[ { "question_id": "19M.3.SL.TZ1.1", "Question": "
\n

In 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.

\n

\n

The 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 \na\n=\ng\n\n\n\n\nm\n1\n\n\n−\n\n\nm\n2\n\n\n\n\n\n\nm\n1\n\n\n+\n\n\nm\n2\n\n\n\n\n and this can be re-arranged to give \ng\n=\na\n\n\n\n\nm\n1\n\n\n+\n\n\nm\n2\n\n\n\n\n\n\nm\n1\n\n\n−\n\n\nm\n2\n\n\n\n\n.

\n

In 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.

\n
\n

Calculate the percentage error in the measured value of g.

\n
[3]
\n
a.i.
\n
\n

Deduce the value of g and its absolute uncertainty for this experiment.

\n
[2]
\n
a.ii.
\n
\n

There is an advantage and a disadvantage in using two masses that are almost equal.

\n

State and explain the advantage with reference to the magnitude of the acceleration that is obtained.

\n
[2]
\n
b.i.
\n
\n

There is an advantage and a disadvantage in using two masses that are almost equal.

\n

State and explain the disadvantage with reference to your answer to (a)(ii).

\n
[2]
\n
b.ii.
\n
", "Markscheme": "
\n

error in m1 + m2 is 1 % OR error in m1 − m2 is 40 % OR error in a is 1 % ✔

\n

adds percentage errors ✔

\n

so error in g is 42 % OR 40 % OR 41.8 % ✔

\n

Allow answer 0.42 or 0.4 or 0.418.

\n

Award [0] for comparing the average value with a known value, e.g. 9.81 m s-2.

\n
a.i.
\n
\n

g = 9.996 «m s−2» OR Δg = 4.20 «m s−2» ✔

\n

g = (10 ± 4) «m s−2»

\n

OR

\n

g = (10.0 ± 4.2) «m s−2» ✔

\n

Award [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 » .

\n
a.ii.
\n
\n

the acceleration would be small/the time of fall would be large ✔

\n

easier to measure /a longer time of fall reduces the % error in the time of fall and «hence acceleration» ✔

\n

Do not accept ideas related to the mass/moment of inertia of the pulley.

\n
b.i.
\n
\n

the percentage error in the difference of the masses is large ✔

\n

leading to a large percentage error/uncertainty in g/of the experiment ✔

\n

Do not accept ideas related to the mass/moment of inertia of the pulley.

\n
b.ii.
\n
", "Examiners report": "
\n

Atwoods 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.

\n
a.i.
\n
\n

Many 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.

\n
a.ii.
\n
\n

Q 1 b) was quite well answered. Only the weakest candidates presented difficulty in understanding simple mechanics.

\n
b.i.
\n
\n

In 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.

\n
b.ii.
\n
", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-1-3-controlling-variables", "inquiry-1-exploring-and-designing", "tool-3-mathematics" ] }, { "question_id": "19M.3.SL.TZ2.1", "Question": "
\n

A student investigates the electromotive force (emf) ε and internal resistance r of a cell.

\n

\n

The current I and the terminal potential difference V are measured.

\n

For this circuit V = ε - Ir .

\n

The table shows the data collected by the student. The uncertainties for each measurement
are shown.

\n

\n

The graph shows the data plotted.

\n

\n
\n

The student has plotted error bars for the potential difference. Outline why no error bars are shown for the current.

\n
[1]
\n
a.
\n
\n

Determine, using the graph, the emf of the cell including the uncertainty for this value. Give your answer to the correct number of significant figures.

\n
[3]
\n
b.
\n
\n

Outline, without calculation, how the internal resistance can be determined from this graph.

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

ΔI is too small to be shown/seen

\n

OR

\n

Error bar of negligible size compared to error bar in V ✔

\n
a.
\n
\n

evidence that ε can be determined from the y-intercept of the line of best-fit or lines of min and max gradient ✔

\n

states ε=1.59 OR 1.60 OR 1.61V«» ✔

\n

states uncertainty in ε is 0.02 V«» OR 0.03«V» ✔

\n
b.
\n
\n

determine the gradient «of the line of best-fit» ✔

\n

r is the negative of this gradient ✔

\n
c.
\n
", "Examiners report": "
\n

Almost 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.

\n
a.
\n
\n

The 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.

\n
b.
\n
\n[N/A]\n
c.
\n
", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-2-1-collecting-data", "inquiry-2-collecting-and-processing-data", "tool-3-mathematics" ] }, { "question_id": "19M.3.SL.TZ2.3", "Question": "
\n

A 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.

\n

\n

The distance D from the double slits to the screen is measured using a ruler with a smallest scale division of 1 mm.

\n

The fringe separation s is measured with uncertainty ± 0.1 mm.

\n

The slit separation d has negligible uncertainty.

\n

The wavelength is calculated using the relationship  \nλ\n=\n\n\ns\nd\n\nD\n\n.

\n
\n

When d = 0.200 mm, s = 0.9 mm and D = 280 mm, determine the percentage uncertainty in the wavelength.

\n
[2]
\n
a.
\n
\n

Explain how the student could use this apparatus to obtain a more reliable value for λ. 

\n
[2]
\n
b.
\n
", "Markscheme": "
\n

Evidence of \n\n\nΔ\ns\n\ns\n\n\nAND\n\n\n\nΔ\nD\n\nD\n\n used   ✔

\n

«add fractional/% uncertainties»

\n

obtains 11 % (or 0.11) OR 10 % (or 0.1) ✔

\n
a.
\n
\n

ALTERNATIVE 1:

\n

measure the combined width for several fringes

\n

OR

\n

repeat measurements ✓

\n

take the average

\n

OR

\n

so the «percentage» uncertainties are reduced ✓

\n

ALTERNATIVE 2:

\n

increase D «hence s»

\n

OR

\n

Decrease d ✓

\n

so the «percentage» uncertainties are reduced ✓

\n

Do not accept answers which suggest using different apparatus.

\n
b.
\n
", "Examiners report": "
\n

A 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.

\n
a.
\n
\n

The 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.

\n
b.
\n
", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-3-2-evaluating", "inquiry-3-concluding-and-evaluating", "tool-3-mathematics" ] }, { "question_id": "19M.2.HL.TZ1.2", "Question": "
\n

A beam of electrons each of de Broglie wavelength 2.4 × 10–15 m is incident on a thin film of silicon-30  \n\n(\n\n\n\n\n\n14\n\n\n30\n\n\n\nSi\n\n\n)\n\n. The variation in the electron intensity of the beam with scattering angle is shown.

\n

\n
\n

Use the graph to show that the nuclear radius of silicon-30 is about 4 fm.

\n
[3]
\n
a.i.
\n
\n

Estimate, using the result from (a)(i), the nuclear radius of thorium-232 \n\n(\n\n\n\n\n\n90\n\n\n232\n\n\n\nTh\n\n\n)\n\n.

\n
[2]
\n
a.ii.
\n
\n

Suggest 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.

\n
[1]
\n
a.iii.
\n
\n

Outline why deviations from Rutherford scattering are observed when high-energy alpha particles are incident on nuclei.

\n
[2]
\n
a.iv.
\n
", "Markscheme": "
\n

read off between 17 and 19 «deg» ✔

\n

correct use of d = \n\nλ\n\nsin\n⁡\nθ\n\n\n = 7.8 × 10−15 «m» ✔

\n

so radius = \n\n7.8\n2\n\n «fm» = 3.9 «fm» ✔

\n

Award ecf for wrong angle in MP1.

\n

Answer for MP3 must show at least 2 sf.

\n
a.i.
\n
\n

RTh = Rsi  \n\n\n\n(\n\n\n\n\n\nA\n\n\nTh\n\n\n\n\n\n\n\n\nA\n\n\nSi\n\n\n\n\n\n\n\n)\n\n\n\n1\n3\n\n\n\n\n or substitution ✔

\n

7.4 «fm» ✔

\n
a.ii.
\n
\n

electron wavelength shorter than alpha particles (thus increased resolution)
OR
electron is not subject to strong nuclear force ✔

\n

 

\n
a.iii.
\n
\n

nuclear forces act ✔

\n

nuclear recoil occurs ✔

\n

significant penetration into nucleus / probing internal structure of individual nucleons ✔

\n

incident particles are relativistic ✔

\n
a.iv.
\n
", "Examiners report": "
\n

This 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.

\n
a.i.
\n
\n

This 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.

\n
a.ii.
\n
\n

This 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.

\n
a.iii.
\n
\n

Candidates 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.

\n
a.iv.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-3-wave-phenomena" ] }, { "question_id": "19M.2.HL.TZ1.3", "Question": "
\n

A beam of microwaves is incident normally on a pair of identical narrow slits S1 and S2.

\n

\n

When 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.

\n
\n

Explain why intensity maxima are observed at X and Y.

\n
[2]
\n
a.
\n
\n

The distance from S1 to Y is 1.243 m and the distance from S2 to Y is 1.181 m.

\n

Determine the frequency of the microwaves.

\n
[3]
\n
b.
\n
\n

Outline one reason why the maxima observed at W, X and Y will have different intensities from each other.

\n
[1]
\n
c.
\n
\n

The 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.

\n

\n

The 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.

\n

\n
[2]
\n
d.
\n
", "Markscheme": "
\n

two waves superpose/mention of superposition/mention of «constructive» interference ✔

\n

they arrive in phase/there is a path length difference of an integer number of wavelengths ✔

\n
a.
\n
\n

path difference = 0.062 «m»✔

\n

so wavelength = 0.031 «m»✔

\n

frequency = 9.7 × 109 «Hz»✔

\n

Award [2 max] for 4.8 x 109 Hz.

\n
b.
\n
\n

intensity is modulated by a single slit diffraction envelope OR

\n

intensity varies with distance OR points are different distances from the slits ✔

\n
c.
\n
\n

\n

cos2 variation shown ✔

\n

with zero at 90° (by eye) ✔

\n

Award [1 max] for an inverted curve with maximum at 90°.

\n
d.
\n
", "Examiners report": "
\n

Many 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.

\n
a.
\n
\n

Many 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.

\n
b.
\n
\n

This 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.

\n
c.
\n
\n

This 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.

\n
d.
\n
", "topics": [ "a--space-time-and-motion", "c-wave-behaviour" ], "subtopics": [ "a-5-galilean-and-special-relativity", "c-2-wave-model", "c-3-wave-phenomena" ] }, { "question_id": "19M.2.HL.TZ1.5", "Question": "
\n

The moon Phobos moves around the planet Mars in a circular orbit.

\n
\n

Outline the origin of the force that acts on Phobos.

\n
[1]
\n
a.i.
\n
\n

Outline why this force does no work on Phobos.

\n
[1]
\n
a.ii.
\n
\n

The orbital period T of a moon orbiting a planet of mass M is given by

\n

\n\n\n\n\nR\n3\n\n\n\n\n\n\nT\n2\n\n\n\n\n=\nk\nM\n

\n

where R is the average distance between the centre of the planet and the centre of the moon.

\n

Show that \nk\n=\n\nG\n\n4\n\n\nπ\n2\n\n\n\n\n

\n
[3]
\n
b.i.
\n
\n

The following data for the Mars–Phobos system and the Earth–Moon system are available:

\n

Mass of Earth = 5.97 × 1024 kg

\n

The Earth–Moon distance is 41 times the Mars–Phobos distance.

\n

The orbital period of the Moon is 86 times the orbital period of Phobos.

\n

Calculate, in kg, the mass of Mars.

\n
[2]
\n
b.ii.
\n
\n

The 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.

\n

\n

Determine, using the graph, the mass of the Moon.

\n
[3]
\n
c.
\n
", "Markscheme": "
\n

gravitational attraction/force/field «of the planet/Mars» ✔

\n

Do not accept “gravity”.

\n
a.i.
\n
\n

the force/field and the velocity/displacement are at 90° to each other OR

\n

there is no change in GPE of the moon/Phobos ✔

\n
a.ii.
\n
\n

ALTERNATE 1

\n

«using fundamental equations»

\n

use of Universal gravitational force/acceleration/orbital velocity equations ✔

\n

equating to centripetal force or acceleration. ✔

\n

rearranges to get \nk\n=\n\nG\n\n4\n\n\nπ\n2\n\n\n\n\n  ✔

\n

ALTERNATE 2

\n

«starting with \n\n\n\n\nR\n3\n\n\n\n\n\n\nT\n2\n\n\n\n\n=\nk\nM\n»

\n

substitution of proper equation for T from orbital motion equations ✔

\n

substitution of proper equation for M OR R from orbital motion equations ✔

\n

rearranges to get \nk\n=\n\nG\n\n4\n\n\nπ\n2\n\n\n\n\n  ✔

\n
b.i.
\n
\n

\n\n\nm\n\n\nMars\n\n\n\n\n=\n\n\n\n(\n\n\n\n\n\nR\n\n\nMars\n\n\n\n\n\n\n\n\nR\n\n\nEarth\n\n\n\n\n\n\n\n)\n\n3\n\n\n\n\n\n(\n\n\n\n\n\nT\n\n\nEarth\n\n\n\n\n\n\n\n\nT\n\nM\na\nr\ns\n\n\n\n\n\n\n)\n\n2\n\n\n\n\nm\n\n\nEarth\n\n\n\n\n or other consistent re-arrangement ✔

\n

6.4 × 1023 «kg» ✔

\n

 

\n
b.ii.
\n
\n

read off separation at maximum potential 0.9 ✔

\n

equating of gravitational field strength of earth and moon at that location OR ✔

\n

7.4 × 1022 «kg» ✔

\n

Allow ECF from MP1

\n
c.
\n
", "Examiners report": "
\n

This 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.

\n
a.i.
\n
\n

Some 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.

\n
a.ii.
\n
\n

This 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.

\n
b.i.
\n
\n

This 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.

\n
b.ii.
\n
\n

This 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.

\n
c.
\n
", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "a-3-work-energy-and-power", "d-1-gravitational-fields" ] }, { "question_id": "19M.2.HL.TZ1.6", "Question": "
\n

A small metal pendulum bob is suspended at rest from a fixed point with a length of thread of negligible mass. Air resistance is negligible.

\n

The pendulum begins to oscillate. Assume that the motion of the system is simple harmonic, and in one vertical plane.

\n

The graph shows the variation of kinetic energy of the pendulum bob with time.

\n

\n
\n

When 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.

\n

\n
\n

Calculate, in m, the length of the thread. State your answer to an appropriate number of significant figures.

\n
[3]
\n
a.i.
\n
\n

Label on the graph with the letter X a point where the speed of the pendulum is half that of its initial speed.

\n
[1]
\n
a.ii.
\n
\n

The mass of the pendulum bob is 75 g. Show that the maximum speed of the bob is about 0.7 m s–1.

\n
[2]
\n
a.iii.
\n
\n

Calculate the speed of the combined masses immediately after the collision.

\n
[1]
\n
b.i.
\n
\n

Show that the collision is inelastic.

\n
[3]
\n
b.ii.
\n
\n

Sketch, 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.

\n

\n
[2]
\n
b.iii.
\n
\n

The 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′.

\n

Explain why f and f′ are different.

\n
[3]
\n
b.iv.
\n
", "Markscheme": "
\n

identifies T as 2.25 s ✔

\n

L = 1.26 m ✔

\n

1.3 / 1.26 «m» ✔

\n

Accept any number of s.f. for MP2.

\n

Accept any answer with 2 or 3 s.f. for MP3.

\n
a.i.
\n
\n

X labels any point on the curve where EK   \n\n1\n4\n\n of maximum/5 mJ ✔

\n
a.ii.
\n
\n

\n\n1\n2\n\n mv2 = 20 × 10−3 seen OR \n\n1\n2\n\n × 7.5 × 10-2 × v2 = 20 × 10-3 ✔

\n

0.73 «m s−1» ✔

\n

Must see at least 2 s.f. for MP2.

\n
a.iii.
\n
\n

0.40 «m s-1» ✔

\n
b.i.
\n
\n

initial energy 24 mJ and final energy 12 mJ ✔

\n

energy is lost/unequal /change in energy is 12 mJ ✔

\n

inelastic collisions occur when energy is lost ✔

\n
b.ii.
\n
\n

graph with same period but inverted ✔

\n

amplitude one half of the original/two boxes throughout (by eye) ✔

\n
b.iii.
\n
\n

mention of Doppler effect ✔

\n

there is a change in the wavelength of the reflected wave ✔

\n

because the wave speed is constant, there is a change in frequency ✔

\n
b.iv.
\n
", "Examiners report": "
\n

This 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.

\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n

This 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.

\n
a.iii.
\n
\n[N/A]\n
b.i.
\n
\n

Candidates 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.

\n
b.ii.
\n
\n

Many 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.

\n
b.iii.
\n
\n

This 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.

\n
b.iv.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "19M.2.HL.TZ1.7", "Question": "
\n
\n (a)\n
\n
\n

\n Show that during an adiabatic expansion of an ideal monatomic gas the temperature\n \n \n \n \n T\n \n \n \n \n and volume\n \n \n \n \n V\n \n \n \n \n are given by\n

\n

\n \n \n \n \n T\n \n \n \n \n V\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n \n \n \n \n = constant\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Calculate the efficiency of the cycle.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n The work done during the isothermal expansion A → B is 540 J. Calculate the thermal energy that leaves the gas during one cycle.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.iii)\n
\n
\n

\n Calculate the ratio\n \n \n \n \n \n \n \n \n V\n \n \n C\n \n \n \n \n \n \n \n \n V\n \n \n B\n \n \n \n \n \n \n \n \n where\n \n V\n \n C\n \n \n is the volume of the gas at C and\n \n V\n \n B\n \n \n is the volume at B.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c.i)\n
\n
\n

\n Calculate the change in the entropy of the gas during the change A to B.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c.ii)\n
\n
\n

\n Explain, 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).\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n substitution of\n \n \n \n \n P\n \n \n =\n \n \n \n \n n\n \n \n R\n \n \n T\n \n \n \n V\n \n \n \n \n in\n \n \n \n \n \n P\n \n \n X\n \n \n \n \n \n V\n \n \n X\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n =\n \n \n \n \n P\n \n \n Y\n \n \n \n \n \n V\n \n \n Y\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n \n \n ✔\n

\n

\n manipulation to get result ✔\n

\n
\n
\n (b.i)\n
\n

\n e « = 1 −\n \n \n \n \n \n \n \n \n T\n \n \n c\n \n \n \n \n \n \n \n \n T\n \n \n h\n \n \n \n \n \n \n \n \n = 1 −\n \n \n \n \n 340\n \n \n 620\n \n \n \n \n » = 0.45 ✔\n

\n
\n
\n (b.ii)\n
\n

\n heat into gas «is along AB» and equals\n

\n

\n \n Q\n \n in\n \n \n «= Δ\n \n U\n \n +\n \n W\n \n = 0 + 540» = 540 «J» ✔\n

\n

\n heat out is (1−\n \n e\n \n )\n \n Q\n \n in =\n \n \n (1−0.45) × 540 = 297 «J» ≈ 3.0 × 10\n \n 2\n \n «J» ✔\n

\n

\n \n Award\n \n [2]\n \n for bald correct answer.\n \n

\n
\n
\n (b.iii)\n
\n

\n \n \n \n \n \n \n T\n \n \n B\n \n \n \n \n \n V\n \n \n B\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n \n =\n \n \n \n \n T\n \n \n C\n \n \n \n \n \n V\n \n \n C\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n \n ⇒\n \n \n \n \n \n \n V\n \n \n C\n \n \n \n \n \n \n \n \n V\n \n \n B\n \n \n \n \n \n \n =\n \n \n \n \n \n (\n \n \n \n \n \n \n \n T\n \n \n B\n \n \n \n \n \n \n \n \n T\n \n \n C\n \n \n \n \n \n \n \n )\n \n \n \n \n \n 3\n \n \n 2\n \n \n \n \n \n \n \n \n ✔\n

\n

\n \n \n \n \n \n \n \n \n V\n \n \n C\n \n \n \n \n \n \n \n \n V\n \n \n B\n \n \n \n \n \n \n =\n \n \n \n \n \n (\n \n \n \n \n \n 620\n \n \n \n \n 340\n \n \n \n \n \n )\n \n \n \n \n \n 3\n \n \n 2\n \n \n \n \n \n \n =\n \n \n 2.5\n \n \n \n \n ✔\n

\n

\n \n Award\n \n [2]\n \n for bald correct answer.\n \n

\n
\n
\n (c.i)\n
\n

\n Δ\n \n S\n \n «=\n \n \n \n \n Q\n \n \n T\n \n \n \n \n =\n \n \n \n \n 540\n \n \n 620\n \n \n \n \n »= 0.87 «JK\n \n −1\n \n » ✔\n

\n
\n
\n (c.ii)\n
\n

\n the Carnot cycle has the maximum efficiency «for heat engines operating between two given temperatures »✔\n

\n

\n real engine can not work at Carnot cycle/ideal cycle ✔\n

\n

\n the second law of thermodynamics says that it is impossible to convert all the input heat into mechanical work ✔\n

\n

\n a real engine would have additional losses due to friction\n \n etc\n \n ✔\n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n The algebraic manipulation required for this question was well mastered by only the better-prepared candidates. Many candidates tried to find the required formula via randomly selected equations from the data booklet.\n

\n
\n
\n (b.i)\n
\n

\n The efficiency of the Carnot cycle was well calculated by most of the candidates, but the thermal energy that leaves the gas was well calculated only by the best candidates. Many candidates were able to establish that the heat added to the gas was 540 J - but struggled to then link this with the efficiency to determine the thermal energy leaving the gas. In iii) the better candidates used the formula given at i) to appropriately calculate the ratio but many were not able to manipulate the expressions to achieve the desired outcome. The reciprocal of 2.5 (0.4) was a common error in the final result and so was the incorrect use of the formula for isobaric expansion. The change in the entropy was well calculated by many candidates with ECF being very prominent from part biii). A complete and proper explanation in ii) was well formulated only by the better candidates. However, many were able to recognize that the Carnot engine was ideal and not real. Many answers referred to heat being lost to the environment but not to “additional losses” that make the engine less than its ideal capacity.\n

