0 cm
\nC. 18 cm
\nD. 90 cm
\nC
\nThree possible features of an atomic model are
\nI. orbital radius
\nII. quantized energy
\nIII. quantized angular momentum.
\nWhich of these are features of the Bohr model for hydrogen?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II, and III
\nD
\nPhotons of discrete energy are emitted during gamma decay. This is evidence for
A. atomic energy levels.
B. nuclear energy levels.
C. pair annihilation.
D. quantum tunneling.
\nB
\nTwo charges Q1 and Q2, each equal to 2 nC, are separated by a distance 3 m in a vacuum. What is the electric force on Q2 and the electric field due to Q1 at the position of Q2?
\nA
\nWhat is the unit of electrical potential difference expressed in fundamental SI units?
A. kg m s-1 C-1
B. kg m2 s-2 C-1
C. kg m2 s-3 A-1
D. kg m2 s-1 A
\nC
\nThe most popular answer was B giving a low discrimination index for this question. It should be a relatively straightforward question provided the candidate can remember which of ‘C’ or ‘A’ is the fundamental unit.
\nA boat with an output engine power of 15 kW moves through water at a speed of 10 m s-1. What is the resistive force acting on the boat?
A. 0.15 kN
B. 0.75 kN
C. 1.5 kN
D. 150 kN
\nC
\nTwo conductors S and T have the V/I characteristic graphs shown below.
\nWhen the conductors are placed in the circuit below, the reading of the ammeter is 6.0 A.
\nWhat is the emf of the cell?
\nA. 4.0 V
\nB. 5.0 V
\nC. 8.0 V
\nD. 13 V
\nA
\nTwo forces of magnitude 12 N and 24 N act at the same point. Which force cannot be the resultant of these forces?
A. 10 N
B. 16 N
C. 19 N
D. 36 N
\nA
\nFor a real cell in a circuit, the terminal potential difference is at its closest to the emf when
\nA. the internal resistance is much smaller than the load resistance.
\nB. a large current flows in the circuit.
\nC. the cell is not completely discharged.
\nD. the cell is being recharged.
\nA
\nIn nuclear magnetic resonance imaging (NMR) a patient is exposed to a strong external magnetic field so that the spin of the protons in the body align parallel or antiparallel to the magnetic field. A pulse of a radio frequency (RF) electromagnetic wave is then directed at the patient.
\nDescribe the effect of the RF signal on the protons in the body.
\nOutline the measurement that needs to be made after the RF signal is turned off.
\nDescribe how the measurement in (b) provides diagnostic information for the doctor.
\nprotons spin direction changes
\nOR
\nproton energy state changes ✔
\nRelaxation time «of signal/proton spin» ✔
\nLocation/time delay of the emitted RF signal ✔
\nRelaxation time gives information on tissue type/density/health/OWTTE✔
\nLocation information provides 3D image/OWTTE✔
\nThe ratio = 1.5.
\nShow that the intensity of solar radiation at the orbit of Mars is about 600 W m–2.
\nDetermine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.
\nThe atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the greenhouse effect is not significant on Mars.
\nuse of «1.36 × 103 × » ✔
\n604 «W m–2» ✔
\nuse of for mean intensity ✔
\ntemperature/K = «» 230 ✔
\nrecognize the link between molecular density/concentration and pressure ✔
\nlow pressure means too few molecules to produce a significant heating effect
\nOR
\nlow pressure means too little radiation re-radiated back to Mars ✔
\nThe data for the star Eta Aquilae A are given in the table.
\nis the luminosity of the Sun and is the mass of the Sun.
\nShow by calculation that Eta Aquilae A is not on the main sequence.
\nEstimate, in , the distance to Eta Aquilae A using the parallax angle in the table.
\nEstimate, in , the distance to Eta Aquilae A using the luminosity in the table, given that .
\nSuggest why your answers to (b)(i) and (b)(ii) are different.
\nEta Aquilae A is a Cepheid variable. Explain why the brightness of Eta Aquilae A varies.
\nEta Aquilae A was on the main sequence before it became a variable star. Compare, without calculation, the time Eta Aquilae A spent on the main sequence to the total time the Sun is likely to spend on the main sequence.
\n✓
\nthe luminosity of Eta (2630) is very different «so it is not main sequence» ✓
\nAllow calculation of to give so not main sequence
\nOWTTE
✓
\nUse of ✓
\n✓
\n✓
\n\n
Award [3] marks for a bald correct answer between and
\nparallax angle in milliarc seconds/very small/at the limits of measurement ✓
\nuncertainties/error in measuring οr or ✓
\nvalues same order of magnitude, so not significantly different ✓
\n\n
Accept answers where MP1 and MP2 both refer to parallax angle
\nOWTTE
\nreference to change in size ✓
reference to change in temperature ✓
reference to periodicity of the process ✓
reference to transparency / opaqueness ✓
shorter time ✓
\nstar more massive and mass related to luminosity
OR
star more massive and mass related to time in main sequence
OR
position on HR diagram to the left and above shows that will reach red giant region sooner ✓
Most candidates were successful in using the mass luminosity relationship.
\nThe conversion to parsecs proved to be a very well known skill.
\nVery well answered by most candidates.
\nCandidates related the difference in the two methods for finding d to the large uncertainty in finding parallax angle at this distance (>400pc). Fewer also spotted that the luminosity of the star is also error prone unless its distance is already known.
\nCandidates were clearly familiar with Cepheids and the process leading to its variability in brightness.
\nCandidates showed clear ideas and were able to explain successfully why Eta Aquilae A lifetime as a main sequence star is much shorter than the one expected for the Sun.
\nAn astronaut is moving at a constant velocity in the absence of a gravitational field when he throws a tool away from him.
What is the effect of throwing the tool on the total kinetic energy of the astronaut and the tool and the total momentum of the astronaut and the tool?
\nD
\nAn electron enters the space inside a current-carrying solenoid. The velocity of the electron is parallel to the solenoid’s axis. The electron is
\nA. slowed down.
\nB. speeded up.
\nC. undeflected.
\nD. deflected outwards.
\nA
\nThe graph shows the variation of velocity of a body with time along a straight line.
\nWhat is correct for this graph?
A. The maximum acceleration is at P.
B. The average acceleration of the body is given by the area enclosed by the graph and time axis.
C. The maximum displacement is at Q.
D. The total displacement of the body is given by the area enclosed by the graph and time axis.
\nD
\nA beam of microwaves is incident normally on a pair of identical narrow slits S1 and S2.
\nWhen a microwave receiver is initially placed at W which is equidistant from the slits, a maximum in intensity is observed. The receiver is then moved towards Z along a line parallel to the slits. Intensity maxima are observed at X and Y with one minimum between them. W, X and Y are consecutive maxima.
\nExplain why intensity maxima are observed at X and Y.
\nThe distance from S1 to Y is 1.243 m and the distance from S2 to Y is 1.181 m.
Determine the frequency of the microwaves.
Outline one reason why the maxima observed at W, X and Y will have different intensities from each other.
\ntwo waves superpose/mention of superposition/mention of «constructive» interference ✔
\nthey arrive in phase/there is a path length difference of an integer number of wavelengths ✔
\nIgnore references to nodes/antinodes.
\npath difference = 0.062 «m» ✔
\nso wavelength = 0.031 «m» ✔
\nfrequency = 9.7 × 109 «Hz» ✔
\nIf no unit is given, assume the answer is in Hz. Accept other prefixes (eg 9.7 GHz)
\nAward [2 max] for 4.8 x 109 Hz.
\nintensity varies with distance OR points are different distances from the slits ✔
\nAccept “Intensity is modulated by a single slit diffraction envelope”.
\nMany candidates were able to discuss the interference that is taking place in this question, but few were able to fully describe the path length difference. That said, the quality of responses on this type of question seems to have improved over the last few examination sessions with very few candidates simply discussing the crests and troughs of waves.
\nMany candidates struggled with this question. Few were able to calculate a proper path length difference, and then use that to calculate the wavelength and frequency. Many candidates went down blind paths of trying various equations from the data booklet, and some seemed to believe that the wavelength is just the reciprocal of the frequency.
\nThis is one of many questions on this paper where candidates wrote vague answers that did not clearly connect to physics concepts or include key information. There were many overly simplistic answers like “they are farther away” without specifying what they are farther away from. Candidates should be reminded that their responses should go beyond the obvious and include some evidence of deeper understanding.
\nThe average temperature of ocean surface water is 289 K. Oceans behave as black bodies.
\nShow that the intensity radiated by the oceans is about 400 W m-2.
\nExplain why some of this radiation is returned to the oceans from the atmosphere.
\n5.67 × 10−8 × 2894
\nOR
\n= 396 «W m−2» ✔
\n«≈ 400 W m−2»
\n\n
«most of the radiation emitted by the oceans is in the» infrared ✔
«this radiation is» absorbed by greenhouse gases/named greenhouse gas in the atmosphere ✔
«the gases» reradiate/re-emit ✔
partly back towards oceans/in all directions/awareness that radiation in other directions is also present ✔
\nThis was well answered with candidates scoring the mark for either a correct substitution or an answer given to at least one more sf than the show that value. Some candidates used 298 rather than 289.
\nFor many this was a well-rehearsed answer which succinctly scored full marks. For others too many vague terms were used. There was much talk about energy being trapped or reflected and the ozone layer was often included. The word ‘albedo’ was often written down with no indication of what it means and ‘the albedo effect also featured.
\nA container holds 20 g of argon-40() and 40 g of neon-20 () .
What is in the container?
A. 0.25
B. 0.5
C. 2
D. 4
\nA
\nA child stands on a horizontal rotating platform that is moving at constant angular speed. The centripetal force on the child is provided by
\nA. the gravitational force on the child.
\nB. the friction on the child’s feet.
\nC. the tension in the child’s muscles.
\nD. the normal reaction of the platform on the child.
\nB
\nMuons are created at a height of 3230 m above the Earth’s surface. The muons move vertically downward at a speed of 0.980 c relative to the Earth’s surface. The gamma factor for this speed is 5.00. The half-life of a muon in its rest frame is 2.20 μs.
\nEstimate in the Earth frame the fraction of the original muons that will reach the Earth’s surface before decaying according to Newtonian mechanics.
\nEstimate in the Earth frame the fraction of the original muons that will reach the Earth’s surface before decaying according to special relativity.
\nDemonstrate how an observer moving with the same velocity as the muons accounts for the answer to (a)(ii).
\ntime of travel is «» = 1.10 × 10−5 «s» ✔
\nwhich is «» = 5.0 half-lives ✔
\nso fraction arriving as muons is « » =
\nOR
\n3 % ✔
\nAward [3] for a bald correct answer.
\n\n
time of travel corresponds to «» = 1.0 half-life ✔
\nso fraction arriving as muons is
\nOR
\n50 % ✔
\nAward [2] for a bald correct answer.
\n\n
observer measures the distance to the surface to be shorter «by a factor of 5.0»/ length contraction occurs ✔
\nso time of travel again corresponds to «»= 1.0 half-life ✔
\nMuons. The problem of muons created above the Earth surface moving almost at the speed of light, explicitly mentioned in the Guide, was solved well by prepared candidates. In i) many correctly calculated the time of travel in the earth’s frame though some struggled to recognize and apply the decay half-life aspect of this problem.
\nStudents who correctly answered part i) did so in ii).
\nSome candidates struggled in b), with only the best candidates identified length contraction. Importantly, the command term here is “demonstrate” which means students must make their response clear by reasoning or evidence. Many who identified length contraction did not provide adequate reasoning to gain full marks.
\nThe Hubble constant is 2.3 × 10-18 s-1.
\nA galaxy is 1.6 × 108 ly from Earth. Show that its recessional speed as measured from Earth is about 3.5 × 106 m s-1.
\nA line in the hydrogen spectrum when measured on Earth has a wavelength of 486 nm. Calculate, in nm, the wavelength of the same hydrogen line when observed in the galaxy’s emission spectrum.
\nOutline how observations of spectra from distant galaxies provide evidence that the universe is expanding.
\nd = «1.6 × 108 × 9.46 × 1015 =» 1.51 × 1024 «m»✔
v = «H0d = 2.3 × 10−18 ×1.5 × 1024 =» 3.48 × 106 «m s–1» ✔
Answer given, correct working required or at least 3sf needed for MP2.
\n» 5.64«nm» ✔
observed λ = «486 + 5.64 =» 492 «nm»✔
\nall distant galaxies exhibit red-shift ✔
\nOWTTE
\nThis very simple application of Hubble’s law was answered correctly by the vast majority of candidates.
\nMany candidates subtracted the change in wavelength and obtained a blue shift. Others were unsure which wavelength λo is in the data book equation. But correct answers were common.
\nNearly all candidates were able to mention redshift as the evidence for galaxy recession and the universe expansion.
\nThe temperature of a fixed mass of an ideal gas changes from 200 °C to 400 °C.
What is ?
A. 0.50
B. 0.70
C. 1.4
D. 2.0
\nB
\nMost candidates chose A having forgotten to convert from oC to K.
\nA beam of ultrasound of intensity I0 enters a layer of muscle of thickness 4.1 cm.
\nThe fraction of the intensity that is reflected at a boundary is
\n\n
where Z1 and Z2 are the acoustic impedances of the two media at the boundary. After travelling a distance x in a medium the intensity of ultrasound is reduced by a factor e–μx where μ is the absorption coefficient.
\nThe following data are available.
\nAcoustic impedance of muscle = 1.7 × 106 kg m–2 s–1
\nAcoustic impedance of bone = 6.3 × 106 kg m–2 s–1
\nAbsorption coefficient of muscle = 23 m–1
\nDetermine, in terms of I0, the intensity of ultrasound that is incident on the muscle–bone boundary.
\nDetermine, in terms of I0, the intensity of ultrasound that is reflected at the muscle–bone boundary.
\nDetermine, in terms of I0, the intensity of ultrasound that returns to the muscle–gel boundary.
\nI0e−23 × 0.041 ✔
\n= 0.39 I0 ✔
\nR = « =» 0.33 ✔
\nso reflected intensity is 0.33 × 0.39I0 = 0.13I0 ✔
\n0.13I0 × 0.39 = 0.05I0 ✔
\nMonochromatic light travelling upwards in glass is incident on a boundary with air. The path of the refracted light is shown.
\nA layer of liquid is then placed on the glass without changing the angle of incidence on the glass. The refractive index of the glass is greater than the refractive index of the liquid and the refractive index of the liquid is greater than that of air.
\nWhat is the path of the refracted light when the liquid is placed on the glass?
\nD
\nA low discrimination index with most candidates choosing C. They have deduced, correctly, that the ray moves away from the normal on entering the denser medium but have apparently forgotten that the stem of the question has shown them that it reaches the glass-air boundary at an angle greater than the critical angle.
\nA simple model of an atom has three energy levels. The differences between adjacent energy levels are shown below.
\nWhat are the two smallest frequencies in the emission spectrum of this atom?
\nA. 0.5 × 1015 Hz and 1.0 × 1015 Hz
\nB. 0.5 × 1015 Hz and 1.5 × 1015 Hz
\nC. 1.0 × 1015 Hz and 2.0 × 1015 Hz
\nD. 1.0 × 1015 Hz and 3.0 × 1015 Hz
\nC
\nThe diagram shows space and time axes and ct for an observer at rest with respect to a galaxy. A spacecraft moving through the galaxy has space and time axes ′ and ct′.
\nA rocket is launched towards the right from the spacecraft when it is at the origin of the axes. This is labelled event 1 on the spacetime diagram. Event 2 is an asteroid exploding at = 100 ly and ct = 20 ly.
\nPlot, on the axes, the point corresponding to event 2.
\nSuggest whether the rocket launched by the spacecraft might be the cause of the explosion of the asteroid.
\nShow that the value of the invariant spacetime interval between events 1 and 2 is 9600 ly2.
\nAn observer in the spacecraft measures that events 1 and 2 are a distance of 120 ly apart. Determine, according to the spacecraft observer, the time between events 1 and 2.
\nUsing the spacetime diagram, determine which event occurred first for the spacecraft observer, event 1 or event 2.
\nDetermine, using the diagram, the speed of the spacecraft relative to the galaxy.
\npoint as shown ✔
\nALTERNATIVE 1
the rocket would have to travel faster than the speed of light ✔
so impossible ✔
\nALTERNATIVE 2
drawing of future lightcone at origin ✔
and seeing that the asteroid explodes outside the lightcone so impossible ✔
\nALTERNATIVE 3
the event was observed at +20 years, but its distance (stationary) is 100 ly ✔
so the asteroid event happened 80 years before t = 0 for the galactic observer ✔
\n1002 − 202 = 9600 «ly2» ✔
\nAlso accept 98 (the square root of 9600).
\nAllow negative value.
\n9600 = 1202 − c2t2 ✔
\nct = «−» 69.3 «ly» / t = «−» 69.3 «y» ✔
\nAllow approach with Lorentz transformation.
\nline from event 2 parallel to ’ axis intersects ct’ axis at a negative value ✔
\nevent 2 occurred first ✔
\nuse of tan θ = with the angle between the time axes ✔
\nto get (0.70 ± 0.02)c ✔
\nAlmost all candidates were able to plot the event on the diagram.
\nMost of the candidates identified, that the spacecraft was launched after the asteroid explosion and better candidates were also able to explain their reasoning with a drawing of light cones on the spacetime graph being the most popular response type.
\nMost of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference.
\nMost of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference.
\nMost of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference. Most candidates well determined that event 2 occurred first in d ii), with a lesser number showing this correctly via the spacetime diagram.
\nMany candidates determined the speed using the spacetime diagram. However, some experienced difficulties reading accurately from the graph, and though their approach was correct, they failed to gain a result within the accepted range of values.
\nWhat is the relation between the value of the unified atomic mass unit in grams and the value of Avogadro’s constant in mol−1?
\nA. Their ratio is 1.
\nB. Their product is 1.
\nC. Their sum is 1.
\nD. Their difference is 0.
\nB
\nObject P moves vertically with simple harmonic motion (shm). Object Q moves in a vertical circle with a uniform speed. P and Q have the same time period T. When P is at the top of its motion, Q is at the bottom of its motion.
\nWhat is the interval between successive times when the acceleration of P is equal and opposite to the acceleration of Q?
A.
B.
C.
D. T
B
\nIn a hydrogen atom, the sum of the masses of a proton and of an electron is larger than the mass of the atom. Which interaction is mainly responsible for this difference?
\nA. Electromagnetic
\nB. Strong nuclear
\nC. Weak nuclear
\nD. Gravitational
\nA
\nThe resistance of component X decreases when the intensity of light incident on it increases. X is connected in series with a cell of negligible internal resistance and a resistor of fixed resistance. The ammeter and voltmeter are ideal.
What is the change in the reading on the ammeter and the change in the reading on the voltmeter when the light incident on X is increased?
\nA
\nA spaceship moves away from the Earth in the direction of a nearby planet. An observer on the Earth determines the planet is 4 ly from the Earth. The spacetime diagram for the Earth’s reference frame shows the worldline of the spaceship. Assume the clock on the Earth, the clock on the planet, and the clock on the spaceship were all synchronized when ct = 0.
\nShow, using the spacetime diagram, that the speed of the spaceship relative to the Earth is 0.80c.
\nLabel, with the letter E, the event of the spaceship going past the planet.
\nDetermine, according to an observer on the spaceship as the spaceship passes the planet, the time shown by the clock on the spaceship.
\nDetermine, according to an observer on the spaceship as the spaceship passes the planet, the time shown by the clock on the planet.
\nOn passing the planet a probe containing the spaceship’s clock and an astronaut is sent back to Earth at a speed of 0.80c relative to Earth. Suggest, for this situation, how the twin paradox arises and how it is resolved.
\nEvidence of finding 1/gradient such as:
use of any correct coordinate pair to find v - eg or .
OR
measures tan of angle between ct and ct’ as about 39° AND tan 39 ≈ 0.8 ✔
\nAnswer 0.8c given, so check coordinate values carefully.
E labelled at = 4, ct = 5 ✔
\nCheck that E is placed on the worldline of S.
\nOR
\nAllow solutions involving the use of Lorentz equations.
\nt = 5 years OR ct = 5 ly ✔
\nOn return to Earth the astronaut will have aged less than Earthlings «by 4 years»
OR
time passed on Earth is greater than time passed for the astronaut «by 4 years» ✔
astronaut accelerated/changed frames but Earth did not
OR
for astronaut the Earth clock jumps forward at turn-around ✔
\nOWTTE
Treat as neutral any mention of both the Earth and astronaut seeing each other’s clock as running slow.
\nMost candidates could show that the velocity of the spacecraft was 0.8c.
\nEvent E was usually correctly labelled on the space-time diagram.
\nA very simple time dilation question which most candidates got wrong at SL but the question was better answered at HL.
\nMany candidates tried to use time dilation again without realising that the clock on P must also read 5 years at event E because that is the time on the Earth clock in P’s frame for the event.
\nThe twin paradox is now well understood and there were some good quality answers. Some candidates even knew that the Earth clock jumps forward when the Astronaut turns around.
\nWhich Feynman diagram describes the annihilation of an electron and its antiparticle?
\nA
\nIn 2017, two neutron stars were observed to merge, forming a black hole. The material released included chemical elements produced by the r process of neutron capture. Describe two characteristics of the elements produced by the r process.
\nhigher atomic number than iron ✓
\nexcess of neutrons ✓
\nradioactive/undergoing beta decay ✓
\n\n
Allow heavier than iron for MP1
\nThe rapid process proved to be known by many although fewer candidates were able to provide two characteristics.
\nA planet orbits at a distance d from a star. The power emitted by the star is P. The total surface area of the planet is A.
\nExplain why the power incident on the planet is
\n\n
The albedo of the planet is . The equilibrium surface temperature of the planet is T. Derive the expression
\n\nwhere e is the emissivity of the planet.
\nOn average, the Moon is the same distance from the Sun as the Earth. The Moon can be assumed to have an emissivity e = 1 and an albedo = 0.13. The solar constant is 1.36 × 103 W m−2. Calculate the surface temperature of the Moon.
\nis the power received by the planet/at a distance d «from star» ✓
\nis the projected area/cross sectional area of the planet ✓
\n\n
use of ✓
\nwith correct manipulation to show the result ✓
\n\n
✓
\nT = 268.75 «K» ≅ 270 «K» ✓
\nThe cosmic microwave background (CMB) radiation is observed to have anisotropies.
\nState the nature of the anisotropies observed in the CMB radiation.
\nIdentify two possible causes of the anisotropies in (a).
\nthe temperature/«peak» wavelength/intensity «of the CMBR» varies «slightly» / is not constant in different directions ✓
\nquantum fluctuations «that have expanded» ✓
\ndensity perturbations «that resulted in galaxies and clusters of galaxies» ✓
\ndipole distortion «due to the motion of the Earth» ✓
\nThis was not the best answered question in the Option. Candidates oscillated between correct identification of characteristics of the CMB radiation and more generic explanations about the Big Bang.
\nThose who correctly identified specific characteristics of the CMB radiation were able to quote causes for this during the early Big Bang.
\nA horizontal wire PQ lies perpendicular to a uniform horizontal magnetic field.
\nA length of 0.25 m of the wire is subject to a magnetic field strength of 40 mT. A downward magnetic force of 60 mN acts on the wire.
\nWhat is the magnitude and direction of the current in the wire?
\nB
\nWhich graph shows the relationship between gravitational force F between two point masses and their separation r?
\nD
\nThe positions of stable nuclei are plotted by neutron number n and proton number p. The graph indicates a dotted line for which n = p. Which graph shows the line of stable nuclides and the shaded region where unstable nuclei emit beta minus (β-) particles?
\nD
\nThis question proved challenging, a low discrimination index and a relatively even spread of answers suggests that maybe guesswork was responsible for the candidates choice.
\nWhich Feynman diagram shows the emission of a photon by a charged antiparticle?
\nC
\nThree methods for the production of electrical energy are
I. wind turbine
\nII. photovoltaic cell
\nIII. fossil fuel power station.
\nWhich methods involve the use of a primary energy source?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nD
\nThis question seems to have prompted some discussion among teachers and slightly more candidates chose response A than the others. Primary energy is defined as coming from a natural resource so whereas fossil fuels are non-renewable they are a primary energy resource. Also, a photovoltaic cell produces electricity, defined as a secondary energy source from a primary energy source, the sun. The clue is given in the question ‘involve the USE of a primary energy source’.
\nA student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of 64 m s–1. The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms.
\nThe student strikes the tennis ball at point P. The tennis ball is initially directed at an angle of 7.00° to the horizontal.
\nThe following data are available.
Height of P = 2.80 m
Distance of student from net = 11.9 m
Height of net = 0.910 m
Initial speed of tennis ball = 64 m s-1
\nCalculate the average force exerted by the racquet on the ball.
\nCalculate the average power delivered to the ball during the impact.
\nCalculate the time it takes the tennis ball to reach the net.
\nShow that the tennis ball passes over the net.
\nDetermine the speed of the tennis ball as it strikes the ground.
\nThe student models the bounce of the tennis ball to predict the angle θ at which the ball leaves a surface of clay and a surface of grass.
\nThe model assumes
• during contact with the surface the ball slides.
• the sliding time is the same for both surfaces.
• the sliding frictional force is greater for clay than grass.
• the normal reaction force is the same for both surfaces.
Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.
\n✔
\n= 148«»≈150«» ✔
\n\n
ALTERNATIVE 1
\n✔
\n» ✔
\n\n
ALTERNATIVE 2
\n✔
\n» ✔
\n\n
horizontal component of velocity is 64.0 × cos7° = 63.52 «ms−1» ✔
\n» ✔
\nDo not award BCA. Check working.
Do not award ECF from using 64 m s-1.
\nALTERNATIVE 1
uy = 64 sin7/7.80 «ms−1»✔
decrease in height = 7.80 × 0.187 + × 9.81 × 0.1872/1.63 «m» ✔
final height = «2.80 − 1.63» = 1.1/1.2 «m» ✔
«higher than net so goes over»
ALTERNATIVE 2
vertical distance to fall to net «= 2.80 − 0.91» = 1.89 «m»✔
time to fall this distance found using «=1.89 = 7.8t + × 9.81 ×t2»
t = 0.21 «s»✔
0.21 «s» > 0.187 «s» ✔
«reaches the net before it has fallen far enough so goes over»
Other alternatives are possible
ALTERNATIVE 1
Initial KE + PE = final KE /
\n × 0.058 × 642 + 0.058 × 9.81 × 2.80 = × 0.058 × v2 ✔
v = 64.4 «ms−1» ✔
ALTERNATIVE 2
» ✔
\n« »
\n» ✔
\n\n
so horizontal velocity component at lift off for clay is smaller ✔
normal force is the same so vertical component of velocity is the same ✔
so bounce angle on clay is greater ✔
At both HL and SL many candidates scored both marks for correctly answering this. A straightforward start to the paper. For those not gaining both marks it was possible to gain some credit for calculating either the change in momentum or the acceleration. At SL some used 64 ms-1 as a value for a and continued to use this value over the next few parts to the question.
\nThis was well answered although a significant number of candidates approached it using P = Fv but forgot to divide v by 2 to calculated the average velocity. This scored one mark out of 2.
\nThis question scored well at HL but less so at SL. One common mistake was to calculate the direct distance to the top of the net and assume that the ball travelled that distance with constant speed. At SL particularly, another was to consider the motion only when the ball is in contact with the racquet.
\nThere were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.
\nA common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.
\nThis proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.
\nAs the command term in this question is ‘predict’ a bald answer of clay was acceptable for one mark. This was a testing question that candidates found demanding but there were some very well-reasoned answers. The most common incorrect answer involved suggesting that the greater frictional force on the clay court left the ball with less kinetic energy and so a smaller angle. At SL many gained the answer that the angle on clay would be greater with the argument that frictional force is greater and so the distance the ball slides is less.
\nA tube of constant circular cross-section, sealed at one end, contains an ideal gas trapped by a cylinder of mercury of length 0.035 m. The whole arrangement is in the Earth’s atmosphere. The density of mercury is 1.36 × 104 kg m–3.
\nWhen the mercury is above the gas column the length of the gas column is 0.190 m.
\nThe tube is slowly rotated until the gas column is above the mercury.
\nThe length of the gas column is now 0.208 m. The temperature of the trapped gas does not change during the process.
\nA solid cylinder of height h and density ρ rests on a flat surface.
\nShow that the pressure pc exerted by the cylinder on the surface is given by pc = ρgh.
\nShow that (po + pm) × 0.190 = where
\npo = atmospheric pressure
\npm = pressure due to the mercury column
\nT = temperature of the trapped gas
\nn = number of moles of the trapped gas
\nA = cross-sectional area of the tube.
\nDetermine the atmospheric pressure. Give a suitable unit for your answer.
\nOutline why the gas particles in the tube hit the mercury surface less often after the tube has been rotated.
\nweight of cylinder = Ahg ρ ✔
\npressure = = ✔
\nAllow use of A = in MP1.
\nuse of PV = nRT and V = Area × (0.190) seen ✔
\nsubstitution of P = po + pm «re-arrangement to give answer»✔
\nrecognition that is constant OR 190po + 190pm = 208po − 208pm
\nOR po = pm ✔
\npressure due to mercury pm = 0.035 × 1.36 × 104 × 9.81(= 4.67 × 103 Pa) ✔
\n1.03 × 105 ✔
\nPa OR Nm-2 OR kgm-1s-2 ✔
\nDo not award for a bald correct answer. Working must be shown to award MP3.
\nAward MP4 for any correct unit of pressure (eg “mm of mercury / Hg”).
\nsame number of particles to collide with a larger surface area OR greater volume with constant rms speed decreases collision frequency ✔
\nLook for a correct statement that connects pressure to molecular movement/collisions.
\nThis question was fairly well answered with a not insignificant number of candidates making clearly wrong substitutions (such as F=mgh) to make the equation work out. As a “show that” question the derivation should be neatly laid out with the fundamental equations written out and the substitutions/cancelations clearly shown.
\nThis question was generally well answered. Most candidates took the time to show the set up and substitutions they used to derive the given expression. A small number of candidates attempted to “show that” by making unit substitutions - this is not acceptable for a question like this.
\nThis question was left blank by many candidates. Of those who attempted a solution, few appreciated the difference made to the derived equation from 4bi when the tube was rotated. It should be noted that this was also the “unit question” on this exam, and a candidate could have been awarded a mark for clearly writing any correct unit of pressure without doing any calculations. Candidates should be reminded to keep an eye out for this opportunity and to at least write a unit even if the question seems unapproachable.
\nThis was generally poorly answered with the candidates split between answers that generally demonstrated some good understanding of physics (such as connecting the increase in volume AND constant rms speed of particles with the rate of collisions with the mercury) and answers that did not (such as gases want to rise so the gas it will hit the mercury less).
\nWhich quantity has the same units as those for energy stored per unit volume?
\nA. Density
\nB. Force
\nC. Momentum
\nD. Pressure
\nD
\nA list of four physical quantities is
\nHow many scalar quantities are in this list?
\nA. 1
\nB. 2
\nC. 3
\nD. 4
\nC
\nBurning one litre of gasoline produces more energy than burning one kilogram of coal, and the density of gasoline is smaller than 1 g cm−3. What can be deduced from this information?
\nA. Energy density is greater for gasoline.
\nB. Specific energy is greater for gasoline.
\nC. Energy density is greater for coal.
\nD. Specific energy is greater for coal.
\nB
\nWhich is correct for the tangential acceleration of a simple pendulum at small amplitudes?
\nA. It is inversely proportional to displacement.
\nB. It is proportional to displacement.
\nC. It is opposite to displacement.
\nD. It is proportional and opposite to displacement
\nD
\nThe diagram shows the diffraction pattern for light passing through a single slit.
\nWhat is
\n\n
A. 0.01
\nB. 0.02
\nC. 1
\nD. 2
\nA
\nLiquid oxygen at its boiling point is stored in an insulated tank. Gaseous oxygen is produced from the tank when required using an electrical heater placed in the liquid.
\nThe following data are available.
\nMass of 1.0 mol of oxygen = 32 g
\nSpecific latent heat of vaporization of oxygen = 2.1 × 105 J kg–1
\nAn oxygen flow rate of 0.25 mol s–1 is needed.
\nDistinguish between the internal energy of the oxygen at the boiling point when it is in its liquid phase and when it is in its gas phase.
\nCalculate, in kW, the heater power required.
\nCalculate the volume of the oxygen produced in one second when it is allowed to expand to a pressure of 0.11 MPa and to reach a temperature of 260 K.
\nState one assumption of the kinetic model of an ideal gas that does not apply to oxygen.
\nInternal energy is the sum of all the PEs and KEs of the molecules (of the oxygen) ✔
\nPE of molecules in gaseous state is zero ✔
\n(At boiling point) average KE of molecules in gas and liquid is the same ✔
\ngases have a higher internal energy ✔
\n\n
Molecules/particles/atoms must be included once, if not, award [1 max]
\nALTERNATIVE 1:
\nflow rate of oxygen = 8 «g s−1» ✔
\n«2.1 ×105 × 8 × 10−3» = 1.7 «kW» ✔
\n\n
ALTERNATIVE 2:
\nQ = «0.25 × 32 ×10−3 × 2.1 × 105 =» 1680 «J» ✔
\npower = «1680 W =» 1.7 «kW» ✔
\nV = «» 4.9 × 10−3 «m3»
\nideal gas has point objects ✔
\nno intermolecular forces ✔
\nnon liquefaction ✔
\nideal gas assumes monatomic particles ✔
\nthe collisions between particles are elastic ✔
\n\n
Allow the opposite statements if they are clearly made about oxygen eg oxygen/this can be liquified
\nThe Hertzsprung-Russell (HR) diagram shows several star types. The luminosity of the Sun is L☉.
\nIdentify, on the HR diagram, the position of the Sun. Label the position S.
\nSuggest the conditions that will cause the Sun to become a red giant.
\nOutline why the Sun will maintain a constant radius after it becomes a white dwarf.
\nDuring its evolution, the Sun is likely to be a red giant of surface temperature 3000 K and luminosity 104 L☉. Later it is likely to be a white dwarf of surface temperature 10 000 K and luminosity 10-4 L☉. Calculate the .
\nthe letter S should be in the region of the shaded area ✔
\nthe fusion of hydrogen in the core eventually stops
OR
core contracts ✔
the hydrogen in a layer around the core will begin to fuse ✔
Sun expands AND the surface cools ✔
helium fusion begins in the core ✔
Sun becomes more luminous/brighter✔
\nIgnore any mention of the evolution past the red giant stage
\nelectron degeneracy <<prevents further compression>> ✔
\nIgnore mention of the Chandrasekhar limit.