\n
\n
\n (b.ii)\n
\n

\n The efficiency of the Carnot cycle was well calculated by most of the candidates, but the thermal energy that leaves the gas was well calculated only by the best candidates. Many candidates were able to establish that the heat added to the gas was 540 J - but struggled to then link this with the efficiency to determine the thermal energy leaving the gas. In iii) the better candidates used the formula given at i) to appropriately calculate the ratio but many were not able to manipulate the expressions to achieve the desired outcome. The reciprocal of 2.5 (0.4) was a common error in the final result and so was the incorrect use of the formula for isobaric expansion. The change in the entropy was well calculated by many candidates with ECF being very prominent from part biii). A complete and proper explanation in ii) was well formulated only by the better candidates. However, many were able to recognize that the Carnot engine was ideal and not real. Many answers referred to heat being lost to the environment but not to “additional losses” that make the engine less than its ideal capacity.\n

\n
\n
\n (b.iii)\n
\n

\n The efficiency of the Carnot cycle was well calculated by most of the candidates, but the thermal energy that leaves the gas was well calculated only by the best candidates. Many candidates were able to establish that the heat added to the gas was 540 J - but struggled to then link this with the efficiency to determine the thermal energy leaving the gas. In iii) the better candidates used the formula given at i) to appropriately calculate the ratio but many were not able to manipulate the expressions to achieve the desired outcome. The reciprocal of 2.5 (0.4) was a common error in the final result and so was the incorrect use of the formula for isobaric expansion. The change in the entropy was well calculated by many candidates with ECF being very prominent from part biii). A complete and proper explanation in ii) was well formulated only by the better candidates. However, many were able to recognize that the Carnot engine was ideal and not real. Many answers referred to heat being lost to the environment but not to “additional losses” that make the engine less than its ideal capacity.\n

\n
\n
\n (c.i)\n
\n

\n The efficiency of the Carnot cycle was well calculated by most of the candidates, but the thermal energy that leaves the gas was well calculated only by the best candidates. Many candidates were able to establish that the heat added to the gas was 540 J - but struggled to then link this with the efficiency to determine the thermal energy leaving the gas. In iii) the better candidates used the formula given at i) to appropriately calculate the ratio but many were not able to manipulate the expressions to achieve the desired outcome. The reciprocal of 2.5 (0.4) was a common error in the final result and so was the incorrect use of the formula for isobaric expansion. The change in the entropy was well calculated by many candidates with ECF being very prominent from part biii). A complete and proper explanation in ii) was well formulated only by the better candidates. However, many were able to recognize that the Carnot engine was ideal and not real. Many answers referred to heat being lost to the environment but not to “additional losses” that make the engine less than its ideal capacity.\n

\n
\n
\n (c.ii)\n
\n

\n The efficiency of the Carnot cycle was well calculated by most of the candidates, but the thermal energy that leaves the gas was well calculated only by the best candidates. Many candidates were able to establish that the heat added to the gas was 540 J - but struggled to then link this with the efficiency to determine the thermal energy leaving the gas. In iii) the better candidates used the formula given at i) to appropriately calculate the ratio but many were not able to manipulate the expressions to achieve the desired outcome. The reciprocal of 2.5 (0.4) was a common error in the final result and so was the incorrect use of the formula for isobaric expansion. The change in the entropy was well calculated by many candidates with ECF being very prominent from part biii). A complete and proper explanation in ii) was well formulated only by the better candidates. However, many were able to recognize that the Carnot engine was ideal and not real. Many answers referred to heat being lost to the environment but not to “additional losses” that make the engine less than its ideal capacity.\n

\n
\n", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "19M.2.HL.TZ2.10", "Question": "
\n

A small magnet is dropped from rest above a stationary horizontal conducting ring. The south (S) pole of the magnet is upwards.

\n

\n
\n

While the magnet is moving towards the ring, state why the magnetic flux in the ring is increasing.

\n
[1]
\n
a.
\n
\n

While the magnet is moving towards the ring, sketch, using an arrow on Diagram 2, the direction of the induced current in the ring.

\n
[1]
\n
b.
\n
\n

While the magnet is moving towards the ring, deduce the direction of the magnetic force on the magnet.

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

the magnetic field at the position of the ring is increasing «because the magnet gets closer to the ring» ✔

\n
a.
\n
\n

the current must be counterclockwise «in diagram 2» ✔

\n

eg:

\n

\n
b.
\n
\n

since the induced magnetic field is upwards

\n

OR

\n

by Lenz law the change «of magnetic field/flux» must be opposed

\n

OR

\n

by conservation of energy the movement of the magnet must be opposed ✔

\n

therefore the force is repulsive/upwards ✔

\n
c.
\n
", "Examiners report": "
\n

This was well-answered.

\n
a.
\n
\n

Answers here were reasonably evenly split between clockwise and anti-clockwise, with the odd few arrows pointing left or right.

\n
b.
\n
\n

The 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.

\n
c.
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "19M.2.HL.TZ2.11", "Question": "
\n

In 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.

\n

\n
\n

A pure sample of copper-64 has a mass of 28 mg. The decay constant of copper-64 is 5.5 × 10-2 hour–1.

\n
\n

Suggest 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.

\n
[2]
\n
a.
\n
\n

Estimate, using the graph, the radius of a carbon-12 nucleus.

\n
[2]
\n
bi.
\n
\n

The ratio \n\n\n\nvolume of a nucleaus of mass number \n\nA\n\n\n\nvolume of a nucleon\n\n\n\n is approximately A.

\n

Comment on this observation by reference to the strong nuclear force.

\n
[2]
\n
bii.
\n
\n

Estimate, in Bq, the initial activity of the sample.

\n
[2]
\n
ci.
\n
\n

Calculate, in hours, the time at which the activity of the sample has decreased to one-third of the initial activity.

\n
[2]
\n
cii.
\n
", "Markscheme": "
\n

«de Broglie’s hypothesis states that the» electron is represented by a wave ✔

\n

therefore it cannot be localized/it is spread out/it does not have a definite position ✔

\n

Award MP1 for any mention of wavelike property of an electron.

\n
a.
\n
\n

«\nd\nsin\n⁡\nθ\n=\nλ\n⇒\n» \nd\n=\n\n\n1.6\n×\n\n\n\n10\n\n\n−\n15\n\n\n\n\n\nsin\n⁡\n\n17\n∘\n\n\n\n\n/\n\n5.47\n×\n\n\n10\n\n−\n15\n\n\n\n «m» ✔

\n

\nR\n=\n\nd\n2\n\n≈\n2.7\n\n/\n\n2.8\n×\n\n\n10\n\n−\n15\n\n\n\n «m» ✔

\n
bi.
\n
\n

this implies that the nucleons are very tightly packed/that there is very little space in between the nucleons ✔

\n

because the nuclear force is stronger than the electrostatic force ✔

\n
bii.
\n
\n

number of nuclei is \n\n\n28\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n64\n\n\n×\n6.02\n×\n\n\n10\n\n23\n\n\n\n\n/\n\n2.63\n×\n\n\n10\n\n20\n\n\n\n ✔

\n

\nA\n=\n«\nλ\nN\n=\n2.63\n×\n\n\n10\n\n20\n\n\n\n×\n\n\n5.5\n×\n\n\n\n10\n\n\n−\n2\n\n\n\n\n\n3600\n\n\n» \n4.0\n×\n\n\n10\n\n15\n\n\n\n «Bq» ✔

\n
ci.
\n
\n

\n\n\n1\n\n\n3\n\n\n=\n\n\ne\n\n−\nλ\nt\n\n\n\n ✔

\n

t = 20«hr» ✔

\n
cii.
\n
", "Examiners report": "", "topics": [ "c-wave-behaviour", "e-nuclear-and-quantum-physics" ], "subtopics": [ "c-3-wave-phenomena", "e-1-structure-of-the-atom", "e-2-quantum-physics", "e-3-radioactive-decay" ] }, { "question_id": "19M.2.HL.TZ2.4", "Question": "
\n

Three 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.

\n

\n
\n

When fully charged the space between the plates of the capacitor is filled with a dielectric with double the permittivity of a vacuum.

\n
\n

The switch S is initially open. Calculate the total power dissipated in the circuit.

\n
[2]
\n
a.
\n
\n

The switch is now closed. State, without calculation, why the current in the cell will increase.

\n
[1]
\n
bi.
\n
\n

The switch is now closed. \n\nDeduce the ratio \n\n\n\n\npower dissipated in Y with S open\n\n\n\n\npower dissipated in Y with S closed\n\n\n\n.

\n

 

\n
[2]
\n
bii.
\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.

\n

\n

The capacitance of the capacitor is 6.0 μF and the reading of the voltmeter is 12 V.

\n

Calculate the energy stored in the capacitor.

\n
[1]
\n
c.
\n
\n

Calculate the change in the energy stored in the capacitor.

\n
[3]
\n
di.
\n
\n

Suggest, in terms of conservation of energy, the cause for the above change.

\n
[1]
\n
dii.
\n
", "Markscheme": "
\n

total resistance of circuit is 8.0 «Ω» ✔

\n

\nP\n=\n\n\n\n\n\n12\n\n2\n\n\n\n\n8.0\n\n\n=\n18\n«W» ✔

\n
a.
\n
\n

«a resistor is now connected in parallel» reducing the total resistance

\n

OR

\n

current through YZ unchanged and additional current flows through X ✔

\n
bi.
\n
\n

evidence in calculation or statement that pd across Y/current in Y is the same as before ✔

\n

so ratio is 1 ✔

\n
bii.
\n
\n

\nE\n=\n\n«\n\n\n1\n2\n\nC\n\n\nV\n2\n\n\n=\n\n1\n2\n\n×\n6\n×\n\n\n10\n\n−\n6\n\n\n\n×\n\n\n12\n2\n\n\n=\n\n»\n\n4.3\n×\n\n\n10\n\n−\n4\n\n\n\n«\n\nJ\n\n» ✔

\n
c.
\n
\n

ALTERNATIVE 1

\n

capacitance doubles and voltage halves ✔

\n

since \nE\n=\n\n1\n2\n\nC\n\n\nV\n2\n\n\n energy halves   ✔

\n

so change is «–»2.2×10–4 «J»  ✔

\n

 

\n

ALTERNATIVE 2

\n

\nE\n=\n\n1\n2\n\nC\n\n\nV\n2\n\n\n\n and \n\nQ\n=\nC\nV\n\n so \n\nE\n=\n\n\n\n\nQ\n2\n\n\n\n\n2\nC\n\n\n  ✔ 

\n

capacitance doubles and charge unchanged so energy halves ✔

\n

so change is «−»2.2 × 10−4 «J» ✔

\n
di.
\n
\n

it is the work done when inserting the dielectric into the capacitor ✔

\n
dii.
\n
", "Examiners report": "
\n

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.

\n
a.
\n
\n

Most 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.

\n
bi.
\n
\n

Very 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.

\n
bii.
\n
\n

Most answered this correctly.

\n
c.
\n
\n

By 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.

\n
di.
\n
\n

Very few scored on this question.

\n
dii.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-5-galilean-and-special-relativity" ] }, { "question_id": "19M.2.HL.TZ2.6", "Question": "
\n
\n (a)\n
\n
\n

\n \n Define\n \n proper length\n \n .\n \n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (bi)\n
\n
\n

\n \n In the reference frame of the train a ball travels with speed 0.50\n \n c\n \n 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.\n \n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (bii)\n
\n
\n

\n \n In the reference frame of the train a ball travels with speed 0.50\n \n c\n \n 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.\n \n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n the length measured «in a reference frame» where the object is at rest ✔\n
\n \n

\n
\n
\n (bi)\n
\n

\n \n \n
\n \n

\n
\n
\n (bii)\n
\n

\n \n \n \n ALTERNATIVE 1:\n \n \n \n \n
\n \n

\n

\n \n \n \n

\n

\n \n \n \n \n ALTERNATIVE 2:\n \n \n \n \n

\n

\n \n \n \n \n v\n \n of ball is 0.846\n \n c\n \n for platform ✔\n
\n \n \n \n

\n

\n \n \n \n length of train is 68m for platform ✔\n \n \n \n

\n

\n \n \n \n \n \n \n \n

\n

\n \n \n \n \n \n ALTERNATIVE 3:\n \n \n \n \n \n

\n

\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n Proper length is quite well understood. A common mistake is to mention that it is the length measured by a reference frame at rest.\n

\n
\n
\n (bi)\n
\n

\n Because 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.\n

\n
\n
\n (bii)\n
\n

\n Almost 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.\n

\n
\n", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-5-galilean-and-special-relativity" ] }, { "question_id": "19M.2.HL.TZ2.8", "Question": "
\n

Monochromatic 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.

\n

\n

Initially 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.

\n
\n

The width of each slit is increased to 0.030 mm. D, d and λ remain the same.

\n
\n

Deduce, in W m-2, the intensity at M.

\n
[3]
\n
a.
\n
\n

P is the first maximum of intensity on one side of M. The following data are available.

\n

d = 0.12 mm

\n

D = 1.5 m

\n

Distance MP = 7.0 mm

\n

Calculate, in nm, the wavelength λ of the light.

\n
[2]
\n
b.
\n
\n

Suggest why, after this change, the intensity at P will be less than that at M.

\n
[1]
\n
ci.
\n
\n

Show that, due to single slit diffraction, the intensity at a point on the screen a distance of 28 mm from M is zero.

\n
[2]
\n
cii.
\n
", "Markscheme": "
\n

there is constructive interference at M

\n

OR

\n

the amplitude doubles at M ✔

\n

intensity is «proportional to» amplitude2 ✔

\n

88 «W m−2» ✔

\n
a.
\n
\n

«\ns\n=\n\n\nλ\nD\n\nd\n\n⇒\n\n»\n\nλ\n=\n\n\ns\nd\n\nD\n\n\n/\n\n\n\n0.12\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n×\n7.0\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n1.5\n\n\n    ✔

\n

\nλ\n=\n560\n\n«\n\n\nnm\n\n»  ✔

\n

 

\n
b.
\n
\n

«the interference pattern will be modulated by»

\n

single slit diffraction ✔

\n

«envelope and so it will be less»

\n
ci.
\n
\n

ALTERNATIVE 1

\n

the angular position of this point is \nθ\n=\n\n\n28\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n1.5\n\n\n=\n0.01867\n«rad»  ✔

\n

which coincides with the first minimum of the diffraction envelope

\n

\nθ\n=\n\nλ\nb\n\n=\n\n\n560\n×\n\n\n\n10\n\n\n−\n9\n\n\n\n\n\n0.030\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n=\n0.01867\n «rad» ✔

\n

«so intensity will be zero»

\n

 

\n

ALTERNATIVE 2

\n

the first minimum of the diffraction envelope is at \nθ\n=\n\nλ\nb\n\n=\n\n\n560\n×\n\n\n\n10\n\n\n−\n9\n\n\n\n\n\n0.030\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n=\n0.01867\n«rad»   ✔

\n

distance on screen is \ny\n=\n1.50\n×\n0.01867\n=\n28\n«mm»  ✔

\n

«so intensity will be zero»

\n

 

\n
cii.
\n
", "Examiners report": "
\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.

\n
a.
\n
\n

This 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.

\n
b.
\n
\n

Many recognised the significance of the single slit diffraction envelope.

\n
ci.
\n
\n

Credit 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.

\n
cii.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum", "a-4-rigid-body-mechanics" ] }, { "question_id": "19M.2.HL.TZ2.9", "Question": "
\n

A 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.

\n
\n

Show that the total energy of the planet is given by the equation shown.

\n

\nE\n=\n\n1\n2\n\nm\nV\n

\n
[2]
\n
ai.
\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.

\n
[2]
\n
aii.
\n
\n

The diagram shows some of the electric field lines for two fixed, charged particles X and Y.

\n

\n

The magnitude of the charge on X is \nQ\n and that on Y is \nq\n. The distance between X and Y is 0.600 m. The distance between P and Y is 0.820 m.

\n

At P the electric field is zero. Determine, to one significant figure, the ratio \n\nQ\nq\n\n.

\n
[2]
\n
b.
\n
", "Markscheme": "
\n

\nE\n=\n\n1\n2\n\nm\n\n\nG\nM\n\nr\n\n−\n\n\nG\nM\nm\n\nr\n\n=\n−\n\n1\n2\n\n\n\nG\nM\nm\n\nr\n\n  ✔

\n

comparison with \nV\n=\n−\n\n\nG\nM\n\nr\n\n   ✔

\n

«to give answer»

\n

 

\n
ai.
\n
\n

ALTERNATIVE 1

\n

«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 ✔

\n

the potential will not change and so the energy will not change ✔

\n

ALTERNATIVE 2

\n

r / distance between the centres of the objects / orbital radius remains unchanged ✔

\n

since \n\n\nE\n\nT\no\nt\na\nl\n\n\n\n=\n−\n\n1\n2\n\n\n\nG\nM\nm\n\nr\n\n, energy will not change ✔

\n

 

\n
aii.
\n
\n

\n\n\nk\nQ\n\n\n\n\n\n(\n0.600\n+\n0.820\n)\n\n2\n\n\n\n\n=\n\n\nk\nq\n\n\n\n\n\n0.820\n\n2\n\n\n\n\n  ✔

\n

\n\nQ\nq\n\n=\n\n«\n\n\n\n\n\n\n(\n0.600\n+\n0.820\n)\n\n2\n\n\n\n\n\n\n\n0.820\n\n2\n\n\n\n\n=\n2.9988\n≈\n\n»\n\n3\n  ✔

\n

 

\n
b.
\n
", "Examiners report": "
\n

This was generally well answered but with candidates sometimes getting in to trouble over negative signs but otherwise producing well-presented answers.

\n
ai.
\n
\n

A large number of candidates thought that the total energy of the planet would change, mostly double.

\n
aii.
\n
\n

The 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.

\n
b.
\n
", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-4-rigid-body-mechanics", "d-1-gravitational-fields", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "19M.2.SL.TZ1.3", "Question": "
\n

A beam of microwaves is incident normally on a pair of identical narrow slits S1 and S2.

\n

\n

When 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.

\n
\n

Explain why intensity maxima are observed at X and Y.

\n
[2]
\n
a.
\n
\n

The 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.

\n
[3]
\n
b.
\n
\n

Outline one reason why the maxima observed at W, X and Y will have different intensities from each other.

\n
[1]
\n
c.
\n
", "Markscheme": "
\n

two waves superpose/mention of superposition/mention of «constructive» interference ✔

\n

they arrive in phase/there is a path length difference of an integer number of wavelengths ✔

\n

Ignore references to nodes/antinodes.

\n
a.
\n
\n

path difference = 0.062 «m» ✔

\n

so wavelength = 0.031 «m» ✔

\n

frequency = 9.7 × 109 «Hz» ✔

\n

If no unit is given, assume the answer is in Hz. Accept other prefixes (eg 9.7 GHz)

\n

Award [2 max] for 4.8 x 109 Hz.

\n
b.
\n
\n

intensity varies with distance OR points are different distances from the slits ✔

\n

Accept “Intensity is modulated by a single slit diffraction envelope”.

\n
c.
\n
", "Examiners report": "
\n

Many 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.

\n
a.
\n
\n

Many 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.

\n
b.
\n
\n

This 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.

\n
c.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-3-wave-phenomena" ] }, { "question_id": "19M.2.SL.TZ1.5", "Question": "
\n

A 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.

\n

At 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.

\n

\n
\n

Calculate the speed of the combined masses immediately after the collision.

\n
[1]
\n
a.
\n
\n

Show that the collision is inelastic.

\n
[3]
\n
b.
\n
\n

Describe the changes in gravitational potential energy of the oscillating system from t = 0 as it oscillates through one cycle of its motion.

\n
[1]
\n
c.
\n
", "Markscheme": "
\n

0.40 «m s−1» ✔

\n
a.
\n
\n

initial energy 24 mJ and final energy 12 mJ ✔

\n

energy is lost/unequal /change in energy is 12 mJ ✔

\n

inelastic collisions occur when energy is lost ✔

\n
b.
\n
\n

maximum GPE at extremes, minimum in centre ✔

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n

Candidates 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.

\n
b.
\n
\n

This 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.

\n
c.
\n
", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "a-3-work-energy-and-power", "d-1-gravitational-fields" ] }, { "question_id": "19M.2.SL.TZ1.6", "Question": "
\n

The 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.

\n

\n
\n

A 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.

\n
[2]
\n
a.i.
\n
\n

Another black body is on the Moon’s surface at point B.

\n

Outline, without calculation, why the aximum temperature of the black body at point B is less than at point A.

\n
[2]
\n
a.ii.
\n
\n

The 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.

\n
[1]
\n
b.
\n
\n

Outline why a force acts on the Moon.

\n
[1]
\n
c.i.
\n
\n

Outline why this force does no work on the Moon.

\n
[1]
\n
c.ii.
\n
", "Markscheme": "
\n

T = \n\n\n\n(\n\n\n\n1360\n\nσ\n\n\n)\n\n\n0.25\n\n\n\n   ✔

\n

390 «K» ✔

\n

Must see 1360 (from data booklet) used for MP1.

\n

Must see at least 2 s.f.

\n
a.i.
\n
\n

energy/Power/Intensity lower at B ✔

\n

connection made between energy/power/intensity and temperature of blackbody ✔

\n
a.ii.
\n
\n

(28 %) of sun’s energy is scattered/reflected by earth’s atmosphere OR only 72 % of incident energy gets absorbed by blackbody ✔

\n

Must be clear that the energy is being scattered by the atmosphere.

\n

Award [0] for simple definition of “albedo”.