Award [0] for answer mentioning radiation pressure or fusion reactions.
\nLocating the Sun’s position on the HR diagram was correctly done by most candidates, although a few were unsure of the surface temperature of the Sun.
\nThe evolution of a main sequence star to the red giant region is reasonably well understood. However many struggled to find three different facts to describe the changes. Answers were often too vague, when writing about a change in temperature or size of a star, the candidates are expected to mention whether they are referring to the core or the surface/outer layer. A surprising number of candidates wrote that the Sun must be less than eight solar masses.
\nThe mention of electron degeneracy pressure was fairly common, but incorrect answers were even more common at SL.
\nCalculating the ratio of the radius of a white dwarf to a red giant star was done quite well by most candidates. However quite a few candidates made POT errors or forgot to take the final square root.
\nA proton has a total energy 1050 MeV after being accelerated from rest through a potential difference V.
\nDefine total energy.
\nDetermine the momentum of the proton.
\nDetermine the speed of the proton.
\nCalculate the potential difference V.
\ntotal energy is the sum of the rest energy and the kinetic energy ✔
\n«p2 c2 = 10502–9382 therefore» p=472«MeVc–1»✔
\n✔
\n\n
OR
\n«ms–1» ✔
\n\n
V = 112 «MV» ✔
\nGenerally well answered by most candidates. A common mistake was to define the total energy in the context of classical mechanics.
\nMost candidates seemed to have the right starting points but mistakes were often made in attempting to convert units. The energy-momentum equation is generally best answered using only ‘MeV’ based units.
\nAn easy calculation, that was generally well answered.
\nVery few candidates realised that this question required a simple calculation using eV= KE = (γ-1)E.
\nA solid cylinder of mass M and radius R is free to rotate about a fixed horizontal axle. A rope is tied around the cylinder and a block of mass is attached to the end of the rope.
\nThe system is initially at rest and the block is released. The moment of inertia of the cylinder about the axle is MR2
\nShow that the angular acceleration of the cylinder is
\nShow that the tension T in the string is
\nThe block falls a distance 0.50 m after its release before hitting the ground. Show that the block hits the ground 0.55 s after release.
\nCalculate, for the cylinder, at the instant just before the block hits the ground the angular momentum.
\nCalculate, for the cylinder, at the instant just before the block hits the ground the kinetic energy.
\nequations of motion are: TR = MR2 and − T =
\nOR
\ngR = MR2 + Ra
\nuse of a = R ✔
\ncombine equations to get result ✔
\nAllow energy conservation use.
\nThis is a show that question, so look for correct working.
\nDo not allow direct use of tension from a ii)
\nuse of T = MR to find T = MR × ✔
\n«cancelling to show final answer»
\na = 3.27 «ms−2» / a = g/3 ✔
\n✔
\n= 0.55 «s»
\nDo not apply ECF from MP1 to MP2 if for a=g, giving answer 0.32 s.
\nALTERNATIVE 1
ΔL «= ΓΔt = TRΔt » = ✔
\nΔL = 2.2J «Js»✔
ALTERNATIVE 2
\nω =<Δt = Δt = = > 8.99 «rads−1» ✔
\nΔL «=Iω» × 12 × 0.202 × 8.99 = 2.2 «Js»
\nAward [2] for a bald correct answer.
\nω =<Δt = Δt = = > 8.99 «rads−1» ✔
\nEk = « Iω2 = MR2 ω2 = × 12 × 0.202 × 8.992 = » 9.7 «J»
\nAward [2] for a bald correct answer.
\nThe diagram shows a Carnot cycle for an ideal monatomic gas.
\nThe highest temperature in the cycle is 620 K and the lowest is 340 K.
\nShow that during an adiabatic expansion of an ideal monatomic gas the temperature and volume are given by
\n= constant
\nCalculate the efficiency of the cycle.
\nThe work done during the isothermal expansion A → B is 540 J. Calculate the thermal energy that leaves the gas during one cycle.
\nCalculate the ratio where VC is the volume of the gas at C and VB is the volume at B.
\nCalculate the change in the entropy of the gas during the change A to B.
\nExplain, by reference to the second law of thermodynamics, why a real engine operating between the temperatures of 620 K and 340 K cannot have an efficiency greater than the answer to (b)(i).
\nsubstitution of in ✔
\nmanipulation to get result ✔
\ne « = 1 − = 1 − » = 0.45 ✔
\nheat into gas «is along AB» and equals
\nQin «= ΔU + W = 0 + 540» = 540 «J» ✔
\nheat out is (1−e) Qin = (1−0.45) × 540 = 297 «J» ≈ 3.0 × 102 «J» ✔
\nAward [2] for bald correct answer.
\n✔
\n✔
\nAward [2] for bald correct answer.
\nΔS «= = »= 0.87 «JK−1» ✔
\nthe Carnot cycle has the maximum efficiency «for heat engines operating between two given temperatures »✔
\nreal engine can not work at Carnot cycle/ideal cycle ✔
\nthe second law of thermodynamics says that it is impossible to convert all the input heat into mechanical work ✔
\na real engine would have additional losses due to friction etc ✔
\nPlane wavefronts in air are incident on the curved side of a transparent semi-circular block of refractive index 2.0.
\nPart of wavefront XY outside the block is shown.
\nDraw, on the diagram, the wavefront inside the block.
\nsmooth curve of correct curvature continuous at the boundary as shown ✔
\nwavelength must be half the one in air; judge by eye ✔
\nLight is incident on a diffraction grating. The wavelength lines 600.0 nm and 601.5 nm are just resolved in the second order spectrum. How many slits of the diffraction grating are illuminated?
\nA. 20
\nB. 40
\nC. 200
\nD. 400
\nC
\nAn object O is placed in front of concave mirror. The centre of the mirror is labelled with the letter C.
\nLabel the focal point of the mirror with the letter F.
\nSketch two appropriate rays on the diagram to show the formation of the image. Label the image with the letter I.
\nThe upper half of the mirror is blackened so it cannot reflect any light. State the effect of this, if any, on the image.
\nA concave mirror of radius 3.0 m is used to form the image of the full Moon. The distance from the mirror to the Moon is 3.8 × 108 m and the diameter of the Moon is 3.5 × 106 m.
\nDetermine the diameter of the image of the Moon.
\nF half-way between C and mirror vertex and on the principal axis ✔
\none correct ray ✔
\nsecond correct ray that allows the image to be located ✔
\nimage drawn ✔
\nimage will be less bright / dimmer ✔
\n«image distance is ie» v = 1.5 «m» ✔
\n✔
\nimage diameter is × 3.5 × 106 = 1.4 «cm» ✔
\nAward [3] for bald correct answer.
\nA container of volume 3.2 × 10-6 m3 is filled with helium gas at a pressure of 5.1 × 105 Pa and temperature 320 K. Assume that this sample of helium gas behaves as an ideal gas.
\nA helium atom has a volume of 4.9 × 10-31 m3.
\nThe molar mass of helium is 4.0 g mol-1. Show that the mass of a helium atom is 6.6 × 10-27 kg.
\nEstimate the average speed of the helium atoms in the container.
\nShow that the number of helium atoms in the container is about 4 × 1020.
\nCalculate the ratio .
\nExplain, using your answer to (d)(i) and with reference to the kinetic model, why this sample of helium can be assumed to be an ideal gas.
\n«kg»
OR
6.64 × 10−27 «kg» ✔
\n\n
✔
\nv = 1.4 × 103 «ms−1» ✔
\n\n
OR
\n✔
\nN = 3.7 × 1020 ✔
\n« ✔
\n«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas
OR
«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas
OR
«For an ideal gas» particles are assumed to be point objects ✔
calculation/ratio/result in (d)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔
\nThe mark was awarded for a clear substitution or an answer to at least 3sf. Many gained the mark for a clear substitution with a conversion from g to kg somewhere in their response. Fewer gave the answer to the correct number of sf.
\nAt HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.
\nAgain at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by NA to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by NA.
\nThis was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.
\nIn general this was poorly answered at SL. Many other non-related gas properties given such as no / negligible intermolecular forces, low pressure, high temperature. Some candidates interpreted the ratio as meaning it is a low density gas. At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.
\nOn approaching a stationary observer, a train sounds its horn and decelerates at a constant rate. At time t the train passes by the observer and continues to decelerate at the same rate. Which diagram shows the variation with time of the frequency of the sound measured by the observer?
\nD
\nThe diagram shows a light ray incident from air into the core of an optic fibre. The angle of incidence is θ. Values of refractive indices are shown on the diagram.
\nCalculate the critical angle at the core–cladding boundary.
\nShow that the maximum value of θ for which total internal reflection will take place at the core–cladding boundary is about 90°.
\nComment on your answer to part (a)(ii).
\nA signal consists of two rays that enter the core at angle of incidence θ = 0 and θ = θmax. Identify a disadvantage of this fibre for transmitting this signal.
\nOutline the significance of optic fibres in modern communications.
\n«sin θc = » sin θc = ✔
\nθc = 51.97° ✔
\nAward [2] for bald correct answer.
\n«angle of refraction at air-core boundary is 90°−θc «=90.00° − 51.97° = 38.03°»✔
\n1.000 × sinθmax =1.620 × sin 38.03° ✔
\nθmax = 86.41° ✔
\n«θmax is almost 90° which means that» a ray entering the core almost at any angle will be totally internally reflected/will not escape ✔
\nrays will follow very different paths in the core ✔
\nleading to waveguide dispersion/different arrival times/pulse overlap ✔
\nReference to 2 of:
\nsecure/encrypted transfer of data ✔
\nhigh bandwidth/volume of data transferred ✔
\nhigh quality/minimal noise in transmission ✔
\nfree from cross talk ✔
\nlow «specific» attenuation ✔
\nThe Hertzsprung–Russell (HR) diagram shows the Sun and a main sequence star X.
\nThe following data are available for the mass and radius of star X where M is the mass of the Sun and R is the radius of the Sun:
\nMX = 5.0 M
\nRX = 3.2 R
\nThe parallax angle for star X is 0.125 arc-second.
\nStar X will evolve to become a white dwarf star D.
\nShow that the luminosity of star X is about 280 times greater than the luminosity of the Sun L.
\nDetermine the ratio .
\nOutline how the parallax angle of a star can be measured.
\nShow that the distance to star X is 1.6 × 106 AU.
\nThe apparent brightness of the Sun is 1400 Wm–2. Calculate, in Wm–2, the apparent brightness of star X.
\nLabel, on the HR diagram, the region of white dwarf stars.
\nOutline the condition that prevents star D from collapsing further.
\nStar D emits energy into space in the form of electromagnetic radiation. State the origin of this energy.
\nPredict the change in luminosity of star D as time increases.
\n✔
\nCorrect working or answer to 4 sig figs required.
\n✔
\n✔
\nAward [2] for bald correct answer.
\nthe position of the star is recorded 6 months apart
OR
the radius/diameter of the Earth orbit clearly labelled on a diagram ✔
the parallax is measured from the shift of the star relative to the background of the distant stars ✔
\nFor MP2 accept a correctly labelled parallax angle on a diagram.
\nAward MP2 only if background distance stars are mentioned.
\nd = = 8.0 «pc» ✔
\nd = 8.0 × 3.26 × «AU» ✔
\n«= 1.64 × 106 AU»
\nALTERNATIVE 1
\n\n
OR
\n\n
✔
\n«W m–2» ✔
\n\n
ALTERNATIVE 2
\nOR OR ✔
\n✔
\nAward [2] for bald correct answer.
\nAllow ECF from MP1 to MP2
\n ✔
Allow any region with L below Sun and left to the main sequence.
\nan electron degeneracy «pressure develops that opposes gravitation»/reference to Pauli principle ✔
\nthermal energy/internal energy ✔
\n«temperature decreases so» luminosity decreases ✔
\nExplain the formation of a type I a supernova which enables the star to be used as a standard candle.
\nDescribe the r process which occurs during type II supernovae nucleosynthesis.
\nwhite dwarf attracts mass from another star ✔
\nexplodes/becomes supernova when mass equals/exceeds the Chandrasekhar limit / 1.4MSUN ✔
\nhence luminosity of all type I a supernova is the same ✔
\nOWTTE
\n«successive» rapid neutron capture ✔
faster than «β» decay can occur ✔
results in formation of heavier/neutron rich isotopes ✔
\nOWTTE
\nThe formation of Type 1a supernovae was well known by most candidates but few were able to explain how this process resulted in a standard candle.
\nMany candidates could describe the r process correctly but quite a large number of candidates seemed completely at a loss and could not relate the r process to neutron capture.
\nTwo equal positive fixed point charges Q = +44 μC and point P are at the vertices of an equilateral triangle of side 0.48 m.
\nPoint P is now moved closer to the charges.
\nA point charge q = −2.0 μC and mass 0.25 kg is placed at P. When x is small compared to d, the magnitude of the net force on q is F ≈ 115x.
\nAn uncharged parallel plate capacitor C is connected to a cell of emf 12 V, a resistor R and another resistor of resistance 20 MΩ.
\nShow that the magnitude of the resultant electric field at P is 3 MN C−1
\nState the direction of the resultant electric field at P.
\nExplain why q will perform simple harmonic oscillations when it is released.
\nCalculate the period of oscillations of q.
\nAt t = 0, the switch is connected to X. On the axes, draw a sketch graph to show the variation with time of the voltage VR across R.
\nThe switch is then connected to Y and C discharges through the 20 MΩ resistor. The voltage Vc drops to 50 % of its initial value in 5.0 s. Determine the capacitance of C.
\n«electric field at P from one charge is »
\nOR
\n«NC−1» ✓
\n
« net field is » «NC−1» ✓
directed vertically up «on plane of the page» ✓
\n\n
Allow an arrow pointing up on the diagram.
\nforce «on q» is proportional to the displacement ✓
\nand opposite to the displacement / directed towards equilibrium ✓
\n✓
\n✓
\n\n
Award [2] marks for a bald correct answer.
\nAllow ECF for MP2.
\ndecreasing from 12 ✓
\ncorrect shape as shown ✓
\n\n
Do not penalize if the graph does not touch the t axis.
\n✓
\n«F» ✓
\n\n
Award [2] for a bald correct answer.
\nA student measures the radius R of a circular plate to determine its area. The absolute uncertainty in R is ΔR.
\nWhat is the fractional uncertainty in the area of the plate?
\nA.
\nB.
\nC.
\nD.
\n\n
A
\nLight from a distant galaxy observed on Earth shows a redshift of 0.15.
\nOutline what is meant by redshift.
\nDetermine the distance to this galaxy assuming a Hubble constant of H0 = 72 km s–1 Mpc–1.
\nThe cosmic microwave background (CMB) radiation provides strong evidence for the Big Bang model. State the two main pieces of this evidence.
\nThe graph shows the variation of the intensity I of the CMB with wavelength λ.
\nDetermine, using the graph, the temperature of the CMB.
\n«the received» wavelength is longer than that emitted ✔
\nAllow context of Doppler redshift as well as cosmological redshift.
\nv = zc = 0.15 × 3.0 × 105 = 4.5 × 104 «km s−1» ✔
\nd = = 625 «Mpc» ✔
\n\n
Award [2] for bald correct answer.
\nAccept in other units, eg, 1.95 x 1025m.
\nthe radiation has a black body spectrum/it is black body radiation ✔
\nthe radiation is highly isotropic/uniform ✔
\nmatched the «predicted» wavelength/temperature if the Big Bang had increased/cooled by expansion ✔
\npeak wavelength read off graph as (1.1±0.05)«mm» ✔
\nsubstitution into Wien’s law to get T = (2.5 to 2.8)«K» ✔
\nA student is verifying the equation
\n\n
The percentage uncertainties are:
\n\n
What is the percentage uncertainty in x?
\nA. 5 %
\nB. 15 %
\nC. 25 %
\nD. 30 %
\nB
\nThis question was well answered by candidates, as shown by a high difficulty index.
\nA proton has momentum 10-20 N s and the uncertainty in the position of the proton is 10-10 m. What is the minimum fractional uncertainty in the momentum of this proton?
A. 5 × 10-25
B. 5 × 10-15
C. 5 × 10-5
D. 2 × 104
\nC
\nOver 100 candidates left this blank. It is testing fractional uncertainty and also involves the Heisenberg uncertainty principle.
\nA rocket is accelerating upwards at 9.8 m s-2 in deep space. A photon of energy 14.4 keV is emitted upwards from the bottom of the rocket and travels to a detector in the tip of the rocket 52.0 m above.
\nExplain why a change in frequency is expected for the photon detected at the top of the rocket.
\nCalculate the frequency change.
\nALTERNATIVE 1
detector accelerates/moves away from point of photon emission ✔
so Doppler effect / redshift ✔
so f decreases ✔
ALTERNATIVE 2
equivalent to stationary rocket on earth’s surface ✔
photons lose «gravitational» energy as they move upwards ✔
h f OR f decreases ✔
\nMost candidates answered this question correctly using the equivalence principle and could show that the frequency would decrease.
\nArithmetic mistakes were common at the different stages of the calculations even when the process used was correct.
\nA boy throws a ball horizontally at a speed of 15 m s-1 from the top of a cliff that is 80 m above the surface of the sea. Air resistance is negligible.
\nWhat is the distance from the bottom of the cliff to the point where the ball lands in the sea?
\nA. 45 m
\nB. 60 m
\nC. 80 m
\nD. 240 m
\nB
\nA book is at rest on a table. What is a pair of action–reaction forces for this situation according to Newton’s third law of motion?
\nC
\nA uniform ladder of weight 50.0 N and length 4.00 m is placed against a frictionless wall making an angle of 60.0° with the ground.
\nOutline why the normal force acting on the ladder at the point of contact with the wall is equal to the frictional force F between the ladder and the ground.
\nCalculate F.
\nThe coefficient of friction between the ladder and the ground is 0.400. Determine whether the ladder will slip.
\n«translational equilibrium demands that the» resultant force in the horizontal direction must be zero✔
«hence NW = F»
\nEquality of forces is given, look for reason why.
\n«clockwise moments = anticlockwise moments»
50 × 2cos 60 = NW × 4sin 60 ✔
\n«»
\nF = 14.4«N» ✔
\nmaximum friction force = «0.4 × 50N» = 20«N» ✔
14.4 < 20 AND so will not slip ✔
\nMany candidates stated that the resultant of all forces must be zero but failed to mention the fact that horizontal forces must balance in this particular question.
\nVery few candidates could take moments about any point and correct answers were rare both at SL and HL.
\nThe question about the slipping of the ladder was poorly answered. The fact that the normal reaction on the floor was 50N was not known to many.
\nAn object has a weight of 6.10 × 102 N. What is the change in gravitational potential energy of the object when it moves through 8.0 m vertically?
A. 5 kJ
B. 4.9 kJ
C. 4.88 kJ
D. 4.880 kJ
\nB
\nAt SL, more candidates chose C with B the second most popular response. This question was about significant figures and candidates should be reminded that on the multiple choice paper they are not expected to perform detailed calculations. In this case 6.10 (to 3 sig figs) times 8.0 (to 2 sig figs) produces an answer to 2 sig figs giving B as the correct response. All answers are equivalent from a numerical point of view with the difference being the number of sig figs used.
\nThe graph shows the variation of momentum with time for an object.
\nWhat net force acts on the object for the first 2.0 s and for the second 2.0 s of the motion?
\nA
\nA small metal pendulum bob of mass 75 g is suspended at rest from a fixed point with a length of thread of negligible mass. Air resistance is negligible. The bob is then displaced to the left.
\nAt time t = 0 the bob is moving horizontally to the right at 0.8 m s–1. It collides with a small stationary object also of mass 75 g. Both objects then move together with motion that is simple harmonic.
\nCalculate the speed of the combined masses immediately after the collision.
\nShow that the collision is inelastic.
\nDescribe the changes in gravitational potential energy of the oscillating system from t = 0 as it oscillates through one cycle of its motion.
\n0.40 «m s−1» ✔
\ninitial energy 24 mJ and final energy 12 mJ ✔
\nenergy is lost/unequal /change in energy is 12 mJ ✔
\ninelastic collisions occur when energy is lost ✔
maximum GPE at extremes, minimum in centre ✔
\nCandidates fell into some broad categories on this question. This was a “show that” question, so there was an expectation of a mathematical argument. Many were able to successfully show that the initial and final kinetic energies were different and connect this to the concept of inelastic collisions. Some candidates tried to connect conservation of momentum unsuccessfully, and some simply wrote an extended response about the nature of inelastic collisions and noted that the bobs stuck together without any calculations. This approach was awarded zero marks.
\nThis straightforward question had surprisingly poorly answers. Candidate answers tended to be overly vague, such as “as the bob went higher the GPE increased and as it fell the GPE decreased.” Candidates needed to specify when GPE would be at maximum and minimum values. Some candidates mistakenly assumed that at t=0 the pendulum bob was at maximum height despite being told otherwise in the question stem.
\nA sports car is accelerated from 0 to 100 km per hour in 3 s. What is the acceleration of the car?
\nA. 0.1 g
\nB. 0.3 g
\nC. 0.9 g
\nD. 3 g
\nC
\nResponse D was the most common (but incorrect) response, with candidates neglecting to convert km/h to m/s.
\nThe diagram shows the direction of a sound wave travelling in a metal sheet.
\nThe frequency of the sound wave in the metal is 250 Hz.
\nParticle P in the metal sheet performs simple harmonic oscillations. When the displacement of P is 3.2 μm the magnitude of its acceleration is 7.9 m s-2. Calculate the magnitude of the acceleration of P when its displacement is 2.3 μm.
\nThe wave is incident at point Q on the metal–air boundary. The wave makes an angle of 54° with the normal at Q. The speed of sound in the metal is 6010 m s–1 and the speed of sound in air is 340 m s–1. Calculate the angle between the normal at Q and the direction of the wave in air.
\nState the frequency of the wave in air.
\nDetermine the wavelength of the wave in air.
\nThe sound wave in air in (c) enters a pipe that is open at both ends. The diagram shows the displacement, at a particular time T, of the standing wave that is set up in the pipe.
\nOn the diagram, at time T, label with the letter C a point in the pipe that is at the centre of a compression.
\nExpression or statement showing acceleration is proportional to displacement ✔
so «» ✔
\n ✔
θ = 2.6° ✔
\nf = 250 «Hz» OR Same OR Unchanged ✔
» ✔
any point labelled C on the vertical line shown below ✔
eg:
This was well answered at both levels.
\nMany scored full marks on this question. Common errors were using the calculator in radian mode or getting the equation upside down.
\nMany used a ratio of the speeds to produce a new frequency of 14Hz (340 x 250/6010). It would have helped candidates if they had been aware that the command term ‘state’ means ‘give a specific name, value or other brief answer without explanation or calculation.’
\nThis was answered well at both levels.
\nAn object of mass moving at velocity collides with a stationary object of mass . The objects stick together after the collision. What is the final speed and the change in total kinetic energy immediately after the collision?
\nB
\nA small object is placed at a distance of 2.0 cm from the objective lens of an optical compound microscope in normal adjustment.
\nThe following data are available.
\nMagnification of the microscope = 70
Focal length of the eyepiece = 3.0 cm
Near point distance = 24 cm
State what is meant by normal adjustment when applied to a compound microscope.
\nCalculate, in cm, the distance between the eyepiece and the image formed by the objective lens.
\nDetermine, in cm, the focal length of the objective lens.
\n«the final» image is formed at the near point of the eye ✔
\n«image is virtual so» «cm» ✔
\n«» so «cm» ✔
\nAND ✔
\n✔
\nso ✔
\nNOTE: MP1 allow
\nA spaceship is travelling at , away from Earth. It launches a probe away from Earth, at relative to the spaceship. An observer on the probe measures the length of the probe to be .
\nAn object of mass is thrown downwards from a height of . The initial speed of the object is .
The object hits the ground at a speed of . Assume . What is the best estimate of the energy transferred from the object to the air as it falls?
A.
\nB.
\nC.
\nD.
\nCalculate the speed of the probe in terms of , relative to Earth.
\nB
\n✓
\n✓
\nAllow all negative signs for velocities
Award [2] marks for a bald correct answer
\n\n
A girl throws an object horizontally at time t = 0. Air resistance can be ignored. At t = 0.50 s the object travels horizontally a distance in metres while it falls vertically through a distance in metres.
\nWhat is the initial velocity of the object and the vertical distance fallen at t = 1.0 s?
\nD
\nThe correct response (D) was the most common selection by a minority of candidates, with incorrect responses being roughly equally distributed among the remaining options. This question has one of the highest discrimination indexes.
\nPlutonium-238 (Pu) decays by alpha (α) decay into uranium (U).
\nThe following data are available for binding energies per nucleon:
\nplutonium 7.568 MeV
\nuranium 7.600 MeV
\nalpha particle 7.074 MeV
\nThe energy in b(i) can be transferred into electrical energy to run the instruments of a spacecraft. A spacecraft carries 33 kg of pure plutonium-238 at launch. The decay constant of plutonium is 2.50 × 10−10 s−1.
\nSolar radiation falls onto a metallic surface carried by the spacecraft causing the emission of photoelectrons. The radiation has passed through a filter so it is monochromatic. The spacecraft is moving away from the Sun.
\nState what is meant by the binding energy of a nucleus.
\nDraw, on the axes, a graph to show the variation with nucleon number of the binding energy per nucleon, . Numbers are not required on the vertical axis.
\nIdentify, with a cross, on the graph in (a)(ii), the region of greatest stability.
\nSome unstable nuclei have many more neutrons than protons. Suggest the likely decay for these nuclei.
\nShow that the energy released in this decay is about 6 MeV.
\nThe plutonium nucleus is at rest when it decays.
\nCalculate the ratio .
\nEstimate the power, in kW, that is available from the plutonium at launch.
\nThe spacecraft will take 7.2 years (2.3 × 108 s) to reach a planet in the solar system. Estimate the power available to the spacecraft when it gets to the planet.
\nState and explain what happens to the kinetic energy of an emitted photoelectron.
\nState and explain what happens to the rate at which charge leaves the metallic surface.
\nthe energy needed to «completely» separate the nucleons of a nucleus
\nOR
\nthe energy released when a nucleus is assembled from its constituent nucleons ✓
\n\n
Accept reference to protons and neutrons.
\ncurve rising to a maximum between 50 and 100 ✓
\ncurve continued and decreasing ✓
\n\n
Ignore starting point.
Ignore maximum at alpha particle.
\nAt a point on the peak of their graph ✓
\nbeta minus «decay» ✓
\ncorrect mass numbers for uranium (234) and alpha (4) ✓
\n«MeV» ✓
\nenergy released 5.51 «MeV» ✓
\n\n
Ignore any negative sign.
\n«» OR ✓
\n«» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAccept for MP2.
\nnumber of nuclei present ✓
\ninitial activity is ✓
\npower is «kW» ✓
\n\n
Allow a final answer of 20 kW if 6 MeV used.
\nAllow ECF from MP1 and MP2.
\navailable power after time t is ✓
\n«kW» ✓
\n\n
MP1 may be implicit.
\nAllow ECF from (c)(i).
\nAllow 17.4 kW from unrounded power from (c)(i).
\nAllow 18.8 kW from 6 MeV.
\nstays the same ✓
\nas energy depends on the frequency of light ✓
\n\n
Allow reference to wavelength for MP2.
\nAward MP2 only to answers stating that KE decreases due to Doppler effect.
\ndecreases ✓
\nas number of photons incident decreases ✓
\nA sample of vegetable oil, initially in the liquid state, is placed in a freezer that transfers thermal energy from the sample at a constant rate. The graph shows how temperature of the sample varies with time .
\nThe following data are available.
\nMass of the sample
Specific latent heat of fusion of the oil
Rate of thermal energy transfer
Calculate the thermal energy transferred from the sample during the first minutes.
\nEstimate the specific heat capacity of the oil in its liquid phase. State an appropriate unit for your answer.
\nThe sample begins to freeze during the thermal energy transfer. Explain, in terms of the molecular model of matter, why the temperature of the sample remains constant during freezing.
\nCalculate the mass of the oil that remains unfrozen after minutes.
\n✓
\n\n
OR ✓
\nOR ✓
\n
Allow any appropriate unit that is
«intermolecular» bonds are formed during freezing ✓
\n
bond-forming process releases energy
OR
«intermolecular» PE decreases «and the difference is transferred as heat» ✓
«average random» KE of the molecules does not decrease/change ✓
temperature is related to «average» KE of the molecules «hence unchanged» ✓
\n
To award MP3 or MP4 molecules/particles/atoms must be mentioned.
\nmass of frozen oil ✓
\nunfrozen mass ✓
\nA waiter carrying a tray is accelerating to the right as shown in the image.
\nWhat is the free-body diagram of the forces acting on the tray?
\nD
\nResponse D was the most common response, with the free-body diagram in response A providing a significant distractor for roughly a third of candidates. Most candidates recognized that the only upward vector would be one perpendicular to the tray.
\nThe moment of inertia of a solid sphere is where m is the mass of the sphere and r is the radius.
\nShow that the total kinetic energy Ek of the sphere when it rolls, without slipping, at speed v is .
\n\n
A solid sphere of mass 1.5 kg is rolling, without slipping, on a horizontal surface with a speed of 0.50 m s-1. The sphere then rolls, without slipping, down a ramp to reach a horizontal surface that is 45 cm lower.
\nCalculate the speed of the sphere at the bottom of the ramp.
\nEk = Ek linear + Ek rotational
\nOR
\n✔
\n✔
\n«»
\n\n
Answer is given in the question so check working is correct at each stage.
\nInitial «=0.26J» ✔
\nFinal «=6.88J» ✔
\n«m s–1» ✔
\n\n
Other solution methods are possible.
\n\n
The derivation of the formula for the total kinetic energy of a rolling ball was well answered.
\nAlthough there were many correct answers, many candidates forgot to include the initial kinetic energy of the ball at the top of the ramp. The process followed to obtain the answer was too often poorly presented, candidates are encouraged to explain what is being calculated rather than just writing numbers.
\nAn object of mass is falling vertically through the air. The drag force acting on the object is . What is the best estimate of the acceleration of the object?
\nA. Zero
\nB.
\nC.
\nD.
\nB
\nThe Moon has no atmosphere and orbits the Earth. The diagram shows the Moon with rays of light from the Sun that are incident at 90° to the axis of rotation of the Moon.
\nA black body is on the Moon’s surface at point A. Show that the maximum temperature that this body can reach is 400 K. Assume that the Earth and the Moon are the same distance from the Sun.
\nAnother black body is on the Moon’s surface at point B.
\nOutline, without calculation, why the aximum temperature of the black body at point B is less than at point A.
\nThe albedo of the Earth’s atmosphere is 0.28. Outline why the maximum temperature of a black body on the Earth when the Sun is overhead is less than that at point A on the Moon.
\nOutline why a force acts on the Moon.
\nOutline why this force does no work on the Moon.
\nT = ✔
\n390 «K» ✔
\nMust see 1360 (from data booklet) used for MP1.
\nMust see at least 2 s.f.
\nenergy/Power/Intensity lower at B ✔
\nconnection made between energy/power/intensity and temperature of blackbody ✔
\n(28 %) of sun’s energy is scattered/reflected by earth’s atmosphere OR only 72 % of incident energy gets absorbed by blackbody ✔
\nMust be clear that the energy is being scattered by the atmosphere.
\nAward [0] for simple definition of “albedo”.
\ngravitational attraction/force/field «of the planet/Moon» ✔
\nDo not accept “gravity”.
\nthe force/field and the velocity/displacement are at 90° to each other OR there is no change in GPE of the moon ✔
\nAward [0] for any mention of no net force on the satellite.
\nDo not accept acceleration is perpendicular to velocity.
\nMany candidates struggled with this question. A significant portion attempted to apply Wein’s Law and simply stated that a particular wavelength was the peak and then used that to determine the temperature. Some did use the solar constant from the data booklet and were able to calculate the correct temperature. As part of their preparation for the exam candidates should thoroughly review the data booklet and be aware of what constants are given there. As with all “show that” questions candidates should be reminded to include an unrounded answer.
\nThis is question is another example of candidates not thinking beyond the obvious in the question. Many simply said that point B is farther away, or that it is at an angle. Some used vague terms like “the sunlight is more spread out” rather than using proper physics terms. Few candidates connected the lower intensity at B with the lower temperature of the blackbody.
\nThis question was assessing the understanding of the concept of albedo. Many candidates were able to connect that an albedo of 0.28 meant that 28 % of the incident energy from the sun was being reflected or scattered by the atmosphere before reaching the black body.
\nThis was generally well answered, although some candidates simply used the vague term “gravity” rather than specifying that it is a gravitational force or a gravitational field. Candidates need to be reminded about using proper physics terms and not more general, “every day” terms on the exam.
\nSome candidates connected the idea that the gravitational force is perpendicular to the velocity (and hence the displacement) for the mark. It was also allowed to discuss that there is no change in gravitational potential energy, so therefore no work was being done. It was not acceptable to simply state that the net displacement over one full orbit is zero. Unfortunately, some candidates suggested that there is no net force on the moon so there is no work done, or that the moon is so much smaller so no work could be done on it.
\nA glass block of refractive index 1.5 is immersed in a tank filled with a liquid of higher refractive index. Light is incident on the base of the glass block. Which is the correct diagram for rays incident on the glass block at an angle greater than the critical angle?
\nD
\nResponse D was the most common response, with response A providing a significant distractor for roughly a third of candidates unsure about refraction beyond the critical angle.
\nIn an experiment to determine the speed of sound in air, a tube that is open at the top is filled with water and a vibrating tuning fork is held over the tube as the water is released through a valve.
\nAn increase in intensity in the sound is heard for the first time when the air column length is . The next increase is heard when the air column length is .
\nWhich expressions are approximately correct for the wavelength of the sound?
\nI. 4
\nII. 4
\nIII.
\nA. I and II
\nB. I and III
\nC. II and III
\nD. I, II and III
\nB
\nThe question was well answered by students.
\nTwo parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively charged plate to the positively charged plate. The charge gains kinetic energy W. The distance between the plates is doubled and the potential difference between them is halved. What is the kinetic energy gained by an identical charge moving between these plates?
\nA.
\nB. W
\nC. 2W
\nD. 4W
\nA
\nThe correct response (A) was the most common from candidates, however a significant number of candidates appeared unsure of the impact of the distance between plates and (incorrectly) selected response B.
\nA resistor of resistance R is connected to a fully charged cell of negligible internal resistance. A constant power P is dissipated in the resistor and the cell discharges in time t. An identical cell is connected in series with two identical resistors each of resistance R.
\nWhat is the power dissipated in each resistor and the time taken to discharge the cell?
\nB
\nThis question was not well answered, with fewer than 25 % of candidates correctly selecting response B. Furthermore, the discrimination index for this question was remarkably low, suggesting this question would provide rich classroom discussion.
\nTwo currents of 3 A and 1 A are established in the same direction through two parallel straight wires R and S.
\nWhat is correct about the magnetic forces acting on each wire?
\nA. Both wires exert equal magnitude attractive forces on each other.
\nB. Both wires exert equal magnitude repulsive forces on each other.
\nC. Wire R exerts a larger magnitude attractive force on wire S.
\nD. Wire R exerts a larger magnitude repulsive force on wire S.
\nA
\nThis question had the lowest difficulty index on the HL paper, with roughly 10 % of candidates selecting response A. Responses C and D were roughly equally common candidate answers, with students not recognizing the applicability of Newton’s 3rd law.
\nCommunication signals are transmitted over long distances through optic fibres.
\nA signal is transmitted along an optic fibre with attenuation per unit length of 0.40 dB km–1. The signal must be amplified when the power of the signal has fallen to 0.02 % of the input power.
\nDescribe why a higher data transfer rate is possible in optic fibres than in twisted pair cables.
\nState one cause of attenuation in the optic fibre.
\nDetermine the distance at which the signal must be amplified.
\nfibres have broader bandwidth than cables ✔
therefore can carry multiple signals simultaneously ✔
absorption/scattering of light
OR
impurities in the «glass core of the» fibre ✔
attenuation = «dB» ✔
amplification required after or 93 «km» ✔
NOTE: Allow ECF from mp1 for wrong dB value.(eg: 42 km if % symbol ignored).
P and Q leave the same point, travelling in the same direction. The graphs show the variation with time of velocity for both P and Q.
\nWhat is the distance between P and Q when ?
\nA.
\nB.
\nC.
\nD.
\nB
\nA heat pump is modelled by the cycle A→B→C→A.
\nThe heat pump transfers thermal energy to the interior of a building during processes C→A and A→B and absorbs thermal energy from the environment during process B→C. The working substance is an ideal gas.
\nShow that the work done on the gas for the isothermal process C→A is approximately 440 J.
\nCalculate the change in internal energy of the gas for the process A→B.
\nCalculate the temperature at A if the temperature at B is −40°C.
\nDetermine, using the first law of thermodynamics, the total thermal energy transferred to the building during the processes C→A and A→B.
\nSuggest why this cycle is not a suitable model for a working heat pump.
\nevidence of work done equals area between AC and the Volume axis ✓
reasonable method to estimate area giving a value 425 to 450 J ✓
\nAnswer 440 J is given, check for valid working.
Examples of acceptable methods for MP2:
- estimates 17 to18 small squares x 25 J per square = 425 to 450 J.
- 250 J for area below BC plus a triangle of dimensions 5 × 3, 3 × 5, or 4 × 4 small square edges giving 250 J + 187.5 J or 250 J + 200 J.
Accurate integration value is 438 J - if method seen award [2].
\n«use of and to give»
\n✔
\n«»
\n=«–»375«J» ✔
\nAnother method is possible: eg realisation that ΔU for BC has same magnitude, so ΔU = 3/2 PΔV.
\nTA = 816«K» OR 543«°C»✔
\nfor CA ΔU = 0 so Q = W = −440 «J» ✔
for AB W = 0 so Q = ΔU = −375 «J» ✔
815 «J» transferred to the building ✔
\nMust use the first law of thermodynamics for MP1 and MP2.
\nthe temperature changes in the cycle are too large ✔
the cycle takes too long «because it contains an isothermal stage» ✔
energy/power output would be too small ✔
\nAt SL, Correct answers were rare and very few candidates used the fact the work done was area under the curve, and even fewer could estimate this area. At HL, the question was better answered. Candidates used a range of methods to estimate the area including counting the squares, approximating the area using geometrical shapes and on a few occasions using integral calculus.
\nNot very many candidates seem to know the generalised formula ΔU =1.5(P2V2 -P1V1) however many correct answers were seen.
\nThe temperature at A was found correctly by most candidates.
\nThe main problem here was deciding whether each Q was positive or negative. But the question was quite well answered.
\nBecause the question was about a heat pump rather than a heat engine very few answers were correct. Only a very small number of candidates mentioned the fact that the isothermal change would take an impracticably long time.
\nA horizontal electrical cable carries a steady current out of the page. The Earth’s magnetic field exerts a force on the cable.
\nWhich arrow shows the direction of the force on the cable due to the Earth’s magnetic field?
\nB
\nThe correct answer was well answered by candidates, with a relatively high discrimination index.
\nTwo loudspeakers, A and B, are driven in phase and with the same amplitude at a frequency of . Point P is located from A and from B. The speed of sound is .
\nDeduce that a minimum intensity of sound is heard at P.
\nA microphone moves along the line from P to Q. PQ is normal to the line midway between the loudspeakers.
\n\n
The intensity of sound is detected by the microphone. Predict the variation of detected intensity as the microphone moves from P to Q.
\nWhen both loudspeakers are operating, the intensity of sound recorded at Q is . Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to .
\nEstimate .
\nwavelength✓
path difference ✓
OR «half-wavelengths» ✓
waves meet in antiphase «at P»
OR
destructive interference/superposition «at P» ✓
\n
Allow approach where path length is calculated in terms of number of wavelengths; along path A () and
path B () for MP2, hence path difference wavelengths for MP3
«equally spaced» maxima and minima ✓
\na maximum at Q ✓
\nfour «additional» maxima «between P and Q» ✓
\nthe amplitude of sound at Q is halved ✓
«intensity is proportional to amplitude squared hence» ✓
Three forces act on a block which is sliding down a slope at constant speed. is the weight, is the reaction force at the surface of the block and is the friction force acting on the block.
\nIn this situation
\nA. there must be an unbalanced force down the plane.
\nB. .
\nC. .
\nD. the resultant force on the block is zero.
\nD
\nThe diagram shows the emission spectrum of an atom.
\nWhich of the following atomic energy level models can produce this spectrum?
\nA
\nWith a low difficulty index, most candidate responses were divided between (incorrect) responses C and D. Students appeared to select more familiar energy level diagrams rather than the diagram that best correlated with the emission spectrum given.
\nIn an experiment to measure the acceleration of free fall a student ties two different blocks of masses m1 and m2 to the ends of a string that passes over a frictionless pulley.
\nThe student calculates the acceleration a of the blocks by measuring the time taken by the heavier mass to fall through a given distance. Their theory predicts that and this can be re-arranged to give .
\nIn a particular experiment the student calculates that a = (0.204 ±0.002) ms–2 using m1 = (0.125 ±0.001) kg and m2 = (0.120 ±0.001) kg.
\nCalculate the percentage error in the measured value of g.
\nDeduce the value of g and its absolute uncertainty for this experiment.
\nThere is an advantage and a disadvantage in using two masses that are almost equal.
\nState and explain the advantage with reference to the magnitude of the acceleration that is obtained.
\nThere is an advantage and a disadvantage in using two masses that are almost equal.
\nState and explain the disadvantage with reference to your answer to (a)(ii).
\nerror in m1 + m2 is 1 % OR error in m1 − m2 is 40 % OR error in a is 1 % ✔
\nadds percentage errors ✔
\nso error in g is 42 % OR 40 % OR 41.8 % ✔
\nAllow answer 0.42 or 0.4 or 0.418.
\nAward [0] for comparing the average value with a known value, e.g. 9.81 m s-2.
\ng = 9.996 «m s−2» OR Δg = 4.20 «m s−2» ✔
\ng = (10 ± 4) «m s−2»
\nOR
\ng = (10.0 ± 4.2) «m s−2» ✔
\nAward [1] max for not proper significant digits or decimals use, such as: 9.996±4.178 or 10±4.2 or 10.0±4 or 10.0±4.18« m s−2 » .
\nthe acceleration would be small/the time of fall would be large ✔
\neasier to measure /a longer time of fall reduces the % error in the time of fall and «hence acceleration» ✔
\nDo not accept ideas related to the mass/moment of inertia of the pulley.
\nthe percentage error in the difference of the masses is large ✔
\nleading to a large percentage error/uncertainty in g/of the experiment ✔
\nDo not accept ideas related to the mass/moment of inertia of the pulley.
\nAtwoods machine a) is a quite straightforward question that tests the ability to propagate uncertainties through calculations. Almost all candidates proved the ability to add percentages or relative calculations, however, many weaker candidates failed in the percentage uncertainty when subtracting the two masses.
\nMany average candidates did not use the correct number of significant figures and wrote the answers inappropriately. Only the best candidates rounded out and wrote the proper answer of 10±4 ms−2. Some candidates did not propagate uncertainties and only compared the average calculated value with the known value 9.81 ms−2.
\nQ 1 b) was quite well answered. Only the weakest candidates presented difficulty in understanding simple mechanics.
\nIn part ii) many were able to appreciate that the resultant percentage error in “g” was relatively large however linking this with what caused the large uncertainty (that is, the high % error from the small difference in masses) proved more challenging.
\nA table-tennis ball of mass 3 g is fired with a speed of 10 m s-1 from a stationary toy gun of mass 0.600 kg. The gun and ball are an isolated system.
What are the recoil speed of the toy gun and the total momentum of the system immediately after the gun is fired?
\nA
\nThis question gives good discrimination at both levels with the correct response, A, being the most popular at HL. Response B was second most popular at HL and most popular by a small margin at SL, however a significant number of candidates chose the other responses at both levels. Realising the gun and ball are initially at rest and momentum must be conserved leads to a zero momentum after firing, immediately removing options B and D.
\nA balloon rises at a steady vertical velocity of . An object is dropped from the balloon at a height of above the ground. Air resistance is negligible. What is the time taken for the object to hit the ground?
\nA.
\nB.
\nC.
\nD.
\nC
\nEven though over half the candidates are choosing the correct response it has a low discrimination index. Many are choosing D indicating that they forgot to take the velocity upward as negative.
\nThe graph shows how current varies with potential difference across a component X.
\nComponent X and a cell of negligible internal resistance are placed in a circuit.
\nA variable resistor R is connected in series with component X. The ammeter reads .
\nComponent X and the cell are now placed in a potential divider circuit.
\n\n
Outline why component X is considered non-ohmic.
\nDetermine the resistance of the variable resistor.
\nCalculate the power dissipated in the circuit.
\nState the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.
\nDescribe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).
\ncurrent is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓
ALTERNATIVE 1
\nvoltage across X ✓
\nvoltage across R ✓
\nresistance of variable resistor ✓
\n\n
ALTERNATIVE 2
\noverall resistance ✓
\nresistance of X ✓
\nresistance of variable resistor ✓
\npower ✓
\nfrom to ✓
\nallows zero current through component X / potential divider arrangement ✓
\nprovides greater range «of current through component X» ✓
\nThe carbon isotope C is radioactive. It decays according to the equation
\nC → N + X + Y
\n\n
What are X and Y?
\nB
\nThis question was well answered by candidates, with a high discrimination index.
\nThe four pendulums shown have been cut from the same uniform sheet of board. They are attached to the ceiling with strings of equal length.
\nWhich pendulum has the shortest period?
\nD
\nCandidate answers were almost equally divided between responses B and D (correct). This question indirectly assesses experimental skills; how do we determine the effective length of a pendulum?
\nA block of weight W slides down a ramp at constant velocity. A friction force F acts between the bottom of the block and the surface of the ramp. A normal reaction N acts between the ramp and the block. What is the free-body diagram for the forces that act on the block?
\nD
\nEta Cassiopeiae A and B is a binary star system located in the constellation Cassiopeia.
\nDistinguish between a constellation and a stellar cluster.
\nstars in a cluster are gravitationally bound OR in constellation are not ✔
stars in a cluster are the same/similar age OR in constellation are not ✔
stars in a cluster are close in space/the same distance away OR in constellation are not ✔
stars in a cluster originate from same gas cloud OR in constellation do not ✔
stars in a cluster appear much closer in night sky than in a constellation ✔
NOTE: Take care to reward only 1 comment from a given marking point for MP1 to MP5.
\nThree identical light bulbs, X, Y and Z, each of resistance 4.0 Ω are connected to a cell of emf 12 V. The cell has negligible internal resistance.
\nThe switch S is initially open. Calculate the total power dissipated in the circuit.
\nThe switch is now closed. State, without calculation, why the current in the cell will increase.
\nThe switch is now closed. Deduce the ratio .
\n\n
\n
total resistance of circuit is 8.0 «Ω» ✔
P = =18 «W» ✔
\n\n
«a resistor is now connected in parallel» reducing the total resistance
OR
current through YZ unchanged and additional current flows through X ✔
\nevidence in calculation or statement that pd across Y/current in Y is the same as before ✔
so ratio is 1 ✔
\nMost candidates scored both marks. ECF was awarded for those who didn’t calculate the new resistance correctly. Candidates showing clearly that they were attempting to calculate the new total resistance helped examiners to award ECF marks.
\nMost recognised that this decreased the total resistance of the circuit. Answers scoring via the second alternative were rare as the statements were often far too vague.
\nVery few gained any credit for this at both levels. Most performed complicated calculations involving the total circuit and using 12V – they had not realised that the question refers to Y only.
\nAn object of mass strikes a vertical wall horizontally at speed . The object rebounds from the wall horizontally at speed .
\nWhat is the magnitude of the change in the momentum of the object?
\nA.
\nB.
\nC.
\nD.
\nD
\nIn an investigation a student folds paper into cylinders of the same diameter D but different heights. Beginning with the shortest cylinder they applied the same fixed load to each of the cylinders one by one. They recorded the height H of the first cylinder to collapse.
\nThey then repeat this process with cylinders of different diameters.
\nThe graph shows the data plotted by the student and the line of best fit.
\nTheory predicts that H = where c is a constant.
\nSuggest why the student’s data supports the theoretical prediction.
\nDetermine c. State an appropriate unit for c.
\nDetermine c. State an appropriate unit for c.
\nIdentify one factor that determines the value of c.
\ntheory «» predicts that H3 ∝ D2 ✔
\ngraph «of H3 vs D2 » is a straight line through the origin/graph of proportionality ✔
\nAllow gives H3 = c3D2 for MP1.
\nDo not award MP2 for “the graph is linear” without mention of origin.
\nevidence of gradient calculation to give gradient = 3.0 ✔
\nc3 = 3.0 ⇒ c = 1.4 ✔
\n\n
evidence of gradient calculation to give gradient = 3.0 ✔
\nc3 = 3.0 ⇒ c = 1.4 ✔
\n\n
the load/the thickness of paper/the type of paper/ the number of times the paper is rolled to form a cylinder ✔
\nLoad on a cylinder. The question was successful for candidates well prepared to write conclusions in their lab reports. Theory in the stem of the question predicts a directly proportional relationship between H and D2/3, which graphed are H3 and D2. Well prepared candidates were able to identify, that the theory predicts that H3 should be directly proportional to D2 and that this proportionality can be seen from the graph. Many candidates were able to mention that the relationship was linear and passed through the origin (as an alternative to proportional). However, a common response mentioned only linear or linear regression which is not sufficient to fully demonstrate proportionality.
\nPart b) was the most difficult part of section A, but still accessible. Many candidates only calculated the slope of the graph and did not realise that the third root of the slope is the constant c. Some students who were able to achieve the numerical value of c=1.4 struggled to establish the correct unit - perhaps lacking confidence or familiarity with the notion that a unit could be raised to a fractional index.
\nPart b) was the most difficult part of section A, but still accessible. Many candidates only calculated the slope of the graph and did not realise that the third root of the slope is the constant c. Some students who were able to achieve the numerical value of c=1.4 struggled to establish the correct unit - perhaps lacking confidence or familiarity with the notion that a unit could be raised to a fractional index.
\nIn c) most of the candidates well identified the load or the type of the paper as possible controlled variables. A common mistake here was answer discussing the height or the diameter of the cylinders.
\nA solid sphere is released from rest below the surface of a fluid and begins to fall.
\nDraw and label the forces acting on the sphere at the instant when it is released.
\nExplain why the sphere will reach a terminal speed.
\nThe weight of the sphere is 6.16 mN and the radius is 5.00 × 10-3 m. For a fluid of density 8.50 × 102 kg m-3, the terminal speed is found to be 0.280 m s-1. Calculate the viscosity of the fluid.
\nBoth forces must be suitably labeled.
\nDo not accept just ‘gravity’
\nAward [0] if a third force is shown.
\n«as the ball falls» there is a drag force ✔
when drag force+buoyant force/upthrust =«-» weight
OR
When net/resultant force =0 ✔
«terminal speed occurs»
\nOWTTE
Terminal speed is mentioned in the question, so no additional marks for reference to it.
\nany evidence (numerical or algebraic) of a realisation that
\n✔
\n«»
\n«Pas»✔
\n\n
The question was generally well answered but many candidates did not realise that the drag force would only be present when the ball starts moving.
\nMany candidates could explain correctly that the drag force would increase as the speed increases and that the weight would be balanced by the buoyant force and the drag force.
\nWhen the condition for forces in equilibrium was correctly formed, many candidates managed to obtain the correct answer. The working was often poorly presented making it difficult to mark or award marks for the process.
\nA substance changes from the solid phase to the gas phase without becoming a liquid and without a change in temperature.
What is true about the internal energy of the substance and the total intermolecular potential energy of the substance when this phase change occurs?
\nC
\nThis question has a low discrimination index at SL with more candidates choosing response D rather than the correct C. Candidates should remember that all information given in the question is important and the clue here is ‘without a change in temperature’. Thus the kinetic energy does not change so internal energy and potential energy will both have the same change and in addition energy must be provided to change the state of a solid.
\nIn a Young’s double-slit experiment, the distance between fringes is too small to be observed.
\nWhat change would increase the distance between fringes?
\nA. Increasing the frequency of light
\nB. Increasing the distance between slits
\nC. Increasing the distance from the slits to the screen
\nD. Increasing the distance between light source and slits
\nC
\nSatellite X is in orbit around the Earth. An identical satellite Y is in a higher orbit. What is correct for the total energy and the kinetic energy of the satellite Y compared with satellite X?
\nB
\nWe accept the comment from G2 forms that the wording of this question could be improved. The correct answer (B) considers the total and kinetic energies of satellite X the most popular answer.
\nA horizontal force acts on a sphere. A horizontal resistive force acts on the sphere where is the speed of the sphere and is a constant. What is the terminal velocity of the sphere?
\nA.
\nB.
\nC.
\nD.
\nD
\nThe escape speed from a planet of radius R is vesc. A satellite orbits the planet at a distance R from the surface of the planet. What is the orbital speed of the satellite?
\nA.
\nB.
\nC.
\nD.
\nA
\nThis had a very low discrimination index with the majority of candidates choosing B, followed by C. Response A, the correct answer, was third in popularity. The candidates missed that the satellite orbits at a distance of R from the surface of a planet of radius R so the total distance to be considered was 2R.
\nThe headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer whose eye has an aperture of 4.0 mm. The observer can just distinguish the headlights as separate images. What is the distance between the observer and the headlights?
\nA. 8 km
\nB. 10 km
\nC. 15 km
\nD. 20 km
\nB
\nResponse B was the most common (correct) response, with responses A and C as equally significant distractors.
\nA liquid of mass m and specific heat capacity c cools. The rate of change of the temperature of the liquid is k. What is the rate at which thermal energy is transferred from the liquid?
\nA.
\nB.
\nC.
\nD. kmc
\nD
\nA square loop of side 5.0 cm enters a region of uniform magnetic field at t = 0. The loop exits the region of magnetic field at t = 3.5 s. The magnetic field strength is 0.94 T and is directed into the plane of the paper. The magnetic field extends over a length 65 cm. The speed of the loop is constant.
\nShow that the speed of the loop is 20 cm s−1.
\nSketch, on the axes, a graph to show the variation with time of the magnetic flux linkage in the loop.
\nSketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.
\nThere are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.
\nThe resistance of the loop is 2.4 Ω. Calculate the magnitude of the magnetic force on the loop as it enters the region of magnetic field.
\nShow that the energy dissipated in the loop from t = 0 to t = 3.5 s is 0.13 J.
\nThe mass of the wire is 18 g. The specific heat capacity of copper is 385 J kg−1 K−1. Estimate the increase in temperature of the wire.
\n✓
\nshape as above ✓
\nshape as above ✓
\n\n
Vertical lines not necessary to score.
\nAllow ECF from (b)(i).
\nALTERNATIVE 1
\nmaximum flux at «» «Wb» ✓
\nemf = «» «V» ✓
\n
ALTERNATIVE 2
emf induced in one turn = BvL = «V» ✓
\nemf «V» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAllow ECF from MP1.
\nOR «A» ✓
\n«N» ✓
\n\n
Allow ECF from (c)(i).
\nAward [2] marks for a bald correct answer.
\nEnergy is being dissipated for 0.50 s ✓
\n
« J»
OR
\n« J» ✓
\n\n
Allow ECF from (b) and (c).
\nWatch for candidates who do not justify somehow the use of 0.5 s and just divide by 2 their answer.
\n✓
\n«K» ✓
\n\n
Allow [2] marks for a bald correct answer.
\nAward [1] for a POT error in MP1.
\nThe equation = constant is applied to a real gas where p is the pressure of the gas, V is its volume and T is its temperature.
What is correct about this equation?
A. It is empirical.
B. It is theoretical.
C. It cannot be tested.
D. It cannot be disproved.
\nA
\nCylinder X has a volume and contains 3.0 mol of an ideal gas. Cylinder Y has a volume and contains 2.0 mol of the same gas.
The gases in X and Y are at the same temperature . The containers are joined by a valve which is opened so that the temperatures do not change.
\nWhat is the change in pressure in X?
\nA.
\nB.
\nC.
\nD.
\nA
\nA transparent liquid film of refractive index 1.5 coats the outside of a glass lens of higher refractive index. The liquid film is used to eliminate reflection from the lens at wavelength λ in air.
\nWhat is the minimum thickness of the liquid film coating and the phase change at the liquid–glass interface?
\nD
\nHalf of candidates (incorrectly) selected response B, suggesting that while they recognized the phase change of π, determining the minimum thickness of the thin film was challenging. The discrimination index was very low for this question.
\nA railway track passes over a bridge that has a span of 20 m.
\nThe bridge is subject to a periodic force as a train crosses, this is caused by the weight of the train acting through the wheels as they pass the centre of the bridge.
\nThe wheels of the train are separated by 25 m.
\nThe graph shows the variation of the amplitude of vibration A of the bridge with driving frequency fD, when the damping of the bridge system is small.
\nShow that, when the speed of the train is 10 m s-1, the frequency of the periodic force is 0.4 Hz.
\nOutline, with reference to the curve, why it is unsafe to drive a train across the bridge at 30 m s-1 for this amount of damping.
\nThe damping of the bridge system can be varied. Draw, on the graph, a second curve when the damping is larger.
\ntime period
\nT = «» = 2.5 s AND f =
\nOR
\nevidence of f = ✔
\nAnswer 0.4 Hz is given, check correct working is shown.
\n30 m s–1 corresponds to f = 1.2 Hz ✔
the amplitude of vibration is a maximum for this speed
OR
corresponds to the resonant frequency ✔
\nsimilar shape with lower amplitude ✔
maximum shifted slightly to left of the original curve ✔
\nAmplitude must be lower than the original, but allow the amplitude to be equal at the extremes.
\nThe question was correctly answered by almost all candidates.
\nThe answers to this question were generally well presented and a correct argument was presented by almost all candidates. Resonance was often correctly referred to.
\nA correct curve, with lower amplitude and shifted left, was drawn by most candidates.
\nThe graph shows the variation of the displacement of a wave with distance along the wave.
The wave speed is 0.50 m s-1.
\nWhat is the period of the wave?
A. 0.33 s
B. 1.5 s
C. 3.0 s
D. 6.0 s
\nC
\nAn ideal gas of constant mass is heated in a container of constant volume.
\nWhat is the reason for the increase in pressure of the gas?
\nA. The average number of molecules per unit volume increases.
\nB. The average force per impact at the container wall increases.
\nC. Molecules collide with each other more frequently.
\nD. Molecules occupy a greater fractional volume of the container.
\nB
\nMany candidates chose option C which is a typical misconception that collision between molecules has something to do with pressure.
\nOne possible fission reaction of uranium-235 (U-235) is
\n\nMass of one atom of U-235
Binding energy per nucleon for U-235
Binding energy per nucleon for Xe-140
Binding energy per nucleon for Sr-94
A nuclear power station uses U-235 as fuel. Assume that every fission reaction of U-235 gives rise to of energy.
\nA sample of waste produced by the reactor contains of strontium-94 (Sr-94). Sr-94 is radioactive and undergoes beta-minus () decay into a daughter nuclide X. The reaction for this decay is
\n.
\n\n
The graph shows the variation with time of the mass of Sr-94 remaining in the sample.
\nState what is meant by binding energy of a nucleus.
\nOutline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units.
\nShow that the energy released in the reaction is about .
\nEstimate, in , the specific energy of U-235.
\nThe power station has a useful power output of and an efficiency of . Determine the mass of U-235 that undergoes fission in one day.
\nWrite down the proton number of nuclide X.
\nState the half-life of Sr-94.
\nCalculate the mass of Sr-94 remaining in the sample after minutes.
\nenergy required to «completely» separate the nucleons
OR
energy released when a nucleus is formed from its constituent nucleons ✓
Allow protons AND neutrons.
the values «in SI units» would be very small ✓
\nOR ✓
\nsee AND ✓
\n✓
\nenergy produced in one day ✓
\nmass ✓
\n✓
\n
Do not allow unless the proton number is indicated.
✓
\nALTERNATIVE 1
\n✓
\nmass remaining ✓
\n\n
ALTERNATIVE 2
\ndecay constant ✓
\nmass remaining ✓
\nA negative charge Q is to be moved within an electric field E, to equidistant points from its position, as shown.
\nWhich path requires the most work done?
\nD
\nThe most common answer was A, suggesting that students missed the prompt that Q is a negative charge.
\nThe escape velocity for an object at the surface of the Earth is vesc. The diameter of the Moon is 4 times smaller than that of the Earth and the mass of the Moon is 81 times smaller than that of the Earth. What is the escape velocity of the object on the Moon?
\nA. vesc
\nB. vesc
\nC. vesc
\nD. vesc
\nC
\nThis question was well answered by candidates.
\nA student places an object 5.0 cm from a converging lens of focal length 10.0 cm.
\nThe student mounts the same lens on a ruler and light from a distant object is incident on the lens.
\nConstruct rays, on the diagram, to locate the image of this object formed by the lens. Label this with the letter I.
\nDetermine, by calculation, the linear magnification produced in the above diagram.
\nSuggest an application for the lens used in this way.
\nIdentify, with a vertical line, the position of the focussed image. Label the position I.
\nThe image at I is the object for a second converging lens. This second lens forms a final image at infinity with an overall angular magnification for the two lens arrangement of 5. Calculate the distance between the two converging lenses.
\nA new object is placed a few meters to the left of the original lens. The student adjusts spacing of the lenses to form a virtual image at infinity of the new object. Outline, without calculation, the required change to the lens separation.
\nany two correct rays with extensions ✔
extensions converging to locate an upward virtual image labelled I with position within shaded region around focal point on diagram ✔
\nv = «–» 10«cm» ✔
\nM «= –=–» = «+» 2 ✔
\n\n
\n
magnifying glass
OR
Simple microscope
OR
eyepiece lens ✔
\nI labelled at 25 cm mark ✔
\nthe second lens has «cm» ✔
«so for telescope image to be at infinity»
the second lens is placed at 27 «cm»
OR
separation becomes 12 «cm» ✔
\nimage formed by 10 cm lens is greater than 10 cm/further to the right of the first lens ✔
so second lens must also move to the right OR lens separation increases ✔
\nAward [1 max] for bald “separation increases”.
The simple ray diagram was constructed well by most candidates, especially compared to previous years.
\nThe very simple calculation of magnification was done well by nearly everybody.
\nUsing a converging lens as a magnifying glass was the most common correct answer.
\nAnother very easy and well answered ray diagram question.
\nOnly candidates who realised that a simple telescope was being constructed were able to answer the question correctly. Most candidates realised that the focal lenses need to be added but few found the focal lens of the second lens correctly.
\nMany candidates did not read the question carefully and provided totally incorrect answers. It does not seem to be generally well known that if a distant object is moved to the right, for a converging lens, then the real image must also move to the right.
\nAn object at the end of a spring oscillates vertically with simple harmonic motion (shm). The graph shows the variation with time of the displacement of the object.
\nWhat is the velocity of the object?
\nA.
\nB.
\nC.
\nD.
\nB
\nA spherical soap bubble is made of a thin film of soapy water. The bubble has an internal air pressure and is formed in air of constant pressure . The theoretical prediction for the variation of is given by the equation
\n\nwhere is a constant for the thin film and is the radius of the bubble.
\nData for and were collected under controlled conditions and plotted as a graph showing the variation of with .
Suggest whether the data are consistent with the theoretical prediction.
\nShow that the value of is about 0.03.
\nIdentify the fundamental units of .
\nIn order to find the uncertainty for , a maximum gradient line would be drawn. On the graph, sketch the maximum gradient line for the data.
\nThe percentage uncertainty for is . State , with its absolute uncertainty.
\nThe expected value of is . Comment on your result.
\n«theory suggests» is proportional to ✓
\n
graph/line of best fit is straight/linear «so yes»
OR
graph/line of best fit passes through the origin «so yes» ✓
\n
MP1: Accept ‘linear’
MP2 do not award if there is any contradiction
eg: graph not proportional, does not pass through origin.
gradient «»
OR
use of equation with coordinates of a point ✓
✓
\n
MP1 allow gradients in range to
MP2 allow a range to for
\n\n
✓
\n\n
Accept
\n\n
straight line, gradient greater than line of best fit, and within the error bars ✓
\n« of » =
OR
« of » = ✓
rounds uncertainty to 1sf
OR
✓
\n
Allow ECF from (b)(i)
Award [2] marks for a bald correct answer
Experimental value matches this/correct, as expected value within the range ✓
OR
experimental value does not match/incorrect, as it is not within range ✓
A satellite in a circular orbit around the Earth needs to reduce its orbital radius.
\nWhat is the work done by the satellite rocket engine and the change in kinetic energy resulting from this shift in orbital height?
\nC
\nThis question was generally well answered, however a significant number of students (incorrectly) selected response A suggesting a lack of clarity around the work done as a result of changes in orbital height.
\nThe graph below shows the variation with time of the magnetic flux through a coil.
\nWhich of the following gives three times for which the magnitude of the induced emf is a maximum?
\nA. 0, ,
\nB. 0, , T
\nC. 0, , T
\nD. , ,
\nB
\nThis question was extremely well done, with the highest difficulty index seen on this paper.
\nOutline how the light spectra of distant galaxies are used to confirm hypotheses about the expansion of the universe.
\nLight from a hydrogen source in a laboratory on Earth contains a spectral line of wavelength 122 nm. Light from the same spectral line reaching Earth from a distant galaxy has a wavelength of 392 nm. Determine the ratio of the present size of the universe to the size of the universe when the light was emitted by the galaxy.
\nEstimate the age of the universe in seconds using the Hubble constant H0 = 70 km s–1 Mpc–1.
\nOutline why the estimate made in (b)(i) is unlikely to be the actual age of the universe.
\nspectra of galaxies are redshifted «compared to spectra on Earth» ✔
\nredshift/longer wavelength implies galaxies recede/ move away from us
OR
redshift is interpreted as cosmological expansion of space ✔
«hence universe expands»
\nNOTE: Universe expansion is given, so no mark for repeating this.
Do not accept answers based on CMB radiation.
ALTERNATIVE 1
\n✔
\n✔
\n\n
ALTERNATIVE 2
\n✔
\n=3.21 ✔
\n✔
\n✔
\nbecause estimate assumes the «present» constant rate of expansion ✔
it is unlikely that the expansion rate of the universe was ever constant ✔
there is uncertainty in the value of H0 ✔
NOTE: OWTTE
\nA mass on a spring is displaced from its equilibrium position. Which graph represents the variation of acceleration with displacement for the mass after it is released?
\nD
\nAn object of mass 2.0 kg rests on a rough surface. A person pushes the object in a straight line with a force of 10 N through a distance d.
\nThe resultant force acting on the object throughout d is 6.0 N.
\nWhat is the value of the sliding coefficient of friction between the surface and the object and what is the acceleration a of the object?
\nA
\nThere is no evidence that candidates were disadvantaged by the use of sliding friction rather than dynamic friction with the correct option being the most popular.
\nTwo capacitors of 3 μF and 6 μF are connected in series and charged using a 9 V battery.
\nWhat charge is stored on each capacitor?
\nA
\nResponse C was an effective distractor as the most common answer selected by candidates, suggesting that candidates are not fully understanding capacitor rules in series and parallel.
\nThe circuit diagram shows a capacitor that is charged by the battery after the switch is connected to terminal X. The cell has emf V and internal resistance r. After the switch is connected to terminal Y the capacitor discharges through the resistor of resistance R.
\nWhat is the nature of the current and magnitude of the initial current in the resistor after the switch is connected to terminal Y?
\nC
\nThe most common (incorrect) answer was response D, suggesting that candidates were uncertain of initial current magnitude while understanding that current would decrease upon discharge.
\nA metallic surface is first irradiated with infrared radiation and photoelectrons are emitted from the surface. The infrared radiation is replaced by ultraviolet radiation of the same intensity.
\nWhat will be the change in the kinetic energy of the photoelectrons and the rate at which they are ejected?
\nA
\nWith a low difficulty index, fewer than 15 % of candidates correctly selected response A. The large majority of candidates selected response C. These candidates likely did not recognize that since intensity stays constant, there must be fewer ultraviolet photons ejected for the power per unit area to remain constant. The discrimination index was very low for this question.
\nUnpolarized light is incident on two polarizers. The axes of polarization of both polarizers are initially parallel. The second polarizer is then rotated through 360° as shown.
\nWhich graph shows the variation of intensity with angle θ for the light leaving the second polarizer?
\nB
\nThe half-life of a radioactive nuclide is 8.0 s. The initial activity of a pure sample of the nuclide is 10 000 Bq. What is the approximate activity of the sample after 4.0 s?