\n
b.
\n
\n

gravitational attraction/force/field «of the planet/Moon» ✔

\n

Do not accept “gravity”.

\n
c.i.
\n
\n

the force/field and the velocity/displacement are at 90° to each other OR there is no change in GPE of the moon ✔

\n

Award [0] for any mention of no net force on the satellite.

\n

Do not accept acceleration is perpendicular to velocity.

\n
c.ii.
\n
", "Examiners report": "
\n

Many 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.

\n
a.i.
\n
\n

This 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.

\n
a.ii.
\n
\n

This 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.

\n
b.
\n
\n

This 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.

\n
c.i.
\n
\n

Some 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.

\n
c.ii.
\n
", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "a-3-work-energy-and-power", "b-2-greenhouse-effect", "d-1-gravitational-fields" ] }, { "question_id": "19M.2.SL.TZ2.1", "Question": "
\n

A 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.

\n
\n

The student strikes the tennis ball at point P. The tennis ball is initially directed at an angle of 7.00° to the horizontal.

\n

\n

The following data are available.

\n

Height of P = 2.80 m

\n

Distance of student from net = 11.9 m

\n

Height of net = 0.910 m

\n

Initial speed of tennis ball = 64 m s-1

\n
\n

Calculate the average force exerted by the racquet on the ball.

\n
[2]
\n
ai.
\n
\n

Calculate the average power delivered to the ball during the impact.

\n
[2]
\n
aii.
\n
\n

Calculate the time it takes the tennis ball to reach the net.

\n
[2]
\n
bi.
\n
\n

Show that the tennis ball passes over the net.

\n
[3]
\n
bii.
\n
\n

Determine the speed of the tennis ball as it strikes the ground.

\n
[2]
\n
biii.
\n
\n

The 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.

\n

\n

The 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.

\n

Predict for the student’s model, without calculation, whether θ is greater for a clay surface or for a grass surface.

\n
[3]
\n
c.
\n
", "Markscheme": "
\n

\nF\n=\n\n\nΔ\nm\nv\n\n\nΔ\nt\n\n\n\n/\n\nm\n\n\nΔ\nv\n\n\nΔ\nt\n\n\n\n/\n\n\n\n0.058\n×\n64.0\n\n\n25\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n  ✔

\n

\nF\n = 148«\n\nN\n\n»≈150«\n\nN\n\n»  ✔

\n

 

\n
ai.
\n
\n

ALTERNATIVE 1

\n

\nP\n=\n\n\n\n1\n2\n\nm\n\n\nv\n2\n\n\n\nt\n\n\n/\n\n\n\n\n1\n2\n\n×\n0.058\n×\n\n\n\n64.0\n\n2\n\n\n\n\n25\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n  ✔

\n

\nP\n=\n4700\n\n/\n\n4800\n\n«\n\n\nW\n\n»  ✔

\n

 

\n

ALTERNATIVE 2

\n

\nP\n=\n\naverage\n\nF\nv\n\n/\n\n148\n×\n\n\n64.0\n\n2\n\n  ✔

\n

\nP\n=\n4700\n\n/\n\n4800\n\n«\n\n\nW\n\n»  ✔

\n

 

\n
aii.
\n
\n

horizontal component of velocity is 64.0 × cos7° = 63.52 «ms−1» ✔

\n

\nt\n=\n\n«\n\n\n\n11.9\n\n\n63.52\n\n\n=\n\n»\n\n0.187\n\n/\n\n0.19\n\n«\n\n\ns\n\n»   ✔

\n

Do not award BCA. Check working.

\n

Do not award ECF from using 64 m s-1.

\n
bi.
\n
\n

ALTERNATIVE 1

\n

uy = 64 sin7/7.80 «ms−1»✔

\n

decrease in height = 7.80 × 0.187 + \n\n1\n2\n\n × 9.81 × 0.1872/1.63 «m» ✔

\n

final height = «2.80 − 1.63» = 1.1/1.2 «m» ✔

\n

«higher than net so goes over»

\n

ALTERNATIVE 2

\n

vertical distance to fall to net «= 2.80 − 0.91» = 1.89 «m»✔

\n

time to fall this distance found using «=1.89 = 7.8t + \n\n1\n2\n\n × 9.81 ×t2»

\n

t = 0.21 «s»✔

\n

0.21 «s» > 0.187 «s» ✔

\n

«reaches the net before it has fallen far enough so goes over»

\n

Other alternatives are possible

\n
bii.
\n
\n

ALTERNATIVE 1

\n

Initial KE + PE = final KE /

\n

\n\n1\n2\n\n × 0.058 × 642 + 0.058 × 9.81 × 2.80 = \n\n1\n2\n\n × 0.058 × v2 ✔

\n

v = 64.4 «ms−1» ✔

\n

ALTERNATIVE 2

\n

\n\n\nv\nv\n\n\n=\n\n«\n\n\n\n\n\n7.8\n\n2\n\n\n+\n2\n×\n9.81\n×\n2.8\n\n\n»\n\n=\n10.8\n\n«\n\n\nm\n\n\n\n\ns\n\n\n−\n1\n\n\n\n»  ✔ 

\n

« \nv\n=\n\n\n\n\n63.5\n\n2\n\n\n+\n\n\n\n10.8\n\n2\n\n\n\n »

\n

\nv\n=\n64.4\n\n«\n\n\nm\n\n\n\n\ns\n\n\n−\n1\n\n\n\n»   ✔

\n

 

\n
biii.
\n
\n

so horizontal velocity component at lift off for clay is smaller ✔

\n

normal force is the same so vertical component of velocity is the same ✔

\n

so bounce angle on clay is greater ✔

\n
c.
\n
", "Examiners report": "
\n

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.

\n
ai.
\n
\n

This 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.

\n
aii.
\n
\n

This 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.

\n
bi.
\n
\n

There 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.

\n

A 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.

\n
bii.
\n
\n

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

\n
biii.
\n
\n

As 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.

\n
c.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power" ] }, { "question_id": "19M.2.SL.TZ2.14", "Question": "
\n
\n (a)\n
\n
\n

\n \n Show that, when the speed of the train is 10 m s-1, the frequency of the periodic force is 0.4 Hz.\n \n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n \n \n Outline, with reference to the curve, why it is unsafe to drive a train across the bridge at 30 m s\n \n -1\n \n for this amount of damping.\n \n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n \n The damping of the bridge system can be varied. Draw, on the graph, a second curve when the damping is larger.\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n time period\n \n

\n

\n \n \n T = «\n \n \n \n \n 25\n \n \n 10\n \n \n \n \n »\n \n \n \n = 2.5 s\n \n \n \n \n AND\n \n f =\n \n \n \n \n 1\n \n \n T\n \n \n \n \n \n \n

\n

\n \n \n \n OR\n \n \n \n

\n

\n \n \n evidence of f =\n \n \n \n \n 10\n \n \n 25\n \n \n \n \n ✔\n \n \n

\n

\n \n \n \n Answer 0.4 Hz is given, check correct working is shown.\n \n \n \n

\n
\n
\n (b)\n
\n

\n \n 30 m s\n \n –1\n \n corresponds to\n \n f\n \n = 1.2 Hz ✔\n
\n \n

\n

\n \n the amplitude of vibration is a maximum for this speed\n
\n \n

\n

\n \n \n \n OR\n \n \n
\n \n

\n

\n \n corresponds to the resonant frequency ✔\n \n

\n
\n
\n (c)\n
\n

\n \n similar shape with lower amplitude ✔\n
\n \n

\n

\n \n maximum shifted slightly to left of the original curve ✔\n \n

\n

\n \n \n \n Amplitude must be lower than the original, but allow the amplitude to be equal at the extremes.\n \n \n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n The question was correctly answered by almost all candidates.\n

\n
\n
\n (b)\n
\n

\n The answers to this question were generally well presented and a correct argument was presented by almost all candidates. Resonance was often correctly referred to.\n

\n
\n
\n (c)\n
\n

\n A correct curve, with lower amplitude and shifted left, was drawn by most candidates.\n

\n
\n", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-4-standing-waves-and-resonance" ] }, { "question_id": "19M.2.SL.TZ2.15", "Question": "
\n
\n (a)\n
\n
\n

\n \n Identify, on the HR diagram, the position of the Sun. Label the position S.\n \n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n \n During its evolution, the Sun is likely to be a red giant of surface temperature 3000 K and luminosity 10\n \n 4\n \n L\n \n \n \n ☉\n \n . Later it is likely to be a white dwarf of surface temperature 10 000 K and luminosity 10-4 L\n \n \n \n ☉\n \n . Calculate the\n \n \n \n \n \n \n radius of the Sun as a white dwarf\n \n \n \n \n \n \n radius of the Sun as a red giant\n \n \n \n \n \n \n .\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n the letter S should be in the region of the shaded area\n \n \n \n ✔\n \n \n

\n

\n \n \n \n

\n
\n
\n (d)\n
\n

\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n Locating 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.\n

\n
\n
\n (d)\n
\n

\n Calculating 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.\n

\n
\n", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-5-fusion-and-stars" ] }, { "question_id": "19M.2.SL.TZ2.2", "Question": "
\n

A 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

A helium atom has a volume of 4.9 × 10-31 m3.

\n
\n

The molar mass of helium is 4.0 g mol-1. Show that the mass of a helium atom is 6.6 × 10-27 kg.

\n
[1]
\n
a.
\n
\n

Estimate the average speed of the helium atoms in the container.

\n
[2]
\n
b.
\n
\n

Show that the number of helium atoms in the container is about 4 × 1020.

\n
[2]
\n
c.
\n
\n

Calculate the ratio  \n\n\n\ntotal volume of helium atoms\n\n\n\n\nvolume of helium gas\n\n\n\n.

\n
[1]
\n
di.
\n
\n

Explain, 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
[2]
\n
dii.
\n
", "Markscheme": "
\n

\nm\n=\n\n\n4.0\n×\n\n\n\n10\n\n\n−\n3\n\n\n\n\n\n6.02\n×\n\n\n\n10\n\n\n23\n\n\n\n\n\n«kg»

\n

OR

\n

6.64 × 10−27 «kg» ✔

\n

 

\n
a.
\n
\n

\n\n1\n2\n\nm\n\n\nv\n2\n\n\n=\n\n3\n2\n\nk\nT\n\n/\n\nv\n=\n\n\n\n3\nk\nT\n\nm\n\n\n\n/\n\n\n\n\n3\n×\n1.38\n×\n\n\n\n10\n\n\n−\n23\n\n\n\n×\n320\n\n\n6.6\n×\n\n\n\n10\n\n\n−\n27\n\n\n\n\n\n\n  ✔

\n

v = 1.4 × 103 «ms−1» ✔ 

\n
b.
\n
\n

\nN\n=\n\n\np\nV\n\n\nk\nT\n\n\n\n/\n\n\n\n5.1\n×\n\n\n\n10\n\n5\n\n\n×\n3.2\n×\n\n\n\n10\n\n\n−\n6\n\n\n\n\n\n1.38\n×\n\n\n\n10\n\n\n−\n23\n\n\n\n×\n320\n\n\n

\n

OR

\n

\nN\n=\n\n\np\nV\n\n\nN\nA\n\n\n\n\nR\nT\n\n\n\n/\n\n\n\n5.1\n×\n\n\n\n10\n\n5\n\n\n×\n3.2\n×\n\n\n\n10\n\n\n−\n6\n\n\n\n×\n6.02\n×\n\n\n\n10\n\n\n23\n\n\n\n\n\n8.31\n×\n320\n\n\n   ✔

\n

N = 3.7 × 1020 ✔

\n
c.
\n
\n

«\n\n\n4\n×\n\n\n\n10\n\n\n20\n\n\n\n×\n4.9\n×\n\n\n\n10\n\n\n−\n31\n\n\n\n\n\n3.2\n×\n\n\n\n10\n\n\n−\n6\n\n\n\n\n\n=\n\n»\n\n6\n×\n\n\n10\n\n−\n5\n\n\n\n  ✔

\n
di.
\n
\n

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

\n

OR

\n

«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

\n

OR

\n

«For an ideal gas» particles are assumed to be point objects ✔

\n

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» ✔

\n
dii.
\n
", "Examiners report": "
\n

The 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.

\n
a.
\n
\n

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

\n
b.
\n
\n

Again 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.

\n
c.
\n
\n

This 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.

\n
di.
\n
\n

In 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.

\n
dii.
\n
", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter" ], "subtopics": [ "a-3-work-energy-and-power", "b-1-thermal-energy-transfers", "b-3-gas-laws" ] }, { "question_id": "19M.2.SL.TZ2.7", "Question": "
\n

The average temperature of ocean surface water is 289 K. Oceans behave as black bodies.

\n
\n

Show that the intensity radiated by the oceans is about 400 W m-2.

\n
[1]
\n
a.
\n
\n

Explain why some of this radiation is returned to the oceans from the atmosphere.

\n
[3]
\n
b.
\n
", "Markscheme": "
\n

5.67 × 10−8 × 2894

\n

OR

\n

= 396 «W m−2» ✔

\n

«≈ 400 W m−2»

\n

 

\n
a.
\n
\n

«most of the radiation emitted by the oceans is in the» infrared ✔

\n

«this radiation is» absorbed by greenhouse gases/named greenhouse gas in the atmosphere ✔

\n

«the gases» reradiate/re-emit ✔

\n

partly back towards oceans/in all directions/awareness that radiation in other directions is also present ✔

\n
b.
\n
", "Examiners report": "
\n

This 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.

\n
a.
\n
\n

For 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.

\n
b.
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-2-greenhouse-effect" ] }, { "question_id": "19N.3.SL.TZ0.1", "Question": "
\n

A 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.

\n

\n
\n

Suggest, by reference to the graph, why it is unlikely that the relationship between T and v is linear.

\n
[1]
\n
a.
\n
\n

Determine the fractional uncertainty in v when T = 2.115 s, correct to one significant figure.

\n
[2]
\n
b.
\n
\n

The 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.

\n

\n

Determine b, giving an appropriate unit for b.

\n
[3]
\n
c.
\n
\n

The lines of the minimum and maximum gradient are shown.

\n

\n

Estimate the absolute uncertainty in a.

\n
[2]
\n
d.
\n
", "Markscheme": "
\n

a 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 ✔

\n

NOTE: Do not allow “a line cannot be drawn through all error bars” without specifying “straight”.

\n
a.
\n
\n

v=1.15 «m  s-1»  AND  ∆v=0.05 «m  s-1» ✔

\n

«0.051.15=»0.04 ✔

\n

NOTE: Accept 4 %

\n
b.
\n
\n

use of 2 correct points on the line with Δv2 > 2 ✔

\n

b in range 0.012 to 0.013 ✔

\n

s3 m–2 ✔

\n
c.
\n
\n

amax=2.101 «s» ±0.001 «s» AND amin=2.095«s» ±0.001 «s» ✔

\n

«2.101-2.0952=» 0.003 «s» ✔

\n
d.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.
\n
", "topics": [ "tools" ], "subtopics": [ "tool-3-mathematics" ] }, { "question_id": "19N.2.HL.TZ0.11", "Question": "
\n

Monochromatic light of very low intensity is incident on a metal surface. The light causes the emission of electrons almost instantaneously. Explain how this observation

\n
\n

In an experiment to demonstrate the photoelectric effect, light of wavelength 480 nm is incident on a metal surface.

\n

\n

The graph shows the variation of the current I in the ammeter with the potential V of the cathode.

\n

\n
\n

does not support the wave nature of light.

\n
[2]
\n
a(i).
\n
\n

does support the photon nature of light.

\n
[2]
\n
a(ii).
\n
\n

Calculate, in eV, the work function of the metal surface.

\n
[3]
\n
b(i).
\n
\n

The 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 I with potential V after this change.

\n
[2]
\n
b(ii).
\n
", "Markscheme": "
\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 ✔

\n

NOTE: OWTTE

\n
a(i).
\n
\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 ✔

\n

NOTE: Reference to photon-electron collision scores MP1

\n
a(ii).
\n
\n

ϕ=hcλ-EK ✔

\n

EK=1.5«eV» ✔

\n

ϕ=«1.24×10-6480×10-9-1.5=» 1.1 «eV»✔

\n

NOTE: Allow reading from the graph of Ek=1.4leading to an answer of 1.2 «eV».

\n
b(i).
\n
\n

similar curve lower than original ✔

\n

with same horizontal intercept ✔

\n

\n
b(ii).
\n
", "Examiners report": "
\n[N/A]\n
a(i).
\n
\n[N/A]\n
a(ii).
\n
\n[N/A]\n
b(i).
\n
\n[N/A]\n
b(ii).
\n
", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-2-quantum-physics" ] }, { "question_id": "19N.2.HL.TZ0.6", "Question": "
\n
\n (a(i))\n
\n
\n

\n \n Show that the pressure at B is about 130 kPa.\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (a(ii))\n
\n
\n

\n \n Calculate the ratio\n \n \n \n \n V\n \n \n A\n \n \n \n \n V\n \n \n C\n \n \n \n \n .\n \n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b(i))\n
\n
\n

\n \n determine the thermal energy removed from the system.\n \n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b(ii))\n
\n
\n

\n \n explain why the entropy of the gas decreases.\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b(iii))\n
\n
\n

\n \n state and explain whether the second law of thermodynamics is violated.\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a(i))\n
\n

\n \n \n \n P\n \n \n B\n \n \n \n =\n \n \n \n \n 250\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n 1\n \n \n .\n \n \n \n 5\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n \n \n \n \n «\n \n \n from\n \n \n \n \n \n P\n \n \n B\n \n \n \n \n \n \n 1\n \n \n .\n \n \n 5\n \n \n \n \n \n V\n \n \n A\n \n \n \n \n \n \n 5\n \n \n 3\n \n \n \n \n =\n \n \n 250\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n \n \n V\n \n \n A\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n »\n \n \n \n ✔\n \n

\n

\n = 127 kPa\n \n ✔\n \n

\n
\n
\n (a(ii))\n
\n

\n \n \n «\n \n \n 127\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 5\n \n \n \n \n \n V\n \n \n A\n \n \n \n =\n \n \n 250\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n V\n \n \n C\n \n \n \n »\n \n \n

\n

\n \n 1.31 ✔\n \n

\n
\n
\n (b(i))\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n work done\n \n \n ∆\n \n \n W\n \n \n =\n \n \n «\n \n \n -\n \n \n »\n \n \n 250\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n =\n \n \n «\n \n \n -\n \n \n »\n \n \n 375\n \n \n \n \n «\n \n \n J\n \n \n »\n \n \n \n ✔\n \n

\n

\n change in internal energy\n \n \n ∆\n \n \n U\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 300\n \n \n ×\n \n \n 8\n \n \n .\n \n \n 31\n \n \n ×\n \n \n \n \n -\n \n \n 150\n \n \n \n \n =\n \n \n «\n \n \n -\n \n \n »\n \n \n 561\n \n \n \n \n «\n \n \n J\n \n \n »\n \n \n
\n \n \n OR\n
\n \n \n \n \n \n \n ∆\n \n \n U\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n P\n \n \n ∆\n \n \n V\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n ×\n \n \n 375\n \n \n =\n \n \n «\n \n \n -\n \n \n »\n \n \n 563\n \n \n \n \n «\n \n \n J\n \n \n »\n \n \n \n \n \n ✔\n \n

\n

\n thermal energy removed\n \n \n ∆\n \n \n Q\n \n \n =\n \n \n 375\n \n \n +\n \n \n 561\n \n \n =\n \n \n 936\n \n \n \n \n «\n \n \n J\n \n \n »\n \n \n
\n \n \n OR\n
\n \n \n \n \n \n \n ∆\n \n \n Q\n \n \n =\n \n \n 375\n \n \n +\n \n \n 563\n \n \n =\n \n \n 938\n \n \n \n \n «\n \n \n J\n \n \n »\n \n \n \n \n \n ✔\n \n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n \n ∆\n \n \n Q\n \n \n =\n \n \n «\n \n \n n\n \n \n C\n \n \n p\n \n \n ∆\n \n \n T\n \n \n =\n \n \n »\n \n \n \n 5\n \n \n 2\n \n \n \n ×\n \n \n n\n \n \n R\n \n \n T\n \n \n \n ✔\n \n

\n

\n thermal energy removed\n \n \n ∆\n \n \n Q\n \n \n =\n \n \n 0\n \n \n .\n \n \n 300\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 5\n \n \n ×\n \n \n 8\n \n \n .\n \n \n 31\n \n \n ×\n \n \n 150\n \n \n \n ✔\n \n

\n

\n \n \n \n =\n \n \n 935\n \n \n \n \n «\n \n \n J\n \n \n »\n \n \n \n \n ✔\n \n

\n
\n
\n (b(ii))\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n \n «from b(i)»\n \n \n ∆\n \n \n Q\n \n \n is negative\n \n ✔\n \n \n

\n

\n \n \n \n ∆\n \n \n S\n \n \n =\n \n \n \n \n ∆\n \n \n Q\n \n \n \n T\n \n \n \n \n \n AND\n \n \n ∆\n \n \n S\n \n \n \n \n is negative\n \n ✔\n \n \n

\n

\n

\n

\n \n ALTERNATIVE 2\n \n

\n

\n \n T\n \n and/or\n \n V\n \n decreases ✔\n

\n

\n less disorder/more order «so\n \n S\n \n decreases» ✔\n

\n

\n

\n

\n \n ALTERNATIVE 3\n \n

\n

\n \n T\n \n decreases ✔\n

\n

\n \n \n ∆\n \n \n S\n \n \n =\n \n \n K\n \n \n ×\n \n \n ln\n \n \n \n \n \n T\n \n \n 2\n \n \n \n \n T\n \n \n 1\n \n \n \n \n \n <\n \n \n 0\n \n \n ✔\n

\n

\n \n \n

\n

\n \n NOTE: Answer given, look for a valid reason that S decreases.\n \n

\n
\n
\n (b(iii))\n
\n

\n \n not violated ✔\n \n

\n

\n \n the entropy of the surroundings must have increased\n
\n \n \n OR\n \n \n
\n the overall entropy of the system and the surroundings is the same or increased ✔\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "19N.2.SL.TZ0.1", "Question": "
\n

The graph shows the variation with time t of the horizontal force F exerted on a tennis ball by a racket.

\n

\n

The 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.

\n
\n

Calculate the speed of the ball as it leaves the racket.

\n
[2]
\n
a.
\n
\n

Show that the average force exerted on the ball by the racket is about 50 N.

\n
[2]
\n
b.
\n
\n

Determine, 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.

\n
[3]
\n
c.
\n
\n

Draw 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.

\n

\n
[2]
\n
d.
\n
", "Markscheme": "
\n

links 0.84 to Δp ✔

\n

v=«0.845.8×10-2=» 14.5 «m s–1»✔

\n

NOTE: Award [2] for bald correct answer

\n
a.
\n
\n

use of Δt = «(28 – 12) × 10–3 =» 16 × 10–3 «s» ✔

\n

F¯ =«∆p∆t=» 0.8416×10-3 OR  53 «N» ✔

\n

NOTE: Accept a time interval from 14 to 16 ms
Allow ECF from incorrect time interval

\n
b.
\n
\n

Ek = 12 × 5.8 × 10–2 × 14.52 ✔

\n

Ek = W ✔

\n

s = «WF=12×5.8×10-2×14.5253=» 0.12 « m » ✔

\n

 

\n

Allow ECF from (a) and (b)

\n

Allow ECF from MP1

\n

Award [2] max for a calculation without reference to work done, eg: average velocity × time

\n
c.
\n
\n

\n

graph must show increasing speed from an initial of zero all the time ✔
overall correct curvature ✔

\n
d.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power" ] }, { "question_id": "19N.2.SL.TZ0.3", "Question": "
\n

The solid line in the graph shows the variation with distance x of the displacement y 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.

\n

\n
\n

One 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.

\n

\n

Point X (not shown) is a point on the string at a distance of 10 cm from the oscillator.

\n
\n

Calculate, in m s–1, the speed for this wave.

\n
[1]
\n
a(i).
\n
\n

Calculate, in Hz, the frequency for this wave.

\n
[2]
\n
a(ii).
\n
\n

The 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.

\n
[2]
\n
b.
\n
\n

State the number of all other points on the string that have the same amplitude and phase as X.

\n
[1]
\n
c(i).
\n
\n

The frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string.

\n

\n
[1]
\n
c(ii).
\n
", "Markscheme": "
\n

v = «0.050.20×10-3=» 250 «m s–1»✔

\n
a(i).
\n
\n

λ = 0.30 «m» ✔
f = «2500.30=» 830 «Hz» ✔

\n

NOTE: Allow ECF from (a)(i)
Allow ECF from wrong wavelength for MP2

\n
a(ii).
\n
\n

Q ✔
acceleration is proportional to displacement «and Q has larger displacement» ✔

\n
b.
\n
\n

3 «points» ✔

\n
c(i).
\n
\n

first harmonic mode drawn ✔

\n

\n

NOTE: Allow if only one curve drawn, either solid or dashed.

\n
c(ii).
\n
", "Examiners report": "
\n[N/A]\n
a(i).
\n
\n[N/A]\n
a(ii).
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c(i).
\n
\n[N/A]\n
c(ii).
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-4-standing-waves-and-resonance" ] }, { "question_id": "19N.2.SL.TZ0.4", "Question": "
\n

A 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.

\n

\n
\n

The speed of the proton is 2.0 × 106 m s–1 and the magnetic field strength B is 0.35 T.

\n
\n

Explain why the path of the proton is a circle.

\n
[2]
\n
a.
\n
\n

Show that the radius of the path is about 6 cm.

\n
[2]
\n
b(i).
\n
\n

Calculate the time for one complete revolution.

\n
[2]
\n
b(ii).
\n
\n

Explain why the kinetic energy of the proton is constant.

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

magnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔

\n

force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔

\n

force is centripetal/towards the centre ✔

\n

NOTE: Accept reference to acceleration instead of force

\n
a.
\n
\n

qvB=mv2R✔

\n

R=1.67×10-27×2.0×1061.6×10-19×0.35 OR 0.060 « m »

\n

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

\n
b(i).
\n
\n

T=2πRv✔

\n

T=«2π×0.062.0×106=1.9×10-7 « s » ✔

\n

NOTE: Award [2] for bald correct answer

\n
b(ii).
\n
\n

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

\n

NOTE: Award [2] for a reference to work done is zero hence Ek remains constant

\n

 

\n

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

\n

NOTE: Accept mention of speed or velocity indistinctly in MP2

\n

 

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b(i).
\n
\n[N/A]\n
b(ii).
\n
\n[N/A]\n
c.
\n
", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "a-3-work-energy-and-power", "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "19N.2.SL.TZ0.5", "Question": "
\n

An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.

\n

 

\n

\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.

\n
[2]
\n
a.
\n
\n

Calculate the magnitude of the initial acceleration of the electron.

\n
[2]
\n
b(i).
\n
\n

Describe the subsequent motion of the electron.

\n
[3]
\n
b(ii).
\n
", "Markscheme": "
\n

E=k×qr2 ✔

\n

E=8.99×109×6.0×10-30.42  OR  E=3.37×108 «N C-1» ✔

\n

NOTE: Ignore any negative sign.

\n
a.
\n
\n

F=q×E  OR  F=1.6×10-19×3.4×108=5.4×10-11 «N» ✔

\n

a=«5.4×10-119.1×10-31=» 5.9×1019« m s-2»✔

\n

NOTE: Ignore any negative sign.
Award [1] for a calculation leading to a=« m s-2»
Award [2] for bald correct answer

\n

 

\n
b(i).
\n
\n

the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔

\n

NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase

\n
b(ii).
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b(i).
\n
\n[N/A]\n
b(ii).
\n
", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "19N.2.SL.TZ0.7", "Question": "
\n

A stationary nucleus of uranium-238 undergoes alpha decay to form thorium-234.

\n

The following data are available.

\n

Energy released in decay                         4.27 MeV
Binding energy per nucleon for helium      7.07 MeV
Binding energy per nucleon for thorium    7.60 MeV

\n
\n

Radioactive decay is said to be “random” and “spontaneous”. Outline what is meant by each of these terms.

\n

Random: 

\n

Spontaneous:

\n
[2]
\n
a.
\n
\n

Calculate the binding energy per nucleon for uranium-238.

\n
[3]
\n
b(i).
\n
\n

Calculate the ratio kinetic energy of alpha particlekinetic energy of thorium nucleus.

\n
[2]
\n
b(ii).
\n
", "Markscheme": "
\n

random:
it cannot be predicted which nucleus will decay
OR
it cannot be predicted when a nucleus will decay ✔

\n

NOTE: OWTTE

\n

 

\n

spontaneous:
the decay cannot be influenced/modified in any way ✔

\n

NOTE: OWTTE

\n
a.
\n
\n

234 × 7.6  OR  4 × 7.07 ✔

\n

BEU =« 234 × 7.6 + 4 × 7.07 – 4.27 =» 1802 « MeV » ✔

\n

BEUA=«1802238=» 7.57 « MeV » ✔

\n

NOTE: Allow ECF from MP2
Award [3] for bald correct answer
Allow conversion to J, final answer is 1.2 × 10–12

\n
b(i).
\n
\n

states or applies conservation of momentum ✔

\n

ratio is «EkαEkTh=p22mαp22mTh=2344» 58.5 ✔

\n

NOTE: Award [2] for bald correct answer

\n
b(ii).
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b(i).
\n
\n[N/A]\n
b(ii).
\n
", "topics": [ "a--space-time-and-motion", "e-nuclear-and-quantum-physics" ], "subtopics": [ "a-3-work-energy-and-power", "e-3-radioactive-decay" ] }, { "question_id": "20N.3.SL.TZ0.1", "Question": "
\n

A spherical soap bubble is made of a thin film of soapy water. The bubble has an internal air pressure Pi and is formed in air of constant pressure Po. The theoretical prediction for the variation of Pi-Po is given by the equation

\n

(Pi-Po)=4gR

\n

where γ is a constant for the thin film and R is the radius of the bubble.

\n

Data for Pi-Po and R  were collected under controlled conditions and plotted as a graph showing the variation of Pi-Po with 1R.

\n

\n
\n

Suggest whether the data are consistent with the theoretical prediction.

\n
[2]
\n
a.
\n
\n

Show that the value of γ  is about 0.03.

\n
[2]
\n
b(i).
\n
\n

Identify the fundamental units of γ.

\n
[1]
\n
b(ii).
\n
\n

In order to find the uncertainty for γ, a maximum gradient line would be drawn. On the graph, sketch the maximum gradient line for the data.

\n
[1]
\n
b(iii).
\n
\n

The percentage uncertainty for γ is 15%. State γ, with its absolute uncertainty.

\n
[2]
\n
b(iv).
\n
\n

The expected value of γ is 0.027. Comment on your result.

\n
[1]
\n
b(v).
\n
", "Markscheme": "
\n

«theory suggests» Pi-Po is proportional to 1R ✓

\n


graph/line of best fit is straight/linear «so yes»
OR
graph/line of best fit passes through the origin «so yes» ✓

\n

 

\n

MP1: Accept ‘linear’

\n

MP2 do not award if there is any contradiction
eg: graph not proportional, does not pass through origin.

\n
a.
\n
\n

gradient = «4γ» =0.10
OR
use of equation with coordinates of a point ✓

\n


γ=0.025 ✓

\n

 

\n

MP1 allow gradients in range 0.098 to 0.102

\n

MP2 allow a range 0.024 to 0.026 for γ

\n

 

\n
b(i).
\n
\n

kg s-2 ✓

\n

 

\n

Accept kgs2

\n

 

\n
b(ii).
\n
\n

straight line, gradient greater than line of best fit, and within the error bars ✓

\n

\n
b(iii).
\n
\n

«15% of 0.025» = 0.00375
OR
«15% of 0.030» = 0.0045 ✓

\n

rounds uncertainty to 1sf
±0.004
OR
±0.005 ✓

\n

 

\n

Allow ECF from (b)(i)
Award [2] marks for a bald correct answer

\n
b(iv).
\n
\n

Experimental value matches this/correct, as expected value within the range ✓
OR
experimental value does not match/incorrect, as it is not within range ✓

\n
b(v).
\n
", "Examiners report": "", "topics": [ "tools" ], "subtopics": [ "tool-3-mathematics" ] }, { "question_id": "20N.3.SL.TZ0.2", "Question": "
\n

A student studies the relationship between the centripetal force applied to an object undergoing circular motion and its period T.

\n

The object (mass m) is attached by a light inextensible string, through a tube, to a weight W 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 T of the motion until the radius r is constant. The radius of the circle and the mass of the object are measured and remain constant for the entire experiment.

\n

\n

© International Baccalaureate Organization 2020.

\n

The student collects the measurements of T five times, for weight W. The weight is then doubled (2W) and the data collection repeated. Then it is repeated with 3W and 4W. The results are expected to support the relationship

\n

W=4π2mrT2.

\n
\n

In reality, there is friction in the system, so in this case W is less than the total centripetal force in the system. A suitable graph is plotted to determine the value of mr experimentally. The value of mr was also calculated directly from the measured values of m and r.

\n
\n

State why the experiment is repeated with different values of W.

\n
[1]
\n
a.
\n
\n

Predict from the equation whether the value of mr found experimentally will be larger, the same or smaller than the value of mr calculated directly.

\n
[2]
\n
b.
\n
\n

The measurements of T were collected five times. Explain how repeated measurements of T reduced the random error in the final experimental value of mr.

\n
[2]
\n
c(i).
\n
\n

Outline why repeated measurements of T would not reduce any systematic error in T.

\n
[1]
\n
c(ii).
\n
", "Markscheme": "
\n

In order to draw a graph « of W versus 1T2 »
OR

\n

to confirm proportionality between «W and T-2 »

\n

OR

\n

to confirm relationship between «W and T »

\n

OR

\n

because W is the independent variable in the experiment ✓

\n

 

\n

OWTTE

\n
a.
\n
\n

ALTERNATIVE 1

\n

W+friction=4π2mrT2

\n

OR

\n

centripetal force is larger «than W» / W is smaller «than centripetal» ✓

\n

«so» experimental mr is smaller «than calculated value» ✓

\n

 

\n

ALTERNATIVE 2 (refers to graph)

\n

reference to «friction force is» a systematic error «and does not affect gradient» ✓

\n

«so» mr is the same ✓

\n

 

\n

MP2 awarded only with correct justification.
Candidates can gain zero, MP1 alone or full marks.

\n

OWTTE

\n
b.
\n
\n

mention of mean/average value «of T» ✓

\n

this reduces uncertainty in T / result
OR
more accurate/precise ✓

\n

 

\n

Reference to “random errors average out” scores MP1

\n

Accept “closer to true value”, “more reliable value” OWTTE for MP2

\n

 

\n
c(i).
\n
\n

systematic errors «usually» constant/always present/ not influenced by repetition ✓

\n

 

\n

OWTTE

\n
c(ii).
\n
", "Examiners report": "
\n

Most candidates scored. Different wording was used to express the aim of confirming the relationship.

\n
a.
\n
\n

Most 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.

\n
b.
\n
\n

Most 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.

\n
c(i).
\n
\n

Usually very well answered acknowledging that systematic errors are constant and present throughout all  measurements.

\n
c(ii).
\n
", "topics": [ "inquiry" ], "subtopics": [ "i-1-2-designing", "i-2-3-interpreting-results", "inquiry-1-exploring-and-designing", "inquiry-2-collecting-and-processing-data" ] }, { "question_id": "20N.2.HL.TZ0.10", "Question": "
\n

The de Broglie wavelength λ of a particle accelerated close to the speed of light is approximately

\n

λ≈hcE

\n

where E is the energy of the particle.
A beam of electrons of energy 4.2×108 eV is produced in an accelerator.

\n
\n

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 θ relative to the direction of the incident beam.

\n

\n

The graph shows the variation of the intensity of electrons with θ. There is a minimum of intensity for θ=θ0.

\n

\n
\n

Show that the wavelength of an electron in the beam is about 3×10-15 m.

\n
[1]
\n
a.
\n
\n

Discuss how the results of the experiment provide evidence for matter waves.

\n
[2]
\n
b(i).
\n
\n

The accepted value of the diameter of the carbon-12 nucleus is 4.94×10-15 m. Estimate the angle θ0 at which the minimum of the intensity is formed.

\n
[2]
\n
b(ii).
\n
\n

Outline why electrons with energy of approximately 107 eV would be unsuitable for the investigation of nuclear radii.

\n
[2]
\n
b(iii).
\n
\n

Experiments with many nuclides suggest that the radius of a nucleus is proportional to A13, where A is the number of nucleons in the nucleus. Show that the density of a nucleus remains approximately the same for all nuclei.

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

λ=6.63×10-34×3×1081.60×10-19×4.2×108 OR =2.96×10-15«m» ✓

\n


Answer to at least 2 s.f. (i.e. 3.0)

\n
a.
\n
\n

«the shape of the graph suggests that» electrons undergo diffraction «with carbon nuclei» ✓

\n

only waves diffract ✓

\n
b(i).
\n
\n

sinθ0=2.96×10-154.94×10-15«=0.599» ✓

\n

37 «degrees» OR 0.64/0.65 «rad» ✓

\n
b(ii).
\n
\n

the 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 ✓

\n
b(iii).
\n
\n

volume of a nucleus proportional to A133=A AND mass proportional to A ✓

\n

the ratio massvolume independent of A «hence density the same for all nuclei» ✓

\n

 

\n

Both needed for MP1

\n
c.
\n
", "Examiners report": "
\n

An easy calculation with only one energy conversion to consider and a 'show' answer to help. 

\n
a.
\n
\n

This 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.

\n
b(i).
\n
\n

This was answered reasonably well with only the odd omission of the sine in the equation.

\n
b(ii).
\n
\n

Candidates 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.

\n
b(iii).
\n
\n

This 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.

\n
c.
\n
", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-1-structure-of-the-atom", "e-2-quantum-physics" ] }, { "question_id": "20N.2.HL.TZ0.6", "Question": "
\n

One possible fission reaction of uranium-235 (U-235) is

\n

U92235+n01→Xe54140+Sr3894+2n01

\n

Mass of one atom of U-235 =235 u
Binding energy per nucleon for U-235 =7.59 MeV
Binding energy per nucleon for Xe-140 =8.29 MeV
Binding energy per nucleon for Sr-94 =8.59 MeV

\n
\n

A nuclear power station uses U-235 as fuel. Assume that every fission reaction of U-235 gives rise to 180 MeV of energy.

\n
\n

A sample of waste produced by the reactor contains 1.0 kg of strontium-94 (Sr-94). Sr-94 is radioactive and undergoes beta-minus (β-) decay into a daughter nuclide X. The reaction for this decay is

\n

Sr3894→X+v¯e+e.

\n

 

\n
\n

The graph shows the variation with time of the mass of Sr-94 remaining in the sample.

\n

\n
\n

State what is meant by binding energy of a nucleus.

\n
[1]
\n
a(i).
\n
\n

Outline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units.

\n
[1]
\n
a(ii).
\n
\n

Show that the energy released in the reaction is about 180 MeV.

\n
[1]
\n
a(iii).
\n
\n

Estimate, in J kg-1, the specific energy of U-235.

\n
[2]
\n
b(i).
\n
\n

The power station has a useful power output of 1.2 GW and an efficiency of 36 %. Determine the mass of U-235 that undergoes fission in one day.

\n
[2]
\n
b(ii).
\n
\n

The specific energy of fossil fuel is typically 30 MJ kg–1. Suggest, with reference to your answer to (b)(i), one advantage of U-235 compared with fossil fuels in a power station.

\n
[1]
\n
b(iii).
\n
\n

Write down the proton number of nuclide X.

\n
[1]
\n
c(i).
\n
\n

State the half-life of Sr-94.

\n
[1]
\n
c(ii).
\n
\n

Calculate the mass of Sr-94 remaining in the sample after 10 minutes.

\n
[2]
\n
c(iii).
\n
", "Markscheme": "
\n

energy required to «completely» separate the nucleons
OR
energy released when a nucleus is formed from its constituent nucleons ✓

\n


Allow protons AND neutrons.

\n
a(i).
\n
\n

the values «in SI units» would be very small ✓

\n
a(ii).
\n
\n

140×8.29+94×8.59-235×7.59 OR 184 «MeV» ✓

\n
a(iii).
\n
\n

see «energy=» 180×106×1.60×10-19 AND «mass=» 235×1.66×10-27 ✓

\n

7.4×1013 «J kg-1» ✓

\n
b(i).
\n
\n

energy produced in one day =1.2×109×24×36000.36=2.9×1014 «J» ✓

\n

mass =2.9×10147.4×1013=3.9 «kg» ✓

\n
b(ii).
\n
\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 kg
OR
less fuel can be used to create the same amount of energy ✓

\n
b(iii).
\n
\n

39 ✓

\n


Do not allow X3994 unless the proton number is indicated.

\n
c(i).
\n
\n

75 «s» ✓

\n
c(ii).
\n
\n

ALTERNATIVE 1

\n

10 min=8 t1/2 ✓

\n

mass remaining=1.0×128=3.9×10-3«kg» ✓

\n

 

\n

ALTERNATIVE 2

\n

decay constant =«ln275=»9.24×10-3 «s-1» ✓

\n

mass remaining =1.0×e-9.24×10-3×600=3.9×10-3 «kg» ✓

\n
c(iii).
\n
", "Examiners report": "
\n

Generally, 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'. 

\n
a(i).
\n
\n

Well answered with some candidates describing the values as too large or small.

\n
a(ii).
\n
\n

Well answered.

\n
a(iii).
\n
\n

This 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.

\n
b(i).
\n
\n

Most candidates had the right idea, but common problems included forgetting the efficiency or not converting to days.

\n
b(ii).
\n
\n

HL only. This was well answered.

\n
b(iii).
\n
\n

Most candidates answered this correctly.

\n
c(i).
\n
\n

Most candidates answered this correctly.

\n
c(ii).
\n
\n

This was answered well with most candidates (even at HL) going down the number of half-lives route rather than the exponential calculation route.

\n
c(iii).
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "20N.2.HL.TZ0.7", "Question": "
\n

A vertical solid cylinder of uniform cross-sectional area A 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 x below the water surface.

\n

\n

At time t=0 the cylinder is released. The resultant vertical force F on the cylinder is related to the displacement x of the mark by

\n

F=-ρAgx

\n

where ρ is the density of water.

\n
\n

The cylinder was initially pushed down a distance x=0.250 m.

\n
\n

Outline why the cylinder performs simple harmonic motion when released.

\n
[1]
\n
a.
\n
\n

The mass of the cylinder is 118 kg and the cross-sectional area of the cylinder is 2.29×10-1 m2. The density of water is 1.03×103 kg m-3. Show that the angular frequency of oscillation of the cylinder is about 4.4  rad s-1.

\n
[2]
\n
b.
\n
\n

Determine the maximum kinetic energy Ekmax of the cylinder.

\n
[2]
\n
c(i).
\n
\n

Draw, on the axes, the graph to show how the kinetic energy of the cylinder varies with time during one period of oscillation T.

\n

\n
[2]
\n
c(ii).
\n
", "Markscheme": "
\n

the «restoring» force/acceleration is proportional to displacement ✓

\n


Allow use of symbols i.e. F∝-x or a∝-x

\n
a.
\n
\n

Evidence of equating mω2x=ρAgx «to obtain ρAgm=ω2» ✓

\n

 

\n

ω=1.03×103×2.29×10-1×9.81118 OR 4.43«rad s-1» ✓

\n

 

\n

Answer to at least 3 s.f.