\nA. 2500 Bq
\nB. 5000 Bq
\nC. 7100 Bq
\nD. 7500 Bq
\nC
\nRoughly half of candidates (incorrectly) selected response D, without recognizing that the change in activity over time is not linear.
\nA rocket has just been launched vertically from Earth. The image shows the free-body diagram of the rocket. F1 represents a larger force than F2.
\nWhich force pairs with F1 and which force pairs with F2, according to Newton’s third law?
\nB
\nAn object is pushed from rest by a constant net force of 100 N. When the object has travelled 2.0 m the object has reached a velocity of 10 m s−1.
\nWhat is the mass of the object?
\nA. 2 kg
\nB. 4 kg
\nC. 40 kg
\nD. 200 kg
\nB
\nA particle is confined within a nucleus. What is the order of magnitude of the uncertainty in the momentum of the particle?
\nA. 10–10 N s
\nB. 10–15 N s
\nC. 10–20 N s
\nD. 10–25 N s
\nC
\nResponse B was an effective distractor for over a third of candidates.
A particle with charge −2.5 × 10−6 C moves from point X to point Y due to a uniform electrostatic field. The diagram shows some equipotential lines of the field.
\nWhat is correct about the motion of the particle from X to Y and the magnitude of the work done by the field on the particle?
\nD
\nTwo blocks of different masses are released from identical springs of elastic constant k = 100 Nm−1, initially compressed a distance Δx = 0.1 m. Block X has a mass of 1 kg and block Y has a mass of 0.25 kg.
\nWhat are the velocities of the blocks when they leave the springs?
\nC
\nMost candidates chose the correct answer confirming this was not problematic.
\nThe particle omega minus () decays at rest into a neutral pion () and the xi baryon () according to
\n\n
The pion momentum is 289.7 MeV c–1.
\nThe rest masses of the particles are:
\n: 1672 MeV c–2
\n: 135.0 MeV c–2
\n: 1321 MeV c–2
\nShow that energy is conserved in this decay.
\nCalculate the speed of the pion.
\nmomentum of xi baryon is also 289.7«MeVc−1» ✔
\ntotal energy of xi baryon and pion is «MeV» ✔
\nwhich equals the rest energy of the omega ✔
\nAllow a backwards argument, assuming the energy is equal.
\n✔
\n✔
\nAward [2] for bald correct answer.
\nEnergy conservation in a decay. This relativistic mechanics question was very challenging. Only a few candidates were able to calculate the total energy of the baryon and pion, despite being able to recognize the correct momentum of the baryon.
\nEnergy conservation in a decay. This relativistic mechanics question was very challenging. Only a few candidates were able to calculate the total energy of the baryon and pion, despite being able to recognize the correct momentum of the baryon. The same for the speed of the pion in b), most of the attempts demonstrated by candidates were not relevant.
\nA Σ+ particle decays from rest into a neutron and another particle X according to the reaction
\nΣ+ → n + X
\nThe following data are available.
\nRest mass of Σ+ = 1190 MeV c–2
Momentum of neutron = 185 MeV c–1
Calculate, for the neutron,
\nthe total energy.
\nthe speed.
\nDetermine the rest mass of X.
\nneutron energy «MeV» ✔
NOTE: Allow 1.5 × 10–10 «J»
\nALTERNATIVE 1
\n«use of »
\n« so» ✔
\nv = 0.193c ✔
\n\n
ALTERNATIVE 2
\n«use of »
\n✔
\nv = 0.193c ✔
\n\n
ALTERNATIVE 3
\n«use of »
\n✔
\n✔
\n\n
NOTE: Allow v = 5.8 × 107 «ms–1»
\nmomentum of X = 185 «MeV c–1»✔
\nenergy of X = 1190 – 958 = 232 «MeV»✔
\n«MeV c–2» ✔
\nNOTE: Allow mass in kg - gives 2.5 × 10–28 «kg»
\nA company delivers packages to customers using a small unmanned aircraft. Rotating horizontal blades exert a force on the surrounding air. The air above the aircraft is initially stationary.
\nThe air is propelled vertically downwards with speed . The aircraft hovers motionless above the ground. A package is suspended from the aircraft on a string. The mass of the aircraft is and the combined mass of the package and string is . The mass of air pushed downwards by the blades in one second is .
\nState the value of the resultant force on the aircraft when hovering.
\nOutline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.
\nDetermine . State your answer to an appropriate number of significant figures.
\nCalculate the power transferred to the air by the aircraft.
\nThe package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft.
\nzero ✓
\nBlades exert a downward force on the air ✓
\n
air exerts an equal and opposite force on the blades «by Newton’s third law»
OR
air exerts a reaction force on the blades «by Newton’s third law» ✓
Downward direction required for MP1.
«lift force/change of momentum in one second» ✓
\n✓
\nAND answer expressed to sf only ✓
\n
Allow from .
ALTERNATIVE 1
\npower ✓
\n✓
\n\n
ALTERNATIVE 2
\nPower ✓
\n✓
\nvertical force = lift force – weight OR OR ✓
\nacceleration ✓
\nThis was generally answered well with the most common incorrect answer being the weight of the aircraft and package. The question uses the command term 'state' which indicates that the answer requires no working.
\nThe question required candidates to apply Newton's third law to a specific situation. Candidates who had learned the 'action and reaction' version of Newton's third law generally did less well than those who had learned a version describing 'object A exerting a force on object B' etc. Some answers lacked detail of what was exerting the force and in which direction.
\nThis was answered well with many getting full marks. A small number gave the wrong number of significant figures and some attempted to answer using kinematics equations or kinetic energy.
\nHL only. It was common to see answers that neglected to average the velocity and consequently arrived at an answer twice the size of the correct one. This was awarded 1 of the 2 marks.
\nWell done by a good number of candidates. Many earned a mark by simply using the correct mass to find an acceleration even though the force was incorrect.
\nWhich is a correct unit for gravitational potential?
\nA. m2 s−2
\nB. J kg
\nC. m s−2
\nD. N m−1 kg−1
\nA
\nTitan is a moon of Saturn. The Titan-Sun distance is 9.3 times greater than the Earth-Sun distance.
\nThe molar mass of nitrogen is 28 g mol−1.
\nShow that the intensity of the solar radiation at the location of Titan is 16 W m−2.
\nTitan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m−2.
\nShow that the equilibrium surface temperature of Titan is about 90 K.
\nThe mass of Titan is 0.025 times the mass of the Earth and its radius is 0.404 times the radius of the Earth. The escape speed from Earth is 11.2 km s−1. Show that the escape speed from Titan is 2.8 km s−1.
\nThe orbital radius of Titan around Saturn is and the period of revolution is .
\nShow that where is the mass of Saturn.
\nThe orbital radius of Titan around Saturn is 1.2 × 109 m and the orbital period is 15.9 days. Estimate the mass of Saturn.
\nShow that the mass of a nitrogen molecule is 4.7 × 10−26 kg.
\nEstimate the root mean square speed of nitrogen molecules in the Titan atmosphere. Assume an atmosphere temperature of 90 K.
\nDiscuss, by reference to the answer in (b), whether it is likely that Titan will lose its atmosphere of nitrogen.
\nincident intensity OR «W m−2» ✓
\n\n
Allow the use of 1400 for the solar constant.
\nexposed surface is ¼ of the total surface ✓
\nabsorbed intensity = (1−0.22) × incident intensity ✓
\n0.78 × 0.25 × 15.7 OR 3.07 «W m−2» ✓
\n\n
Allow 3.06 from rounding and 3.12 if they use 16 W m−2.
\nσT 4 = 3.07
\nOR
\nT = 86 «K» ✓
\nOR
\n«km s−1» ✓
\ncorrect equating of gravitational force / acceleration to centripetal force / acceleration ✓
\ncorrect rearrangement to reach the expression given ✓
\n\n
Allow use of for MP1.
\n«s» ✓
\n«kg» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAllow ECF from MP1.
\nOR
\n«kg» ✓
\n✓
\n«ms−1» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAllow 282 from a rounded mass.
\nno, molecular speeds much less than escape speed ✓
\n\n
Allow ECF from incorrect (d)(ii).
\nA student studies the relationship between the centripetal force applied to an object undergoing circular motion and its period .
\nThe object (mass ) is attached by a light inextensible string, through a tube, to a weight which hangs vertically. The string is free to move through the tube. A student swings the mass in a horizontal, circular path, adjusting the period of the motion until the radius is constant. The radius of the circle and the mass of the object are measured and remain constant for the entire experiment.
\n© International Baccalaureate Organization 2020.
\nThe student collects the measurements of five times, for weight . The weight is then doubled () and the data collection repeated. Then it is repeated with and . The results are expected to support the relationship
\n\nIn reality, there is friction in the system, so in this case is less than the total centripetal force in the system. A suitable graph is plotted to determine the value of experimentally. The value of was also calculated directly from the measured values of and .
\nState why the experiment is repeated with different values of .
\nPredict from the equation whether the value of found experimentally will be larger, the same or smaller than the value of calculated directly.
\nThe measurements of were collected five times. Explain how repeated measurements of reduced the random error in the final experimental value of .
\nOutline why repeated measurements of would not reduce any systematic error in .
\nIn order to draw a graph « of versus »
OR
to confirm proportionality between « and »
OR
to confirm relationship between « and »
OR
because W is the independent variable in the experiment ✓
\n\n
OWTTE
\nALTERNATIVE 1
\n\n
OR
\ncentripetal force is larger «than » / is smaller «than centripetal» ✓
\n«so» experimental is smaller «than calculated value» ✓
\n\n
ALTERNATIVE 2 (refers to graph)
reference to «friction force is» a systematic error «and does not affect gradient» ✓
\n«so» is the same ✓
\n\n
MP2 awarded only with correct justification.
Candidates can gain zero, MP1 alone or full marks.
OWTTE
\nmention of mean/average value «of » ✓
\nthis reduces uncertainty in / result
OR
more accurate/precise ✓
\n
Reference to “random errors average out” scores MP1
\nAccept “closer to true value”, “more reliable value” OWTTE for MP2
\n\n
systematic errors «usually» constant/always present/ not influenced by repetition ✓
\n\n
OWTTE
\nMost candidates scored. Different wording was used to express the aim of confirming the relationship.
\nMost successful candidates chose to consider a single point then concluding that the calculated mr would be smaller than the real value as W < centripetal force, or even went into analysing the dependence of the frictional force with W. Many were able to deduce this. Some candidates thought that a graph would still have the same gradient (if friction was constant) and mentioned systematic error, so mr was not changed which was also accepted.
\nMost candidates stated that the mean of 5 values of T was used to obtain an answer closer to the true value if there were no systematic errors. Some just repeated the question.
\nUsually very well answered acknowledging that systematic errors are constant and present throughout all measurements.
\nA black hole has a Schwarzschild radius R. A probe at a distance of 0.5R from the event horizon of the black hole emits radio waves of frequency that are received by an observer very far from the black hole.
\nExplain why the frequency of the radio waves detected by the observer is lower than .
\nThe probe emits 20 short pulses of these radio waves every minute, according to a clock in the probe. Calculate the time between pulses as measured by the observer.
\nALTERNATIVE 1
as the photons move away from the black hole, they lose energy in the gravitational field ✔
since «the detected frequency is lower than the emitted frequency» ✔
\n
ALTERNATIVE 2
if the observer was accelerating away from the probe, radio waves would undergo Doppler shift towards lower frequency ✔
by the equivalence principle, the gravitational field has the same effect as acceleration ✔
\n
ALTERNATIVE 3
due to gravitational time dilation, time between arrivals of wavefronts is greater for the observer ✔
since , «the detected frequency is lower than the emitted frequency» ✔
\n
NOTE: The question states that received frequency is lower so take care not to award a mark for simply re-stating this, a valid explanation must be given.
\ntime between pulses = 3 s according to the probe ✔
\n«s» ✔
\nA quantity of an ideal gas is at a temperature T in a cylinder with a movable piston that traps a length L of the gas. The piston is moved so that the length of the trapped gas is reduced to and the pressure of the gas doubles.
\nWhat is the temperature of the gas at the end of the change?
\n
A.
B.
C.
D.
C
\nSome comments queried that the Laws of Thermodynamics are not on the syllabus. This question was set as a test of Thermal Physics, topic 3, with option A coming from Mechanics, topic 2, not Thermodynamics.
\nA planet is in a circular orbit around a star. The speed of the planet is constant.
\nExplain why a centripetal force is needed for the planet to be in a circular orbit.
\nState the nature of this centripetal force.
\nDetermine the gravitational field of the planet.
\nThe following data are given:
\nMass of planet kg
Radius of the planet m.
«circular motion» involves a changing velocity ✓
\n«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓
\nthere must be a force/acceleration towards centre/star ✓
\nwithout a centripetal force the planet will move in a straight line ✓
\ngravitational force/force of gravity ✓
\nuse of ✓
\n6.4 «Nkg−1 or ms−2» ✓
\nA planet has radius R. The escape speed from the surface of the planet is v. At what distance from the surface of the planet is the orbital speed 0.5v?
\nA. 0.5R
\nB. R
\nC. 2R
\nD. 4R
\nB
\nWhat is true for an ideal gas?
\n
A. nRT = NkBT
B. nRT = kBT
\nC. RT = NkBT
\nD. RT = kBT
\nA
\nA conducting ring encloses an area of 2.0 cm2 and is perpendicular to a magnetic field of strength 5.0 mT. The direction of the magnetic field is reversed in a time 4.0 s. What is the average emf induced in the ring?
\nA. 0
\nB. 0.25 μV
\nC. 0.40 μV
\nD. 0.50 μV
\nD
\nTwo parallel current-carrying wires have equal currents in the same direction. There is an attractive force between the wires.
\nMaxwell’s equations led to the constancy of the speed of light. Identify what Maxwell’s equations describe.
\nState a postulate that is the same for both special relativity and Galilean relativity.
\nIdentify the nature of the attractive force recorded by an observer stationary with respect to the wires.
\nA second observer moves at the drift velocity of the electron current in the wires. Discuss how this observer accounts for the force between the wires.
\nmention of electric AND magnetic fields ✓
OR
mention of electromagnetic radiation/wave/fields ✓
the laws of physics are the same in all «inertial» frames of reference/for all «inertial» observers ✓
\n\n
OWTTE
\nmagnetic ✓
\n«In observer frame» protons «in the two wires» move in same/parallel direction ✓
\nthese moving protons produce magnetic attraction ✓
\nthere is also a smaller electrostatic repulsion due to wires appearing positive due to length contraction «of proton spacing» ✓
\n\n
OWTTE
\nEasy introduction fairly well answered by most candidates.
\nWith a few exceptions referring to Newton's first, this was very well answered.
\nMost scored by recognizing the force as magnetic.
\nMany failed to recognize that the magnetic force would still be present due to the current produced by the relative motion of the protons in both wires, and only focused on the repulsive electrostatic due to length contraction.
\nA box is in free fall in a uniform gravitational field. Observer X is at rest inside the box. Observer Y is at rest relative to the gravitational field. A light source inside the box emits a light ray that is initially parallel to the floor of the box according to both observers.
\nState the equivalence principle.
\nState and explain the path of the light ray according to observer X.
\nState and explain the path of the light ray according to observer Y.
\na freely falling frame in a gravitational field is equivalent to an inertial frame
OR
a frame accelerating in free space is equivalent to a frame at rest in a gravitational field ✔
X is in an inertial frame ✔
\nso light will follow a straight line path «parallel to the floor of the box» ✔
\nALTERNATIVE 1
light must hit right wall of box at same place as determined by X ✔
«but box is accelerating» so path must be curved downward ✔
\nALTERNATIVE 2
light is affected by gravity «for the observer at rest to the ground» ✔
so the path is curved downward/toward the ground ✔
\nFreefall in a gravitational field. Most of the candidates answered well.
\nFreefall in a gravitational field. Most of the candidates answered well but in b) omitted to mention that X is inertial.
\nQuite a high number of candidates, in b ii), stated incorrectly, that the light ray will be curved upward.
\nThe conservation of which quantity explains Lenz’s law?
\nA. Charge
\nB. Energy
\nC. Magnetic field
\nD. Mass
\nB
\nWhich assumption is part of the molecular kinetic model of ideal gases?
\n
A. The work done on a system equals the change in kinetic energy of the system.
B. The volume of a gas results from adding the volume of the individual molecules.
\nC. A gas is made up of tiny identical particles in constant random motion.
\nD. All particles in a gas have kinetic and potential energy.
\nC
\nA Pitot tube shown in the diagram is used to determine the speed of air flowing steadily in a horizontal wind tunnel. The narrow tube between points A and B is filled with a liquid. At point B the speed of the air is zero.
\nExplain why the levels of the liquid are at different heights.
\nThe density of the liquid in the tube is 8.7 × 102 kg m–3 and the density of air is 1.2 kg m–3. The difference in the level of the liquid is 6.0 cm. Determine the speed of air at A.
\nair speed at A greater than at B/speed at B is zero
OR
total/stagnation pressure «PB» – static pressure «PA» = dynamic pressure ✔
so PA is less than at PB (or vice versa) «by Bernoulli effect» ✔
\nheight of the liquid column is related to «dynamic» pressure difference «hence lower height in arm B» ✔
\n«»
difference in pressure «Pa» ✔
correct substitution into the Bernoulli equation, eg: ✔
«ms–1» ✔
\nGasoline of density 720 kg m–3 flows in a pipe of constant diameter.
\nState one condition that must be satisfied for the Bernoulli equation
\nρv2 + ρgz + ρ = constant
\nto apply
\nOutline why the speed of the gasoline at X is the same as that at Y.
\nCalculate the difference in pressure between X and Y.
\nThe diameter at Y is made smaller than that at X. Explain why the pressure difference between X and Y will increase.
\nflow must be laminar/steady/not turbulent ✔
\nfluid must be incompressible/have constant density ✔
\nfluid must be non viscous ✔
\n«continuity equation says» Av = constant «and the areas are the same» ✔
\nBernoulli: « + 0 + Px = + pgH + Py » gives Px − Py = pgH ✔
\nPx − Py = 720 × 9.81 × 1.2 = 8.5 «kPa» ✔
\nAward [2] for bald correct answer.
\nWatch for POT mistakes.
\nthe fluid speed at Y will be greater «than that at X» ✔
\nreducing the pressure at Y
OR
the formula used to show that the difference is increased ✔
Gasoline in a pipe. In a), most of the candidates well noted that for the Bernoulli equation, the fluid must be” non-viscous”, some noted, “laminar” and a few, “incompressible”. Some students stated vaguer and less concrete responses such as “the fluid must be ideal”.
\nIn b) most candidates well noted and understood the application of the continuity equation.
\nIn b) most candidates well noted and understood the application of the continuity equation and successfully went on to correctly calculate the pressure difference.
\nSub-question iii) well discriminated between the better and weaker candidates. As weaker candidates often wrote that “lower diameter means higher pressure” without a direct reference to the greater speed at Y implying reduced pressure.
\nA resistor designed for use in a direct current (dc) circuit is labelled “50 W, 2 Ω”. The resistor is connected in series with an alternating current (ac) power supply of peak potential difference 10 V. What is the average power dissipated by the resistor in the ac circuit?
\nA. 25 W
\nB. 35 W
\nC. 50 W
\nD. 100 W
\nA
\nSystem X is at a temperature of 40 °C. Thermal energy is provided to system X until it reaches a temperature of 50 °C. System Y is at a temperature of 283 K. Thermal energy is provided to system Y until it reaches a temperature of 293 K.
\nWhat is the difference in the thermal energy provided to both systems?
\nA. Zero
\nB. Larger for X
\nC. Larger for Y
\nD. Cannot be determined with the data given
\nD
\nThis question gives good discrimination although slightly more candidates chose option A instead of the correct option D. It is unusual that the correct response is 'cannot be determined' but the lack of mass or specific heat capacity in the data should have alerted candidates that they were not able to work out or compare how much thermal energy was supplied.
\nA capacitor of capacitance X is connected to a power supply of voltage V. At time t = 0, the capacitor is disconnected from the supply and discharged through a resistor of resistance R. What is the variation with time of the charge on the capacitor?
\n\n\n\n\nD
\nA particle is moving in a straight line with an acceleration proportional to its displacement and opposite to its direction. What are the velocity and the acceleration of the particle when it is at its maximum displacement?
\nD
\nWhich is a vector quantity?
\nA. Acceleration
\nB. Energy
\nC. Pressure
\nD. Speed
\nA
\nWhat is a consequence of the uncertainty principle?
\nA. The absorption spectrum of hydrogen atoms is discrete.
\nB. Electrons in low energy states have short lifetimes.
\nC. Electrons cannot exist within nuclei.
\nD. Photons do not have momentum.
\nC
\nThree statements about electromagnetic waves are:
\nI. They can be polarized.
II. They can be produced by accelerating electric charges.
III. They must travel at the same velocity in all media.
Which combination of statements is true?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nA ball of mass (50 ± 1) g is moving with a speed of (25 ± 1) m s−1. What is the fractional uncertainty in the momentum of the ball?
\n
A. 0.02
B. 0.04
\nC. 0.06
\nD. 0.08
\nC
\nA wave travels along a string. Graph M shows the variation with time of the displacement of a point X on the string. Graph N shows the variation with distance of the displacement of the string. PQ and RS are marked on the graphs.
\nWhat is the speed of the wave?
\nA.
B.
C.
D.
C
\nA pendulum bob is displaced until its centre is 30 mm above its rest position and then released. The motion of the pendulum is lightly damped.
\nDescribe what is meant by damped motion.
\nAfter one complete oscillation, the height of the pendulum bob above the rest position has decreased to 28 mm. Calculate the Q factor.
\nThe point of suspension now vibrates horizontally with small amplitude and frequency 0.80 Hz, which is the natural frequency of the pendulum. The amount of damping is unchanged.
\nWhen the pendulum oscillates with a constant amplitude the energy stored in the system is 20 mJ. Calculate the average power, in W, delivered to the pendulum by the driving force.
\na situation in which a resistive force opposes the motion
OR
amplitude/energy decreases with time ✔
✔
\n✔
\npower added = «W» ✔
\nIn a photoelectric effect experiment, a beam of light is incident on a metallic surface W in a vacuum.
\nThe graph shows how the current varies with the potential difference V when three different beams X, Y, and Z are incident on W at different times.
\n I. X and Y have the same frequency.
II. Y and Z have different intensity.
III. Y and Z have the same frequency.
Which statements are correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nThe refractive index of glass is and the refractive index of water is . What is the critical angle for light travelling from glass to water?
\nA.
B.
C.
D.
D
\nThe graph shows the variation with time t of the velocity of an object.
\nWhat is the variation with time t of the acceleration of the object?
A
\nThe graphs show the variation with time of the activity and the number of remaining nuclei for a sample of a radioactive nuclide.
\nWhat is the decay constant of the nuclide?
\nA.
\nB.
\nC.
\nD.
\nA
\nThe graph shows the variation with time t of the total energy E of a damped oscillating system.
\nThe Q factor for the system is 25. Determine the period of oscillation for this system.
\nAnother system has the same initial total energy and period as that in (a) but its Q factor is greater than 25. Without any calculations, draw on the graph, the variation with time of the total energy of this system.
\nALTERNATIVE 1
\n« ✔
\n«mJ» ✔
\nreading off the graph, period is 0.48 «s» ✔
\nAllow correct use of any value of E0, not only at the time = 0.
\nAllow answer from interval 0.42−0.55 s
\nALTERNATIVE 2
\nuse of ✔
\nenergy stored = 12 «mJ» AND power loss = 5.6 «mJ/s»✔
\n«f = 1.86 s so» period is 0.54 «s» ✔
\nAllow answer from interval 0.42−0.55 s.
\nAward [3] for bald correct answer.
\nsimilar shape graph starting at 12 mJ and above the original ✔
\nQ factor. Most of the candidates attempted to find the period of the damped system by using the correct formula.
\nMany thus went on to establish the correct period within the range given. Some candidates made POT errors not recognizing or identifying the unit used in this question.
\nA ball is thrown vertically downwards with an initial speed of 4.0 m s−1. The ball hits the ground with a speed of 16 m s−1. Air resistance is negligible. What is the time of fall and what is the distance travelled by the ball?
\nD
\nOutline how ultrasound, in a medical context, is produced.
\nSuggest the advantage in medical diagnosis of using ultrasound of frequency 1 MHz rather than 0.1 MHz.
\nUltrasound can be used to measure the dimensions of a blood vessel. Suggest why a B scan is preferable to an A scan for this application.
\nmention of AC voltage OR to piezo-electric crystal ✔
\ncrystal vibrates «at its resonant frequency» ✔
\n1 MHz waves have shorter wavelength than 0.1 MHz ✔
\ncan probe smaller size areas of organs/have higher resolution ✔
\na B scan is a computer generated combination of a large number of A scans ✔
\nallowing a measurement in different directions/two dimensional image ✔
\nProduction of ultrasound. This is not well known by many candidates. Quite a high number wrote generally correct information or statements about ultrasound, but these were not related or well considered responses directed to the question, for example, “use of gel in ultrasound medical applications”.
\nIn (b) the higher resolution was mentioned by only the best candidates. Some candidates mentioned the notion of “smaller penetration at higher frequencies” – correct, though again not well related to what the question was asking.
\nPart c) was also difficult to answer for most of the candidates.
\nUnpolarized light with an intensity of 320 W m−2 goes through a polarizer and an analyser, originally aligned parallel.
\nThe analyser is rotated through an angle θ = 30°. Cos 30° = .
\nWhat is the intensity of the light emerging from the analyser?
\nA. 120 W m−2
\nB. W m−2
\nC. 240 W m−2
\nD. W m−2
\nA
\nThis question proved a little challenging for SL candidates with many choosing incorrect answers especially C or D instead of correct A. At HL option C also was a popular option although a high discrimination index shows that the question discriminated well between the candidates. It appears that candidates are forgetting that 50% of the intensity is lost when unpolarised light passes through a polarizer, and then more is lost at the analyser according to Malus' Law.
\nWhat was a reason to postulate the existence of neutrinos?
\nA. Nuclear energy levels had a continuous spectrum.
\nB. The photon emission spectrum only contained specific wavelengths.
\nC. Some particles were indistinguishable from their antiparticle.
\nD. The energy of emitted beta particles had a continuous spectrum.
\nD
\nA vertical solid cylinder of uniform cross-sectional area floats in water. The cylinder is partially submerged. When the cylinder floats at rest, a mark is aligned with the water surface. The cylinder is pushed vertically downwards so that the mark is a distance below the water surface.
\nAt time the cylinder is released. The resultant vertical force on the cylinder is related to the displacement of the mark by
\n\nwhere is the density of water.
\nThe cylinder was initially pushed down a distance .
\nOutline why the cylinder performs simple harmonic motion when released.
\nThe mass of the cylinder is and the cross-sectional area of the cylinder is . The density of water is . Show that the angular frequency of oscillation of the cylinder is about .
\nDetermine the maximum kinetic energy of the cylinder.
\nDraw, on the axes, the graph to show how the kinetic energy of the cylinder varies with time during one period of oscillation .
\nthe «restoring» force/acceleration is proportional to displacement ✓
\n
Allow use of symbols i.e. or
Evidence of equating «to obtain » ✓
\n\n
OR ✓
\n\n
Answer to at least s.f.
\n« is a maximum when hence» ✓
\n
✓
energy never negative ✓
\ncorrect shape with two maxima ✓
\nThis was well answered with candidates gaining credit for answers in words or symbols.
\nAgain, very well answered.
\nA straightforward calculation with the most common mistake being missing the squared on the omega.
\nMost candidates answered with a graph that was only positive so scored the first mark.
\nAn X-ray beam, of intensity , is used to examine the flow of blood through an artery in the leg of a patient. The beam passes through an equal thickness of blood and soft tissue.
\nThe thickness of blood and tissue is 5.00 mm. The intensity of the X-rays emerging from the tissue is and the intensity emerging from the blood is .
\nThe following data are available.
\nMass absorption coefficient of tissue = 0.379 cm2 g–1
Mass absorption coefficient of blood = 0.385 cm2 g–1
Density of tissue = 1.10 × 103 kg m–3
Density of blood = 1.06 × 103 kg m–3
Show that the ratio is close to 1.
\nState and explain, with reference to you answer in (a)(i), what needs to be done to produce a clear image of the leg artery using X-rays.
\nIn nuclear magnetic resonance (NMR) protons inside a patient are made to emit a radio frequency electromagnetic radiation. Outline the mechanism by which this radiation is emitted by the protons.
\nAND ✔
\n✔
\n✔
\nthe difference between intensities is negligible so no contrast ✔
modifying the blood is easier than modifying the soft tissue ✔
increase absorption of X-rays in the blood ✔
by injecting/introducing a liquid/chemical/contrast medium ✔
with large mass absorption coefficient/nontoxic/higher density ✔
\n«a uniform» magnetic field is applied to align proton spins ✔
\nproton spins are excited by an «external» radio frequency signal/field
OR
protons change from spin-up to spin-down state due to «external» RF signal/field ✔
«radio frequency» radiation is emitted as the protons relax ✔
\n\n
NOTE: For MP3 do not allow simplistic “protons emit RF radiation” as this is given in the question
\nWhich lists one scalar and two vector quantities?
\nA. Mass, momentum, potential difference
\nB. Mass, power, velocity
\nC. Power, intensity, velocity
\nD. Power, momentum, velocity
\nD
\nA conducting sphere has radius 48 cm. The electric potential on the surface of the sphere is 3.4 × 105 V.
\nThe sphere is connected by a long conducting wire to a second conducting sphere of radius 24 cm. The second sphere is initially uncharged.
\n
Show that the charge on the surface of the sphere is +18 μC.
\nDescribe, in terms of electron flow, how the smaller sphere becomes charged.
\nPredict the charge on each sphere.
\nOR
\n«μC» ✓
\nelectrons leave the small sphere «making it positively charged» ✓
\n✓
\n✓
\nso «μC», «μC» ✓
\n\n
Award [3] marks for a bald correct answer.
\nA parallel beam of X-rays travels through 7.8 cm of tissue to reach the bowel surface. Calculate the fraction of the original intensity of the X-rays that reach the bowel surface. The linear attenuation coefficient for tissue is 0.24 cm–1.
\nThe fluid in the bowel has a similar linear attenuation coefficient as the bowel surface. Gases have much lower linear attenuation coefficients than fluids. Explain why doctors will fill the bowel with air before taking an X-ray image.
\nI0e−0.24 × 7.8 ✔
\n0.15I0 ✔
\nAward [2] for bald correct answer.
\nto produce an X-ray image there must be constrast/a difference in the intensity of the beam transmitted through tissue and the bowel ✔
\nintroduction of air will produce contrast ✔
\nAir as a contrast medium. Part (a) was well calculated by most of the prepared candidates.
\nIn (b) only the best candidates well identified the importance of the change in contrast of the image (resulting from different attenuation values) needed to locate the organ.
\nThe root mean square (rms) current in the primary coil of an ideal transformer is 2.0 A. The rms voltage in the secondary coil is 50 V. The average power transferred from the secondary coil is 20 W.
\nWhat is and what is the average power transferred from the primary coil?
\nC
\nAn elevator (lift) and its load accelerate vertically upwards.
\nWhich statement is correct in this situation?
\n
A. The net force on the load is zero.
B. The tension in the cable is equal but opposite to the combined weight of the elevator and its load.
\nC. The normal reaction force on the load is equal but opposite to the force on the elevator from the load.
\nD. The elevator and its load are in translational equilibrium.
\nC
\nA ball undergoes an elastic collision with a vertical wall. Which of the following is equal to zero?
\nA. The change of the magnitude of linear momentum of the ball
\nB. The magnitude of the change of linear momentum of the ball
\nC. The rate of change of linear momentum of the ball
\nD. The impulse of the force on the ball
\nA
\nA charge Q is at a point between two electric charges Q1 and Q2. The net electric force on Q is zero. Charge Q1 is further from Q than charge Q2.
\nWhat is true about the signs of the charges Q1 and Q2 and their magnitudes?
\nA
\nTwo initially uncharged capacitors X and Y are connected in series to a cell as shown.
\nWhat is ?
\n
A.
B.
\nC.
\nD.
\nC
\nX and Y are two objects on a frictionless table connected by a string. The mass of X is 2 kg and the mass of Y is 4 kg. The mass of the string is negligible. A constant horizontal force of 12 N acts on Y.
\nWhat are the acceleration of Y and the magnitude of the tension in the string?
\nA
\nTwo forces act on an object in different directions. The magnitudes of the forces are 18 N and 27 N. The mass of the object is 9.0 kg. What is a possible value for the acceleration of the object?
\nA. 0 m s−2
\nB. 0.5 m s−2
\nC. 2.0 m s−2
\nD. 6.0 m s−2
\nC
\nA battery of negligible internal resistance is connected to a lamp. A second identical lamp is added in series. What is the change in potential difference across the first lamp and what is the change in the output power of the battery?
\nA
\nTwo identical boxes are stored in a warehouse as shown in the diagram. Two forces acting on the top box and two forces acting on the bottom box are shown.
\nWhich is a force pair according to Newton’s third law?
\nA. 1 and 2
\nB. 3 and 4
\nC. 2 and 3
\nD. 2 and 4
\nC
\nAn object of mass 1.0 kg hangs at rest from a spring. The spring has a negligible mass and the spring constant k is 20 N m−1
\nWhat is the elastic potential energy stored in the spring?
\n
A. 1.0 J
B. 2.5 J
\nC. 5.0 J
\nD. 10 J
\nB
\nEvidence from the Planck space observatory suggests that the density of matter in the universe is about 32 % of the critical density of the universe.
\nOutline how the light spectra of distant galaxies are used to confirm hypotheses about the expansion of the universe.
\nLight from a hydrogen source in a laboratory on Earth contains a spectral line of wavelength 122 nm. Light from the same spectral line reaching Earth from a distant galaxy has a wavelength of 392 nm. Determine the ratio of the present size of the universe to the size of the universe when the light was emitted by the galaxy.
\nState what is meant by the critical density.
\nCalculate the density of matter in the universe, using the Hubble constant 70 km s–1 Mpc–1.
\nIt is estimated that less than 20 % of the matter in the universe is observable. Discuss how scientists use galactic rotation curves to explain this.
\nspectra of galaxies are redshifted «compared to spectra on Earth» ✔
\nredshift/longer wavelength implies galaxies recede/ move away from us
OR
redshift is interpreted as cosmological expansion of space ✔
«hence universe expands»
\nNOTE: Universe expansion is given, so no mark for repeating this.