\n
b.
\n
\n

«EK is a maximum when x=0 hence» EK, max=12×118×4.420.2502-02 ✓

\n


71.4 «J» ✓

\n
c(i).
\n
\n

energy never negative ✓

\n

correct shape with two maxima ✓

\n

\n
c(ii).
\n
", "Examiners report": "
\n

This was well answered with candidates gaining credit for answers in words or symbols.

\n
a.
\n
\n

Again, very well answered.

\n
b.
\n
\n

A straightforward calculation with the most common mistake being missing the squared on the omega.

\n
c(i).
\n
\n

Most candidates answered with a graph that was only positive so scored the first mark.

\n
c(ii).
\n
", "topics": [ "a--space-time-and-motion", "c-wave-behaviour" ], "subtopics": [ "a-4-rigid-body-mechanics", "c-1-simple-harmonic-motion" ] }, { "question_id": "20N.2.HL.TZ0.8", "Question": "
\n

The diagram shows the electric field lines of a positively charged conducting sphere of radius R and charge Q.

\n

\n

Points A and B are located on the same field line.

\n
\n

A 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 1.7×10-16 J. Point A is at a distance of 5.0×10-2 m from the centre of the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere.

\n
\n

Explain why the electric potential decreases from A to B.

\n
[2]
\n
a.
\n
\n

Draw, on the axes, the variation of electric potential V with distance r from the centre of the sphere.

\n

\n
[2]
\n
b.
\n
\n

Calculate the electric potential difference between points A and B.

\n
[1]
\n
c(i).
\n
\n

Determine the charge Q of the sphere.

\n
[2]
\n
c(ii).
\n
\n

The 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.

\n
[1]
\n
d.
\n
", "Markscheme": "
\n

ALTERNATIVE 1
work done on moving a positive test charge in any outward direction is negative ✓
potential difference is proportional to this work «so V decreases from A to B» ✓

\n

 

\n

ALTERNATIVE 2
potential gradient is directed opposite to the field so inwards ✓
the gradient indicates the direction of increase of V «hence V increases towards the centre/decreases from A to B» ✓

\n

 

\n

ALTERNATIVE 3
V=kQR so as r increases V decreases ✓
V is positive as Q is positive ✓

\n

 

\n

ALTERNATIVE 4
the work done per unit charge in bringing a positive charge from infinity ✓
to point B is less than point A ✓

\n
a.
\n
\n

curve decreasing asymptotically for r>R ✓

\n

non - zero constant between 0 and R ✓

\n

\n
b.
\n
\n

«Wq=1.7×10-161.60×10-19=»1.1×103 «V» ✓

\n
c(i).
\n
\n

8.99×109×Q×15.0×10-2-11.0×10-1=1.1×103 ✓

\n


Q=1.2×10-8 «C» ✓

\n
c(ii).
\n
\n

to highlight similarities between «different» fields ✓

\n
d.
\n
", "Examiners report": "
\n

The 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.

\n
a.
\n
\n

This was well answered.

\n
b.
\n
\n

Most 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.

\n
c(i).
\n
\n

Even 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.

\n
c(ii).
\n
\n

This was generally well answered with only a small number of answers suggesting greater international cooperation.

\n
d.
\n
", "topics": [ "a--space-time-and-motion", "d-fields", "nature-of-science" ], "subtopics": [ "a-4-rigid-body-mechanics", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "20N.2.HL.TZ0.9", "Question": "
\n

The diagram shows an alternating current generator with a rectangular coil rotating at a constant frequency in a uniform magnetic field.

\n

\n
\n

The graph shows how the generator output voltage V varies with time t.

\n

\n

Electrical power produced by the generator is delivered to a consumer some distance away.

\n
\n

Explain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.

\n
[3]
\n
a.
\n
\n

The average power output of the generator is 8.5×105 W. Calculate the root mean square (rms) value of the generator output current.

\n
[2]
\n
b(i).
\n
\n

The 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 2.5×102.

\n
[1]
\n
b(ii).
\n
\n

The 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.

\n
[2]
\n
b(iii).
\n
", "Markscheme": "
\n

there is a magnetic flux «linkage» in the coil / coil cuts magnetic field ✓

\n

this 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
a.
\n
\n

Vrms=25×1032«=17.7×103  V» ✓

\n

 

\n

Irms=8.5×10517.7×103=48 «A» ✓

\n
b(i).
\n
\n

«power loss proportional to I2 hence the step-up factor is 2.5×102» 16 ✓

\n
b(ii).
\n
\n

peak emf doubles ✓

\n

T halves ✓

\n


Must show at least 1 cycle.

\n
b(iii).
\n
", "Examiners report": "
\n

This question was well answered with the majority discussing changes in flux rather than wires cutting field lines, which was good to see.

\n
a.
\n
\n

Generally well answered.

\n
b(i).
\n
\n

This 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.

\n
b(ii).
\n
\n

Well answered, with the majority of candidates scoring at least 1 mark.

\n
b(iii).
\n
", "topics": [ "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "b-4-thermodynamics", "d-4-induction" ] }, { "question_id": "20N.2.SL.TZ0.1", "Question": "
\n

A 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.

\n

\n

The air is propelled vertically downwards with speed v. The aircraft hovers motionless above the ground. A package is suspended from the aircraft on a string. The mass of the aircraft is 0.95 kg and the combined mass of the package and string is 0.45 kg. The mass of air pushed downwards by the blades in one second is 1.7 kg.

\n
\n

State the value of the resultant force on the aircraft when hovering.

\n
[1]
\n
a(i).
\n
\n

Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.

\n
[2]
\n
a(ii).
\n
\n

Determine v. State your answer to an appropriate number of significant figures.

\n
[3]
\n
a(iii).
\n
\n

The 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.

\n
[2]
\n
b.
\n
", "Markscheme": "
\n

zero ✓

\n
a(i).
\n
\n

Blades 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» ✓

\n


Downward direction required for MP1.

\n
a(ii).
\n
\n

«lift force/change of momentum in one second» =1.7v ✓

\n

1.7v=0.95+0.45×9.81 ✓

\n

v=8.1«ms-1» AND answer expressed to 2 sf only ✓

\n

Allow 8.2 from g=10 ms-2.

\n
a(iii).
\n
\n

vertical force = lift force – weight OR =0.45×9.81 OR =4.4 «N» ✓

\n

acceleration=0.45×9.810.95=4.6 «ms-2» ✓

\n
b.
\n
", "Examiners report": "
\n[N/A]\n
a(i).
\n
\n[N/A]\n
a(ii).
\n
\n[N/A]\n
a(iii).
\n
\n[N/A]\n
b.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power" ] }, { "question_id": "20N.2.SL.TZ0.12", "Question": "
\n
\n (b)\n
\n
\n

\n Determine the terminal velocity of the sphere.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c(i))\n
\n
\n

\n Determine the force exerted by the spring on the sphere when the sphere is at rest.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (b)\n
\n

\n radius of sphere\n \n \n =\n \n \n 0\n \n \n .\n \n \n 012\n \n \n \n \n «\n \n \n m\n \n \n »\n \n \n ✓\n

\n

\n

\n

\n weight of sphere\n \n \n =\n \n \n 6\n \n \n π\n \n \n η\n \n \n r\n \n \n v\n \n \n +\n \n \n ρ\n \n \n V\n \n \n g\n \n \n

\n

\n \n \n OR\n \n \n

\n

\n \n \n v\n \n \n =\n \n \n \n \n \n \n 1\n \n \n .\n \n \n 26\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n -\n \n \n 915\n \n \n ×\n \n \n 7\n \n \n .\n \n \n 24\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n ×\n \n \n 9\n \n \n .\n \n \n 81\n \n \n \n \n 6\n \n \n π\n \n \n ×\n \n \n 37\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n v\n \n \n =\n \n \n 6\n \n \n .\n \n \n 84\n \n \n \n \n «\n \n \n m\n \n \n \n \n \n s\n \n \n \n –\n \n \n 1\n \n \n \n \n »\n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n Accept use of\n \n \n g\n \n \n =\n \n \n 10\n \n \n leading to\n \n \n v\n \n \n =\n \n \n 7\n \n \n .\n \n \n 0\n \n \n \n \n «\n \n \n m\n \n \n \n \n \n s\n \n \n \n –\n \n \n 1\n \n \n \n \n »\n \n \n \n \n

\n

\n \n \n Allow implicit calculation of radius for MP1\n \n \n

\n

\n \n \n Do not allow ECF for MP3 if buoyant force omitted.\n \n \n

\n
\n
\n (c(i))\n
\n

\n \n \n F\n \n \n =\n \n \n m\n \n \n g\n \n \n -\n \n \n ρ\n \n \n V\n \n \n g\n \n \n

\n

\n \n \n OR\n \n \n

\n

\n \n \n \n \n F\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 0126\n \n \n ×\n \n \n 9\n \n \n .\n \n \n 81\n \n \n \n \n -\n \n \n \n \n 915\n \n \n ×\n \n \n 7\n \n \n .\n \n \n 24\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n ×\n \n \n 9\n \n \n .\n \n \n 81\n \n \n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n \n F\n \n \n =\n \n \n 5\n \n \n .\n \n \n 86\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n «\n \n \n N\n \n \n »\n \n \n ✓\n \n

\n

\n

\n

\n \n \n Accept use of\n \n \n g\n \n \n =\n \n \n 10\n \n \n leading to\n \n \n F\n \n \n =\n \n \n 6\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n N\n \n \n \n \n

\n
\n", "Examiners report": "
\n (b)\n
\n

\n Only those candidates who forgot to include the buoyant force missed marks here.\n

\n
\n
\n (c(i))\n
\n

\n Continuing from b, most candidates scored full marks.\n

\n
\n", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "20N.2.SL.TZ0.15", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the apparent brightness\n \n \n b\n \n \n ∝\n \n \n \n \n A\n \n \n \n T\n \n \n 4\n \n \n \n \n \n d\n \n \n 2\n \n \n \n \n , where\n \n \n d\n \n \n is the distance of the object from Earth,\n \n \n T\n \n \n is the surface temperature of the object and\n \n \n A\n \n \n is the surface area of the object.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Two of the brightest objects in the night sky seen from Earth are the planet Venus and the star Sirius. Explain why the equation\n \n \n b\n \n \n ∝\n \n \n \n \n A\n \n \n \n T\n \n \n 4\n \n \n \n \n \n d\n \n \n 2\n \n \n \n \n is applicable to Sirius but not to Venus.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n substitution of\n \n \n L\n \n \n =\n \n \n σ\n \n \n A\n \n \n \n T\n \n \n 4\n \n \n \n into\n \n \n b\n \n \n =\n \n \n \n L\n \n \n \n 4\n \n \n \n πd\n \n \n 2\n \n \n \n \n \n giving\n \n \n b\n \n \n =\n \n \n \n \n σ\n \n \n A\n \n \n \n T\n \n \n 4\n \n \n \n \n \n 4\n \n \n \n πd\n \n \n 2\n \n \n \n \n \n

\n

\n

\n

\n \n \n Removal of constants\n \n \n σ\n \n \n and\n \n \n 4\n \n \n π\n \n \n is optional\n \n \n

\n
\n
\n (b)\n
\n

\n \n equation applies to Sirius/stars that are luminous/emit light «from fusion»\n \n \n ✓\n \n

\n

\n \n but Venus reflects the Sun’s light/does not emit light\n \n \n «\n \n \n from fusion\n \n \n »\n \n \n ✓\n \n

\n

\n

\n

\n \n \n OWTTE\n \n \n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-5-fusion-and-stars" ] }, { "question_id": "20N.2.SL.TZ0.3", "Question": "
\n

A 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 T of the sample varies with time t.

\n

\n

The following data are available.

\n

Mass of the sample =0.32 kg
Specific latent heat of fusion of the oil =130 kJ kg-1
Rate of thermal energy transfer =15 W

\n
\n

Calculate the thermal energy transferred from the sample during the first 30 minutes.

\n
[1]
\n
a(i).
\n
\n

Estimate the specific heat capacity of the oil in its liquid phase. State an appropriate unit for your answer.

\n
[2]
\n
a(ii).
\n
\n

The 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.

\n
[3]
\n
b.
\n
\n

Calculate the mass of the oil that remains unfrozen after 60 minutes.

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

«15×30×60»=27000 «J» ✓

\n

 

\n
a(i).
\n
\n

27×103=0.32×c×290-250 OR 2100 ✓

\n

J kg-1 K-1 OR J kg-1  0C-1 ✓

\n


Allow any appropriate unit that is energymass×termperature

\n
a(ii).
\n
\n

«intermolecular» bonds are formed during freezing ✓

\n


bond-forming process releases energy
OR
«intermolecular» PE decreases «and the difference is transferred as heat» ✓

\n


«average random» KE of the molecules does not decrease/change ✓

\n


temperature is related to «average» KE of the molecules «hence unchanged» ✓

\n

 

\n

To award MP3 or MP4 molecules/particles/atoms must be mentioned.

\n
b.
\n
\n

mass of frozen oil «=27×103130×103»=0.21 «kg» ✓

\n

unfrozen mass «=0.32-0.21»=0.11 «kg» ✓

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a(i).
\n
\n[N/A]\n
a(ii).
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "20N.2.SL.TZ0.5", "Question": "
\n

The graph shows how current I varies with potential difference V across a component X.

\n

\n
\n

Component X and a cell of negligible internal resistance are placed in a circuit.

\n

A variable resistor R is connected in series with component X. The ammeter reads 20 mA.

\n

\n
\n

Component X and the cell are now placed in a potential divider circuit.

\n

\n

 

\n
\n

Outline why component X is considered non-ohmic.

\n
[1]
\n
a.
\n
\n

Determine the resistance of the variable resistor.

\n
[3]
\n
b(i).
\n
\n

Calculate the power dissipated in the circuit.

\n
[1]
\n
b(ii).
\n
\n

State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.

\n
[1]
\n
c(i).
\n
\n

Describe, by reference to your answer for (c)(i), the advantage of the potential divider arrangement over the arrangement in (b).

\n
[2]
\n
c(ii).
\n
", "Markscheme": "
\n

current is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓

\n
a.
\n
\n

ALTERNATIVE 1

\n

voltage across X=2.3 «V» ✓

\n

voltage across R«=4.0-2.3»=1.7 «V» ✓

\n

resistance of variable resistor «=1.70.020»=85 «Ω» ✓

\n

 

\n

ALTERNATIVE 2

\n

overall resistance «=4.00.020»=200 «Ω» ✓

\n

resistance of X «=2.30.020»=115 «Ω» ✓

\n

resistance of variable resistor «=200-115»=85 «Ω» ✓

\n
b(i).
\n
\n

power «=4.0×0.020»=0.080 «W» ✓

\n
b(ii).
\n
\n

from 0 to 60 mA ✓

\n
c(i).
\n
\n

allows zero current through component X / potential divider arrangement ✓

\n

provides greater range «of current through component X» ✓

\n
c(ii).
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b(i).
\n
\n[N/A]\n
b(ii).
\n
\n[N/A]\n
c(i).
\n
\n[N/A]\n
c(ii).
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-5-current-and-circuits" ] }, { "question_id": "20N.2.SL.TZ0.7", "Question": "
\n
\n (a)\n
\n
\n

\n Outline why the cylinder performs simple harmonic motion when released.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The mass of the cylinder is\n \n \n 118\n \n \n \n \n kg\n \n \n and the cross-sectional area of the cylinder is\n \n \n 2\n \n \n .\n \n \n 29\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 1\n \n \n \n \n \n \n \n m\n \n \n 2\n \n \n \n . The density of water is\n \n \n 1\n \n \n .\n \n \n 03\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n kg\n \n \n \n \n \n m\n \n \n \n -\n \n \n 3\n \n \n \n \n . Show that the angular frequency of oscillation of the cylinder is about\n \n \n 4\n \n \n .\n \n \n 4\n \n \n \n \n \n \n rad\n \n \n \n \n \n s\n \n \n \n -\n \n \n 1\n \n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c(ii))\n
\n
\n

\n Draw, on the axes, the graph to show how the kinetic energy of the cylinder varies with time during\n \n one\n \n period of oscillation\n \n \n T\n \n \n .\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n the\n \n \n «\n \n \n restoring\n \n \n »\n \n \n force/acceleration is proportional to displacement\n \n \n ✓\n \n

\n

\n \n \n
\n Allow use of symbols i.e.\n \n \n \n F\n \n \n ∝\n \n \n -\n \n \n x\n \n \n \n or\n \n \n \n a\n \n \n ∝\n \n \n -\n \n \n x\n \n \n \n

\n
\n
\n (b)\n
\n

\n \n Evidence of equating\n \n \n m\n \n \n \n ω\n \n \n 2\n \n \n \n x\n \n \n =\n \n \n ρ\n \n \n A\n \n \n g\n \n \n x\n \n \n «to obtain\n \n \n \n \n ρ\n \n \n A\n \n \n g\n \n \n \n m\n \n \n \n =\n \n \n \n ω\n \n \n 2\n \n \n \n » ✓\n \n

\n

\n

\n

\n \n \n \n ω\n \n \n =\n \n \n \n \n \n 1\n \n \n .\n \n \n 03\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 29\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 1\n \n \n \n \n ×\n \n \n 9\n \n \n .\n \n \n 81\n \n \n \n 118\n \n \n \n \n \n \n OR\n \n \n \n \n 4\n \n \n .\n \n \n 43\n \n \n «\n \n \n rad\n \n \n \n \n \n s\n \n \n \n -\n \n \n 1\n \n \n \n \n »\n \n \n ✓\n \n

\n

\n

\n

\n \n \n Answer to at least\n \n \n 3\n \n \n s.f.\n \n \n

\n
\n
\n (c(ii))\n
\n

\n \n energy never negative\n \n \n ✓\n \n

\n

\n \n correct shape with two maxima\n \n \n ✓\n \n

\n

\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n This was well answered with candidates gaining credit for answers in words or symbols.\n

\n
\n
\n (b)\n
\n

\n Again, very well answered.\n

\n
\n
\n (c(ii))\n
\n

\n Most candidates answered with a graph that was only positive so scored the first mark.\n

\n
\n", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion" ] }, { "question_id": "21M.2.HL.TZ1.10", "Question": "
\n

In 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.

\n

\n
\n

Describe the photoelectric effect.

\n
[2]
\n
a.
\n
\n

Show that the maximum velocity of the photoelectrons is 700 km  s-1.

\n
[2]
\n
b.
\n
\n

The photoelectrons are emitted from a sodium surface. Sodium has a work function of 2.3 eV.

\n

Calculate the wavelength of the radiation incident on the sodium. State an appropriate unit for your answer.

\n
[3]
\n
c.
\n
", "Markscheme": "
\n

electrons are ejected from the surface of a metal ✓

\n

after gaining energy from photons/electromagnetic radiation ✓

\n

there is a minimum «threshold» energy/frequency
OR
maximum «threshold» wavelength ✓

\n
a.
\n
\n

«eV = 12 mv2» and manipulation to get v ✓

\n

v=2×1.6×10-19×1.49.1×10-31  OR  702«km s-1»✓

\n

Must see either complete substitution or calculation to at least 3 s.f. for MP2

\n
b.
\n
\n

E=2.3+1.4 ✓

\n

λ=6.63×10-34×3×1083.7×1.6×10-19 ✓

\n

=3.4×10-7 m OR 340 nm ✓

\n


Must see an appropriate unit to award MP3.

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-5-current-and-circuits", "e-2-quantum-physics" ] }, { "question_id": "21M.2.HL.TZ1.2", "Question": "
\n

A planet is in a circular orbit around a star. The speed of the planet is constant. The following data are given:

\n

Mass of planet                                      =8.0×1024 kg
Mass of star                                          =3.2×1030 kg
Distance from the star to the planet R  =4.4×1010 m.

\n
\n

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.

\n
\n

Explain why a centripetal force is needed for the planet to be in a circular orbit.

\n
[2]
\n
a.
\n
\n

Calculate the value of the centripetal force.

\n
[1]
\n
b.
\n
\n

Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 109 J kg−1.

\n
[3]
\n
c.i.
\n
\n

Estimate the escape speed of the spacecraft from the planet–star system.

\n
[2]
\n
c.ii.
\n
", "Markscheme": "
\n

«circular motion» involves a changing velocity ✓

\n

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

\n

there must be a force/acceleration towards centre/star ✓

\n

without a centripetal force the planet will move in a straight line ✓

\n
a.
\n
\n

F=(6.67×10-11)(8×1024)(3.2×1030)(4.4×1010)2=8.8×1023 «N» ✓

\n
b.
\n
\n

Vplanet = «−»(6.67×10-11)(8×1024)9.1×106=«−» 5.9 × 107 «J kg−1» ✓

\n

Vstar = «−»(6.67×10-11)(3.2×1030)4.4×1010=«−» 4.9 × 109 «J kg−1» ✓

\n

Vplanet + Vstar = «−» 4.9 «09» × 109 «J kg−1» ✓

\n


Must see substitutions and not just equations.

\n
c.i.
\n
\n

use of vesc = 2V ✓

\n

v = 9.91 × 104 «m s−1» ✓

\n

 

\n

 

\n
c.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.i.
\n
\n[N/A]\n
c.ii.
\n
", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "d-1-gravitational-fields" ] }, { "question_id": "21M.2.HL.TZ1.8", "Question": "
\n

On 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

                                                       

\n
\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.

\n
\n

Outline how a standing wave is produced on the string.

\n
[2]
\n
a.
\n
\n

Show that the speed of the wave on the string is about 240 m s−1.

\n
[2]
\n
b.i.
\n
\n

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

\n

                 

\n
[1]
\n
b.ii.
\n
\n

Calculate, in m s−1, the maximum velocity of vibration of point P when it is vibrating with a frequency of 195 Hz.

\n
[2]
\n
b.iii.
\n
\n

Calculate, in terms of g, the maximum acceleration of P.