Do not accept answers based on CMB radiation.
ALTERNATIVE 1
\n✔
\n✔
\nALTERNATIVE 2
\n✔
\n= 3.21 ✔
\ndensity of flat/Euclidean universe
OR
density for which universe has zero curvature
OR
density resulting in universe expansion rate tending to zero ✔
✔
\n✔
\n✔
\nNOTE: MP1 for conversion of H to base units.
Allow ECF from MP1, but NOT if H is left as 70.
rotation speed of galaxies is larger than expected away from the centre ✔
there must be more mass «at the edges» than is visually observable «indicating the presence of dark matter» ✔
\nA circuit consists of a cell of emf E = 3.0 V and four resistors connected as shown. Resistors R1 and R4 are 1.0 Ω and resistors R2 and R3 are 2.0 Ω.
\nWhat is the voltmeter reading?
\nA. 0.50 V
\nB. 1.0 V
\nC. 1.5 V
\nD. 2.0 V
\nB
\nThere were some comments from teachers that the circuit is unfamiliar, however it is basically a series and parallel circuit and can be solved by considering the parallel sections individually either by calculating the current through each and then the voltages across the individual resistors or by considering the resistors as a potential divider. It has a low discrimination index at HL with many choosing option C (B correct) and very poor discrimination at SL, again with option C the most popular choice.
\nIn a photoelectric experiment a stopping voltage required to prevent photoelectrons from flowing across the photoelectric cell is measured for light of two frequencies and . The results obtained are shown.
The ratio is an estimate of
\n
A.
B.
\nC.
\nD.
\nD
\nA rectangular coil of wire RSTU is connected to a battery and placed in a magnetic field Z directed to the right. Both the plane of the coil and the magnetic field direction are in the same plane.
\nWhat is true about the magnetic force acting on the sides RS and ST?
\nB
\nAn electron has a linear momentum of 4.0 × 10−25 kg m s−1. What is the order of magnitude of the kinetic energy of the electron?
\nA. 10−50 J
\nB. 10−34 J
\nC. 10−19 J
\nD. 106 J
\nC
\nA net force acts on an object of mass that is initially at rest. The object moves in a straight line. The variation of with the distance is shown.
\nWhat is the speed of the object at the distance ?
\n
A.
B.
\nC.
\nD.
\nB
\nDescribe the mechanism of formation of type I a supernovae.
\nDescribe the mechanism of formation of type II supernovae.
\nSuggest why type I a supernovae were used in the study that led to the conclusion that the expansion of the universe is accelerating.
\nALTERNATIVE 1
a white dwarf star in a binary system accretes mass from the companion star ✔
when the white dwarf star mass reaches the Chandrasekhar limit the star explodes «due to fusion reactions»✔
\nALTERNATIVE 2
it can be formed in the collision of two white dwarf stars ✔
where shock waves from the collision rip both stars apart ✔
\na red supergiant star explodes when its core collapses ✔
\n«it was necessary» to measure the distance «of very distant objects more accurately» ✔
\ntype I a are standard candles/objects of known luminosity ✔
\nSupernova. The mechanism of the formation of supernovae Ia was well described by many candidates.
\nIn (ii), for the description of the mechanism of type II supernovae, many responses lacked detail and did not make mention of core collapse.
\nPart b) was well answered by most of the prepared candidates with a good understanding that these stars behave as “standard candles”.
\nA satellite is orbiting Earth in a circular path at constant speed. Three statements about the resultant force on the satellite are:
\nI. It is equal to the gravitational force of attraction on the satellite.
II. It is equal to the mass of the satellite multiplied by its acceleration.
III. It is equal to the centripetal force on the satellite.
Which combination of statements is correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nThis was a good discriminator at HL although many candidates chose option B (D correct). Option B was just the most popular choice at SL. Candidates appear not to realise that although this is circular motion F = ma still applies.
\nThree statements about Newton’s law of gravitation are:
\nI. It can be used to predict the motion of a satellite.
II. It explains why gravity exists.
III. It is used to derive the expression for gravitational potential energy.
Which combination of statements is correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nComments suggested that 'gravitational potential' is more suitable for an HL question. However, candidates should have realised that statement II is incorrect so option B is the only possibility and this proved the most popular answer. The wording will be altered to 'gravitational potential energy' for publication.
\nA ball rolls on the floor towards a wall and rebounds with the same speed and at the same angle to the wall.
\nWhat is the direction of the impulse applied to the ball by the wall?
\nD
\nThe graph shows the variation with diffraction angle of the intensity of light after it has passed through four parallel slits.
\nThe number of slits is increased but their separation and width stay the same. All slits are illuminated.
\nState what is meant by the Doppler effect.
\nA plate performs simple harmonic oscillations with a frequency of 39 Hz and an amplitude of 8.0 cm.
\nShow that the maximum speed of the oscillating plate is about 20 m s−1.
\nSound of frequency 2400 Hz is emitted from a stationary source towards the oscillating plate in (b). The speed of sound is 340 m s−1.
\nDetermine the maximum frequency of the sound that is received back at the source after reflection at the plate.
\n\n
State what will happen to the angular position of the primary maxima.
\nState what will happen to the width of the primary maxima.
\nState what will happen to the intensity of the secondary maxima.
\nthe change in the observed frequency ✓
\nwhen there is relative motion between the source and the observer ✓
\n\n
Do not award MP1 if they refer to wavelength.
\nuse of ✓
\nmaximum speed is «m s−1» ✓
\n\n
Award [2] for a bald correct answer.
\nfrequency at plate «Hz»
\nat source «Hz» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAward [1] mark when the effect is only applied once.
\nstays the same ✓
\ndecreases ✓
\ndecreases ✓
\nA quantity of 2.00 mol of an ideal gas is maintained at a temperature of 127 ºC in a container of volume 0.083 m3. What is the pressure of the gas?
\nA. 8 kPa
\nB. 25 kPa
\nC. 40 kPa
\nD. 80 kPa
\nD
\nA beam of electrons moving in the direction shown is incident on a rectangular slit of width .
\nThe component of momentum of the electrons in direction after passing through the slit is . The uncertainty in is
\n
A. proportional to
B. proportional to
\nC. proportional to
\nD. zero
\nB
\nThree statements about electrons are:
\nI. Electrons interact through virtual photons.
II. Electrons interact through gluons.
III. Electrons interact through particles W and Z.
Which statements identify the particles mediating the forces experienced by electrons?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nA liquid is vaporized to a gas at a constant temperature.
\nThree quantities of the substance are the
\nI. total intermolecular potential energy
II. root mean square speed of the molecules
III. average distance between the molecules.
Which quantities are greater for the substance in the gas phase compared to the liquid phase?
\n
A. I and II only
B. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nTwo identical waves, each with amplitude X0 and intensity I, interfere constructively. What are the amplitude and intensity of the resultant wave?
\nD
\nThe energy levels of an atom are shown. How many photons of energy greater than 1.9 eV can be emitted by this atom?
\n
A. 1
B. 2
\nC. 3
\nD. 4
\nD
\nThe diagram shows the electric field lines of a positively charged conducting sphere of radius and charge .
\nPoints A and B are located on the same field line.
\nA proton is placed at A and released from rest. The magnitude of the work done by the electric field in moving the proton from A to B is . Point A is at a distance of from the centre of the sphere. Point B is at a distance of from the centre of the sphere.
\nExplain why the electric potential decreases from A to B.
\nDraw, on the axes, the variation of electric potential with distance from the centre of the sphere.
\nCalculate the electric potential difference between points A and B.
\nDetermine the charge of the sphere.
\nThe concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common terminology to describe different types of fields.
\nALTERNATIVE 1
work done on moving a positive test charge in any outward direction is negative ✓
potential difference is proportional to this work «so decreases from A to B» ✓
\n
ALTERNATIVE 2
potential gradient is directed opposite to the field so inwards ✓
the gradient indicates the direction of increase of «hence increases towards the centre/decreases from A to B» ✓
\n
ALTERNATIVE 3
so as increases decreases ✓
is positive as is positive ✓
\n
ALTERNATIVE 4
the work done per unit charge in bringing a positive charge from infinity ✓
to point B is less than point A ✓
curve decreasing asymptotically for ✓
\nnon zero constant between and ✓
\n✓
\n✓
\n
✓
to highlight similarities between «different» fields ✓
\nThe majority who answered in terms of potential gained one mark. Often the answers were in terms of work done rather than work done per unit charge or missed the fact that the potential is positive.
\nThis was well answered.
\nMost didn't realise that the key to the answer is the definition of potential or potential difference and tried to answer using one of the formulae in the data booklet, but incorrectly.
\nEven though many were able to choose the appropriate formula from the data booklet they were often hampered in their use of the formula by incorrect techniques when using fractions.
\nThis was generally well answered with only a small number of answers suggesting greater international cooperation.
\nA mass of a liquid of specific heat capacity flows every second through a heater of power . What is the difference in temperature between the liquid entering and leaving the heater?
\n
A.
B.
\nC.
\nD.
\nC
\nSome of the nuclear energy levels of oxygen-14 (14O) and nitrogen-14 (14N) are shown.
\nA nucleus of 14O decays into a nucleus of 14N with the emission of a positron and a gamma ray. What is the maximum energy of the positron and the energy of the gamma ray?
\nA
\nThree quantities used to describe a light wave are
\n I. frequency
II. wavelength
III. speed.
Which quantities increase when the light wave passes from water to air?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nA fixed mass of an ideal gas has a volume of , a pressure of p and a temperature of . The gas is compressed to the volume of and its pressure increases to 12p. What is the new temperature of the gas?
\n
A.
B.
\nC.
\nD.
\nC
\nProxima Centauri is a main sequence star with a mass of 0.12 solar masses.
\nEstimate .
\nDescribe why iron is the heaviest element that can be produced by nuclear fusion processes inside stars.
\nDiscuss one process by which elements heavier than iron are formed in stars.
\nrealization that lifetime ✔
\n✔
\nthe binding energy per nucleon is a maximum for iron ✔
formation of heavier elements than iron by fusion is not energetically possible ✔
\nNOTE: For MP2 some reference to energy is needed
\nALTERNATIVE 1 — s-process
s-process involves «slow» neutron capture ✔
in s-process beta decay occurs before another neutron is captured ✔
s-process occurs in giant stars «AGB stars» ✔
s-process terminates at bismuth/lead/polonium ✔
ALTERNATIVE 2 — r-process
r-process involves «rapid» neutron capture ✔
in r-process further neutrons are captured before the beta decay occurs ✔
r-process occurs in type II supernovae ✔
r-process can lead to elements heavier than bismuth/lead/polonium ✔
NOTE: If the type of the process (r or s/rapid or slow) is not mentioned, award [2 max].
\nA pipe of length L is closed at one end. Another pipe is open at both ends and has length 2L. What is the lowest common frequency for the standing waves in the pipes?
\nA.
\nB.
\nC.
\nD.
\nB
\nWhat statement is not true about radioactive decay?
\n
A. The percentage of radioactive nuclei of an isotope in a sample of that isotope after 7 half-lives is smaller than 1 %.
B. The half-life of a radioactive isotope is the time taken for half the nuclei in a sample of that isotope to decay.
\nC. The whole-life of a radioactive isotope is the time taken for all the nuclei in a sample of that isotope to decay.
\nD. The half-life of radioactive isotopes range between extremely short intervals to thousands of millions of years.
\nC
\nThere was some questioning about the use of the term 'whole-life' from teacher comments. As that option (C) was the correct answer and the most popular it did not confuse the candidates. The statement is clearly incorrect and the use of a non physics specific term that might be used in a general discussion was felt to be acceptable.
\nThe size of a nucleus can be estimated from electron diffraction experiments. What is the order of magnitude of the de Broglie wavelength of the electrons in these experiments?
\n
A. 10−15 m
B. 10−13 m
\nC. 10−11 m
\nD. 10−9 m
\nA
\nThe graph shows the variation with time of the cosmic scale factor R of the universe for the flat model of the universe without dark energy.
\nLight from distant galaxies is redshifted. Explain the cosmological origin of this redshift.
\nDraw, on the axes, a graph to show the variation with time of the cosmic scale factor R for the flat model of the universe with dark energy.
\nCompare and contrast, the variation with time of the temperature of the cosmic background (CMB) radiation, for the two models from the present time onward.
\n«according to general relativity» space expands stretching distances between far away objects ✔
\nwavelengths of photons «received a long time after they were emitted» are thus longer leading to the observed redshift ✔
\nDo not accept references to the Doppler effect.
\n ✔
«since T ∝ » the temperature drops for both models ✔
\nbut in the accelerating model R increases faster and so the temperature drops faster ✔
\nCosmological origin of redshift. The cosmological redshift and variation with time of the cosmic scale factor proved to be a well-mastered concept by many students.
\nCosmological origin of redshift. The cosmological redshift and variation with time of the cosmic scale factor proved to be a well-mastered concept by many students.
\nIn (ii) however, many candidates did not directly answer the question, making little to no reference of temperature.
\nA particle undergoes simple harmonic motion of amplitude and frequency . What is the average speed of the particle during one oscillation?
\n
A.
B.
\nC.
\nD.
\nD
\nA ball of mass 0.250 kg is released from rest at time t = 0, from a height H above a horizontal floor.
\nThe graph shows the variation with time t of the velocity v of the ball. Air resistance is negligible. Take g = −9.80 m s−2. The ball reaches the floor after 1.0 s.
\nDetermine H.
\nLabel the time and velocity graph, using the letter M, the point where the ball reaches the maximum rebound height.
\nState the acceleration of the ball at the maximum rebound height.
\nDraw, on the axes, a graph to show the variation with time of the height of the ball from the instant it rebounds from the floor until the instant it reaches the maximum rebound height. No numbers are required on the axes.
\nEstimate the loss in the mechanical energy of the ball as a result of the collision with the floor.
\nDetermine the average force exerted on the floor by the ball.
\nSuggest why the momentum of the ball was not conserved during the collision with the floor.
\nH = «gt2 =» 4.9 «m» ✓
\n
Accept other methods as area from graph, alternative kinematics equations or conservation of mechanical energy.
Award [1] for a bald correct answer in the range 4.9 - 5.1.
Award [0] if time used is different than 1.0 s.
M at 1.6 s ✓
\n«g =» 9.80 «ms−2» ✓
\n
Accept 9.81, 10 or a plain “g”.
Ignore sign if provided.
concave down parabola as shown «with non-zero initial slope and zero final slope» ✓
\n\n
Award [1] mark if curve starts from a positive time value.
Award [0] if the final slope is negative.
« loss of KE is » «J» ✓
\n
Award [1] mark for an answer in the range 8.7 - 9.5.
✓
\nFnet = «» «N» ✓
\nN «N» ✓
\n
Allow ECF for MP2 and MP3.
there is an external force acting on the ball
\nOR
\nsome momentum is transferred to the floor ✓
\n
Allow references to impulse instead of force.
Do not award references to energy.
The age of the Earth is about 4.5 × 109 years.
\nWhat area of physics provides experimental evidence for this conclusion?
\nA. Newtonian mechanics
\nB. Optics
\nC. Radioactivity
\nD. Electromagnetism
\nC
\nA long straight vertical conductor carries a current I upwards. An electron moves with horizontal speed v to the right.
\nWhat is the direction of the magnetic force on the electron?
\nA. Downwards
\nB. Upwards
\nC. Into the page
\nD. Out of the page
\nA
\nA travelling wave on the surface of a lake has wavelength . Two points along the wave oscillate with the phase difference of . What is the smallest possible distance between these two points?
\n
A.
B.
\nC.
\nD.
\nB
\nThe radius of a circle is measured to be (10.0 ± 0.5) cm. What is the area of the circle?
\nA. (314.2 ± 0.3) cm2
\nB. (314 ± 1) cm2
\nC. (314 ± 15) cm2
\nD. (314 ± 31) cm2
\nD
\nThis question discriminated well at both HL and SL with many candidates choosing the correct option D. However, option B was also a popular choice particularly at SL. Candidates need to be aware that when performing a calculation e.g. the area as here, the uncertainty also has to be propagated - so a 5% uncertainty in the radius becomes a 10% uncertainty in the area. There were some comments on the G2s that the uncertainty should only have been given to 1sf but this is not always correct as uncertainties are given to the precision of the value, depending on the percentage calculated in the propagation.
\nWhich is the definition of gravitational field strength at a point?
\nA. The sum of the gravitational fields created by all masses around the point
\nB. The gravitational force per unit mass experienced by a small point mass at that point
\nC. , where is the mass of a planet and is the distance from the planet to the point
\nD. The resultant force of gravitational attraction on a mass at that point
\nB
\nPhotovoltaic cells and solar heating panels are used to transfer the electromagnetic energy of the Sun’s rays into other forms of energy. What is the form of energy into which solar energy is transferred in photovoltaic cells and solar heating panels?
\nA
\nThe Sankey diagrams for a filament lamp and for an LED bulb are shown below.
\nWhat is the efficiency of the filament lamp and the LED bulb?
\nA
\nHorizontally polarized light is incident on a pair of polarizers X and Y. The axis of polarization of X makes an angle θ with the horizontal. The axis of polarization of Y is vertical.
\nWhat is θ so that the intensity of the light transmitted through Y is a maximum?
\n
A.
B.
\nC.
\nD.
\nB
\nThe dashed line represents the variation with incident electromagnetic frequency of the kinetic energy EK of the photoelectrons ejected from a metal surface. The metal surface is then replaced with one that requires less energy to remove an electron from the surface.
\nWhich graph of the variation of EK with will be observed?
\nA
\nTwo different experiments, P and Q, generate two sets of data to confirm the proportionality of variables and . The graphs for the data from P and Q are shown. The maximum and minimum gradient lines are shown for both sets of data.
\nWhat is true about the systematic error and the uncertainty of the gradient when P is compared to Q?
\nC
\nA ball is thrown upwards at time t = 0. The graph shows the variation with time of the height of the ball. The ball returns to the initial height at time T.
\nWhat is the height h at time t ?
\nA.
B.
C.
D.
D
\nThis question proved challenging, many more candidates chose answer A instead of correct D. Candidates need to be aware that there are useful strategies for answering questions, especially ones they may find difficult. Eliminate choices that are clearly wrong - here A (and therefore B as well) are incorrect as they reflect an equation producing a height that always increases. It is also sometimes helpful to invent numbers to test the equations, e.g. assuming that the height is a known value, testing the possible answers given. Five metres would work here, as the height covered in free-fall from rest during one second. Thrown upwards at 5 m/s, it would take 1 second to go up and another to come back, therefore T = 2s. Only D would give the correct answer of 0 m after 1 s. The question also produced many comments on the G2 due to its difficulty. It must be remembered that questions appear in guide topic order so it is unlikely that harder questions will only appear towards the middle of the paper.
\nThree particles are produced when the nuclide undergoes beta-plus (β+) decay. What are two of these particles?
\nA. and
\nB. and
\nC. and
\nD. and
\nA
\nA ray of monochromatic light is incident on the parallel interfaces between three media. The speeds of light in the media are v1, v2 and v3.
\nWhat is correct about the speeds of light in the media?
\n
A. v3 < v1 < v2
B. v3 < v2 < v1
\nC. v2 < v3 < v1
\nD. v2 < v1 < v3
\nD
\nA radioactive nuclide X decays into a nuclide Y. The graph shows the variation with time of the activity A of X. X and Y have the same nucleon number.
\nWhat is true about nuclide X?
\nA. alpha (α) emitter with a half-life of t
\nB. alpha (α) emitter with a half-life of 2t
\nC. beta-minus (β−) emitter with a half-life of t
\nD. beta-minus (β−) emitter with a half-life of 2t
\nD
\nWhich graph shows a possible probability density function for a given wave function of an electron?
C
\nA book is at rest on a table. One of the forces acting on the book is its weight.
\nWhat is the other force that completes the force pair according to Newton’s third law of motion?
\nA. The pull of the book on Earth
\nB. The pull of Earth on the book
\nC. The push of the table on the book
\nD. The push of the book on the table
\nA
\nThe majority of candidates incorrectly selected option C for this question, resulting in a low difficulty index overall. This question highlights a typical misconception relating to Newton's 3rd law, and emphasises the importance of conceptual physics teaching.
\nA particle reaction is
\n.
\nWhich conservation law is violated by the reaction?
\nA. Baryon number
\nB. Charge
\nC. Lepton number
\nD. Momentum
\nB
\nA string is fixed at both ends. P and Q are two particles on the string.
\nThe first harmonic standing wave is formed in the string. What is correct about the motion of P and Q?
\n
A. P is a node and Q is an antinode.
B. P is an antinode and Q is a node.
\nC. P and Q oscillate with the same amplitude.
\nD. P and Q oscillate with the same frequency.
\nD
\nThree statements about fossil fuels are:
\nI. There is a finite amount of fossil fuels on Earth.
II. The transfer of energy from fossil fuels increases the concentration of CO2 in the atmosphere.
III. The geographic distribution of fossil fuels is uneven and has led to economic inequalities.
Which statements justify the development of alternative energy sources?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nA solid metal ball is dropped from a tower. The variation with time of the velocity of the ball is plotted.
\nA hollow metal ball with the same size and shape is dropped from the same tower. What graph will represent the variation with time of the velocity for the hollow metal ball?
\nC
\nTwo bodies each of equal mass travelling in opposite directions collide head-on.
\nWhat is a possible outcome of the collision?
\nB
\nThis question was well answered by HL candidates. Some students may have answered incorrectly due to consideration of speed rather than velocity.
\nAirboats are used for transport across a river. To move the boat forward, air is propelled from the back of the boat by a fan blade.
\nAn airboat has a fan blade of radius 1.8 m. This fan can propel air with a maximum speed relative to the boat of 20 m s−1. The density of air is 1.2 kg m−3.
\nIn a test the airboat is tied to the river bank with a rope normal to the bank. The fan propels the air at its maximum speed. There is no wind.
\nThe rope is untied and the airboat moves away from the bank. The variation with time t of the speed v of the airboat is shown for the motion.
Outline why a force acts on the airboat due to the fan blade.
\nShow that a mass of about 240 kg of air moves through the fan every second.
\nShow that the tension in the rope is about 5 kN.
\nEstimate the distance the airboat travels to reach its maximum speed.
\nDeduce the mass of the airboat.
\nThe fan is rotating at 120 revolutions every minute. Calculate the centripetal acceleration of the tip of a fan blade.
\nALTERNATIVE 1
\nthere is a force «by the fan» on the air / air is accelerated «to the rear» ✓
\nby Newton 3 ✓
\nthere is an «equal and» opposite force on the boat ✓
\n\n
ALTERNATIVE 2
\nair gains momentum «backward» ✓
\nby conservation of momentum / force is rate of change in momentum ✓
\nboat gains momentum in the opposite direction ✓
\n\n
Accept a reference to Newton’s third law, e.g. N’3, or any correct statement of it for MP2 in ALT 1.
\nAllow any reasonable choice of object where the force of the air is acting on, e.g., fan or blades.
\nOR «mass of air through system per unit time =» seen ✓
\n244 «kg s−1» ✓
\n\n
Accept use of Energy of air per second = 0.5 ρΑv3 = 0.5 mv2 for MP1.
\n«force = Momentum change per sec = = » 244 x 20 OR 4.9 «kN» ✓
\n\n
Allow use of 240
\nrecognition that area under the graph is distance covered ✓
\n«Distance =» 480 - 560 «m» ✓
\n\n
Accept graphical evidence or calculation of correct geometric areas for MP1.
\nMP2 is numerical value within range.
\ncalculation of acceleration as gradient at t = 0 «= 1 m s-2» ✓
\nuse of F=ma OR seen ✓
\n4900 «kg» ✓
\n\n
MP1 can be shown on the graph.
\nAllow an acceleration in the range 1 – 1.1 for MP2 and consistent answer for MP3
\nAllow ECF from MP1.
\nAllow use of average acceleration =
\nor assumption of constant force to obtain 11000 «kg» for [2]
\nAllow use of 4800 or 5000 for MP2
\nALTERNATE 1
\n« ω = » 4 rad s−1 ✓
\n« a = r ω2= » 280 « m s−2 » ✓
\n\n
ALTERNATE 2
\n« » = 22.6 m s−1 ✓
\n« »= 280 « m s−2 » ✓
\n\n
Allow ECF from MP1 for wrong ω (120 gives 2.6 x 104 « m s−2 »)
\nAllow ECF from MP1 for wrong T (2 s gives 18 « m s−2 »)
\nThe majority succeeded in making use of Newton's third law to explain the force on the boat. The question was quite well answered but sequencing of answers was not always ideal. There were some confusions about the air hitting the bank and bouncing off to hit the boat. A small number thought that the wind blowing the fan caused the force on the boat.
\nbi) This was generally well answered with candidates either starting from the wind turbine formula given in the data booklet or with the mass of the air being found using .
\n1bii) Well answered by most candidates. Some creative work to end up with 240 was found in scripts.
\n1ci) Many candidates gained credit here for recognising that the resistive force eventually equalled the drag force and most were able to go on to link this to e.g. zero acceleration. Some had not read the question properly and assumed that the rope was still tied. There was one group of answers that stated something along the lines of \"as there is no rope there is nothing to stop the boat so it can go at max speed.
\n1cii) A slight majority did not realise that they had to find the area under the velocity-time graph, trying equations of motion for non-linear acceleration. Those that attempted to calculate the area under the graph always succeeded in answering within the range.
\n1ciii) Use of the average gradient was common here for the acceleration. However, there also were answers that attempted to calculate the mass via a kinetic energy calculation that made all sorts of incorrect assumptions. Use of average acceleration taken from the gradient of the secant was also common.
\nThe road from city X to city Y is 1000 km long. The displacement is 800 km from X to Y.
\nWhat is the distance travelled from Y to X and the displacement from Y to X?
\nD
\nWhich change produces the largest percentage increase in the maximum theoretical power output of a wind turbine?
\nA. Doubling the area of the blades
\nB. Doubling the density of the fluid
\nC. Doubling the radius of the blades
\nD. Doubling the speed of the fluid
\nD
\nA simple pendulum has a time period on the Earth. The pendulum is taken to the Moon where the gravitational field strength is that of the Earth.
\nWhat is the time period of the pendulum on the Moon?
\nA.
\nB.
\nC.
\nD.
\nA
\nA book of mass m lies on top of a table of mass M that rolls freely along the ground. The coefficient of friction between the book and the table is . A person is pushing the rolling table.
\nWhat is the maximum acceleration of the table so that the book does not slide backwards relative to the table?
\nA.
\nB.
\nC.
\nD.
\nB
\nOver half the candidates incorrectly chose option D. The book is only able to accelerate because of the friction force between the table and the book which depends on μ and the normal reaction force (mg) so independent of M, immediately eliminating options C and D.
\nWater at room temperature is placed in a freezer. The specific heat capacity of water is twice the specific heat capacity of ice. Assume that thermal energy is transferred from the water at a constant rate.
\nWhich graph shows the variation with time of the temperature of the water?
\nB
\nMonochromatic light of wavelength is incident on two slits S1 and S2. An interference pattern is observed on the screen.
\nO is equidistant from S1 and S2. A bright fringe is observed at O and a dark fringe at X.
\nThere are two dark fringes between O and X. What is the path difference between the light arriving at X from the two slits?
\nA.
\nB.
\nC.
\nD.
\nC
\nWith a relatively high discrimination index, this question was well answered by stronger HL candidates. Some students had difficulty recognising that there would be 2.5λ rather than 1.5λ, and as a result option B was a significant distractor.
\nA charge +Q and a charge −2Q are a distance 3x apart. Point P is on the line joining the charges, at a distance x from +Q.
\nThe magnitude of the electric field produced at P by the charge +Q alone is .
\nWhat is the total electric field at P?
\n
A. to the right
B. to the left
\nC. to the right
\nD. to the left
\nC
\nA car accelerates uniformly from rest to a velocity during time . It then continues at constant velocity from to time .
\nWhat is the total distance covered by the car in ?
\nA.
\nB.
C.
D.
D
\nA particle undergoes simple harmonic motion. Which quantities of the motion can be simultaneously zero?
\nA. Displacement and velocity
\nB. Displacement and acceleration
\nC. Velocity and acceleration
\nD. Displacement, velocity and acceleration
\nB
\nIn two different experiments, white light is passed through a single slit and then is either refracted through a prism or diffracted with a diffraction grating. The prism produces a band of colours from M to N. The diffraction grating produces a first order spectrum P to Q.
\nWhat are the colours observed at M and P?
\nA
\nThis has low discrimination and the difficulty index suggests candidates found it hard with the incorrect option C being the most popular. The spreading of colours and formation of a spectrum (or rainbow) is something that is covered during an introductory course in physics and then developed in refraction and diffraction.
\nTwo wires, and , are made of the same material and have equal length. The diameter of is twice that of .
\nWhat is ?
\n\n
A.
\nB.
\nC.
\nD.
\nA
\nA black body at temperature T emits radiation with peak wavelength and power P. What is the temperature of the black body and the power emitted for a peak wavelength of ?
\nD
\nA standing wave is formed on a string. P and Q are adjacent antinodes on the wave. Three statements are made by a student:
\nI. The distance between P and Q is half a wavelength.
II. P and Q have a phase difference of π rad.
III. Energy is transferred between P and Q.
Which statements are correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nThis question was generally well answered by HL candidates. Given the number of candidates who (incorrectly) chose statement III \"Energy is transferred between P and Q\" as true, this question might be a useful review to identify the properties of standing waves.
\nAn object is sliding from rest down a frictionless inclined plane. The object slides 1.0 m during the first second.
\nWhat distance will the object slide during the next second?
\nA. 1.0 m
\nB. 2.0 m
\nC. 3.0 m
\nD. 4.9 m
\nC
\nThe correct response, option C was the most popular chosen at HL but at SL significantly more candidates chose options A or B. The difficulty index of 21 and discrimination index of 0.27 at SL indicates that students found the question to be hard with lower discrimination between stronger and weaker candidates. It is felt that those who chose option A did not realise the block was accelerating down the slope, whereas those choosing B did but were unable to calculate the acceleration correctly.
\nThe coil of a direct current electric motor is turning with a period T. At t = 0 the coil is in the position shown in the diagram. Assume the magnetic field is uniform across the coil.
\nWhich graph shows the variation with time of the force exerted on section XY of the coil during one complete turn?
\nA
\nHas a negative discrimination index with over 80% of candidates choosing the incorrect answer. The difficulty index is also low. The question states that it is about a direct current electric motor, suggesting that C and D are incorrect so by choosing them it would seem that some candidates are confusing an electric motor with a generator.
\nIn the circuit shown, the battery has an emf of 12 V and negligible internal resistance. Three identical resistors are connected as shown. The resistors each have a resistance of 10 Ω.
\nThe resistor L is removed. What is the change in potential at X?
\nA. Increases by 2 V
\nB. Decreases by 2 V
\nC. Increases by 4 V
\nD. Decreases by 4 V
\nB
\nThe majority of HL candidates correctly determined the magnitude of the potential but determining the direction of the change was more problematic. More candidates (incorrectly) selected option A than the correct option B, reinforcing the importance of a conceptual understanding of circuits and potential change.
\nA train is sounding its whistle when approaching a train station. Three statements about the sound received by a stationary observer at the station are:
\nI. The frequency received is higher than the frequency emitted by the train.
II. The wavelength received is longer than the wavelength emitted by the train.
III. The speed of the sound received is not affected by the motion of the train.
Which combination of statements is correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nB
\nAn electric motor of efficiency 0.75 is connected to a power supply with an emf of 20 V and negligible internal resistance. The power output of the motor is 120 W. What is the average current drawn from the power supply?
\n\n
A. 3.1 A
\nB. 4.5 A
\nC. 6.0 A
\nD. 8.0 A
\nD
\nA quantity of 0.24 mol of an ideal gas of constant volume 0.20 m3 is kept at a temperature of 300 K.
\nState what is meant by the internal energy of an ideal gas.
\nCalculate the pressure of the gas.
\nThe temperature of the gas is increased to 500 K. Sketch, on the axes, a graph to show the variation with temperature T of the pressure P of the gas during this change.
\nA container is filled with 1 mole of helium (molar mass 4 g mol−1) and 1 mole of neon (molar mass 20 g mol−1). Compare the average kinetic energy of helium atoms to that of neon atoms.
\nthe total «random» kinetic energy of the molecules/atoms/particles ✓
\n«Pa» ✓
\nstraight line joining (300, 3) and (500, 5) ✓
\ndrawn only in the range from 300 to 500 K ✓
\n\n
Allow ECF from (b)(i) for incorrect initial pressure.
Allow tolerance of ± one grid square for the endpoints.
temperature is the same for both gases ✓
\n«average» kinetic energy is the same «because OR depends on only» ✓
\n
Award [1 max] for a bald statement that kinetic energy is the same.
What is evidence for wave–particle duality?
\nA. Line spectra of elements
\nB. Electron-diffraction experiments
\nC. Rutherford alpha-scattering experiments
\nD. Gamma-ray spectra
\nB
\nThis question was generally well answered by HL candidates.
\nIn a simple climate model for a planet, the incoming intensity is 400 W m−2 and the radiated intensity is 300 W m−2.
\nThe temperature of the planet is constant. What are the reflected intensity from the planet and the albedo of the planet?
\nA
\nTwo cells are connected in parallel as shown below. Each cell has an emf of 5.0 V and an internal resistance of 2.0 Ω. The lamp has a resistance of 4.0 Ω. The ammeter is ideal.
\nWhat is the reading on the ammeter?
\nA. 1.0 A
\nB. 1.3 A
\nC. 2.0 A
\nD. 2.5 A
\nA
\nThe correct option (A) was selected with the lowest frequency of the four possible answers. This question has a low difficulty index, suggesting that the majority of candidates found it challenging. Students were asked to apply the concept of resistors in parallel; omitting the internal resistance in parallel to the external lamp resistance was the most common error here. This question is useful for the revision of resistors in combination.
\nThe decay constant, , of a radioactive sample can be defined as
\nA. the number of disintegrations in the radioactive sample.
\nB. the number of disintegrations per unit time in the radioactive sample.
\nC. the probability that a nucleus decays in the radioactive sample.
\nD. the probability that a nucleus decays per unit time in the radioactive sample.
\nD
\nThis question was well answered by HL candidates.
\nTwo positive and two negative charges are located at the corners of a square as shown. Point X is the centre of the square. What is the value of the electric field E and the electric potential V at X due to the four charges?