\n
[2]
\n
b.iv.
\n
\n

Estimate the displacement needed to double the energy of the string.

\n
[2]
\n
b.v.
\n
\n

The string is made to vibrate in its third harmonic. State the distance between consecutive nodes. 

\n
[1]
\n
c.
\n
", "Markscheme": "
\n

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

\n

superposition/interference of incident and reflected waves ✓

\n

the superposition of the reflections is reinforced only for certain wavelengths ✓ 

\n
a.
\n
\n

λ=2l=2×0.62=«1.24 m» ✓

\n

v=fλ=195×1.24=242 «m s-1» ✓

\n

Answer must be to 3 or more sf or working shown for MP2.

\n
b.i.
\n
\n

straight line through origin with negative gradient ✓

\n
b.ii.
\n
\n

max velocity occurs at x = 0 ✓

\n

v=«(2π)(195)0.0042»=4.9 «m s−1» ✓

\n
b.iii.
\n
\n

a=2π 1952×0.004=6005 «m s−2» ✓

\n

=600 g ✓

\n
b.iv.
\n
\n

use of E∝A2 OR  xo2 ✓

\n

A=0.42=0.57 «cm» ≅ 0.6 «cm» ✓

\n
b.v.
\n
\n

623=21 «cm» ✓

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
b.iii.
\n
\n[N/A]\n
b.iv.
\n
\n[N/A]\n
b.v.
\n
\n[N/A]\n
c.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-4-standing-waves-and-resonance" ] }, { "question_id": "21M.2.HL.TZ2.10", "Question": "
\n

The table gives data for Jupiter and three of its moons, including the radius r of each object.

\n

\n
\n

A spacecraft is to be sent from Io to infinity.

\n
\n

Calculate, for the surface of Io, the gravitational field strength gIo due to the mass of Io. State an appropriate unit for your answer.

\n
[2]
\n
a.
\n
\n

Show that the gravitational potential due to Jupiter at the orbit of Io gravitational potential due to Io at the surface of Io is about 80.

\n
[2]
\n
b.i.
\n
\n

Outline, using (b)(i), why it is not correct to use the equation 2G×mass of Ioradius of Io to calculate the speed required for the spacecraft to reach infinity from the surface of Io.

\n
[1]
\n
b.ii.
\n
\n

An 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
[2]
\n
c.
\n
", "Markscheme": "
\n

«GMr2=6.67×10-11×8.9×10221.8×1062=»1.8 ✓

\n

N kg−1  OR  m s−2  ✓

\n
a.
\n
\n

1.9×10274.9×108  AND  8.9×10221.8×106 seen ✓

\n

«1.9×1027×1.8×1064.9×108×8.9×1022=»78  ✓

\n


For MP1, potentials can be seen individually or as a ratio.

\n
b.i.
\n
\n

«this is the escape speed for Io alone but» gravitational potential / field of Jupiter must be taken into account  ✓

\n


OWTTE

\n
b.ii.
\n
\n

-GMJupiter11.88×109-11.06×109=«5.21×107 J kg-1»  ✓

\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».

\n

Allow ECF from MP1

\n

Award [2] marks for a bald correct answer.

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
c.
\n
", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields" ] }, { "question_id": "21M.2.HL.TZ2.8", "Question": "
\n

Monochromatic 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

  

\n
\n

State the effect on the graph of the variation of sin θ with n of:

\n
\n

State the phase difference between the waves at V and Y.

\n
[1]
\n
a.i.
\n
\n

State, in terms of λ, the path length between points X and Z.

\n
[1]
\n
a.ii.
\n
\n

The separation of adjacent slits is d. Show that for the second-order diffraction maximum 2λ=dsin θ.

\n
[1]
\n
a.iii.
\n
\n

Monochromatic 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.

\n

\n

Determine a mean value for the number of slits per millimetre of the grating.

\n
[4]
\n
b.
\n
\n

using a light source with a smaller wavelength.

\n
[1]
\n
c.i.
\n
\n

increasing the distance between the diffraction grating and the screen.

\n
[1]
\n
c.ii.
\n
", "Markscheme": "
\n

0 OR 2π OR 360° ✓

\n

 

\n
a.i.
\n
\n

4λ ✓

\n
a.ii.
\n
\n

sin θ«=XZVX»=4λ2d✓

\n


Do not award ECF from(a)(ii).

\n
a.iii.
\n
\n

identifies gradient with λd OR use of d sin θ=nλ ✓

\n

gradient = 0.08 OR correct replacement in equation with coordinates of a point ✓

\n

d=633×10-90.080=«7.91×10-6 m» ✓

\n

1.26×102 OR 1.27×102«mm-1» ✓

\n


Allow ECF from MP3

\n
b.
\n
\n

gradient smaller ✓

\n
c.i.
\n
\n

no change ✓

\n
c.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
a.iii.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.i.
\n
\n[N/A]\n
c.ii.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-3-wave-phenomena" ] }, { "question_id": "21M.2.HL.TZ2.9", "Question": "
\n

In 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.

\n

The 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.

\n

\n
\n

Outline the cause of the electron emission for radiation A.

\n
[1]
\n
a.i.
\n
\n

Outline why electrons are never emitted for radiation C.

\n
[1]
\n
a.ii.
\n
\n

Outline why radiation B gives different results.

\n
[1]
\n
a.iii.
\n
\n

Explain why there is no effect on the table of results when the intensity of source B is doubled.

\n
[1]
\n
b.
\n
\n

Photons 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.

\n

Calculate, in eV, the work function of the metal.

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

photon transfers «all» energy to electron ✓

\n
a.i.
\n
\n

photon energy is less than both work functions
OR
photon energy is insufficient «to remove an electron» ✓

\n


Answer must be in terms of photon energy.

\n
a.ii.
\n
\n

Identifies P work function lower than Q work function ✓

\n
a.iii.
\n
\n

changing/doubling intensity «changes/doubles number of photons arriving but» does not change energy of photon ✓

\n
b.
\n
\n

5.1×10-19=1.1×10-18-ϕ ✓

\n

work function = «(11.0-5.1)×10-191.6×10-19= » 3.7 «eV»  ✓

\n


Award [2] marks for a bald correct answer.

\n
c.
\n
", "Examiners report": "
\n[N/A]\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
a.iii.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-2-quantum-physics" ] }, { "question_id": "21M.2.SL.TZ1.1", "Question": "
\n

Two 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.

\n

\n
\n

The 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.

\n
\n

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

\n
[1]
\n
a.
\n
\n

Sketch, 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.

\n

\n
[2]
\n
b.
\n
\n

The 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.

\n

Show that the ball will go over the net.

\n
[3]
\n
c.
\n
\n

Determine the kinetic energy of the ball immediately after the bounce.

\n
[2]
\n
d.i.
\n
\n

Player 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.

\n

\n

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

\n
[3]
\n
d.ii.
\n
", "Markscheme": "
\n

t = «2dg=» 0.22 «s»
OR

\n

t = 2×0.249.8  ✓ 

\n

Answer to 2 or more significant figures or formula with variables replaced by correct values.

\n
a.
\n
\n

increasing straight line from zero up to 0.2 s in x-axis ✓

\n

with gradient = 10 ✓

\n
b.
\n
\n

ALTERNATIVE 1 

\n

t=1.3712=«0.114 s» ✓

\n

y=12×10×0.1142=0.065 m ✓

\n

so (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 = 12×10×t2 so» t = 0.134 s ✓

\n

0.134 × 12 = 1.6 m ✓

\n

1.6 > 1.37 «so ball passed the net already»  ✓

\n

 

\n

Allow use of g = 9.8.

\n
c.
\n
\n

ALTERNATIVE 1 

\n

KE = 12mv2 + mgh = 120.0027 ×10.52 + 0.0027 × 9.8 × 0.18 ✓

\n

0.15 «J» ✓

\n

 

\n

ALTERNATIVE 2

\n

Use of vx = 10.5 AND vy = 1.88 to get v = «10.52 + 1.882» = 10.67 «m s−1» ✓

\n

KE = 12 × 0.0027 × 10.672 = 0.15 «J»  ✓

\n
d.i.
\n
\n

Δv = 21 «m s−1» ✓

\n

F=0.0027 ×210.01

\n

OR

\n

5.67 «N» ✓

\n

any answer to 2 significant figures «N» ✓

\n
d.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.i.
\n
\n[N/A]\n
d.ii.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power" ] }, { "question_id": "21M.2.SL.TZ1.2", "Question": "
\n

A planet is in a circular orbit around a star. The speed of the planet is constant.

\n
\n

Explain why a centripetal force is needed for the planet to be in a circular orbit.

\n
[2]
\n
a.i.
\n
\n

State the nature of this centripetal force.

\n
[1]
\n
a.ii.
\n
\n

Determine the gravitational field of the planet.

\n

The following data are given:

\n

Mass of planet            =8.0×1024 kg
Radius of the planet    =9.1×106 m.

\n
[2]
\n
b.
\n
", "Markscheme": "
\n

«circular motion» involves a changing velocity ✓

\n

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

\n

there must be a force/acceleration towards centre/star ✓

\n

without a centripetal force the planet will move in a straight line ✓

\n
a.i.
\n
\n

gravitational force/force of gravity ✓

\n
a.ii.
\n
\n

use of GMR2 ✓

\n

6.4 «Nkg−1 or ms−2» ✓

\n
b.
\n
", "Examiners report": "
\n[N/A]\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
b.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "21M.2.SL.TZ1.3", "Question": "
\n

A 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.

\n
\n

A mass of 0.86 kg of water remains after it has boiled for 200 s.

\n
\n

The electric heater has two identical resistors connected in parallel.

\n

\n

The circuit transfers 1.6 kW when switch A only is closed. The external voltage is 220 V.

\n
\n

The 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.

\n
[2]
\n
a.i.
\n
\n

State one assumption of the kinetic model of an ideal gas.

\n
[1]
\n
a.ii.
\n
\n

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

\n
[2]
\n
b.i.
\n
\n

Explain why the temperature of water remains at 100 °C during this time.

\n
[1]
\n
b.ii.
\n
\n

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

\n

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

\n

Specific heat capacity of pasta = 1.8 kJ kg−1 K−1
Specific heat capacity of water = 4.2 kJ kg−1 K−1

\n
[3]
\n
c.
\n
\n

Show that each resistor has a resistance of about 30 Ω.

\n
[1]
\n
d.i.
\n
\n

Calculate the power transferred by the heater when both switches are closed.

\n
[2]
\n
d.ii.
\n
", "Markscheme": "
\n

Ek = « 32(1.38×10-23)(373)» = 7.7×10-21 «J» ✓

\n

v = «3×1.38×10-23×6.02×1023×3730.018» = 720 «m s−1» ✓

\n
a.i.
\n
\n

particles can be considered points «without dimensions» ✓

\n

no intermolecular forces/no forces between particles «except during collisions»✓

\n

the volume of a particle is negligible compared to volume of gas ✓

\n

collisions between particles are elastic ✓

\n

time between particle collisions are greater than time of collision ✓

\n

no intermolecular PE/no PE between particles  ✓

\n

 

\n

Accept reference to atoms/molecules for “particle”

\n
a.ii.
\n
\n

«mL = P t» so «L=1600×2000.14» = 2.3 x 106 «J kg-1» ✓

\n

J kg−1 ✓

\n
b.i.
\n
\n

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE ✓

\n

Accept reference to atoms/molecules for “particle” 

\n
b.ii.
\n
\n

use of mcΔT ✓

\n

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10) ✓

\n

Teq = 85.69«°C» ≅ 86«°C» ✓

\n

Accept Teq in Kelvin (359 K).

\n
c.
\n
\n

P=v2R so 22021600 so R=30.25 «Ω» ✓

\n

Must see either the substituted values OR a value for R to at least three s.f.

\n

 

\n
d.i.
\n
\n

use of parallel resistors addition so Req = 15 «Ω» ✓

\n

P = 3200 «W» ✓

\n
d.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.i.
\n
\n[N/A]\n
d.ii.
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers", "b-3-gas-laws", "b-5-current-and-circuits" ] }, { "question_id": "21M.2.SL.TZ1.6", "Question": "
\n

On 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

                                                       

\n
\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.

\n
\n

Outline how a standing wave is produced on the string.

\n
[2]
\n
a.
\n
\n

Show that the speed of the wave on the string is about 240 m s−1.

\n
[2]
\n
b.i.
\n
\n

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

\n

                 

\n
[1]
\n
b.ii.
\n
", "Markscheme": "
\n

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

\n

superposition/interference of incident and reflected waves ✓

\n

the superposition of the reflections is reinforced only for certain wavelengths ✓ 

\n
a.
\n
\n

λ=2l=2×0.62=«1.24 m» ✓

\n

v=fλ=195×1.24=242 «m s-1» ✓

\n

Answer must be to 3 or more sf or working shown for MP2.

\n
b.i.
\n
\n

straight line through origin with negative gradient ✓

\n
b.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-4-standing-waves-and-resonance" ] }, { "question_id": "21M.2.SL.TZ1.7", "Question": "
\n

Conservation of energy and conservation of momentum are two examples of conservation laws.

\n
\n

Outline the significance of conservation laws for physics.

\n
[1]
\n
a.
\n
\n

When 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.

\n
[3]
\n
b.
\n
", "Markscheme": "
\n

they express fundamental principles of nature ✓

\n

allow to model situations ✓

\n

allow to calculate unknown variables ✓

\n

allow to predict possible outcomes ✓

\n

allow to predict missing quantities/particles ✓

\n

allow comparison of different system states ✓

\n
a.
\n
\n

three correct conservation laws listed ✓

\n

at least one conservation law correctly demonstrated ✓

\n

all three conservation laws correctly demonstrated ✓

\n
b.
\n
", "Examiners report": "", "topics": [ "nature-of-science" ], "subtopics": [] }, { "question_id": "21M.2.SL.TZ1.8", "Question": "
\n
\n (a)\n
\n
\n

\n Outline how a standing wave is produced on the string.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Show that the speed of the wave on the string is about 240 m s\n \n −1\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.\n

\n

\n \n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The string is made to vibrate in its third harmonic. State the distance between consecutive nodes.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n «travelling» wave moves along the length of the string and reflects «at fixed end»\n \n ✓\n \n

\n

\n superposition/interference of incident and reflected waves\n \n ✓\n \n

\n

\n the superposition of the reflections is reinforced only for certain wavelengths\n \n ✓\n \n

\n
\n
\n (b.i)\n
\n

\n \n \n λ\n \n \n =\n \n \n 2\n \n \n l\n \n \n =\n \n \n 2\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 62\n \n \n =\n \n \n «\n \n \n 1\n \n \n .\n \n \n 24\n \n \n \n \n m\n \n \n »\n \n \n ✓\n

\n

\n \n \n v\n \n \n =\n \n \n f\n \n \n λ\n \n \n =\n \n \n 195\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 24\n \n \n =\n \n \n 242\n \n \n \n \n «\n \n \n \n m s\n \n \n \n -\n \n \n 1\n \n \n \n \n »\n \n \n ✓\n

\n

\n \n Answer must be to 3 or more sf or working shown for\n \n MP2.\n \n \n

\n
\n
\n (b.ii)\n
\n

\n straight line through origin with negative gradient\n \n ✓\n \n

\n
\n
\n (c)\n
\n

\n \n \n \n 62\n \n \n 3\n \n \n \n =\n \n \n 21\n \n \n \n \n «\n \n \n cm\n \n \n »\n \n \n \n ✓\n \n

\n
\n", "Examiners report": "None", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-4-standing-waves-and-resonance" ] }, { "question_id": "21M.2.SL.TZ2.1", "Question": "
\n

A 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.

\n

\n
\n

The 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.

\n
[2]
\n
a.
\n
\n

The 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.

\n
[2]
\n
b.i.
\n
\n

The top of the wall is 2.4 m above the ground. Deduce whether the ball will hit the wall.

\n
[3]
\n
b.ii.
\n
\n

In 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.

\n
[2]
\n
c.
\n
\n

The player kicks the ball again. It rolls along the ground without sliding with a horizontal velocity of 1.40 m s−1. The radius of the ball is 0.11 m. Calculate the angular velocity of the ball. State an appropriate SI unit for your answer.

\n
[1]
\n
d.
\n
", "Markscheme": "
\n

Δp=0.45×19 OR  a =190.055 ✓ 

\n

«=F=0.45×190.055»160 «N» ✓

\n

Allow [2] marks for a bald correct answer.

\n

Allow ECF for MP2 if 19 sin22 OR 19 cos22 used.

\n
a.
\n
\n

horizontal speed = 19×cos 22 «=17.6 m s-1» ✓ 

\n

time=«distancespeed=1119 cos 22=» 0.62 «s» ✓

\n

Allow ECF for MP2

\n
b.i.
\n
\n

initial vertical speed=19×sin 22 «= 7.1 m s-1» ✓ 

\n

«7.12×0.624-0.5×9.81×0.6242=» 2.5 «m» ✓

\n

ball does not hit wall OR 2.5 «m» > 2.4 «m» ✓

\n


Allow ECF from (b)(i) and from MP1

\n

Allow g = 10 m s−2

\n
b.ii.
\n
\n

air resistance opposes «direction of» motion
OR
air resistance opposes velocity ✓

\n

on the way up «vertical» acceleration is increased OR greater than g ✓

\n

on the way down «vertical» acceleration is decreased OR smaller than g ✓

\n


Allow deceleration/acceleration but meaning must be clear

\n
c.
\n
\n

13 «rad» s-1✓

\n


Unit must be seen for mark

\n

Accept Hz

\n

Accept 4 π «rad» s-1

\n
d.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum" ] }, { "question_id": "21M.2.SL.TZ2.5", "Question": "
\n

A 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.

\n

\n
\n

Describe two ways in which standing waves differ from travelling waves.

\n
[2]
\n
a.
\n
\n

Outline how a standing wave forms in the tube.

\n
[2]
\n
b.i.
\n
\n

The tube is raised until the loudness of the sound reaches a maximum for a second time.

\n

Draw, on the following diagram, the position of the nodes in the tube when the second maximum is heard.

\n

\n
[1]
\n
b.ii.
\n
\n

Between 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
[2]
\n
b.iii.
\n
", "Markscheme": "
\n

energy is not propagated by standing waves ✓

\n

amplitude constant for travelling waves OR amplitude varies with position for standing waves OR standing waves have nodes/antinodes ✓

\n

phase varies with position for travelling waves OR phase constant inter-node for standing waves ✓

\n

travelling waves can have any wavelength OR standing waves have discrete wavelengths ✓

\n


OWTTE

\n
a.
\n
\n

«sound» wave «travels down tube and» is reflected ✓

\n

incident and reflected wave superpose/combine/interfere ✓

\n


OWTTE

\n

Do not award MP1 if the reflection is quoted at the walls/container

\n
b.i.
\n
\n

nodes shown at water surface AND 23way up tube (by eye) ✓

\n


Accept drawing of displacement diagram for correct harmonic without nodes specifically identified.

\n

Award [0] if waveform is shown below the water surface

\n
b.ii.
\n
\n

λ=0.74 «m» ✓

\n

f=«cλ=3200.74=» 430 «Hz» ✓

\n


Allow ECF from MP1

\n
b.iii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
b.iii.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-4-standing-waves-and-resonance" ] }, { "question_id": "21M.2.SL.TZ2.6", "Question": "
\n

A photovoltaic cell is supplying energy to an external circuit. The photovoltaic cell can be modelled as a practical electrical cell with internal resistance.

\n

The 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.

\n

The 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

\n
\n

Explain why the output potential difference to the external circuit and the output emf of the photovoltaic cell are different.

\n

 

\n
[2]
\n
a.
\n
\n

Calculate the internal resistance of the photovoltaic cell for the maximum intensity condition using the model for the cell.

\n

 

\n
[3]
\n
b.
\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.

\n

A measure of the efficiency of a photovoltaic cell is the ratio

\n

energy available every second to the external circuitenergy arriving every second at the photovoltaic cell surface.

\n

Determine the efficiency of this photovoltaic cell when the intensity incident upon it is at a maximum.

\n
[3]
\n
c.
\n
\n

State two reasons why future energy demands will be increasingly reliant on sources such as photovoltaic cells.

\n

\n
[2]
\n
d.
\n
", "Markscheme": "
\n

there is a potential difference across the internal resistance
OR
there is energy/power dissipated in the internal resistance ✓

\n

when there is current «in the cell»/as charge flows «through the cell» ✓

\n


Allow full credit for answer based on V=ε-Ir

\n
a.
\n
\n

ALTERNATIVE 1
pd dropped across cell =6.5 «V»✓

\n

internal resistance =6.50.9 ✓

\n

7.2 «Ω»✓

\n


ALTERNATIVE 2

\n

ε=I(R+r) so ε=V+Ir ✓

\n

21.0=14.5+0.9×r ✓

\n

7.2 «Ω» ✓

\n


Alternative solutions are possible

\n

Award [3] marks for a bald correct answer

\n
b.
\n
\n

power arriving at cell = 680 x 0.35 x 0.45 = «107 W» ✓

\n

power in external circuit = 14.5 x 0.9 = «13.1 W» ✓

\n

efficiency = 0.12 OR 12 % ✓

\n


Award [3] marks for a bald correct answer

\n

Allow ECF for MP3

\n
c.
\n
\n

«energy from Sun/photovoltaic cells» is renewable
OR
non-renewable are running out ✓

\n

non-polluting/clean ✓

\n

no greenhouse gases
OR
does not contribute to global warming/climate change ✓

\n


OWTTE

\n

Do not allow economic aspects (e.g. free energy)

\n
d.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.
\n
", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter" ], "subtopics": [ "a-3-work-energy-and-power", "b-5-current-and-circuits" ] }, { "question_id": "21N.2.HL.TZ0.5", "Question": "
\n

A 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.

\n

\n
\n

Show that the speed of the loop is 20 cm s−1.

\n
[1]
\n
a.
\n
\n

Sketch, on the axes, a graph to show the variation with time of the magnetic flux linkage Φ in the loop.

\n

\n
[1]
\n
b.i.
\n
\n

Sketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.

\n

\n
[1]
\n
b.ii.
\n
\n

There are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.

\n
[2]
\n
c.i.
\n
\n

The 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.

\n
[2]
\n
c.ii.
\n
\n

Show that the energy dissipated in the loop from t = 0 to t = 3.5 s is 0.13 J.

\n
[2]
\n
d.i.
\n
\n

The 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
[2]
\n
d.ii.
\n
", "Markscheme": "
\n

703.5 ✓

\n
a.
\n
\n

\n

shape as above ✓

\n
b.i.
\n
\n

\n

shape as above ✓

\n

 

\n

Vertical lines not necessary to score.

\n

Allow ECF from (b)(i).

\n
b.ii.
\n
\n

ALTERNATIVE 1

\n

maximum flux at «5.0×5.0×10-4×85×0.94» =0.19975≈0.20 «Wb» ✓

\n

emf = «0.200.25=» 0.80 «V» ✓

\n


ALTERNATIVE 2

\n

emf induced in one turn = BvL = 0.94×0.20×0.05=0.0094 «V» ✓

\n

emf =85×0.0094=0.80 «V» ✓

\n

 

\n

Award [2] marks for a bald correct answer.

\n

Allow ECF from MP1.

\n
c.i.
\n
\n

I=«VR=»0.82.4  OR  0.33 «A» ✓

\n

F=«NBIL=85×0.94×0.33×0.05=»=1.3 «N» ✓

\n

 

\n

Allow ECF from (c)(i).

\n

Award [2] marks for a bald correct answer.

\n
c.ii.
\n
\n

Energy is being dissipated for 0.50 s ✓

\n


E=Fvt=1.3×0.20×0.50=«0.13 J»

\n

OR

\n

E=Vlt=0.80×0.33×0.50=«0.13 J» ✓

\n

 

\n

Allow ECF from (b) and (c).

\n

Watch for candidates who do not justify somehow the use of 0.5 s and just divide by 2 their answer.

\n
d.i.
\n
\n

∆T=0.130.018×385 ✓

\n

∆T=1.9×10-2 «K» ✓

\n

 

\n

Allow [2] marks for a bald correct answer.

\n

Award [1] for a POT error in MP1.

\n
d.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
c.i.
\n
\n[N/A]\n
c.ii.
\n
\n[N/A]\n
d.i.
\n
\n[N/A]\n
d.ii.
\n
", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "a-3-work-energy-and-power", "b-1-thermal-energy-transfers", "b-5-current-and-circuits", "d-3-motion-in-electromagnetic-fields", "d-4-induction" ] }, { "question_id": "21N.2.HL.TZ0.7", "Question": "
\n

A conducting sphere has radius 48 cm. The electric potential on the surface of the sphere is 3.4 × 105 V.

\n
\n

The sphere is connected by a long conducting wire to a second conducting sphere of radius 24 cm. The second sphere is initially uncharged.

\n

 

\n
\n

Show that the charge on the surface of the sphere is +18 μC.

\n
[1]
\n
a.
\n
\n

Describe, in terms of electron flow, how the smaller sphere becomes charged.

\n
[1]
\n
b.i.
\n
\n

Predict the charge on each sphere.

\n
[3]
\n
b.ii.
\n
", "Markscheme": "
\n

Q=«VRk=»3.4×105×0.488.99×109

\n

OR

\n

Q=18.2 «μC» ✓

\n
a.
\n
\n

electrons leave the small sphere «making it positively charged» ✓

\n
b.i.
\n
\n

kq148=kq224⇒q1=2q2 ✓

\n

q1+q2=18 ✓

\n

so q1=12 «μC», q2=6.0 «μC» ✓

\n

 

\n

Award [3] marks for a bald correct answer.

\n
b.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
", "topics": [ "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "b-5-current-and-circuits", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "21N.2.HL.TZ0.8", "Question": "
\n

The graph shows the variation with diffraction angle of the intensity of light after it has passed through four parallel slits.

\n

\n

The number of slits is increased but their separation and width stay the same. All slits are illuminated.

\n
\n

State what is meant by the Doppler effect.

\n
[2]
\n
a.
\n
\n

A plate performs simple harmonic oscillations with a frequency of 39 Hz and an amplitude of 8.0 cm.

\n

Show that the maximum speed of the oscillating plate is about 20 m s−1.

\n
[2]
\n
b.
\n
\n

Sound 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.

\n

\n

Determine the maximum frequency of the sound that is received back at the source after reflection at the plate.

\n

 

\n
[2]
\n
c.
\n
\n

State what will happen to the angular position of the primary maxima.

\n
[1]
\n
d.i.
\n
\n

State what will happen to the width of the primary maxima.

\n
[1]
\n
d.ii.
\n
\n

State what will happen to the intensity of the secondary maxima.

\n
[1]
\n
d.iii.
\n
", "Markscheme": "
\n

the change in the observed frequency ✓

\n

when there is relative motion between the source and the observer ✓

\n

 

\n

Do not award MP1 if they refer to wavelength.

\n
a.
\n
\n

use of 2πfA ✓

\n

maximum speed is 2π×39×0.080=19.6 «m s−1» ✓

\n

 

\n

Award [2] for a bald correct answer.

\n
b.
\n
\n

frequency at plate 2400×340+19.6340«=2538 Hz»

\n

at source 2538×340340-19.6=2694≈2700 «Hz» ✓

\n

 

\n

Award [2] marks for a bald correct answer.

\n

Award [1] mark when the effect is only applied once.

\n
c.
\n
\n

stays the same ✓

\n
d.i.
\n
\n

decreases ✓

\n
d.ii.
\n
\n

decreases ✓

\n
d.iii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.
\n
\n[N/A]\n
c.
\n
\n[N/A]\n
d.i.
\n
\n[N/A]\n
d.ii.
\n
\n[N/A]\n
d.iii.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-3-wave-phenomena", "c-5-doppler-effect" ] }, { "question_id": "21N.2.SL.TZ0.3", "Question": "
\n

A 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.

\n

\n
\n

Another wave travels in the medium. The graph shows the variation with time t of the displacement of each wave at the position of P.

\n

\n
\n

A 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 T. The diagram represents the standing wave at t=0 and at t=T8. The wavelength of the wave is 1.20 m. Positive displacements mean displacements to the right.

\n

\n
\n

Calculate the wavelength of the wave.

\n
[2]
\n
a.
\n
\n

State the phase difference between the two waves.

\n
[1]
\n
b.i.
\n
\n

Identify a time at which the displacement of P is zero.

\n
[1]
\n
b.ii.
\n
\n

Estimate the amplitude of the resultant wave.

\n
[1]
\n
b.iii.
\n
\n

Calculate the length of the tube.

\n
[1]
\n
c.i.
\n
\n

A 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 t=T8.

\n
[2]
\n
c.ii.
\n
\n

Draw on the diagram the standing wave at time t=T4.

\n
[1]
\n
c.iii.
\n
", "Markscheme": "
\n

T=4×10-3 «s» or f=250 «Hz» ✓

\n

λ=340×4.0×10-3=1.36≈1.4 «m» ✓

\n

 

\n

Allow ECF from MP1.
Award [2] for a bald correct answer.

\n
a.
\n
\n

«±» π2/90°  OR  3π2/270° ✓

\n
b.i.
\n
\n

1.5 «ms» ✓

\n
b.ii.
\n
\n

8.0 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.

\n
b.iii.
\n
\n

L = «34λ=» 0.90 «m» ✓

\n
c.i.
\n
\n

to the right ✓

\n

displacement is getting less negative

\n

OR

\n

change of displacement is positive ✓

\n
c.ii.
\n
\n

horizontal line drawn at the equilibrium position ✓

\n
c.iii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
b.iii.
\n
\n[N/A]\n
c.i.
\n
\n[N/A]\n
c.ii.
\n
\n[N/A]\n
c.iii.
\n
", "topics": [ "c-wave-behaviour", "d-fields" ], "subtopics": [ "c-1-simple-harmonic-motion", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "21N.2.SL.TZ0.4", "Question": "
\n

A 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.

\n

\n
\n

The diagram shows two parallel wires X and Y that carry equal currents into the page.

\n

\n

Point Q is equidistant from the two wires. The magnetic field at Q due to wire X alone is 15 mT.

\n
\n

The 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.

\n
[1]
\n
a.
\n
\n

Determine the force on Q at the instant it is released.

\n
[2]
\n
b.i.
\n
\n

Describe the motion of Q after release.

\n
[2]
\n
b.ii.
\n
\n

On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone.

\n
[1]
\n
c.i.
\n
\n

Determine the magnitude and direction of the resultant magnetic field at Q.

\n
[2]
\n
c.ii.
\n
", "Markscheme": "
\n

«V=4.50.25=» 18 «V» ✓

\n
a.
\n
\n

F=8.99×109×68×10-6×0.25×10-60.482 ✓

\n

F=0.66 «N» ✓

\n

 

\n

Award [2] marks for a bald correct answer.

\n

Allow symbolic k in substitutions for MP1.

\n

Do not allow ECF from incorrect or not squared distance.

\n
b.i.
\n
\n

Q moves to the right/away from P «along a straight line»

\n

OR

\n

Q is repelled from P ✓

\n


with increasing speed/Q accelerates ✓

\n

acceleration decreases ✓

\n
b.ii.
\n
\n

 

\n

\n

arrow of any length as shown ✓

\n
c.i.
\n
\n

«using components or Pythagoras to get» B = 21 «mT» ✓

\n

directed «horizontally» to the right ✓

\n

 

\n

If no unit seen, assume mT.

\n
c.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
\n[N/A]\n
c.i.
\n
\n[N/A]\n
c.ii.
\n
", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "b-5-current-and-circuits", "d-2-electric-and-magnetic-fields", "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "21N.2.SL.TZ0.5", "Question": "
\n

Plutonium-238 (Pu) decays by alpha (α) decay into uranium (U).

\n

The following data are available for binding energies per nucleon:

\n

plutonium          7.568 MeV

\n

uranium             7.600 MeV

\n

alpha particle     7.074 MeV

\n
\n

State what is meant by the binding energy of a nucleus.

\n
[1]
\n
a.i.
\n
\n

Draw, on the axes, a graph to show the variation with nucleon number A of the binding energy per nucleon, BEA. Numbers are not required on the vertical axis.

\n

\n
[2]
\n
a.ii.
\n
\n

Identify, with a cross, on the graph in (a)(ii), the region of greatest stability.

\n
[1]
\n
a.iii.
\n
\n

Show that the energy released in this decay is about 6 MeV.

\n
[3]
\n
b.i.
\n
\n

The plutonium nucleus is at rest when it decays.

\n

Calculate the ratio kinetic energy of alpha particlekinetic energy of uranium.

\n
[2]
\n
b.ii.
\n
", "Markscheme": "
\n

the energy needed to «completely» separate the nucleons of a nucleus

\n

OR

\n

the energy released when a nucleus is assembled from its constituent nucleons ✓

\n

 

\n

Accept reference to protons AND neutrons.

\n
a.i.
\n
\n

curve rising to a maximum between 50 and 100 ✓

\n

curve continued and decreasing ✓

\n

 

\n

Ignore starting point.

\n

Ignore maximum at alpha particle

\n
a.ii.
\n
\n

At a point on the peak of their graph ✓

\n
a.iii.
\n
\n

correct mass numbers for uranium (234) and alpha (4) ✓

\n

234×7.600+4×7.074-238×7.568 «MeV» ✓

\n

energy released 5.51 «MeV» ✓

\n

 

\n

Ignore any negative sign.

\n
b.i.
\n
\n

«KEαKEU=»p22mαp22mU  OR  mUmα ✓

\n

«2344=» 58.5 ✓

\n

 

\n

Award [2] marks for a bald correct answer.

\n

Accept 1172 for MP2.

\n
b.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
a.iii.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "a-3-work-energy-and-power", "b-1-thermal-energy-transfers", "b-5-current-and-circuits", "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "21N.2.SL.TZ0.6", "Question": "
\n

Titan is a moon of Saturn. The Titan-Sun distance is 9.3 times greater than the Earth-Sun distance.

\n
\n

Show that the intensity of the solar radiation at the location of Titan is 16 W m−2

\n
[1]
\n
a.i.
\n
\n

Titan 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

\n
[3]
\n
a.ii.
\n
\n

Show that the equilibrium surface temperature of Titan is about 90 K.

\n
[1]
\n
a.iii.
\n
\n

The orbital radius of Titan around Saturn is R and the period of revolution is T.

\n

Show that T2=4π2R 3GM where M is the mass of Saturn.

\n
[2]
\n
b.i.
\n
\n

The orbital radius of Titan around Saturn is 1.2 × 109 m and the orbital period is 15.9 days. Estimate the mass of Saturn.

\n
[2]
\n
b.ii.
\n
", "Markscheme": "
\n

incident intensity 13609.32 OR 15.7≈16 «W m−2» ✓

\n

 

\n

Allow the use of 1400 for the solar constant.

\n
a.i.
\n
\n

exposed surface is ¼ of the total surface ✓

\n

absorbed intensity = (1−0.22) × incident intensity ✓

\n

0.78 × 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.

\n
a.ii.
\n
\n

σT 4 = 3.07

\n

OR

\n

T = 86 «K» ✓

\n
a.iii.
\n
\n

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

\n

correct rearrangement to reach the expression given ✓

\n

 

\n

Allow use of GMR=2πRT for MP1.

\n
b.i.
\n
\n

T=15.9×24×3600 «s» ✓

\n

M=4π21.2×10936.67×10-11×15.9×24×36002=5.4×1026 «kg» ✓

\n

 

\n

Award [2] marks for a bald correct answer.

\n

Allow ECF from MP1.

\n
b.ii.
\n
", "Examiners report": "
\n[N/A]\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
a.iii.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
", "topics": [ "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "b-2-greenhouse-effect", "d-1-gravitational-fields" ] }, { "question_id": "22M.2.HL.TZ1.3", "Question": "
\n

Two 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.

\n

\n

B 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.

\n

\n
\n

Explain why the received intensity varies between maximum and minimum values.

\n
[3]
\n
a.
\n
\n

State and explain the wavelength of the sound measured at M.

\n
[2]
\n
b.
\n
\n

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

\n

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

\n

Speed of sound = 340 m s−1

\n
[2]
\n
c.
\n
\n

Loudspeaker 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.

\n

Determine the difference between the frequency detected at M and that emitted by B.

\n
[2]
\n
d.
\n
", "Markscheme": "
\n

movement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓

\n

interference
OR
superposition «of waves» ✓

\n

maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or π » out of phase / path difference = (n+½) x lambda ✓

\n
a.
\n
\n

wavelength = 26 cm ✓

\n


peak to peak distance is the path difference which is one wavelength

\n

OR

\n

this is the distance B moves to be back in phase «with A» ✓

\n

 

\n

Allow 25 – 27 cm for MP1.

\n
b.
\n
\n

«λ2» = 13 cm ✓

\n

f=«cλ=3400.13=» 2.6 «kHz» ✓

\n

 

\n

Allow ½ of wavelength from (b) or data from graph for MP1.

\n

Allow ECF from MP1.

\n
c.
\n
\n

ALTERNATIVE 1
use of f'=fvv+u0 (+ sign must be seen) OR f'= 2987 «Hz» ✓
« Δf» = 13 «Hz» ✓

\n

 

\n

ALTERNATIVE 2
Attempted use of Δff≈ vc

« Δf » = 13 «Hz» ✓

\n
d.
\n
", "Examiners report": "
\n

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.

\n
a.
\n
\n

Many 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.

\n
b.
\n
\n

This 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.

\n
c.
\n
\n

Many 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.

\n
d.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-3-wave-phenomena", "c-5-doppler-effect" ] }, { "question_id": "22M.2.SL.TZ1.1", "Question": "
\n

A 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.

\n

\n

The 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.

\n

The load is released and accelerates downwards.

\n
\n

Outline two differences between the momentum of the box and the momentum of the load at the same instant.

\n
[2]
\n
a.
\n
\n

The vertical acceleration of the load downwards is 2.4 m s−2.

\n

Calculate the tension in the string.

\n
[2]
\n
b.
\n
\n

Show that the speed of the load when it hits the floor is about 2.1 m s−1.

\n
[2]
\n
c.i.
\n
\n

The 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.

\n
[2]
\n
c.ii.
\n
\n

After 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.

\n
[4]
\n
d.
\n
\n

The 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.

\n

\n

Explain, with reference to the frictional force acting, why the box accelerates once it has started to move. 

\n
[3]
\n
e.
\n
", "Markscheme": "
\n

direction of motion is different / OWTTE ✓

\n

mv / magnitude of momentum is different «even though v the same» ✓

\n
a.
\n
\n

use of ma = mg − T «3.5 x 2.4 = 3.5g − T »

\n

OR

\n

T = 3.5(g − 2.4) ✓

\n

26 «N» ✓

\n

 

\n

Accept 27 N from g = 10 m s−2

\n
b.
\n
\n

proper use of kinematic equation ✓

\n

2×2.4×0.95=2.14 «m s−1» ✓

\n

 

\n

Must see either the substituted values OR a value for v to at least three s.f. for MP2.

\n
c.i.
\n
\n

use of ω=vr to give 84 «rad s−1»

\n

OR

\n

ω=2.1/0.025 to give 84 «rad s−1» ✓

\n

 

\n

quoted to 2sf only✓

\n

 

\n
c.ii.
\n
\n

ALTERNATIVE 1

\n

«v2=u2+2as⇒0=2.12-2a×0.35» leading to a = 6.3 «m s-2»

\n

OR

\n

« x=1/2u+vt » leading to t = 0.33 « s » ✓

\n


Fnet = « ma=1.5×6.3 = » 9.45 «N» ✓

\n

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

\n

friction force = net force – weight down ramp = 2.1 «N» ✓

\n

 

\n

ALTERNATIVE 2

\n

kinetic energy initial = work done to stop 0.5 x 1.5 x (2.1)2 = FNET x 0.35 ✓

\n

Fnet = 9.45 «N» ✓

\n

Weight down ramp = 1.5 x 9.8 x sin(30) = 7.4 «N» ✓

\n

friction 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.

\n
d.
\n
\n

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»

\n

OR

\n

reference to net F = ma ✓

\n
e.
\n
", "Examiners report": "
\n

Many 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.

\n
a.
\n
\n

Some 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.

\n
b.
\n
\n

This 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.

\n
c.i.
\n
\n

This SL only question was generally well done. Despite some power of 10 errors, many candidates correctly reported final answer to 2 sf.

\n
c.ii.
\n
\n

Candidates 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.

\n
d.
\n
\n

This 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.

\n
e.
\n
", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum" ] }, { "question_id": "22M.2.SL.TZ1.2", "Question": "
\n

Cold milk enters a small sterilizing unit and flows over an electrical heating element.

\n

\n

The 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.

\n

Specific heat capacity of milk = 3.9 kJ kg−1 K−1

\n
\n

The 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.

\n

\n
\n

Estimate the power input to the heating element. State an appropriate unit for your answer.

\n
[2]
\n
a.
\n
\n

Outline whether your answer to (a) is likely to overestimate or underestimate the power input.

\n
[2]
\n
b.
\n
\n

Discuss, with reference to the molecules in the liquid, the difference between milk at 11 °C and milk at 84 °C.

\n
[2]
\n
c.
\n
\n

State how energy is transferred from the inside of the metal pipe to the outside of the metal pipe.

\n
[1]
\n
d.i.
\n
\n

The 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.

\n
[3]
\n
d.ii.
\n
\n

Describe one other method by which significant amounts of energy can be transferred from the pipe to the surroundings.

\n
[2]
\n
d.iii.
\n
", "Markscheme": "
\n

energy required for milk entering in 1 s = mass x specific heat x 73 ✓

\n

16 kW OR 16000 W ✓

\n

 

\n

MP1 is for substitution into mcΔT regardless of power of ten.

\n

Allow 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.

\n
a.
\n
\n

Underestimate / more energy or power required ✓

\n

because 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.

\n
b.
\n
\n

the 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. ✓

\n
c.
\n
\n

conduction/conducting/conductor «through metal» ✓

\n
d.i.
\n
\n

use of P=eσAT4 where T = 357 K ✓

\n

use of A=2π r l « = 0.236 m2» ✓

\n

P = 87 «W» ✓

\n

 

\n

Allow 85 – 89 W for MP3.

\n

Allow ECF for MP3.

\n
d.ii.
\n
\n

convection «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»

\n

OR

\n

«due to» conduction «of heat or thermal energy» from pipe to air ✓

\n
d.iii.
\n
", "Examiners report": "
\n

Most 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).

\n
a.
\n
\n

Many 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).

\n
b.
\n
\n

This 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.

\n
c.
\n
\n

Many 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.

\n
d.i.
\n
\n

Candidates 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.

\n
d.ii.
\n
\n

A 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).

\n
d.iii.
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers", "b-2-greenhouse-effect" ] }, { "question_id": "22M.2.SL.TZ1.3", "Question": "
\n

Two 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.

\n

\n

B 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.

\n

\n
\n

Explain why the received intensity varies between maximum and minimum values.

\n
[3]
\n
a.
\n
\n

State and explain the wavelength of the sound measured at M.

\n
[2]
\n
b.
\n
\n

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

\n

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

\n

Speed of sound = 340 m s−1

\n
[2]
\n
c.
\n
", "Markscheme": "
\n

movement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓

\n

interference
OR
superposition «of waves» ✓

\n

maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or π » out of phase / path difference = (n+½) x lambda ✓

\n
a.
\n
\n

wavelength = 26 cm ✓

\n


peak to peak distance is the path difference which is one wavelength

\n

OR

\n

this is the distance B moves to be back in phase «with A» ✓

\n

 

\n

Allow 25 − 27 cm for MP1.