\nA
\nCandidates were unsure about this question with almost equal numbers choosing A and C. Electric potential is a scalar quantity so unaffected by the sign of the charge and can only be 0 in this arrangement removing the choice of C.
\nA variable resistor is connected in series to a cell with internal resistance r as shown.
\nThe resistance of the variable resistor is increased. What happens to the power dissipated in the cell and to the terminal potential difference of the cell?
\n
A
\nAn astronaut is orbiting Earth in a spaceship. Why does the astronaut experience weightlessness?
\nA. The astronaut is outside the gravitational field of Earth.
\nB. The acceleration of the astronaut is the same as the acceleration of the spaceship.
\nC. The spaceship is travelling at a high speed tangentially to the orbit.
\nD. The gravitational field is zero at that point.
\nB
\nA student uses a load to pull a box up a ramp inclined at 30°. A string of constant length and negligible mass connects the box to the load that falls vertically. The string passes over a pulley that runs on a frictionless axle. Friction acts between the base of the box and the ramp. Air resistance is negligible.
\nThe load has a mass of 3.5 kg and is initially 0.95 m above the floor. The mass of the box is 1.5 kg.
\nThe load is released and accelerates downwards.
\nOutline two differences between the momentum of the box and the momentum of the load at the same instant.
\nThe vertical acceleration of the load downwards is 2.4 m s−2.
\nCalculate the tension in the string.
\nShow that the speed of the load when it hits the floor is about 2.1 m s−1.
\nThe radius of the pulley is 2.5 cm. Calculate the angular speed of rotation of the pulley as the load hits the floor. State your answer to an appropriate number of significant figures.
\nAfter the load has hit the floor, the box travels a further 0.35 m along the ramp before coming to rest. Determine the average frictional force between the box and the surface of the ramp.
\nThe student then makes the ramp horizontal and applies a constant horizontal force to the box. The force is just large enough to start the box moving. The force continues to be applied after the box begins to move.
\nExplain, with reference to the frictional force acting, why the box accelerates once it has started to move.
\ndirection of motion is different / OWTTE ✓
\nmv / magnitude of momentum is different «even though v the same» ✓
\nuse of ma = mg − T «3.5 x 2.4 = 3.5g − T »
\nOR
\nT = 3.5(g − 2.4) ✓
\n26 «N» ✓
\n\n
Accept 27 N from g = 10 m s−2
\nproper use of kinematic equation ✓
\n«m s−1» ✓
\n\n
Must see either the substituted values OR a value for v to at least three s.f. for MP2.
\nuse of to give 84 «rad s−1»
\nOR
\nto give 84 «rad s−1» ✓
\n\n
quoted to 2sf only✓
\n\n
ALTERNATIVE 1
\n«» leading to a = 6.3 «m s-2»
\nOR
\n« » leading to t = 0.33 « s » ✓
\n
Fnet = « = » 9.45 «N» ✓
Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓
\nfriction force = net force – weight down ramp = 2.1 «N» ✓
\n\n
ALTERNATIVE 2
\nkinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)2 = FNET x 0.35 ✓
\nFnet = 9.45 «N» ✓
\nWeight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓
\nfriction force = net force – weight down ramp = 2.1 «N» ✓
\n\n
Accept 1.95 N from g = 10 m s-2.
Accept 2.42 N from u = 2.14 m s-1.
static coefficient of friction > dynamic/kinetic coefficient of friction / μs > μk ✓
\n«therefore» force of dynamic/kinetic friction will be less than the force of static friction ✓
\n
there will be a net / unbalanced forward force once in motion «which results in acceleration»
OR
\nreference to net F = ma ✓
\nMany students recognized the vector nature of momentum implied in the question, although some focused on the forces acting on each object rather than discussing the momentum.
\nSome students simply calculated the net force acting on the load and did not recognize that this was not the tension force. Many set up a net force equation but had the direction of the forces backwards. This generally resulted from sloppy problem solving.
\nThis was a \"show that\" questions, so examiners were looking for a clear equation leading to a clear substitution of values leading to an answer that had more significant digits than the given answer. Most candidates successfully selected the correct equation and showed a proper substitution. Some candidates started with an energy approach that needed modification as it clearly led to an incorrect solution. These responses did not receive full marks.
\nThis SL only question was generally well done. Despite some power of 10 errors, many candidates correctly reported final answer to 2 sf.
\nCandidates struggled with this question. Very few drew a clear free-body diagram and many simply calculated the acceleration of the box from the given information and used this to calculate the net force on the box, confusing this with the frictional force.
\nThis was an \"explain\" question, so examiners were looking for a clear line of discussion starting with a comparison of the coefficients of friction, leading to a comparison of the relative magnitudes of the forces of friction and ultimately the rise of a net force leading to an acceleration. Many candidates recognized that this was a question about the comparison between static and kinetic/dynamic friction but did not clearly specify which they were referring to in their responses. Some candidates clearly did not read the stem carefully as they referred to the mass being on an incline.
\nWhite light is emitted from a hot filament. The light passes through hydrogen gas at low pressure and then through a diffraction grating onto a screen. A pattern of lines against a background appears on the screen.
\nWhat is the appearance of the lines and background on the screen?
\nD
\nA student measures the length l and width w of a rectangular table top.
\nWhat is the absolute uncertainty of the perimeter of the table top?
\n
A.
\nB.
\nC.
\nD.
\nB
\nThe graph shows the variation with distance r of the electric potential V from a charge Q.
\nWhat is the electric field strength at distance s?
\nA. The area under the graph between s and infinity
\nB. The area under the graph between 0 and s
\nC. The gradient of the tangent at s
\nD. The negative of the gradient of the tangent at s
\nD
\nA mass at the end of a string is moving in a horizontal circle at constant speed. The string makes an angle θ to the vertical.
\nWhat is the magnitude of the acceleration of the mass?
\n
A. g
B. g sin θ
\nC. g cos θ
\nD. g tan θ
\nD
\nWhich two features are necessary for the operation of a transformer?
\nB
\nThe gravitational field strength at the surface of a planet of radius R is . A satellite is moving in a circular orbit a distance R above the surface of the planet. What is the magnitude of the acceleration of the satellite?
\n
A.
B.
\nC.
\nD.
\nB
\nA conductor is placed in a uniform magnetic field perpendicular to the plane of the paper. A force F acts on the conductor when there is a current in the conductor as shown.
\nThe conductor is rotated 30° about the axis of the magnetic field.
\nWhat is the direction of the magnetic field and what is the magnitude of the force on the conductor after the rotation?
\nC
\nThis question requires careful reading by the candidate. Candidates needed to appreciate that the rotation relative to the magnetic field axis still produces a 90 degree angle between the conductor and the field. Option D was a very effective distractor for students.
\nA neutron is absorbed by a nucleus of uranium-235. One possible outcome is the production of two nuclides, barium-144 and krypton-89.
\nHow many neutrons are released in this reaction?
\nA. 0
\nB. 1
\nC. 2
\nD. 3
\nD
\nAnswer C 2 neutrons, was the most popular choice suggesting that candidates failed to read the question properly and missed 'a neutron is adsorbed' at the beginning.
\nWhat is the unit of power expressed in fundamental SI units?
\nA.
\nB.
\nC.
\nD.
\nD
\nA conducting bar with vertices PQRS is moving vertically downwards with constant velocity v through a horizontal magnetic field B that is directed into the plane of the page.
Which side of the bar will have the greatest density of electrons?
A. PQ
\nB. QR
\nC. RS
\nD. SP
\nD
\nA pure sample of radioactive nuclide decays into a stable nuclide .
\nWhat is after two half-lives?
\n
A. 1
B. 2
\nC. 3
\nD. 4
\nC
\nA longitudinal wave travels in a medium with speed 340 m s−1. The graph shows the variation with time t of the displacement x of a particle P in the medium. Positive displacements on the graph correspond to displacements to the right for particle P.
\nAnother wave travels in the medium. The graph shows the variation with time t of the displacement of each wave at the position of P.
\nA standing sound wave is established in a tube that is closed at one end and open at the other end. The period of the wave is . The diagram represents the standing wave at and at . The wavelength of the wave is 1.20 m. Positive displacements mean displacements to the right.
\nCalculate the wavelength of the wave.
\nState the phase difference between the two waves.
\nIdentify a time at which the displacement of P is zero.
\nEstimate the amplitude of the resultant wave.
\nCalculate the length of the tube.
\nA particle in the tube has its equilibrium position at the open end of the tube.
State and explain the direction of the velocity of this particle at time .
Draw on the diagram the standing wave at time .
\n«s» or «Hz» ✓
\n«m» ✓
\n\n
Allow ECF from MP1.
Award [2] for a bald correct answer.
«±» OR ✓
\n1.5 «ms» ✓
\n8.0 OR 8.5 «μm» ✓
\n
From the graph on the paper, value is 8.0. From the calculated correct trig functions, value is 8.49.
L = «» 0.90 «m» ✓
\nto the right ✓
displacement is getting less negative
\nOR
\nchange of displacement is positive ✓
\nhorizontal line drawn at the equilibrium position ✓
\nP and Q are two moons of equal densities orbiting a planet. The orbital radius of P is twice the orbital radius of Q. The volume of P is half that of Q. The force exerted by the planet on P is F. What is the force exerted by the planet on Q?
\nA. F
\nB. 2F
\nC. 4F
\nD. 8F
\nD
\nOption D was the most frequent (correct) answer, however option C was a significant distractor, perhaps for candidates considering only the change in orbital radius. A relatively high discrimination index was seen with this question.
\nA pure sample of iodine-131 decays into xenon with a half-life of 8 days.
\nWhat is after 24 days?
\nA.
\nB.
\nC.
\nD.
\nB
\nThe majority of candidates correctly selected option B. This question had the highest discrimination index on the HL paper.
\nCold milk enters a small sterilizing unit and flows over an electrical heating element.
\nThe temperature of the milk is raised from 11 °C to 84 °C. A mass of 55 g of milk enters the sterilizing unit every second.
\nSpecific heat capacity of milk = 3.9 kJ kg−1 K−1
\nThe milk flows out through an insulated metal pipe. The pipe is at a temperature of 84 °C. A small section of the insulation has been removed from around the pipe.
\nEstimate the power input to the heating element. State an appropriate unit for your answer.
\nOutline whether your answer to (a) is likely to overestimate or underestimate the power input.
\nDiscuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.
\nState how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.
\nThe missing section of insulation is 0.56 m long and the external radius of the pipe is 0.067 m. The emissivity of the pipe surface is 0.40. Determine the energy lost every second from the pipe surface. Ignore any absorption of radiation by the pipe surface.
\nDescribe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.
\nenergy required for milk entering in 1 s = mass x specific heat x 73 ✓
\n16 kW OR 16000 W ✓
\n\n
MP1 is for substitution into mcΔT regardless of power of ten.
\nAllow any correct unit of power (such as J s-1 OR kJ s-1) if paired with an answer to the correct power of 10 for MP2.
\nUnderestimate / more energy or power required ✓
\nbecause energy transferred as heat / thermal energy is lost «to surroundings or electrical components» ✓
\n\n
Do not allow general term “energy” or “power” for MP2.
\nthe temperature has increased so the internal energy / « average » KE «of the molecules» has increased OR temperature is proportional to average KE «of the molecules». ✓
\n«therefore» the «average» speed of the molecules or particles is higher OR more frequent collisions « between molecules » OR spacing between molecules has increased OR average force of collisions is higher OR intermolecular forces are less OR intermolecular bonds break and reform at a higher rate OR molecules are vibrating faster. ✓
\nconduction/conducting/conductor «through metal» ✓
\nuse of where T = 357 K ✓
\nuse of « = 0.236 m2» ✓
\nP = 87 «W» ✓
\n\n
Allow 85 – 89 W for MP3.
\nAllow ECF for MP3.
\nconvection «is likely to be a significant loss» ✓
\n
«due to reduction in density of air near pipe surface» hot air rises «and is replaced by cooler air from elsewhere»
OR
\n«due to» conduction «of heat or thermal energy» from pipe to air ✓
\nMost candidates recognized that this was a specific heat question and set up a proper calculation, but many struggled to match their answer to an appropriate unit. A common mistake was to leave the answer in some form of an energy unit and others did not match the power of ten of the unit to their answer (e.g. 16 W).
\nMany candidates recognized that this was an underestimate of the total energy but failed to provide an adequate reason. Many gave generic responses (such as \"some power will be lost\"/not 100% efficient) without discussing the specific form of energy lost (e.g. heat energy).
\nThis was generally well answered. Most HL candidates linked the increase in temperature to the increase in the kinetic energy of the molecules and were able to come up with a consequence of this change (such as the molecules moving faster). SL candidates tended to focus more on consequences, often neglecting to mention the change in KE.
\nMany candidates recognized that heat transfer by conduction was the correct response. This was a \"state\" question, so candidates were not required to go beyond this.
\nCandidates at both levels were able to recognize that this was a blackbody radiation question. One common mistake candidates made was not calculating the area of a cylinder properly. It is important to remind candidates that they are expected to know how to calculate areas and volumes for basic geometric shapes. Other common errors included the use of T in Celsius and neglecting to raise T ^4. Examiners awarded a large number of ECF marks for candidates who clearly showed work but made these fundamental errors.
\nA few candidates recognized that convection was the third source of heat loss, although few managed to describe the mechanism of convection properly for MP2. Some candidates did not read the question carefully and instead wrote about methods to increase the rate of heat loss (such as removing more insulation or decreasing the temperature of the environment).
\nA circuit consists of three identical capacitors of capacitance C and a battery of voltage V. Two capacitors are connected in parallel with a third in series. The capacitors are fully charged.
\nWhat is the charge stored in capacitors X and Z?
\nC
\nThe mass of a nucleus of iron-56 () is M.
\nWhat is the mass defect of the nucleus of iron-56?
\n\n
A. M − 26mp − 56mn
\nB. 26mp + 30mn − M
\nC. M − 26mp − 56mn − 26me
\nD. 26mp + 30mn + 26me − M
\nB
\nThe minute hand of a clock hanging on a vertical wall has length
\n
The minute hand is observed pointing at 12 and then again 30 minutes later when the minute hand is pointing at 6.
\nWhat is the average velocity and average speed of point P on the minute hand during this time interval?
\nA
\nTwo loudspeakers A and B are initially equidistant from a microphone M. The frequency and intensity emitted by A and B are the same. A and B emit sound in phase. A is fixed in position.
\nB is moved slowly away from M along the line MP. The graph shows the variation with distance travelled by B of the received intensity at M.
\nExplain why the received intensity varies between maximum and minimum values.
\nState and explain the wavelength of the sound measured at M.
\nB is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.
\nShow that the lowest frequency at which the intensity maximum can occur is about 3 kHz.
\nSpeed of sound = 340 m s−1
\nmovement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓
interference
OR
superposition «of waves» ✓
maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or » out of phase / path difference = (n+½) x lambda ✓
wavelength = 26 cm ✓
\n
peak to peak distance is the path difference which is one wavelength
OR
\nthis is the distance B moves to be back in phase «with A» ✓
\n\n
Allow 25 − 27 cm for MP1.
\n«» = 13 cm ✓
\n«» 2.6 «kHz» ✓
\n\n
Allow ½ of wavelength from (b) or data from graph.
\nThis was an \"explain\" questions, so examiners were looking for a clear discussion of the movement of speaker B creating a changing path difference between B and the microphone and A and the microphone. This path difference would lead to interference, and the examiners were looking for a connection between specific phase differences or path differences for maxima or minima. Some candidates were able to discuss basic concepts of interference (e.g. \"there is constructive and destructive interference\"), but failed to make clear connections between the physical situation and the given graph. A very common mistake candidates made was to think the question was about intensity and to therefore describe the decrease in peak height of the maxima on the graph. Another common mistake was to approach this as a Doppler question and to attempt to answer it based on the frequency difference of B.
\nMany candidates recognized that the wavelength was 26 cm, but the explanations were lacking the details about what information the graph was actually providing. Examiners were looking for a connection back to path difference, and not simply a description of peak-to-peak distance on the graph. Some candidates did not state a wavelength at all, and instead simply discussed the concept of wavelength or suggested that the wavelength was constant.
\nThis was a \"show that\" question that had enough information for backwards working. Examiners were looking for evidence of using the wavelength from (b) or information from the graph to determine wavelength followed by a correct substitution and an answer to more significant digits than the given result.
\nA person is standing at rest on the ground and experiences a downward gravitational force W and an upward normal force from the ground N. Which, according to Newton’s third law, is the force that together with W forms a force pair?
\nA. The gravitational force W acting upwards on the ground.
\nB. The gravitational force W acting upwards on the person.
\nC. The normal force N acting upwards on the person.
\nD. The normal force N acting downwards on the ground.
\nA
\nThe diagram shows atomic transitions E1, E2 and E3 when a particular atom changes its energy state. The wavelengths of the photons that correspond to these transitions are , and .
\nWhat is correct for these wavelengths?
\nA.
\nB.
\nC.
\nD.
\nD
\nA fixed mass of an ideal gas is contained in a cylinder closed with a frictionless piston. The volume of the gas is 2.5 × 10−3 m3 when the temperature of the gas is 37 °C and the pressure of the gas is 4.0 × 105 Pa.
\nEnergy is now supplied to the gas and the piston moves to allow the gas to expand. The temperature is held constant.
\nCalculate the number of gas particles in the cylinder.
\nDiscuss, for this process, the changes that occur in the density of the gas.
\nDiscuss, for this process, the changes that occur in the internal energy of the gas.
\nCorrect conversion of T «T = 310 K» seen ✓
\n« use of = to get » 2.3 × 1023 ✓
\n\n
Allow ECF from MP1 i.e., T in Celsius (Result is 2.7 x 1024)
\nAllow use of n, R and NA
\ndensity decreases ✓
\nvolume is increased AND mass/number of particles remains constant ✓
\ninternal energy is constant ✓
\n
internal energy depends on kinetic energy/temperature «only»
OR
\nsince temperature/kinetic energy is constant ✓
\n\n
Do not award MP2 for stating that “temperature is constant” unless linked to the correct conclusion, as that is mentioned in the stem.
\nAward MP2 for stating that kinetic energy remains constant.
\na) This was well answered with the majority converting to K. Quite a few found the number of moles but did not then convert to molecules.
\nbi) Well answered. It was pleasing to see how many recognised the need to state that the mass/number of molecules stayed the same as well as stating that the volume increased. At SL this recognition was less common so only 1 mark was often awarded.
\nbii) This was less successfully answered. A surprising number of candidates said that the internal energy of an ideal gas increases during an isothermal expansion. Many recognised that constant temp meant constant KE but then went on to state that the PE must increase and so the internal energy would increase.
A person with a weight of stands on a scale in an elevator.
\nWhat is the acceleration of the elevator when the scale reads ?
\nA. downwards
\nB. downwards
\nC. upwards
\nD. upwards
\nD
\nThe Higgs boson was discovered in the Large Hadron Collider at CERN. Which statements are correct about the discovery of the Higgs boson?
\nI. It was independent of previous theoretical work.
II. It involved analysing large amounts of experimental data.
III. It was consistent with the standard model of particle physics.
\n
A. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nC
\nCarbon (C-12) and hydrogen (H-1) undergo nuclear fusion to form nitrogen.
\nphoton
\nWhat is the number of neutrons and number of nucleons in the nitrogen nuclide?
\nB
\nThis question was well answered by HL candidates, although a significant number of candidates incorrectly identified the number of neutrons present in the nitrogen nuclide.
\nThree correct statements about the behaviour of electrons are:
\nI. An electron beam is used to investigate the structure of crystals.
II. An electron beam produces a pattern of fringes when sent through two narrow parallel slits.
III. Electromagnetic radiation ejects electrons from the surface of a metal.
Which statements are explained using the wave-like properties of electrons?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nTwo identical boxes containing different masses are sliding with the same initial speed on the same horizontal surface. They both come to rest under the influence of the frictional force of the surface. How do the frictional force and acceleration of the boxes compare?
\nB
\nLight of wavelength is diffracted after passing through a very narrow single slit of width . The intensity of the central maximum of the diffracted light is . The slit width is doubled.
\nWhat is the intensity of central maximum and the angular position of the first minimum?
\nB
\nOption B was the most frequent (correct) selected by candidates. Interestingly, the more able candidates were distracted by option D, who were likely considering the intensity/amplitude relationship. As a result, this would be a good MC question for teaching purposes.
\nA proton collides with an electron. What are the possible products of the collision?
\n\n
A. Two neutrons
\nB. Neutron and positron
\nC. Neutron and antineutrino
\nD. Neutron and neutrino
\nD
\nSamples of two radioactive nuclides X and Y are held in a container. The number of particles of X is half the number of particles of Y. The half-life of X is twice the half-life of Y.
\nWhat is the initial value of ?
\nA.
\nB.
\nC.
\nD.
\nA
\nA loudspeaker emits sound waves of frequency towards a metal plate that reflects the waves. A small microphone is moved along the line from the metal plate to the loudspeaker. The intensity of sound detected at the microphone as it moves varies regularly between maximum and minimum values.
\nThe speed of sound in air is 340 m s−1.
\nExplain the variation in intensity.
\nAdjacent minima are separated by a distance of 0.12 m. Calculate .
\nThe metal plate is replaced by a wooden plate that reflects a lower intensity sound wave than the metal plate.
\nState and explain the differences between the sound intensities detected by the same microphone with the metal plate and the wooden plate.
\n«incident and reflected» waves superpose/interfere/combine ✓
\n«that leads to» standing waves formed OR nodes and antinodes present ✓
\nat antinodes / maxima there is maximum intensity / constructive interference / «displacement» addition / louder sound ✓
\nat nodes / minima there is minimum intensity / destructive interference / «displacement» cancellation / quieter sound ✓
\n\n
OWTTE
\nAllow a sketch of a standing wave for MP2
\nAllow a correct reference to path or phase differences to identify constructive / destructive interference
\nwavelength = 0.24 «m» ✓
\n= «=» 1.4 «kHz» OR 1400 «Hz» ✓
\n\n
Allow ECF from MP1
\nrelates intensity to amplitude ✓
\nantinodes / maximum intensity will be decreased / quieter ✓
\nnodes / minimum will be increased / louder ✓
\ndifference in intensities will be less ✓
\nmaxima and minima are at the same positions ✓
\n\n
OWTTE
\nai) On most occasions it looked like students knew more than they could successfully communicate. Lots of answers talked about interference between the 2 waves, or standing waves being produced but did not go on to add detail. Candidates should take note of how many marks the question part is worth and attempt a structure of the answer that accounts for that. At SL there were problems recognizing a standard question requiring the typical explanation of how a standing wave is established.
\n3aii) By far the most common answer was 2800 Hz, not doubling the value given to get the correct wavelength. That might suggest that some students misinterpreted adjacent minima as two troughs, therefore missing to use the information to correctly determine the wavelength as 0.24 m.
\nb) A question that turned out to be a good high level discriminator. Most candidates went for an answer that generally had everything at a lower intensity and didn't pick up on the relative amount of superposition. Those that did answer it very well, with very clear explanations, succeeded in recognizing that the nodes would be louder and the anti-nodes would be quieter than before.
\nA fuel has mass density and energy density . What mass of the fuel has to be burned to release thermal energy ?
\n\n
A.
\nB.
\nC.
\nD.
\nA
\nAn observer with an eye of pupil diameter views the headlights of a car that emit light of wavelength . The distance between the headlights is .
\nWhat is the greatest distance between the observer and the car at which the images of the headlights will be resolved by the observer’s eye?
\nA.
\nB.
\nC.
\nD.
\nC
\nDespite option C being the most frequent (correct) response, both options A and B were effective distractors. A useful teaching point relating to this question relates to unit analysis; neither solution presented in option A or option D produce the correct units for distance, and could be eliminated from consideration on this basis.
\nTwo identical blocks, each of mass m and speed v, travel towards each other on a frictionless surface.
\nThe blocks undergo a head-on collision. What is definitely true immediately after the collision?
\nA. The momentum of each block is zero.
\nB. The total momentum is zero.
\nC. The momentum of each block is 2mv.
\nD. The total momentum is 2mv.
\nB
\nA power supply is connected to three resistors P, Q and R of fixed value and to an ideal voltmeter. A variable resistor S, formed from a solid cylinder of conducting putty, is also connected in the circuit. Conducting putty is a material that can be moulded so that the resistance of S can be changed by re-shaping it.
\nThe resistance values of P, Q and R are 40 Ω, 16 Ω and 60 Ω respectively. The emf of the power supply is 6.0 V and its internal resistance is negligible.
\nAll the putty is reshaped into a solid cylinder that is four times longer than the original length.
\nCalculate the potential difference across P.
\nThe voltmeter reads zero. Determine the resistance of S.
\nDeduce the resistance of this new cylinder when it has been reshaped.
\nOutline, without calculation, the change in the total power dissipated in Q and the new cylinder after it has been reshaped.
\nALTERNATIVE 1
\nattempt to use potential divider equation or similar method ✓
«»= 2.4 «V» ✓
ALTERNATIVE 2
«current = » = 0.06 «A» ✓
\n40 x 0.06 = 2.4 «V» ✓
\nALTERNATIVE 1
\nPd across Q = 2.4 V so I = 0.15 « A » ✓
\nand pd across S is 6.0 – 2.4 = 3.6 « V » ✓
\n«Ω» ✓
\n\n
ALTERNATIVE 2
\npd at PR junction = pd at QS junction ✓
\nso OR ✓
\n24 «Ω» ✓
\n\n
ALTERNATIVE 3
\n«A» ✓
\nOR ✓
\nR = 24 «Ω» ✓
\n\n
Allow ECF for MP3 from incorrect MP1 or MP2.
\nrecognition that leads to / «» = ✓
\n«because the volume of S is constant new area is» ✓
\n16 x 24 = 384 «Ω» ✓
\n«total» power has decreased ✓
\n
Because current in the branch has decreased «and P=I2R »
OR
\nBecause resistance has increased in branch «and » ✓
\n\n
Allow opposite argument as ECF from (c)(i) (if candidate deduces a lower resistance).
\nAllow “power doesn’t change” if candidate has no change of resistance from (b) to (c)(i).
\nThis was generally well done at the higher level. Some SL candidates struggled to calculate the correct current but earned the second marking point through ECF.
\nMany candidates struggled with this question. A very common mistake was to assume the current was the same in each branch, leading to a resistance of 84 Ohms. The placement of the voltmeter may have caused some confusion for candidates, and they may not have understood what the zero reading was indicating. It is important that candidates understand what voltmeters are actually reading and are familiar with different placements in circuits.
\nMost candidates recognized that increasing the length of the conductive putty would increase the resistance by a factor of four, but very few considered that if the volume of the putty remained constant that the cross-sectional surface area would decrease as well.
\nThis was an item that caused a bit of confusion for candidates. The prompt asks for a comparison of the power in the branch before and after changing the length of the conductive putty. Many candidates correctly identified that the power in Q would decrease, but either did not discuss the power in the whole branch or were not clear that the power in the putty would decrease as well.
\nA charged particle, P, of charge +68 μC is fixed in space. A second particle, Q, of charge +0.25 μC is held at a distance of 48 cm from P and is then released.
\nThe diagram shows two parallel wires X and Y that carry equal currents into the page.
\nPoint Q is equidistant from the two wires. The magnetic field at Q due to wire X alone is 15 mT.
\nThe work done to move a particle of charge 0.25 μC from one point in an electric field to another is 4.5 μJ. Calculate the magnitude of the potential difference between the two points.
\nDetermine the force on Q at the instant it is released.
\nDescribe the motion of Q after release.
\nOn the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone.
\nDetermine the magnitude and direction of the resultant magnetic field at Q.
\n«» 18 «V» ✓
\n✓
\n«N» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAllow symbolic k in substitutions for MP1.
\nDo not allow ECF from incorrect or not squared distance.
\nQ moves to the right/away from P «along a straight line»
\nOR
\nQ is repelled from P ✓
\n
with increasing speed/Q accelerates ✓
acceleration decreases ✓
\n\n
arrow of any length as shown ✓
\n«using components or Pythagoras to get» B = 21 «mT» ✓
\ndirected «horizontally» to the right ✓
\n\n
If no unit seen, assume mT.
\nThe Sankey diagram shows the energy transfers in a nuclear power station.
\nElectrical power output of the power station is 1000 MW.
\nWhat is the thermal power loss in the heat exchanger?
\n\n
A. 500 MW
\nB. 1000 MW
\nC. 1500 MW
\nD. 2500 MW
\nB
\nDescribe the quark structure of a baryon.
\nThe Feynman diagram shows a possible decay of the K+ meson.
\nIdentify the interactions that are involved at points A and B in this decay.
\nThe K+ meson can decay as
\nK+ → μ+ + vμ.
\nState and explain the interaction that is responsible for this decay.
\n3 quarks / example with three quarks «e.g. up up down» ✓
\ninteger / zero / 1 / no fractional «electron» charge
OR
held together by the strong force / gluons
OR
half integer spin
OR
baryon number = 1
OR
colour neutral ✓
A «Decay of the strange antiquark is a» weak «interaction» ✓
\nB «Decay of the u to a gluon and eventually to d and anti-d is a» strong «interaction» ✓
\nweak «interaction» ✓
\nstrangeness is not conserved and this is possible only in weak interactions
OR
the weak interaction allows change of quark flavour
OR
only the weak interaction has a boson / an exchange particle / a W+ to conserve the charge
OR
neutrinos are only produced via the weak interaction ✓
A significant number of candidates recognized that baryons are composed of three quarks. The second mark was for a statement concerning baryons as a result of the quark composition, and not for a general statement about the quarks (e.g. \"the baryon number is 1\" rather than \"each quark has a baryon number of ⅓). It is worth noting that the information about individual quarks is given in the data booklet which is why no marks were awarded for simply copying this information over.
\nMany candidates were able to successfully identify the two interactions in the diagram. Some candidates only described what was happening in the diagram without identifying the actual interaction. A common mistake was identifying the gluon at B as a graviton, and/or suggesting that this was a gravitational interaction. Many candidates also did not make the connection between the term \"interaction\" in the stem and the concept of force.
\nThis was another item where some candidates simply described the particles without specifying the weak interaction. The second marking point was for a justification based on an aspect of this decay that could only be true of the weak nuclear force. A commonly incorrect answer was that this was the only force that acted on quarks and leptons, which was not accepted due to the fact that the gravitational force also acts on these particles as well. Another common incorrect answer among SL candidates was to assume that this was an example of beta negative decay due to the presence of a neutrino.
\nA charged sphere in a gravitational field is initially stationary between two parallel metal plates. There is a potential difference V between the plates.
\nThree changes can be made:
\nI. Increase the separation of the metal plates
II. Increase V
III. Apply a magnetic field into the plane of the paper
What changes made separately will cause the charged sphere to accelerate?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nA
\nOption C was a very successful distractor, selected by the majority of candidates. Most candidates missed that change III (\"Apply a magnetic field into the plane of the paper\") can never be correct if the charge is stationary.
\nA projectile is launched upwards at an angle θ to the horizontal with an initial momentum p0 and an initial energy E0. Air resistance is negligible. What are the momentum and total energy of the projectile at the highest point of the motion?
\nA
\nWhich is correct for a black-body radiator?
\n
A. The power it emits from a unit surface area depends on the temperature only.
B. It has an albedo of 1.
\nC. It emits monochromatic radiation whose wavelength depends on the temperature only.
\nD. It emits radiation of equal intensity at all wavelengths.
\nA
\nThe graph shows the variation with distance of a horizontal force acting on an object. The object, initially at rest, moves horizontally through a distance of .
\nA constant frictional force of opposes the motion. What is the final kinetic energy of the object after it has moved ?
\nA.
\nB.
\nC.
\nD.
\nB
\nA cyclist rides up a hill of vertical height 100 m in 500 s at a constant speed. The combined mass of the cyclist and the bicycle is 80 kg. The power developed by the cyclist is 200 W. What is the efficiency of the energy transfer in this system?
\n\n
A. 8 %
\nB. 20 %
\nC. 60 %
\nD. 80 %
\nD
\nAn object of mass is launched from the surface of the Earth. The Earth has a mass and radius . The acceleration due to gravity at the surface of the Earth is . What is the escape speed of the object from the surface of the Earth?
\nA.
\nB.
\nC.
\nD.
\nB
\nOptions B and C were selected by a roughly equal number of candidates. Again, this is a situation where unit analysis is beneficial; options C and D would not produce units associated with speed (mass is already incorporated in the constant 'g').
\nA sample of oxygen gas with a volume of is at . The gas is heated so that it expands at a constant pressure to a final volume of . What is the final temperature of the gas?
\nA.
\nB.
\nC.
\nD.
\nB
\nA cell is connected to an ideal voltmeter, a switch S and a resistor R. The resistance of R is 4.0 Ω.
\nWhen S is open the reading on the voltmeter is 12 V. When S is closed the voltmeter reads 8.0 V.
\nElectricity can be generated using renewable resources.
\nIdentify the laws of conservation that are represented by Kirchhoff’s circuit laws.
\nState the emf of the cell.
\nDeduce the internal resistance of the cell.
\nThe voltmeter is used in another circuit that contains two secondary cells.
\nCell A has an emf of 10 V and an internal resistance of 1.0 Ω. Cell B has an emf of 4.0 V and an internal resistance of 2.0 Ω.
\nCalculate the reading on the voltmeter.
\nOutline why electricity is a secondary energy source.
\nSome fuel sources are renewable. Outline what is meant by renewable.
\nA fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.
\nThe cell is then re-charged by a rectangular solar panel of dimensions 0.40 m × 0.15 m at a place where the maximum intensity of sunlight is 380 W m−2.
\nThe overall efficiency of the re-charging process is 18 %.
\nCalculate the minimum time required to re-charge the cell fully.
\nOutline why research into solar cell technology is important to society.
\n« conservation of » charge ✓
\n« conservation of » energy ✓
\n\n
Allow [1] max if they explicitly refer to Kirchhoff’ laws linking them to the conservation laws incorrectly.
\n12 V ✓
\nI = 2.0 A OR 12 = I (r +4) OR 4 = Ir OR 8 = 4I ✓
\n«Correct working to get » r = 2.0 «Ω» ✓
\n\n
Allow ECF from (b)(i)
\nLoop equation showing EITHER correct voltages, i.e., 10 – 4 on one side or both emfs positive on different sides of the equation OR correct resistances, i.e. I (1 + 2) ✓
\n10−4 = I (1 + 2) OR I = 2.0 «A» seen✓
\nV = 8.0 «V» ✓
\n\n
Allow any valid method
\nis generated from primary/other sources ✓
\n«a fuel » that can be replenished/replaced within a reasonable time span
\nOR
\n«a fuel» that can be replaced faster than the rate at which it is consumed
\nOR
\nrenewables are limitless/never run out
\nOR
\n«a fuel» produced from renewable sources
\nOR
\ngives an example of a renewable (biofuel, hydrogen, wood, wind, solar, tidal, hydro etc..) ✓
\n\n
OWTTE
\nALTERNATIVE 1
\n«energy output of the panel =» VIt OR 6 x 5 x 0.25 x 3600 OR 27000 «J» ✓
\n«available power =» 380 x 0.4 x 0.15 x 0.18 OR 4.1 «W» ✓
\n«=» 6600 «s» ✓
\n\n
ALTERNATIVE 2
\n«energy needed from Sun =» OR OR 150000 «J» ✓
\n« incident power=» 380 x 0.4 x 0.15 OR 22.8 «W» ✓
\n«=» 6600 «s» ✓
\n\n
Allow ECF for MP3
\nAccept final answer in minutes (110) or hours (1.8).
\ncoherent reason ✓
\ne.g., to improve efficiency, is non-polluting, is renewable, does not produce greenhouse gases, reduce use of fossil fuels
\n\n
Do not allow economic reasons
\na) Most just stated Kirchhoff's laws rather than the underlying laws of conservation of energy and charge, basically describing the equations from the data booklet. When it came to guesses, energy and momentum were often the two, although even a baryon and lepton number conservation was found. It cannot be emphasised enough the importance of correctly identifying the command verb used to introduce the question. In this case, identify, with the specific reference to conservation laws, seem to have been explicit tips not picked up by some candidates.
\nbi) This was probably the easiest question on the paper and almost everybody got it right. 12V. Some calculations were seen, though, that contradict the command verb used. State a value somehow implies that the value is right in front to be read or interpreted suitably.
\nbii) In the end a lot of the answers here were correct. Some obtained 2 ohms and were able to provide an explanation that worked. A very few negative answers were found, suggesting that some candidates work mechanically without properly reflecting in the nature of the value obtained.
\nci) A lot of candidates figured out they had to do some sort of loop here but most had large currents in the voltmeter. Currents of 2 A and 10 A simultaneously were common. Some very good and concise work was seen though, leading to correct steps to show a reading of 8V.
\ncii) This question was cancelled due to an internal reference error. The paper total was adjusted in grade award. This is corrected for publication and future teaching use.
\ndi) The vast majority of candidates could explain why electricity was a secondary energy source.
\ndii) An ideal answer was that renewable fuels can be replenished faster than they are consumed. However, many imaginative alternatives were accepted.
\nei) This question was often very difficult to mark. Working was often scattered all over the answer box. Full marks were not that common, most candidates achieved partial marks. The commonest problem was determining the energy required to charge the battery. It was also common to see a final calculation involving a power divided by a power to calculate the time.
\neii) Almost everybody could give a valid reason why research into solar cells was important. Most answers stated that solar is renewable. There were very few that didn't get a mark due to discussing economic reasons.
\nTwo loudspeakers A and B are initially equidistant from a microphone M. The frequency and intensity emitted by A and B are the same. A and B emit sound in phase. A is fixed in position.
\nB is moved slowly away from M along the line MP. The graph shows the variation with distance travelled by B of the received intensity at M.
\nExplain why the received intensity varies between maximum and minimum values.
\nState and explain the wavelength of the sound measured at M.
\nB is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.
\nShow that the lowest frequency at which the intensity maximum can occur is about 3 kHz.
\nSpeed of sound = 340 m s−1
\nLoudspeaker A is switched off. Loudspeaker B moves away from M at a speed of 1.5 m s−1 while emitting a frequency of 3.0 kHz.
\nDetermine the difference between the frequency detected at M and that emitted by B.
\nmovement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓
interference
OR
superposition «of waves» ✓
maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or » out of phase / path difference = (n+½) x lambda ✓
wavelength = 26 cm ✓
\n
peak to peak distance is the path difference which is one wavelength
OR
\nthis is the distance B moves to be back in phase «with A» ✓
\n\n
Allow 25 – 27 cm for MP1.
\n«» = 13 cm ✓
\n«» 2.6 «kHz» ✓
\n\n
Allow ½ of wavelength from (b) or data from graph for MP1.
\nAllow ECF from MP1.
\nALTERNATIVE 1
use of (+ sign must be seen) OR = 2987 «Hz» ✓
« » = 13 «Hz» ✓
\n
ALTERNATIVE 2
Attempted use of ≈
« Δf » = 13 «Hz» ✓
This was an \"explain\" questions, so examiners were looking for a clear discussion of the movement of speaker B creating a changing path difference between B and the microphone and A and the microphone. This path difference would lead to interference, and the examiners were looking for a connection between specific phase differences or path differences for maxima or minima. Some candidates were able to discuss basic concepts of interference (e.g. \"there is constructive and destructive interference\"), but failed to make clear connections between the physical situation and the given graph. A very common mistake candidates made was to think the question was about intensity and to therefore describe the decrease in peak height of the maxima on the graph. Another common mistake was to approach this as a Doppler question and to attempt to answer it based on the frequency difference of B.
\nMany candidates recognized that the wavelength was 26 cm, but the explanations were lacking the details about what information the graph was actually providing. Examiners were looking for a connection back to path difference, and not simply a description of peak-to-peak distance on the graph. Some candidates did not state a wavelength at all, and instead simply discussed the concept of wavelength or suggested that the wavelength was constant.
\nThis was a \"show that\" question that had enough information for backwards working. Examiners were looking for evidence of using the wavelength from (b) or information from the graph to determine wavelength followed by a correct substitution and an answer to more significant digits than the given result.
\nMany candidates were successful in setting up a Doppler calculation and determining the new frequency, although some missed the second step of finding the difference in frequencies.
\nAn insulated container of negligible mass contains a mass 2M of a liquid. A piece of a metal of mass M is dropped into the liquid. The temperature of the liquid increases by 10 °C and the temperature of the metal decreases by 80 °C in the same time.
\nWhat is ?
\n
A. 2
B. 4
\nC. 8
\nD. 16
\nB
\nThe graph shows the variation of magnetic flux in a coil with time .
\nWhat represents the variation with time of the induced emf across the coil?
\nA
\nCandidate selections were divided across the options presented, with option B as the most common (incorrect answer). This suggests that candidates could use more guidance on how to interpret slope on a ɛ vs. t graph.
\nA block rests on a frictionless horizontal surface. An air rifle pellet is fired horizontally into the block and remains embedded in the block.
\nWhat happens to the total kinetic energy and to the total momentum of the block and pellet system as a result of the collision?
\nC
\nTwo identical containers X and Y each contain an ideal gas. X has N molecules of gas at an absolute temperature of T and Y has 3N molecules of gas at an absolute temperature of What is the ratio of the pressures ?
\nA.
\nB.
\nC.
\nD.
\nC
\nThe molar mass of an ideal gas is . A fixed mass of the gas expands at a constant pressure . The graph shows the variation with temperature T of the gas volume V.
\nWhat is the gradient of the graph?
\n
A.
B.
\nC.
\nD.
\nD
\nPlutonium-238 (Pu) decays by alpha (α) decay into uranium (U).
\nThe following data are available for binding energies per nucleon:
\nplutonium 7.568 MeV
\nuranium 7.600 MeV
\nalpha particle 7.074 MeV
\nState what is meant by the binding energy of a nucleus.
\nDraw, on the axes, a graph to show the variation with nucleon number of the binding energy per nucleon, . Numbers are not required on the vertical axis.
\nIdentify, with a cross, on the graph in (a)(ii), the region of greatest stability.
\nShow that the energy released in this decay is about 6 MeV.
\nThe plutonium nucleus is at rest when it decays.
\nCalculate the ratio .
\nthe energy needed to «completely» separate the nucleons of a nucleus
\nOR
\nthe energy released when a nucleus is assembled from its constituent nucleons ✓
\n\n
Accept reference to protons AND neutrons.
\ncurve rising to a maximum between 50 and 100 ✓
\ncurve continued and decreasing ✓
\n\n
Ignore starting point.
Ignore maximum at alpha particle
\nAt a point on the peak of their graph ✓
\ncorrect mass numbers for uranium (234) and alpha (4) ✓
\n«MeV» ✓
\nenergy released 5.51 «MeV» ✓
\n\n
Ignore any negative sign.
\n«» OR ✓
\n«» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAccept for MP2.
\nThe diagram shows an interference pattern observed on a screen in a double-slit experiment with monochromatic light of wavelength 600 nm. The screen is 1.0 m from the slits.
\nWhat is the separation of the slits?
\n
A. 6.0 × 10−7 m
B. 6.0 × 10−6 m
\nC. 6.0 × 10−5 m
\nD. 6.0 × 10−4 m
\nD
\nA direct current in a lamp dissipates power P. What root mean square (rms) value of an alternating current dissipates average power P through the same lamp?
\nA.
\nB.
\nC.
\nD.
\nC
\nThis question was very challenging for HL candidates, with the lowest difficulty index on the paper. Option B was a very effective distractor; perhaps candidates were selecting their answer based on a connection between rms values and .
\nA piece of metal at a temperature of is dropped into an equal mass of water at a temperature of in a container of negligible mass. The specific heat capacity of water is four times that of the metal. What is the final temperature of the mixture?
\nA.
\nB.
\nC.
\nD.
\nD
\nThe arrangement shows four diodes connected to an alternating current (ac) supply. The output is connected to an external circuit.
\nWhat is the output to the external circuit?
\nA. Full-wave rectified current
\nB. Half-wave rectified current
\nC. Constant non-zero current
\nD. Zero current
\nD
\nThis was another challenging HL question, where candidate responses were divided across the options. Options A and B were most common, due perhaps to the fact that the presentation of this circuit looks like a rectifier. Candidates need to pay attention to the direction of the diodes when considering this type of circuit.
\nA cell has an emf of 3.0 V and an internal resistance of 2.0 Ω. The cell is connected in series with a resistance of 10 Ω.
\nWhat is the terminal potential difference of the cell?
\n
A. 0.5 V
B. 1.5 V
\nC. 2.5 V
\nD. 3.0 V
\nC
\nThe bob of a pendulum has an initial displacement to the right. The bob is released and allowed to oscillate. The graph shows how the displacement varies with time. At which point is the velocity of the bob at its maximum magnitude directed towards the left?
\n\n
C
\nTwo parallel wires carry equal currents in the same direction out of the paper. Which diagram shows the magnetic field surrounding the wires?
\nA
\nThree identical capacitors are connected together as shown.
\nWhat is the order of increasing total capacitance for these arrangements?
\nA. P, S, R, Q
\nB. Q, R, S, P
\nC. P, R, S, Q
\nD. Q, S, R, P
\nC
\nOption C was the most common selection, however all other options were effective distractors. This question would be useful in a review of combined capacitance when capacitors in series and parallel.
\nUnpolarized light of intensity is incident on a polarizer. The light that passes through this polarizer then passes through a second polarizer.
\n
The second polarizer can be rotated to vary the intensity of the emergent light. What is the maximum value of the intensity emerging from the second polarizer?
\nA.
\nB.
\nC.
\nD.
\nB
\nA detector measures the count rate from a sample of a radioactive nuclide. The graph shows the variation with time of the count rate.
\nThe nuclide has a half-life of 20 s. The average background count rate is constant.
\nWhat is the average background count rate?
\n
A. 1 s−1
B. 2 s−1
\nC. 3 s−1
\nD. 4 s−1
\nC
\nLight with photons of energy 8.0 × 10−20 J are incident on a metal surface in a photoelectric experiment.
\nThe work function of the metal surface is 4.8 × 10−20 J . What minimum voltage is required for the ammeter reading to fall to zero?
\nA. 0.2 V
\nB. 0.3 V
\nC. 0.5 V
\nD. 0.8 V
\nA
\nOptions B and C were both effective distractors in this photoelectric effect question. There was a heightened number of blanks (no response) relative to the questions immediately before and after, and the low difficulty index suggests that candidates found this question challenging.
\nTwo wave generators, placed at position P and position Q, produce water waves with a wavelength of. Each generator, operating alone, will produce a wave oscillating with an amplitude of at position R. PR is and RQ is .
\nBoth wave generators now operate together in phase. What is the amplitude of the resulting wave at R?
\n
A.
B.
\nC.
\nD. zero
\nD
\nTitan is a moon of Saturn. The Titan-Sun distance is 9.3 times greater than the Earth-Sun distance.
\nShow that the intensity of the solar radiation at the location of Titan is 16 W m−2
\nTitan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m−2
\nShow that the equilibrium surface temperature of Titan is about 90 K.
\nThe orbital radius of Titan around Saturn is and the period of revolution is .
\nShow that where is the mass of Saturn.
\nThe orbital radius of Titan around Saturn is 1.2 × 109 m and the orbital period is 15.9 days. Estimate the mass of Saturn.
\nincident intensity OR «W m−2» ✓
\n\n
Allow the use of 1400 for the solar constant.
\nexposed surface is ¼ of the total surface ✓
\nabsorbed intensity = (1−0.22) × incident intensity ✓
\n0.78 × 0.25 × 15.7 OR 3.07 «W m−2» ✓
\n\n
Allow 3.06 from rounding and 3.12 if they use 16 W m−2.
\nσT 4 = 3.07
\nOR
\nT = 86 «K» ✓
\ncorrect equating of gravitational force / acceleration to centripetal force / acceleration ✓
\ncorrect rearrangement to reach the expression given ✓
\n\n
Allow use of for MP1.
\n«s» ✓
\n«kg» ✓
\n\n
Award [2] marks for a bald correct answer.
\nAllow ECF from MP1.
\nA glass block has a refractive index in air of ng. The glass block is placed in two different liquids: liquid X with a refractive index of nX and liquid Y with a refractive index of nY.
\nIn liquid X and in liquid Y What is ?
\nA.
\nB.
\nC.
\nD.
\nC
\nThe Feynman diagram shows an interaction between a proton and an electron.
\nWhat is the charge of the exchange particle and what is the lepton number of particle X?
\nB
\nThe diagram shows a simple model of the energy balance in the Earth surface-atmosphere system. The intensities of the radiations are given.
\nWhat is the average intensity radiated by the atmosphere towards the surface?
\n
A. 100 W m−2
B. 150 W m−2
\nC. 240 W m−2
\nD. 390 W m−2
\nB
\nAn experiment is carried out to determine the count rate, corrected for background radiation, when different thicknesses of copper are placed between a radioactive source and a detector. The graph shows the variation of corrected count rate with copper thickness.
\nOutline how the count rate was corrected for background radiation.
\nWhen a single piece of thin copper foil is placed between the source and detector, the count rate is 810 count minute−1. The foil is replaced with one that has three times the thickness. Estimate the new count rate.
\nFurther results were obtained in this experiment with copper and lead absorbers.
\nComment on the radiation detected from this radioactive source.
\nAnother radioactive source consists of a nuclide of caesium that decays to barium .
\nWrite down the reaction for this decay.
\nbackground count rate is subtracted «from each reading» ✓
\n\n
OWTTE
\nthickness is 0.25 «mm» ✓
\n380 «count min−1» ✓
\n\n
MP1 and MP2 can be shown on the graph
\nAllow a range of 0.23 to 0.27 mm for MP1
\nAllow ECF from MP1.
\nAccept a final answer in the range 350 – 420
\nlead better absorber than copper ✓
\nnot alpha ✓
\nas it does not go through the foil / it is easily stopped / it is stopped by paper ✓
\nthere is gamma ✓
\nas it goes through lead ✓
\n\n
ALTERNATIVE 1
\ncan be beta ✓
\nas it is attenuated by «thin» metal / can go through «thin» metal ✓
\n\n
ALTERNATIVE 2
\nnot beta ✓
\nit is stopped by «thin» metal ✓
\n✓
\n✓
\n\n
Accept β or e in MP1.
\nDo not penalize if proton / nucleon numbers or electron subscript in antineutrino are missing.
\na) A majority of candidates were able to say that background radiation count was subtracted from all readings.
\nb) A fairly easy question with most candidates being able to take readings from the graph to get a final count rate of approximately 380 counts per second. Many did not seem to have used a ruler to help their reading.
\nc) This was a bit chaotic with candidates showing all sorts of misconceptions. The first marking point was the one most commonly awarded. The 2 big misconceptions were that the copper and lead were radioactive themselves and produced the radiation, or that the higher the figures the better absorbers they were. Far too many candidates thought that the question was only about the radiation passing through the 3.5 mm of lead and copper. Most of these candidates realised that there must be some gamma radiation in the radiation detected. Far fewer stated that there could not be any alpha. Opinions varied as to whether there was beta, but any sensible answers were given credit.
\nd) This question was generally well answered, with most candidates getting
\nA simple pendulum undergoes simple harmonic motion. The gravitational potential energy of the pendulum is zero at the equilibrium position. How many times during one oscillation is the kinetic energy of the pendulum equal to its gravitational potential energy?
\n
A. 1
B. 2
\nC. 3
\nD. 4
\nD
\nWhen monochromatic light is incident on a single slit a diffraction pattern forms on a screen. The width of the slit is decreased.
\nWhat are the changes in the width and in the intensity of the central maximum of the diffraction pattern?
\nB
\nThe frequency of the first harmonic in a pipe is measured. An adjustment is then made which causes the speed of sound in the pipe to increase. What is true for the frequency and the wavelength of the first harmonic when the speed of sound has increased?
\nA
\nThe diagram shows two cylindrical wires, X and Y. Wire X has a length , a diameter , and a resistivity . Wire Y has a length , a diameter of and a resistivity of .
\nWhat is ?
\nA. 4
\nB. 2
\nC. 0.5
\nD. 0.25
\nD
\nThe diagram shows equipotential lines for an electric field. Which arrow represents the acceleration of an electron at point P?
D
\nMonochromatic light of wavelength in air is incident normally on a thin liquid film of refractive index that is suspended in air. The rays are shown at an angle to the normal for clarity.
\nWhat is the minimum thickness of the film so that the reflected light undergoes constructive interference?
\n
A.
B.
\nC.
\nD.
\nA
\nA satellite of mass orbits a planet of mass in a circular orbit of radius . What is the work that must be done on the satellite to increase its orbital radius to ?
\n
A.
B.
\nC.
\nD.
\nC
\nAirboats are used for transport across a river. To move the boat forward, air is propelled from the back of the boat by a fan blade.
\nAn airboat has a fan blade of radius 1.8 m. This fan can propel air with a maximum speed relative to the boat of 20 m s−1. The density of air is 1.2 kg m−3.
\nIn a test the airboat is tied to the river bank with a rope normal to the bank. The fan propels the air at its maximum speed. There is no wind.
\nThe rope is untied and the airboat moves away from the bank. The variation with time t of the speed v of the airboat is shown for the motion.
Outline why a force acts on the airboat due to the fan blade.
\nShow that a mass of about 240 kg of air moves through the fan every second.
\nShow that the tension in the rope is about 5 kN.
\nExplain why the airboat has a maximum speed under these conditions.
\nEstimate the distance the airboat travels to reach its maximum speed.
\nDeduce the mass of the airboat.
\nALTERNATIVE 1
\nthere is a force «by the fan» on the air / air is accelerated «to the rear» ✓
\nby Newton 3 ✓
\nthere is an «equal and» opposite force on the boat ✓
\n\n
ALTERNATIVE 2
\nair gains momentum «backward» ✓
\nby conservation of momentum / force is rate of change in momentum ✓
\nboat gains momentum in the opposite direction ✓
\n\n
Accept a reference to Newton’s third law, e.g. N’3, or any correct statement of it for MP2 in ALT 1.
\nAllow any reasonable choice of object where the force of the air is acting on, e.g., fan or blades.
\nOR «mass of air through system per unit time =» seen ✓
\n244 «kg s−1» ✓
\n\n
Accept use of Energy of air per second = 0.5 ρΑv3 = 0.5 mv2 for MP1.
\n«force = Momentum change per sec = = » 244 x 20 OR 4.9 «kN» ✓
\n\n
Allow use of 240
\nresistive forces increase with speed OR resistive forces/drag equal forward thrust ✓
\nacceleration/net force becomes zero/speed remains constant ✓
\nrecognition that area under the graph is distance covered ✓
\n«Distance =» 480 - 560 «m» ✓
\n\n
Accept graphical evidence or calculation of correct geometric areas for MP1.
\nMP2 is numerical value within range.
\ncalculation of acceleration as gradient at t = 0 «= 1 m s-2» ✓
\nuse of F=ma OR seen ✓
\n4900 «kg» ✓
\n\n
MP1 can be shown on the graph.
\nAllow an acceleration in the range 1 – 1.1 for MP2 and consistent answer for MP3
\nAllow ECF from MP1.
\nAllow use of average acceleration =
or assumption of constant force to obtain 11000 «kg» for [2]
Allow use of 4800 or 5000 for MP2
\nThe majority succeeded in making use of Newton's third law to explain the force on the boat. The question was quite well answered but sequencing of answers was not always ideal. There were some confusions about the air hitting the bank and bouncing off to hit the boat. A small number thought that the wind blowing the fan caused the force on the boat.
\nbi) This was generally well answered with candidates either starting from the wind turbine formula given in the data booklet or with the mass of the air being found using .
\n1bii) Well answered by most candidates. Some creative work to end up with 240 was found in scripts.
\n1ci) Many candidates gained credit here for recognising that the resistive force eventually equalled the drag force and most were able to go on to link this to e.g. zero acceleration. Some had not read the question properly and assumed that the rope was still tied. There was one group of answers that stated something along the lines of \"as there is no rope there is nothing to stop the boat so it can go at max speed.
\n1cii) A slight majority did not realise that they had to find the area under the velocity-time graph, trying equations of motion for non-linear acceleration. Those that attempted to calculate the area under the graph always succeeded in answering within the range.
\n1ciii) Use of the average gradient was common here for the acceleration. However, there also were answers that attempted to calculate the mass via a kinetic energy calculation that made all sorts of incorrect assumptions. Use of average acceleration taken from the gradient of the secant was also common.
\nTwo charged parallel plates have electric potentials of 10 V and 20 V.
\nA particle with charge +2.0 μC is moved from the 10 V plate to the 20 V plate. What is the change in the electric potential energy of the particle?
\n
A. −20 μJ
B. −10 μJ
\nC. 10 μJ
\nD. 20 μJ
\nD
\nA small magnet is released from rest to drop through a stationary horizontal conducting ring.
\nWhat is the variation with time of the emf induced in the ring?
\nC
\nAn alternating supply is connected to a diode bridge rectification circuit.
\nThe conventional current in the load resistor
\n
A. is a maximum twice during one oscillation of the input voltage.
B. is never zero.
\nC. has a zero average value during one oscillation of the input voltage.
\nD. can only flow from P to Q.
\nA
\nAn experiment to investigate simple harmonic motion consists of a mass oscillating at the end of a vertical spring.
\nThe mass oscillates vertically above a motion sensor that measures the speed of the mass. Test 1 is carried out with a 1.0 kg mass and spring of spring constant . Test 2 is a repeat of the experiment with a 4.0 kg mass and spring of spring constant .
\nThe variation with time of the vertical speed of the masses, for one cycle of the oscillation, is shown for each test.
Calculate the frequency of the oscillation for both tests.
\nDeduce .
\nDetermine the amplitude of oscillation for test 1.
\nIn test 2, the maximum elastic potential energy Ep stored in the spring is 44 J.
\nWhen t = 0 the value of Ep for test 2 is zero.
\nSketch, on the axes, the variation with time of Ep for test 2.
\nThe motion sensor operates by detecting the sound waves reflected from the base of the mass. The sensor compares the frequency detected with the frequency emitted when the signal returns.
\nThe sound frequency emitted by the sensor is 35 kHz. The speed of sound is 340 m s−1.
\nDetermine the maximum frequency change detected by the sensor for test 2.
\n1.3 «Hz» ✓
\nOR ✓
\n0.25 OR ✓
\nvmax = 4.8 «m s−1» ✓
\n«» = 0.61 «m» ✓
\n\n
Allow a range of 4.7 to 4.9 for MP1
\nAllow a range of 0.58 to 0.62 for MP2
\nAllow ECF from (a)(i)
\nAllow ECF from MP1.
\nall energy shown positive ✓
\ncurve starting and finishing at E = 0 with two peaks with at least one at 44 J
OR
curve starting and finishing at E = 0 with one peak at 44 J ✓
\n
Do not accept straight lines or discontinuous curves for MP2
\nread off of 9.4 «m s−1» ✓
\nuse of OR ✓
\nf = 36 «kHz» OR 34 «kHz» ✓
\n«recognition that there are two shifts so» change in f = 2 «kHz» OR f = 37 «kHz» OR 33 «kHz» ✓
\n\n
Allow a range of 9.3 to 9.5 for MP1
\nAllow ECF from MP1.
\nMP4 can also be found by applying the Doppler effect twice.
\nai) The majority managed to answer this question correctly.
\naii) A very well answered question where most worked correctly from the formula for the period of oscillation of a spring.
\naiii) Quite a few answers had vmax from the wrong test.
\naiv) Most common answers were a correct 2 peak curve, a correct 1 peak curve and a sine curve. Several alternatives were included in the MS as the original data provided in the question was inconsistent, i.e. 44 J is not the maximum kinetic energy available for the second test, and that was taken into account not to disadvantage any candidate´s interpretation.
\nb) Many got the first three marks for a correct Doppler shift calculation from the correct speed. . There were very few good correct full answers, might be a question to look at for 6/7 during grading.
\nAn ion moves in a circle in a uniform magnetic field. Which single change would increase the radius of the circular path?
\n
A. Decreasing the speed of the ion
B. Increasing the charge of the ion
\nC. Increasing the mass of the ion
\nD. Increasing the strength of the magnetic field
\nC
\nA beam of light containing two different wavelengths is incident on a diffraction grating. The wavelengths are just resolved in the third order of diffraction.
\nWhat change increases the resolution of the image?
\n
A. Increasing the width of the incident beam
B. Increasing the intensity of light
\nC. Decreasing the number of lines per unit length in the diffraction grating
\nD. Decreasing the order of diffraction
\nA
\nA longitudinal wave travels in a medium with speed 340 m s−1. The graph shows the variation with time t of the displacement x of a particle P in the medium. Positive displacements on the graph correspond to displacements to the right for particle P.
\nAnother wave travels in the medium. The graph shows the variation with time t of the displacement of each wave at the position of P.
\nA standing sound wave is established in a tube that is closed at one end and open at the other end. The period of the wave is . The diagram represents the standing wave at and at . The wavelength of the wave is 1.20 m. Positive displacements mean displacements to the right.
\nCalculate the wavelength of the wave.
\nDetermine, for particle P, the magnitude and direction of the acceleration at t = 2.0 m s.
\nState the phase difference between the two waves.
\nIdentify a time at which the displacement of P is zero.
\nEstimate the amplitude of the resultant wave.
\nCalculate the length of the tube.
\nA particle in the tube has its equilibrium position at the open end of the tube.
State and explain the direction of the velocity of this particle at time .
Draw on the diagram the standing wave at time .
\n«s» or «Hz» ✓
\n«m» ✓
\n\n
Allow ECF from MP1.
Award [2] for a bald correct answer.
OR «s−1» ✓
\n«ms−2» ✓
\n«opposite to displacement so» to the right ✓
\n«±» OR ✓
\n1.5 «ms» ✓
\n8.0 OR 8.5 «μm» ✓
\n
From the graph on the paper, value is 8.0. From the calculated correct trig functions, value is 8.49.
L = «» 0.90 «m» ✓
\nto the right ✓
displacement is getting less negative
\nOR
\nchange of displacement is positive ✓
\nhorizontal line drawn at the equilibrium position ✓
\nA metal sphere is charged positively and placed far away from other charged objects. The electric potential at a point on the surface of the sphere is 53.9 kV.
\nA small positively charged object moves towards the centre of the metal sphere. When the object is 2.8 m from the centre of the sphere, its speed is 3.1 m s−1. The mass of the object is 0.14 g and its charge is 2.4 × 10−8 C.
\nOutline what is meant by electric potential at a point.
\nThe electric potential at a point a distance 2.8 m from the centre of the sphere is 7.71 kV. Determine the radius of the sphere.
\nComment on the angle at which the object meets equipotential surfaces around the sphere.
\nShow that the kinetic energy of the object is about 0.7 mJ.
\nDetermine whether the object will reach the surface of the sphere.
\nthe work done per unit charge ✓
\nIn bringing a small/point/positive/test «charge» from infinity to the point ✓
\n\n
Allow use of energy per unit charge for MP1
\nuse of Vr = constant ✓
\n0.40 m ✓
\n\n
Allow [1] max if r + 2.8 used to get 0.47 m.
\nAllow [2] marks if they calculate Q at one potential and use it to get the distance at the other potential.
\n90° / perpendicular ✓
\nOR 0.67 «mJ» seen ✓
\n«p.d. between point and sphere surface = » (53.9 kV – 7.71) «kV» OR 46.2 «kV» seen ✓
\n«energy required =» VQ « = 46 200 × 2.4 × 10-8» = 1.11 mJ ✓
\nthis is greater than kinetic energy so will not reach sphere ✓
\n\n
MP3 is for a conclusion consistent with the calculations shown.
\nAllow ECF from MP1
\na) Well answered.
\nb) Generally, well answered, but there were quite a few using r + 2.8.
\nci) Very few had problems to recognize the perpendicular angle
\ncii) Good simple calculation
\nciii) Many had a good go at this, but a significant number tried to answer it based on forces.
\nIn the circuits shown, the cells have the same emf and zero internal resistance. All resistors are identical.
\nWhat is the order of increasing power dissipated in each circuit?
\nA
\nA fixed horizontal coil is connected to an ideal voltmeter. A bar magnet is released from rest so that it falls vertically through the coil along the central axis of the coil.
\nThe variation with time t of the emf induced in the coil is shown.
\n\n
Write down the maximum magnitude of the rate of change of flux linked with the coil.
\nState the fundamental SI unit for your answer to (a)(i).
\nExplain why the graph becomes negative.
\nPart of the graph is above the t-axis and part is below. Outline why the areas between the t-axis and the curve for these two parts are likely to be the same.
\nPredict the changes to the graph when the magnet is dropped from a lower height above the coil.
\n«−» 5.0 «mV» OR 5.0 × 10−3 «V» ✓
\n\n
Accept 5.1
\nkg m2 A−1 s−3 ✓
\nALTERNATIVE 1
\nFlux linkage is represented by magnetic field lines through the coil ✓
\nwhen magnet has passed through the coil / is moving away ✓
\nflux «linkage» is decreasing ✓
\nsuitable comment that it is the opposite when above ✓
\nwhen the magnet goes through the midpoint the induced emf is zero ✓
\n\n
ALTERNATIVE 2
\nreference to / states Lenz’s law ✓
\nwhen magnet has passed through the coil / is moving away ✓
\n«coil attracts outgoing S pole so» induced field is downwards ✓
\nbefore «coil repels incoming N pole so» induced field is upwards
OR
induced field has reversed ✓
when the magnet goes through the midpoint the induced emf is zero ✓
\n\n
OWTTE
\narea represents the total change in flux «linkage» ✓
\nthe change in flux is the same going in and out ✓
\n«when magnet is approaching» flux increases to a maximum ✓
\n«when magnet is receding» flux decreases to zero ✓
\n«so areas must be the same»
\nmagnet moves slower ✓
\noverall time «for interaction» will be longer ✓
\npeaks will be smaller ✓
\nareas will be the same as before ✓
\n\n
Allow a graphical interpretation for MP2 as “graph more spread out”
\nai
Well answered, with wrong answers stating 8 for the difference or 3 without realising that the sign does not matter.
aii
Very few candidates managed to get the correct fundamental SI unit for V. All kinds of errors were observed, from power errors to the use of C as a fundamental unit instead of A.
bi) Most scored best by marking using an alternative method introduced to the markscheme in standardisation. There were some confused and vague comments. Clear, concise answers were rare.
\nbii) It was common to see conservation of energy invoked here with suggestions that energy was the area under the graph. Many candidates described the shapes to explain why the areas were the same rather than talking about the physics e.g. one peak is short and fat and the other is tall and thin so they balance out.
\nc) A surprising number didn't pick up on the fact that the magnet would be moving slower. As a result, they discussed everything happening sooner, i.e. the interaction with the magnet and the coil, and that led onto things happening quicker so peaks being bigger.
\nIn an experiment a beam of electrons with energy 440 MeV are incident on oxygen-16 nuclei. The variation with scattering angle of the relative intensity of the scattered electrons is shown.
Identify a property of electrons demonstrated by this experiment.
\nShow that the energy E of each electron in the beam is about 7 × 10−11 J.
\nThe de Broglie wavelength for an electron is given by . Show that the diameter of an oxygen-16 nucleus is about 4 fm.
\nEstimate, using the result in (a)(iii), the volume of a tin-118 nucleus. State your answer to an appropriate number of significant figures.
\nwave properties ✓
\n
Accept reference to diffraction or interference.
440 x 106 x 1.6 x 10-19 OR 7.0 × 10-11 «J» ✓
\nOR OR 2.8 × 10-15 «m» seen ✓
\nread off graph as 46° ✓
\n«Use of =» 3.9 × 10-15 m ✓
\n\n
Accept an angle between 45 and 47 degrees.
\nAllow ECF from MP2
\nALTERNATIVE 1
\nuse of OR ✓
\nvolume of or equivalent working ✓
\n2.3 to 2.5 × 10-43 «m3»✓
\nanswer to 1 or 2sf ✓
\n\n
ALTERNATIVE 2
\nuse of ✓
\nvolume of OR 5.9 x 10-15 seen ✓
\n8.5 × 10-43 «m3»✓
\nanswer to 1 or 2sf ✓
\n\n
Although the question expects candidates to work from the oxygen radius found, allow ALT 2 working from the Fermi radius.
\nMP4 is for any answer stated to 1 or 2 significant figures.
\nai) Well answered.
\naii) Well answered.
\naiii) This was generally well done but quite a few attempted the small angle approximation. Probably worth a mention in the report.
\nb) Most gained credit from the first alternative solution, trying to use the data as the question intended. There were the inevitable slips and calculator mistakes. Most got the fourth mark.
\nThe magnitude of the resultant of two forces acting on a body is 12 N. Which pair of forces acting on the body can combine to produce this resultant?
\nA. 1 N and 2 N
\nB. 1 N and 14 N
\nC. 5 N and 6 N
\nD. 6 N and 7 N
\nD
\nThis question was well answered by HL and SL candidates. There was a higher number of blanks (no response) among SL students than is typical this early in the exam paper.
\nThree identical resistors of resistance R are connected as shown to a battery with a potential difference of and an internal resistance of . A voltmeter is connected across one of the resistors.
\n
What is the reading on the voltmeter?
\nA.
\nB.
\nC.
\nD.
\nC
\nA student measures the time for 20 oscillations of a pendulum. The experiment is repeated four times. The measurements are:
\n10.45 s
\n10.30 s
\n10.70 s
\n10.55 s
\nWhat is the best estimate of the uncertainty in the average time for 20 oscillations?
\nA. 0.01 s
\nB. 0.05 s
\nC. 0.2 s
\nD. 0.5 s
\nC
\nThis question was well answered, although option B was a significant distractor for candidates focusing on the last significant digit.
\nA block moving with initial speed is brought to rest, after travelling a distance d, by a frictional force . A second identical block moving with initial speed u is brought to rest in the same distance d by a frictional force . What is u?
\nA.
\nB.
\nC.
\nD.
\nB
\nWith a lower difficulty index for SL candidates than for HL candidates, this question asked students to recognize the relationship between variables in a kinematics equation. For both groups, option C (incorrect) was most frequently selected, as candidates struggled to show the relationship between U and the change in frictional force. This question would be a useful teaching tool, as results here suggest candidates should spend more time working with equations without numerical substitutions.
\nMagnetic field lines are an example of
\nA. a discovery that helps us understand magnetism.
\nB. a model to aid in visualization.
\nC. a pattern in data from experiments.
\nD. a theory to explain concepts in magnetism.
B
\nA cell is connected to an ideal voltmeter, a switch S and a resistor R. The resistance of R is 4.0 Ω.
\nWhen S is open the reading on the voltmeter is 12 V. When S is closed the voltmeter reads 8.0 V.
\nElectricity can be generated using renewable resources.
\nThe voltmeter is used in another circuit that contains two secondary cells.
\nCell A has an emf of 10 V and an internal resistance of 1.0 Ω. Cell B has an emf of 4.0 V and an internal resistance of 2.0 Ω.
\nIdentify the laws of conservation that are represented by Kirchhoff’s circuit laws.
\nState the emf of the cell.
\nDeduce the internal resistance of the cell.
\nCalculate the reading on the voltmeter.
\nComment on the implications of your answer to (c)(i) for cell B.
\nOutline why electricity is a secondary energy source.
\nSome fuel sources are renewable. Outline what is meant by renewable.
\nA fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.
\nThe cell is then re-charged by a rectangular solar panel of dimensions 0.40 m × 0.15 m at a place where the maximum intensity of sunlight is 380 W m−2.
\nThe overall efficiency of the re-charging process is 18 %.
\nCalculate the minimum time required to re-charge the cell fully.
\nOutline why research into solar cell technology is important to society.
\n« conservation of » charge ✓
\n« conservation of » energy ✓
\n\n
Allow [1] max if they explicitly refer to Kirchhoff’ laws linking them to the conservation laws incorrectly.
\n12 V ✓
\nI = 2.0 A OR 12 = I (r + 4) OR 4 = Ir OR 8 = 4I ✓
\n«Correct working to get » r = 2.0 «Ω» ✓
\n\n
Allow any valid method.
\nAllow ECF from (b)(i)
\nLoop equation showing EITHER correct voltages, i.e., 10 – 4 on one side or both emf’s positive on different sides of the equation OR correct resistances, i.e. I (1 + 2) ✓
\n10−4 = I (1 + 2) OR I = 2.0 «A» seen ✓
\nV = 8.0 «V» ✓
\n\n
Allow any valid method
\nCharge is being driven through the 4.0 V cell OR it is being (re-)charged ✓
\nis generated from primary/other sources ✓
\n«a fuel » that can be replenished/replaced within a reasonable time span
\nOR
\n«a fuel» that can be replaced faster than the rate at which it is consumed
\nOR
\nrenewables are limitless/never run out
\nOR
\n«a fuel» produced from renewable sources
\nOR
\ngives an example of a renewable (biofuel, hydrogen, wood, wind, solar, tidal, hydro etc..) ✓
\n\n
OWTTE
\nALTERNATIVE 1
\n«energy output of the panel =» Vlt OR 6 x 5 x 0.25 x 3600 OR 27000 «J» ✓
\n«available power =» 380 x 0.4 x 0.15 x 0.18 OR 4.1 «W» ✓
\n«=» 6600 «s» ✓
\n\n
ALTERNATIVE 2
\n«energy needed from Sun =» OR OR 150000 «J» ✓
\n« incident power=» 380 x 0.4 x 0.15 OR 22.8 «W» ✓
\n«=» 6600 «s» ✓
\n\n
Allow ECF for MP3
\nAccept final answer in minutes (110) or hours (1.8).
\ncoherent reason ✓
\ne.g., to improve efficiency, is non-polluting, is renewable, does not produce greenhouse gases, reduce use of fossil fuels
\n\n
Do not allow economic reasons
\na) Most just stated Kirchhoff's laws rather than the underlying laws of conservation of energy and charge, basically describing the equations from the data booklet. When it came to guesses, energy and momentum were often the two, although even a baryon and lepton number conservation was found. It cannot be emphasised enough the importance of correctly identifying the command verb used to introduce the question. In this case, identify, with the specific reference to conservation laws, seem to have been explicit tips not picked up by some candidates.
\nbi) This was probably the easiest question on the paper and almost everybody got it right. 12V. Some calculations were seen, though, that contradict the command verb used. State a value somehow implies that the value is right in front to be read or interpreted suitably.
\nbii) In the end a lot of the answers here were correct. Some obtained 2 ohms and were able to provide an explanation that worked. A very few negative answers were found, suggesting that some candidates work mechanically without properly reflecting in the nature of the value obtained.
\nci) A lot of candidates figured out they had to do some sort of loop here but most had large currents in the voltmeter. Currents of 2 A and 10 A simultaneously were common. Some very good and concise work was seen though, leading to correct steps to show a reading of 8V.
\ncii) This question was cancelled due to an internal reference error. The paper total was adjusted in grade award. This is corrected for publication and future teaching use.
\ndi) The vast majority of candidates could explain why electricity was a secondary energy source.
\ndii) An ideal answer was that renewable fuels can be replenished faster than they are consumed. However, many imaginative alternatives were accepted.
\nei) This question was often very difficult to mark. Working was often scattered all over the answer box. Full marks were not that common, most candidates achieved partial marks. The commonest problem was determining the energy required to charge the battery. It was also common to see a final calculation involving a power divided by a power to calculate the time.
\neii) Almost everybody could give a valid reason why research into solar cells was important. Most answers stated that solar is renewable. There were very few that didn't get a mark due to discussing economic reasons.
\nA stone is kicked horizontally at a speed of 1.5 m s−1 from the edge of a cliff on one of Jupiter’s moons. It hits the ground 2.0 s later. The height of the cliff is 4.0 m. Air resistance is negligible.
\nWhat is the magnitude of the displacement of the stone?
\nA. 7.0 m
\nB. 5.0 m
\nC. 4.0 m
\nD. 3.0 m
\nB
\nThis question was generally well answered by both HL and SL candidates and had a mid-range difficulty index (indicating an easier question). Option D was an effective distractor for candidates calculating the horizontal range rather than the displacement. Candidates are encouraged to read the questions carefully to ensure it is clear what each question is asking for.
\nWhat is the order of magnitude of the wavelength of visible light?
\nA. 10−10 m
\nB. 10−7 m
\nC. 10−4 m
\nD. 10−1 m
\nB
\nThis question was correctly answered by the majority of SL candidates.
\nWhich of the formulae represents Newton’s second law?
\nA.
\nB.
\nC.
\nD.
\nC
\nThis question was very well answered by SL candidates, as demonstrated by the high difficulty index.
\nAn object moves in a circle of constant radius. Values of the centripetal force are measured for different values of angular velocity . A graph is plotted with on the -axis. Which quantity plotted on the -axis will produce a straight-line graph?
\nA.
\nB.
\nC.
\nD.
\nA
\nTwo masses and are connected by a string over a frictionless pulley of negligible mass. The masses are released from rest. Air resistance is negligible.
\nMass accelerates downwards at . What is ?
A.
B.
\nC. 2
\nD. 3
\nA
\nWith a low difficulty index for both, this question was challenging for both HL and SL candidates. Option B was the most common (incorrect) answer, and only a small number of candidates correctly selected option A. This question would be a useful teaching tool for students, as they consider the relationship between variables without numeric substitution.
\nA cart travels from rest along a horizontal surface with a constant acceleration. What is the variation of the kinetic energy Ek of the cart with its distance s travelled? Air resistance is negligible.
\nD
\nOption A was the most common (incorrect) response among both HL and SL candidates, suggesting that candidates were looking for a curve representing speed rather than kinetic energy against distance. A low discrimination index suggests that both high and low achieving students were caught by this effective distractor.
\nTwo trolleys of equal mass travel in opposite directions as shown.
\nThe trolleys collide head-on and stick together.
\nWhat is their velocity after the collision?
\nA. 1 m s−1
\nB. 2 m s−1
\nC. 5 m s−1
\nD. 10 m s−1
\nA
\nThe majority of SL candidates selected option B, finding the difference in velocity but neglecting to recognize that mass will have doubled. This question had a relatively high discrimination index suggesting more able candidates had greater success demonstrating this recognition.
\nAn ideal gas is maintained at a temperature of 100 K. The variation of the pressure P and of the gas is shown.
\nWhat is the quantity of the gas?
\nA.
\nB.
\nC.
\nD.
\nC
\nThis question tested candidate understanding of the relationship between the slope of a graph and the ideal gas law. SL candidates found this question more difficult than their HL counterparts, but in both groups of students, option C was the most frequent (and correct) answer.
\nA sphere is suspended from the end of a string and rotates in a horizontal circle. Which freebody diagram, to the correct scale, shows the forces acting on the sphere?
\nD
\nA wave of period 10 ms travels through a medium. The graph shows the variation of particle displacement with distance for the wave.
\nWhat is the average speed of a particle in the medium during one cycle?
\nA. 4.0 m s−1
\nB. 8.0 m s−1
\nC. 16 m s−1
\nD. 20 m s−1
\nC
\nOption D was a very efficient distractor as the most common (incorrect) selection by both HL and SL candidates. The difficulty index was low for this question, suggesting that HL and SL candidates found this question quite challenging. Candidates are again encouraged to read the questions carefully; it is likely that candidates selecting option D were providing the wave speed rather than particle speed.
\nA light source of power P is observed from a distance . The power of the source is then halved.
\nAt what distance from the source will the intensity be the same as before?
\nA.
\nB.
\nC.
\nD.
\nA
\nSL candidate responses were divided across options A, B and C, and so this question would be a useful teaching tool for exploring the relationship between power, intensity and distance.
\nA driver uses the brakes on a car to descend a hill at constant speed. What is correct about the internal energy of the brake discs?
\nA. The internal energy increases.
\nB. The internal energy decreases.
\nC. There is no change in the internal energy.
\nD. The internal energy is zero.
\nA
\nThis question was well answered by HL and SL candidates, although option C did prove to be a distraction for some.
\nAn interference pattern with minima of zero intensity is observed between light waves. What must be true about the frequency and amplitude of the light waves?
\nC
\nThis question was generally well answered by SL candidates with both a high difficulty and discrimination index. Candidates who selected an incorrect answer seems more certain about the correct amplitude than the resulting intensity.
\nA beam of unpolarized light of intensity is incident on a polarizing filter. The polarizing filter is rotated through an angle θ. What is the variation in the intensity of the beam with angle θ after passing through the polarizing filter?
\nD
\nOption B was chosen most frequently among both HL and SL candidates. It seems likely that students selecting this answer were anticipating a question with two polarising filters where the second filter was rotated. Again, careful reading of the questions by candidates is necessary. A high discrimination index was observed for this question. This is a good conceptual question and would be useful in the teaching and/or revision of polarisation.
\nA proton moves along a circular path in a region of a uniform magnetic field. The magnetic field is directed into the plane of the page.
\nLabel with arrows on the diagram the magnetic force F on the proton.
\nLabel with arrows on the velocity vector v of the proton.
\nThe speed of the proton is 2.16 × 106 m s-1 and the magnetic field strength is 0.042 T. For this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.
\nF towards centre ✔
\nv tangent to circle and in the direction shown in the diagram ✔
\n« ✔
R = 0.538 «m»✔
R = 0.54 «m» ✔
Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.
\nExaminers were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.
\nThis was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.
\nTwo blocks, X and Y, are placed in contact with each other. Data for the blocks are provided.
\nX has a mass . What is the mass of Y?
\nA.
\nB.
\nC.
\nD.
\nC
\nThis question was very well answered by candidates, reinforced by the high difficulty index for both HL and SL groups. This is another question that requires the rearrangement of an equation to determine a relationship between variables; interestingly candidates showed greater success on this question than others of this type. This may be due to the fact that there was not an easy distractor included in the response options, requiring candidates to work through equation substitution and rearrangement to reach a final answer.
\nA ray of light is incident on the flat side of a semi-circular glass block placed in paraffin. The ray is totally internally reflected inside the glass block as shown.
\nThe refractive index of glass is and the refractive index of paraffin is .
\nWhat is correct?
\nA.
\nB.
\nC.
\nD.
\nB
\nWhile option A (correct) was the most frequent response from candidates, options B and D were significant distractors. Very few candidates selected option A, recognizing that different frequency and amplitude could not lead to zero intensity through interference.
\nA rocket moving with speed v relative to the ground emits a flash of light in the backward direction.
\nAn observer in the rocket measures the speed of the flash of light to be c.
\nState the speed of the flash of light according to an observer on the ground using Galilean relativity.
\nState the speed of the flash of light according to an observer on the ground using Maxwell’s theory of electromagnetism.
\nState the speed of the flash of light according to an observer on the ground using Einstein’s theory of relativity.
\nc–v ✔
\nc ✔
\nc ✔
\nThe speed of a flash of light from different viewpoints. Most of the prepared candidates well stated the speed of the flash using Galilean relativity, Maxwell’s theory and Einstein’s theory.
\nThe speed of a flash of light from different viewpoints. Most of the prepared candidates well stated the speed of the flash using Galilean relativity, Maxwell’s theory and Einstein’s theory.
\nThe speed of a flash of light from different viewpoints. Most of the prepared candidates well stated the speed of the flash using Galilean relativity, Maxwell’s theory and Einstein’s theory.
\nA standing wave is formed on a rope. The distance between the first and fifth antinode on the standing wave is 60 cm. What is the wavelength of the wave?
\nA. 12 cm
\nB. 15 cm
\nC. 24 cm
\nD. 30 cm
\nD
\nOption D was the most common (correct) answer, however answers A and B proved to be significant distractors. This would be a useful practice question when reviewing standing waves, nodes/antinodes and wavelengths.
\nP and Q are two opposite point charges. The force F acting on P due to Q and the electric field strength E at P are shown.
\nWhich diagram shows the force on Q due to P and the electric field strength at Q?
\nB
\nOption A was the most frequent answer selected by both HL and SL candidates, suggesting that determining the direction of the electric field was more problematic than the direction of the force (Newton 3).
\nThree point charges of equal magnitude are placed at the vertices of an equilateral triangle. The signs of the charges are shown. Point P is equidistant from the vertices of the triangle. What is the direction of the resultant electric field at P?
\nC
\nThis question was very well answered by both HL and SL candidates, reflected in the high difficulty level for both papers.
\nThree identical resistors each of resistance R are connected with a variable resistor X as shown. X is initially set to R. The current in the cell is 0.60 A.
\nThe cell has negligible internal resistance.
\nX is now set to zero. What is the current in the cell?
\nA. 0.45 A
\nB. 0.60 A
\nC. 0.90 A
\nD. 1.80 A
\nC
\nOption C was the most common (correct) answer, however option B was also a frequent response. This question had a relatively high discrimination index, suggesting that more able candidates had less difficulty managing resistance in this combination circuit.
\nTwo cylinders, X and Y, made from the same material, are connected in series.
\nThe cross-sectional area of Y is twice that of X. The drift speed of the electrons in X is and in Y it is .
\nWhat is the ratio ?
\nA. 4
\nB. 2
\nC. 1
\nD.
\nB
\nResponses from SL candidates were split between options B and D. Candidates appeared to recognize the factor of two in the drift speed ratio, but were unclear as to whether it was 2:1 or 1:2. Candidates are encouraged to think if their answer makes sense given the context of the question; it is likely that common sense would help candidates in this instance. Practice manipulating ratios to compare changing variables is a useful skill, and this question could be put to good use in this regard.
\nA ball of mass 0.3 kg is attached to a light, inextensible string. It is rotated in a vertical circle. The length of the string is 0.6 m and the speed of rotation of the ball is 4 m s−1.
\nWhat is the tension when the string is horizontal?
\nA. 5 N
\nB. 8 N
\nC. 11 N
\nD. 13 N
\nB
\nThis question was well answered by both HL and SL candidates with a high difficulty index for each paper.
\nSome transitions between the energy states of a particular atom are shown.
\nEnergy transition E3 gives rise to a photon of green light. Which transition will give rise to a photon of longer wavelength?
\nA. E1
\nB. E2
\nC. E4
\nD. E5
\nC
\nThis question was well answered by SL candidates, although answer A was a distractor for many. This question had the highest discrimination index on this SL paper.
\nThree statements about radioactive decay are:
\nI. The rate of decay is exponential.
II. It is unaffected by temperature and pressure.
III. The decay of individual nuclei cannot be predicted.
Which statements are correct?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nOption B was the most frequent answer by candidates, suggesting that many candidates are unclear about the basic characteristics of radioactive decay.
\nThe background count in a laboratory is 20 counts per second. The initial observed count rate of a pure sample of nitrogen-13 in this laboratory is 180 counts per second. The half-life of nitrogen-13 is 10 minutes. What is the expected count rate of the sample after 30 minutes?
\nA. 20 counts per second
\nB. 23 counts per second
\nC. 40 counts per second
\nD. 60 counts per second
\nC
\nOption B was the most frequent answer, incorrectly selected by candidates who did not consider the background count in the laboratory.
\nundergoes an alpha decay, followed by a beta-minus decay. What is the number of protons and neutrons in the resulting nuclide?
\nC
\nThis question was generally well answered by candidates, however a significant number selected option A (incorrectly) perhaps due to confusion between nuclear mass and the number of neutrons. This question had a relatively high discrimination index.
\nWind of speed flows through a wind generator. The wind speed drops to after passing through the blades. What is the maximum possible efficiency of the generator?
\nA.
\nB.
\nC.
\nD.
\nD
\nBoth HL and SL candidates found this question more challenging, with the large majority of candidates split between the three incorrect options. Option A was the most frequent (incorrect) answer, suggesting that candidates had correctly determined the proportion lost rather than that remaining to produce energy. Candidates should be reminded to consider whether or not their quantitative solutions are realistic; it is highly unlikely that a generator would have maximum efficiency of 1/27 (option A).
\nThree mechanisms that affect the composition of the atmosphere of the Earth are:
\nI. Loss of forests that would otherwise store carbon dioxide – CO2
II. Release of methane – CH4 by the digestive system of grazing animals
III. Increase of nitrous oxide – N2O due to extensive use of fertilizer
Which of these statements describe a process that contributes to global warming?
\nA. I and II only
\nB. I and III only
\nC. II and III only
\nD. I, II and III
\nD
\nThis question was well answered by candidates, although option A was a frequent distractor suggesting candidates may be less clear about the role of nitrous oxide in global warming.
\nThe diagram shows, for a region on the Earth’s surface, the incident, radiated and reflected intensities of the solar radiation.
\nWhat is the albedo of the region?
\nA.
\nB.
\nC.
\nD.
\nA
\nThis question was well answered by both HL and SL candidates.
The intensity of a wave can be defined as the energy per unit area per unit time. What is the unit of intensity expressed in fundamental SI units?
\nA. kg m−2 s−1
\nB. kg m2 s−3
\nC. kg s−2
\nD. kg s−3
\nD
\nThe unit analysis in this question proved tricky for many HL candidates, with option A being the most common (incorrect) answer. The high discrimination index suggests that this question was more problematic for weaker candidates.
\nThe uncertainty in reading a laboratory thermometer is 0.5 °C. The temperature of a liquid falls from 20 °C to 10 °C as measured by the thermometer. What is the percentage uncertainty in the change in temperature?
\nA. 2.5 %
\nB. 5 %
\nC. 7.5 %
\nD. 10 %
\nD
\nMany candidates failed to recognize that the uncertainty in this error propagation question would affect both the initial and final temperature readings. The most common answer (option B) was incorrect, and only a minority of students correctly selected option D.
\nThe homogeneous model of the universe predicts that it may be considered as a spherical cloud of matter of radius r and uniform density ρ. Consider a particle of mass m at the edge of the universe moving with velocity v and obeying Hubble’s law.
\nJustify that the total energy of this particle is .
\nAt critical density there is zero total energy. Show that the critical density of the universe is: .
\nThe accepted value for the Hubble constant is 2.3 × 10−18 s−1. Estimate the critical density of the universe.
\ntotal energy=kinetic energy+potential energy
\nOR
\ntotal energy= ✔
\nsubstitution of M = ✔
\n«Hence answer given»
\nAnswer given so for MP2 look for clear evidence that MUniverse is stated and substituted.
\nsubstitutes H0r for v ✔
\n«total energy = 0»
\n✔
\n«hence ρc = »
\nAnswer given, check working carefully.
\n9.5 × 10−27 « kgm–3» ✔
\nThe vast majority of the candidates could state that the total energy is equal to the sum of the kinetic and potential energies but quite a few did not use the correct formula for the gravitational potential energy. The formula for the mass of the sun was usually correctly substituted.
\nThis was a relatively easy demonstration given the equation in 22a. However many candidates did not show the process followed in a coherent manner that could be understood by examiners.
\nThe question was well answered by many candidates.
\nWhich quantity has the fundamental SI units of kg m–1 s–2?
A. Energy
B. Force
C. Momentum
D. Pressure
D
\nAn object is held in equilibrium by three forces of magnitude F, G and H that act at a point in the same plane.
\nThree equations for these forces are
I. F cos θ = G
II. F = G cos θ + H sin θ
III. F = G + H
Which equations are correct?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nA
\nTwo forces act along a straight line on an object that is initially at rest. One force is constant; the second force is in the opposite direction and proportional to the velocity of the object.
\nWhat is correct about the motion of the object?
\n
A. The acceleration increases from zero to a maximum.
B. The acceleration increases from zero to a maximum and then decreases.
C. The velocity increases from zero to a maximum.
D. The velocity increases from zero to a maximum and then decreases.
\nC
\nThe variation with time t of the acceleration a of an object is shown.
\nWhat is the change in velocity of the object from t = 0 to t = 6 s?
A. 6 m s–1
B. 8 m s–1
C. 10 m s–1
D. 14 m s–1
\nC
\nA climber of mass m slides down a vertical rope with an average acceleration a. What is the average frictional force exerted by the rope on the climber?
A. mg
B. m(g + a)
C. m(g – a)
D. ma
\nC
\nA cube slides down the surface of a ramp at a constant velocity. What is the magnitude of the frictional force that acts on the cube due to the surface?
A. The weight of the cube
B. The component of weight of the cube parallel to the plane
C. The component of weight of the cube perpendicular to the plane
D. The component of the normal reaction at the surface parallel to the plane
\nB
\nA ball is thrown vertically upwards. Air resistance is negligible. What is the variation with time t of the kinetic energy Ek of the ball?
\nD
\nThe tension in a horizontal spring is directly proportional to the extension of the spring. The energy stored in the spring at extension is . What is the work done by the spring when its extension changes from to ?
\nA.
\nB.
\nC.
\nD.
\nD
\nA mass of water is at a temperature of 290 K. The specific heat capacity of water is . Ice, at its melting point, is added to the water to reduce the water temperature to the freezing point. The specific latent heat of fusion for ice is . What is the minimum mass of ice that is required?
\nA.
\nB.
\nC.
\nD.
\nA
\nAn ideal gas is in a closed container. Which changes to its volume and temperature when taken together must cause a decrease in the gas pressure?
\nD
\nTwo flasks P and Q contain an ideal gas and are connected with a tube of negligible volume compared to that of the flasks. The volume of P is twice the volume of Q.
\nP is held at a temperature of 200 K and Q is held at a temperature of 400 K.
\nWhat is mass of ?
\n\n
A.
\nB.
\nC. 4
\nD. 8
\nC
\nThe motion of an object is described by the equation
\nacceleration ∝ − displacement.
\nWhat is the direction of the acceleration relative to that of the displacement and what is the displacement when the speed is a maximum?
\nD
\nA transverse travelling wave is moving through a medium. The graph shows, for one instant, the variation with distance of the displacement of particles in the medium.
\nThe frequency of the wave is 25 Hz and the speed of the wave is 100 m s–1. What is correct for this wave?
\n
A. The particles at X and Y are in phase.
B. The velocity of the particle at X is a maximum.
C. The horizontal distance between X and Z is 3.0 m.
D. The velocity of the particle at Y is 100 m s–1.
\nC
\nMonochromatic light is used to produce double-slit interference fringes on a screen. The fringe separation on the screen is . The distance from the slits to the screen and the separation of the slits are both doubled, and the light source is unchanged. What is the new fringe separation on the screen?
\nA.
\nB.
\nC.
\nD.
\nB
\nUnpolarized light is incident on two polarizing filters X and Y. They are arranged so that light emerging from Y has a maximum intensity. X is fixed and Y is rotated through θ about the direction of the incident beam in its own plane.
\nWhat are the first three successive values of θ for which the final transmitted intensity is a maximum?
\n
A. 90°, 180°, 270°
B. 90°, 270°, 450°
C. 180°, 360°, 540°
D. 180°, 540°, 720°
\n\n
C
\nA pipe is open at both ends. What is correct about a standing wave formed in the air of the pipe?
\nA. The sum of the number of nodes plus the number of antinodes is an odd number.
\nB. The sum of the number of nodes plus the number of antinodes is an even number.
C. There is always a central node.
D. There is always a central antinode.
\nA
\nA negatively charged particle in a uniform gravitational field is positioned mid-way between two charged conducting plates.
\nThe potential difference between the plates is adjusted until the particle is held at rest relative to the plates.
\nWhat change will cause the particle to accelerate downwards relative to the plates?
\n\n
A. Decreasing the charge on the particle
B. Decreasing the separation of the plates
C. Increasing the length of the plates
D. Increasing the potential difference between the plates
\nA
\nA thin copper wire and a thick copper wire are connected in series to an electric cell. Which quantity will be greater in the thin wire?
\nA. Current
B. Number of free charge carriers per unit volume
C. Net number of charge carriers crossing a section of a wire every second
D. Drift speed of the charge carriers
\nD
\nThe diagram shows a resistor network. The potential difference between X and Y is 8.0 V.
\nWhat is the current in the 5Ω resistor?
\nA. 1.0A
B. 1.6A
C. 2.0A
D. 3.0A
\nA
\nWhen a wire with an electric current I is placed in a magnetic field of strength B it experiences a magnetic force F. What is the direction of F?
\nA. In a direction determined by I only
B. In a direction determined by B only
C. In the plane containing I and B
D. At 90° to the plane containing I and B
\nD
\nAn object hangs from a light string and moves in a horizontal circle of radius r.
\nThe string makes an angle θ with the vertical. The angular speed of the object is ω. What is tan θ?
\nA.
\nB.
\nC.
\nD.
\nA
\nAn object of mass m makes n revolutions per second around a circle of radius r at a constant speed. What is the kinetic energy of the object?
\nA. 0
B.
C.
D.
C
\nA satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation?
\nA. The resultant force is the same direction as the satellite’s acceleration.
B. The gravitational force acting on the satellite is negligible.
C. There is no resultant force on the satellite relative to the Earth.
D. The satellite does not exert any force on the Earth.
\nA
\nThe energy levels for an atom are shown to scale.
\nA photon of wavelength λ is emitted because of a transition from E3 to E2. Which transition leads to the emission of a photon of longer wavelength?
A. E4 to E1
B. E4 to E3
C. E3 to E1
D. E2 to E1
\nB
\nA proton, an electron and an alpha particle are at rest. Which particle has the smallest magnitude of ratio of charge to mass and which particle has the largest magnitude of ratio of charge to mass?
\nA
\nX is a radioactive nuclide that decays to a stable nuclide. The activity of X falls to th of its original value in 32 s.
What is the half-life of X?
A. 2 s
B. 4 s
C. 8 s
D. 16 s
\nC
\nWhat is correct about the nature and range of the strong interaction between nuclear particles?
\nA. It is attractive at all particle separations.
B. It is attractive for particle separations between 0.7 fm and 3 fm.
C. It is repulsive for particle separations greater than 3 fm.
D. It is repulsive at all particle separations.
\nB
\nWhat are the units of specific energy and energy density?
\nB
\nWhat is the function of the moderator in a thermal nuclear fission reactor?
\nA. To decrease the kinetic energy of neutrons emitted from fission reactions
B. To increase the kinetic energy of neutrons emitted from fission reactions
C. To decrease the overall number of neutrons available for fission
D. To increase the overall number of neutrons available for fission
\nA
\nWhat is meant by the statement that the average albedo of the Moon is 0.1?
\nA. 10% of the radiation incident on the Moon is absorbed by its surface
B. 10% of the radiation emitted by the Moon is absorbed by its atmosphere
C. 10% of the radiation incident on the Moon is reflected by its surface
D. 10% of the radiation emitted by the Moon is at infrared wavelengths
\nC
\nA ball falls from rest in the absence of air resistance. The position of the centre of the ball is determined at one-second intervals from the instant at which it is released. What are the distances, in metres, travelled by the centre of the ball during each second for the first 4.0 s of the motion?
\nA. 5, 10, 15, 20
B. 5, 15, 25, 35
C. 5, 20, 45, 80
D. 5, 25, 70, 150
\nB
\nAn object is thrown from a cliff at an angle to the horizontal. The ground below the cliff is horizontal.
\nThree quantities are known about this motion.
\nI. The horizontal component of the initial velocity of the object
II. The initial angle between the velocity of the object and the horizontal
III. The height of the cliff
What are the quantities that must be known in order to determine the horizontal distance from the point of projection to the point at which the object hits the ground?
\n\n
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
\nD
\nA nuclear particle has an energy of 108 eV. A grain of sand has a mass of 32 mg. What speed must the grain of sand have for its kinetic energy to equal the energy of the nuclear particle?
\nA. 1 mm s–1
B. 3 mm s–1
C. 10 mm s–1
D. 16 mm s–1
\nA
\nUnder which conditions of pressure and density will a real gas approximate to an ideal gas?
\nD
\nThe graph shows the variation with time for the displacement of a particle in a travelling wave.
\nWhat are the frequency and amplitude for the oscillation of the particle?
\nC
\nA pipe of length 0.6 m is filled with a gas and closed at one end. The speed of sound in the gas is 300 m s–1. What are the frequencies of the first two harmonics in the tube?
\nA. 125 Hz and 250 Hz
B. 125 Hz and 375 Hz
C. 250 Hz and 500 Hz
D. 250 Hz and 750 Hz
\nB
\nTwo power supplies, one of constant emf 24 V and the other of variable emf P, are connected to two resistors as shown. Both power supplies have negligible internal resistances.
\nWhat is the magnitude of P for the reading on the ammeter to be zero?
\nA. Zero
B. 6 V
C. 8 V
D. 18 V
\nB
\nNuclide X can decay by two routes. In Route 1 alpha (α) decay is followed by beta-minus (β–) decay. In Route 2 β– decay is followed by α decay. P and R are the intermediate products and Q and S are the final products.
\nWhich statement is correct?
\n\n
A. Q and S are different isotopes of the same element.
B. The mass numbers of X and R are the same.
C. The atomic numbers of P and R are the same.
D. X and R are different isotopes of the same element.
\nB
\nGamma () radiation
\nA. is deflected by a magnetic field.
B. affects a photographic plate.
C. originates in the electron cloud outside a nucleus.
D. is deflected by an electric field.
\nB
\nThe equations of motion for uniform acceleration
\nA. apply to all accelerations.
B. cannot be proved mathematically.
C. relate force to other quantities in mechanics.
D. were developed through observing the natural world.
\nD
\nAn object undergoes simple harmonic motion (shm) of amplitude 0. When the displacement of the object is , the speed of the object is . What is the speed when the displacement is 0?
\nA. 0
\nB.
\nC.
\nD.
\nA
\nLight of frequency 500 THz is incident on a single slit and forms a diffraction pattern. The first diffraction minimum forms at an angle of 2.4 x 10–3 rad to the central maximum. The frequency of the light is now changed to 750 THz. What is the angle between the first diffraction minimum and the central maximum?
\nA. 1.6 × 10–3 rad
B. 1.8 × 10–3 rad
C. 2.4 × 10–3 rad
D. 3.6 × 10–3 rad
\nA
\nLight of wavelength λ is normally incident on a diffraction grating of spacing 3λ. What is the angle between the two second-order maxima?
\nA.
\nB.
\nC.
\nD. >90° so no second orders appear
\nC
\nSea waves move towards a beach at a constant speed of 2.0 m s–1. They arrive at the beach with a frequency of 0.10 Hz. A girl on a surfboard is moving in the sea at right angles to the wave fronts. She observes that the surfboard crosses the wave fronts with a frequency of 0.40 Hz.
\nWhat is the speed of the surfboard and what is the direction of motion of the surfboard relative to the beach?
\n\n
B
\nThe gravitational potential is at a distance above the surface of a spherical planet of radius and uniform density. What is the gravitational potential a distance above the surface of the planet?
\nA.
\nB.
\nC.
\nD.
\nD
\nThe force acting between two point charges is when the separation of the charges is . What is the force between the charges when the separation is increased to ?
\nA.
\nB.
\nC.
\nD.
\nC
\nAn electron enters a uniform electric field of strength E with a velocity v. The direction of v is not parallel to E. What is the path of the electron after entering the field?
\nA. Circular
B. Parabolic
C. Parallel to E
D. Parallel to v
\nB
\nX and Y are two plane coils parallel to each other that have a common axis. There is a constant direct current in Y.
\nX is first moved towards Y and later is moved away from Y. What, as X moves, is the direction of the current in X relative to that in Y?
\nC
\n![](\"data:image/png;base64,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\"/)
![](\"data:image/png;base64,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\")