\n
b.
\n
\n

«λ2» = 13 cm ✓

\n

f=«cλ=3400.13=» 2.6 «kHz» ✓

\n

 

\n

Allow ½ of wavelength from (b) or data from graph.

\n
c.
\n
", "Examiners report": "
\n

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.

\n
a.
\n
\n

Many 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.

\n
b.
\n
\n

This 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.

\n
c.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-2-wave-model", "c-3-wave-phenomena" ] }, { "question_id": "22M.2.SL.TZ1.6", "Question": "
\n
\n (a)\n
\n
\n

\n Outline\n \n two\n \n reasons why both models predict that the motion is simple harmonic when\n \n \n a\n \n \n is small.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine the time period of the system when\n \n \n a\n \n \n is small.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Outline, without calculation, the change to the time period of the system for the model represented by graph B when\n \n \n a\n \n \n is large.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n The graph shows for model A the variation with\n \n \n x\n \n \n of elastic potential energy\n \n E\n \n \n p\n \n stored in the spring.\n

\n

\n \n

\n

\n Describe the graph for model B.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n For both models:\n \n \n
\n displacement is ∝ to acceleration/force «because graph is straight and through origin» ✓\n

\n

\n displacement and acceleration / force in opposite directions «because gradient is negative»\n
\n \n \n OR\n \n \n
\n acceleration/«restoring» force is always directed to equilibrium ✓\n

\n
\n
\n (b)\n
\n

\n attempted use of\n \n \n \n ω\n \n \n 2\n \n \n \n =\n \n \n \n -\n \n \n \n \n a\n \n \n x\n \n \n \n ✓\n

\n

\n suitable read-offs leading to gradient of line = 28 « s\n \n -2\n \n » ✓\n

\n

\n \n \n T\n \n \n =\n \n \n \n \n 2\n \n \n π\n \n \n \n ω\n \n \n \n «\n \n \n \n \n 2\n \n \n π\n \n \n \n \n 28\n \n \n \n \n » ✓\n

\n

\n \n \n T\n \n \n =\n \n \n 1\n \n \n .\n \n \n 2\n \n \n s ✓\n

\n
\n
\n (c)\n
\n

\n time period increases ✓\n

\n

\n

\n

\n because average ω «for whole cycle» is smaller\n

\n

\n \n \n OR\n \n \n

\n

\n slope / acceleration / force at large x is smaller\n

\n

\n \n \n OR\n \n \n

\n

\n area under graph B is smaller so average speed is smaller. ✓\n

\n
\n
\n (d)\n
\n

\n same curve\n \n \n OR\n \n \n shape for small amplitudes «to about 0.05 m» ✓\n

\n

\n for large amplitudes «outside of 0.05 m»\n \n E\n \n \n p\n \n smaller for model B / values are lower than original / spread will be wider ✓\n \n \n OWTTE\n \n \n

\n

\n

\n

\n \n Accept answers drawn on graph – e.g.\n \n

\n

\n \n \n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n This 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.\n

\n
\n
\n (b)\n
\n

\n This 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.\n

\n
\n
\n (c)\n
\n

\n Many 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.\n

\n
\n
\n (d)\n
\n

\n Candidates 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.\n

\n
\n", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion" ] }, { "question_id": "22M.2.SL.TZ2.2", "Question": "
\n

A 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.

\n
\n

Energy is now supplied to the gas and the piston moves to allow the gas to expand. The temperature is held constant.

\n
\n

Calculate the number of gas particles in the cylinder.

\n
[2]
\n
a.
\n
\n

Discuss, for this process, the changes that occur in the density of the gas.

\n
[2]
\n
b.i.
\n
\n

Discuss, for this process, the changes that occur in the internal energy of the gas.

\n
[2]
\n
b.ii.
\n
", "Markscheme": "
\n

Correct conversion of T «T = 310 K» seen ✓

\n

« use of = N=pVkT to get » 2.3 × 1023 ✓

\n

 

\n

Allow ECF from MP1 i.e., T in Celsius (Result is 2.7 x 1024)

\n

Allow use of n, R and NA

\n
a.
\n
\n

density decreases ✓

\n

volume is increased AND mass/number of particles remains constant ✓

\n
b.i.
\n
\n

internal energy is constant ✓

\n


internal energy depends on kinetic energy/temperature «only»

\n

OR

\n

since 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.

\n

Award MP2 for stating that kinetic energy remains constant.

\n
b.ii.
\n
", "Examiners report": "
\n

a) 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.

\n

bi) 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.

\n

bii) 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.

\n
a.
\n
\n[N/A]\n
b.i.
\n
\n[N/A]\n
b.ii.
\n
", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-3-gas-laws" ] }, { "question_id": "22M.2.SL.TZ2.3", "Question": "
\n

A loudspeaker emits sound waves of frequency f 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.

\n

\n

The speed of sound in air is 340 m s−1.

\n
\n

Explain the variation in intensity.

\n
[3]
\n
a.i.
\n
\n

Adjacent minima are separated by a distance of 0.12 m. Calculate f.

\n
[2]
\n
a.ii.
\n
\n

The metal plate is replaced by a wooden plate that reflects a lower intensity sound wave than the metal plate.

\n

State and explain the differences between the sound intensities detected by the same microphone with the metal plate and the wooden plate.

\n
[3]
\n
b.
\n
", "Markscheme": "
\n

«incident and reflected» waves superpose/interfere/combine ✓

\n

«that leads to» standing waves formed OR nodes and antinodes present ✓

\n

at antinodes / maxima there is maximum intensity / constructive interference / «displacement» addition / louder sound ✓

\n

at nodes / minima there is minimum intensity / destructive interference / «displacement» cancellation / quieter sound ✓

\n

 

\n

OWTTE

\n

Allow a sketch of a standing wave for MP2

\n

Allow a correct reference to path or phase differences to identify constructive / destructive interference

\n
a.i.
\n
\n

wavelength = 0.24 «m» ✓

\n

f = «3400.24=» 1.4 «kHz» OR 1400 «Hz» ✓

\n

 

\n

Allow ECF from MP1

\n
a.ii.
\n
\n

relates intensity to amplitude ✓

\n

antinodes / maximum intensity will be decreased / quieter ✓

\n

nodes / minimum will be increased / louder ✓

\n

difference in intensities will be less ✓

\n

maxima and minima are at the same positions ✓

\n

 

\n

OWTTE

\n
b.
\n
", "Examiners report": "
\n

ai) 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.

\n

3aii) 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.

\n

b) 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.

\n
a.i.
\n
\n[N/A]\n
a.ii.
\n
\n[N/A]\n
b.
\n
", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-4-standing-waves-and-resonance" ] }, { "question_id": "22N.2.HL.TZ0.6", "Question": "
\n
\n (a)\n
\n
\n

\n Outline, by reference to nuclear binding energy, why the mass of a nucleus is less than the sum of the masses of its constituent nucleons.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Calculate, in MeV, the energy released in this decay.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n The polonium nucleus was stationary before the decay.\n

\n

\n Show, by reference to the momentum of the particles, that the kinetic energy of the alpha particle is much greater than the kinetic energy of the lead nucleus.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b.iii)\n
\n
\n

\n In the decay of polonium−210, alpha emissions can be accompanied by the emissions of gamma photons, all of the same wavelength of 1.54 × 10\n \n −12\n \n m.\n

\n

\n Discuss how this observation provides evidence for discrete nuclear energy levels.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n A sample contains 5.0 g of pure polonium-210. The decay constant of polonium-210 is 5.8 × 10\n \n −8\n \n s\n \n −1\n \n . Lead-206 is stable.\n

\n

\n Calculate the mass of lead-206 present in the sample after one year.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n according to Δ\n \n E\n \n = Δ\n \n mc\n \n \n 2\n \n / identifies mass energy equivalence ✓\n

\n

\n

\n

\n energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»\n

\n

\n \n \n OR\n \n \n

\n

\n energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓\n

\n

\n

\n

\n \n Accept protons and neutrons.\n \n

\n
\n
\n (a)\n
\n

\n according to Δ\n \n E\n \n = Δ\n \n mc\n \n \n 2\n \n / identifies mass energy equivalence ✓\n

\n

\n

\n

\n energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»\n

\n

\n \n \n OR\n \n \n

\n

\n energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓\n

\n

\n

\n

\n \n Accept protons and neutrons.\n \n

\n
\n
\n (b.i)\n
\n

\n (\n \n m\n \n \n polonium\n \n −\n \n m\n \n \n lead\n \n −\n \n m\n \n α\n \n \n )\n \n c\n \n \n 2\n \n \n \n OR\n \n \n (209.93676 − 205.92945 − 4.00151)\n

\n

\n \n \n OR\n \n \n

\n

\n mass difference = 5.8 × 10\n \n −3\n \n ✓\n

\n

\n

\n

\n conversion to MeV using 931.5 to give 5.4 «MeV» ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n .\n \n

\n

\n \n Award\n \n [2]\n \n for a\n \n BCA\n \n .\n \n

\n

\n \n Award\n \n [1]\n \n for 8.6 x 10\n \n −13\n \n J.\n \n

\n
\n
\n (b.ii)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n energy ratio expressed in terms of momentum, e.g.\n \n \n \n \n E\n \n \n α\n \n \n \n \n E\n \n \n lead\n \n \n \n \n =\n \n \n \n \n \n p\n \n \n α\n \n \n 2\n \n \n \n /\n \n \n 2\n \n \n \n m\n \n \n α\n \n \n \n \n \n \n p\n \n \n lead\n \n \n 2\n \n \n \n /\n \n \n 2\n \n \n \n m\n \n \n lead\n \n \n \n \n \n ✓\n

\n

\n \n \n \n p\n \n \n α\n \n \n \n =\n \n \n \n p\n \n \n lead\n \n \n \n hence\n \n \n \n \n E\n \n \n α\n \n \n \n \n E\n \n \n lead\n \n \n \n \n =\n \n \n \n \n m\n \n \n lead\n \n \n \n \n m\n \n \n α\n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n \n \n m\n \n \n lead\n \n \n \n \n m\n \n \n α\n \n \n \n \n ≃\n \n \n \n 206\n \n \n 4\n \n \n \n =\n \n \n 51\n \n \n .\n \n \n 5\n \n \n ⇒\n \n \n \n E\n \n \n α\n \n \n \n =\n \n \n 51\n \n \n .\n \n \n 5\n \n \n \n E\n \n \n lead\n \n \n \n «so\n \n \n α\n \n \n has a much greater KE»\n

\n

\n \n \n OR\n \n \n

\n

\n \n \n \n m\n \n \n lead\n \n \n \n «much» greater than\n \n \n \n m\n \n \n alpha\n \n \n \n «so\n \n \n α\n \n \n has a much greater KE» ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n alpha particle and lead particle have equal and opposite momenta ✓\n

\n

\n so their velocities are inversely proportional to mass ✓\n

\n

\n but KE ∝\n \n v\n \n \n 2\n \n «so\n \n \n α\n \n \n has a much greater KE» ✓\n

\n
\n
\n (b.iii)\n
\n

\n photon energy is determined by its wavelength ✓\n

\n

\n
\n photons are emitted when nucleus undergoes transitions between its «nuclear» energy levels\n

\n

\n \n \n OR\n \n \n

\n

\n photon energy equals the difference between «nuclear» energy levels ✓\n

\n

\n
\n photons have the same energy / a fixed value\n

\n

\n \n \n OR\n \n \n

\n

\n energy is quantized / discrete ✓\n

\n
\n
\n (c)\n
\n

\n undecayed mass\n \n \n =\n \n \n 5\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n e\n \n \n \n -\n \n \n 5\n \n \n .\n \n \n 8\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 8\n \n \n \n \n ×\n \n \n 365\n \n \n ×\n \n \n 24\n \n \n ×\n \n \n 60\n \n \n ×\n \n \n 60\n \n \n \n \n «\n \n \n =\n \n \n 0\n \n \n .\n \n \n 8\n \n \n g» ✓\n

\n

\n mass of decayed polonium «\n \n \n =\n \n \n 5\n \n \n .\n \n \n 0\n \n \n -\n \n \n undecayed mass»\n \n \n =\n \n \n 4\n \n \n .\n \n \n 2\n \n \n «g» ✓\n

\n

\n mass of lead «\n \n \n =\n \n \n \n 206\n \n \n 210\n \n \n \n ×\n \n \n 4\n \n \n .\n \n \n 2\n \n \n »\n \n \n =\n \n \n 4\n \n \n .\n \n \n 1\n \n \n «g» ✓\n

\n

\n

\n

\n \n Allow\n \n [2] max\n \n for answers that ignore mass difference between Pb and Po (4.2 g).\n \n

\n

\n \n Allow calculations in number of particles or moles for\n \n MP1\n \n and\n \n MP2\n \n .\n \n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n and\n \n MP2\n \n .\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.\n

\n
\n
\n (a)\n
\n

\n Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.\n

\n
\n
\n (b.i)\n
\n

\n Generally, well answered. There were quite a few who fell into the trap of multiplying by an unnecessary c\n \n 2\n \n as they were not sure of the significance of the unit of u.\n

\n
\n
\n (b.ii)\n
\n

\n Those who answered using the mass often did not get MP3 whereas those who converted to the number of particles or moles before the first calculation did, although that could be considered an unnecessary complication.\n

\n
\n
\n (b.iii)\n
\n

\n Many identified conservation of momentum and consequently the relative velocities but it was common to miss MP3 for correctly relating this to KE.\n

\n
\n
\n (c)\n
\n

\n Several answers referred incorrectly to electron energy levels. Successful candidates managed to score full marks, although it was also common to miss the relationship between energy and wavelength.\n

\n
\n", "topics": [ "a--space-time-and-motion", "e-nuclear-and-quantum-physics" ], "subtopics": [ "a-2-forces-and-momentum", "e-3-radioactive-decay" ] }, { "question_id": "22N.2.HL.TZ0.7", "Question": "
\n
\n (a.i)\n
\n
\n

\n The intensity of light at point O is\n \n \n \n I\n \n \n 0\n \n \n \n . The distance OP is\n \n \n x\n \n \n .\n

\n

\n Sketch, on the axes, a graph to show the variation of the intensity of light with distance from point O on the screen. Your graph should cover the distance range from 0 to 2\n \n \n x\n \n \n .\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n smooth curve decreasing from\n \n \n \n I\n \n \n 0\n \n \n \n to 0 between 0 and\n \n \n x\n \n \n ✓\n

\n

\n secondary maximum correctly placed\n \n \n AND\n \n \n of intensity less than 0.3\n \n \n \n I\n \n \n 0\n \n \n \n ✓\n

\n

\n \n E.g.\n \n \n

\n
\n
\n (a.i)\n
\n

\n smooth curve decreasing from\n \n \n \n I\n \n \n 0\n \n \n \n to 0 between 0 and\n \n \n x\n \n \n ✓\n

\n

\n secondary maximum correctly placed\n \n \n AND\n \n \n of intensity less than 0.3\n \n \n \n I\n \n \n 0\n \n \n \n ✓\n

\n

\n \n E.g.\n \n \n

\n
\n", "Examiners report": "
\n (a.i)\n
\n

\n Most scored MP1. Many candidates scored full marks but it was common to see a maximum at 2x or a secondary maxima too high.\n

\n
\n
\n (a.i)\n
\n

\n Most scored MP1. Many candidates scored full marks but it was common to see a maximum at 2x or a secondary maxima too high.\n

\n
\n", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-3-wave-phenomena" ] }, { "question_id": "22N.2.HL.TZ0.8", "Question": "
\n
\n (a)\n
\n
\n

\n The diagram shows field lines for an electrostatic field. X and Y are two points on the same field line.\n

\n

\n \n

\n

\n Outline which of the two points has the larger electric potential.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Show that the kinetic energy of the satellite in orbit is about 2 × 10\n \n 10\n \n J.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n Determine the minimum energy required to launch the satellite. Ignore the original kinetic energy of the satellite due to Earth’s rotation.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n potential greater at Y ✓\n

\n

\n

\n

\n «from\n \n \n E\n \n \n =\n \n \n -\n \n \n \n \n Δ\n \n \n \n V\n \n \n e\n \n \n \n \n \n Δ\n \n \n r\n \n \n \n \n » the potential increases in the direction opposite to field strength «so from X to Y»\n

\n

\n \n \n OR\n \n \n

\n

\n opposite to the direction of the field lines, «so from X to Y»\n

\n

\n \n \n OR\n \n \n

\n

\n «from\n \n \n W\n \n \n =\n \n \n q\n \n \n ∆\n \n \n \n V\n \n \n e\n \n \n \n » work done to move a positive charge from X to Y is positive «so the potential increases from X to Y» ✓\n

\n
\n
\n (a)\n
\n

\n potential greater at Y ✓\n

\n

\n

\n

\n «from\n \n \n E\n \n \n =\n \n \n -\n \n \n \n \n Δ\n \n \n \n V\n \n \n e\n \n \n \n \n \n Δ\n \n \n r\n \n \n \n \n » the potential increases in the direction opposite to field strength «so from X to Y»\n

\n

\n \n \n OR\n \n \n

\n

\n opposite to the direction of the field lines, «so from X to Y»\n

\n

\n \n \n OR\n \n \n

\n

\n «from\n \n \n W\n \n \n =\n \n \n q\n \n \n ∆\n \n \n \n V\n \n \n e\n \n \n \n » work done to move a positive charge from X to Y is positive «so the potential increases from X to Y» ✓\n

\n
\n
\n (b.i)\n
\n

\n orbital radius\n \n \n =\n \n \n 6\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n +\n \n \n 5\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n «\n \n \n =\n \n \n 6\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n m» ✓\n

\n

\n \n \n KE\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n ×\n \n \n 8\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n 2\n \n \n \n ×\n \n \n \n \n 6\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n ×\n \n \n 6\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n 24\n \n \n \n \n \n 6\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n \n \n \n OR\n \n \n \n \n 2\n \n \n .\n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n 10\n \n \n \n «J» ✓\n

\n

\n

\n

\n \n Award\n \n [1] max\n \n for answers ignoring orbital height (KE = 2.5 × 10\n \n \n \n 10\n \n \n J\n \n ).\n \n

\n
\n
\n (b.ii)\n
\n

\n change in PE\n \n \n =\n \n \n 6\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n ×\n \n \n 6\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n 24\n \n \n \n ×\n \n \n 8\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n 2\n \n \n \n \n \n \n 1\n \n \n \n 6\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n \n -\n \n \n \n 1\n \n \n \n 6\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n \n \n \n =\n \n \n «\n \n \n 3\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n J» ✓\n

\n

\n energy needed = KE + ΔPE =\n \n \n 2\n \n \n .\n \n \n 7\n \n \n ×\n \n \n \n 10\n \n \n 10\n \n \n \n «J» ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from 8(b)(i).\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n A significant majority guessed at X, probably because the field lines are closer together. Those that identified Y were generally successful in their explanation.\n

\n
\n
\n (a)\n
\n

\n A significant majority guessed at X, probably because the field lines are closer together. Those that identified Y were generally successful in their explanation.\n

\n
\n
\n (b.i)\n
\n

\n This question was well done, with only a few missing the height of the satellite.\n

\n
\n
\n (b.ii)\n
\n

\n Generally, this question was not well done. Most carried out a calculation based on the formula for escape velocity. An opportunity to remind candidates of reading back the stem for the sub-question when answering a second or any subsequent part of it.\n

\n
\n", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "22N.2.SL.TZ0.1", "Question": "
\n
\n (a)\n
\n
\n

\n State the initial acceleration of the raindrop.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Explain, by reference to the vertical forces, how the raindrop reaches a constant speed.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c.i)\n
\n
\n

\n Determine the energy transferred to the air during the first 3.0 s of motion. State your answer to an appropriate number of significant figures.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c.ii)\n
\n
\n

\n Describe the energy change that takes place for\n \n t\n \n > 3.0 s.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n g\n \n \n \n OR\n \n \n 9.81 «m s\n \n −2\n \n »\n \n \n OR\n \n \n acceleration of gravity/due to free fall ✓\n

\n

\n

\n

\n \n Accept 10 «\n \n m s\n \n −2\n \n \n ».\n \n

\n

\n \n Ignore sign.\n \n

\n

\n \n Do\n \n not\n \n accept bald “gravity”.\n \n

\n

\n \n Accept answer that indicates tangent of the graph at time t=0.\n \n

\n
\n
\n (a)\n
\n

\n \n g\n \n \n \n OR\n \n \n 9.81 «m s\n \n −2\n \n »\n \n \n OR\n \n \n acceleration of gravity/due to free fall ✓\n

\n

\n

\n

\n \n Accept 10 «\n \n m s\n \n −2\n \n \n ».\n \n

\n

\n \n Ignore sign.\n \n

\n

\n \n Do\n \n not\n \n accept bald “gravity”.\n \n

\n

\n \n Accept answer that indicates tangent of the graph at time t=0.\n \n

\n
\n
\n (b)\n
\n

\n Identification of air resistance/drag force «acting upwards» ✓\n

\n

\n «that» increases with speed ✓\n

\n

\n
\n «until» weight and air resistance cancel out\n

\n

\n \n \n OR\n \n \n

\n

\n net force/acceleration becomes zero ✓\n

\n

\n

\n

\n \n A statement as “air resistance increases with speed” scores\n \n MP1\n \n and\n \n MP2\n \n .\n \n

\n
\n
\n (c.i)\n
\n

\n «loss in» GPE = 3.4 × 10\n \n −5\n \n × 9.81 × 21 «= 7.0 × 10\n \n −3\n \n » «J»\n

\n

\n \n \n OR\n \n \n

\n

\n «gain in» KE = 0.5 × 3.4 × 10\n \n −5\n \n × 9.0\n \n 2\n \n «= 1.4 × 10\n \n −3\n \n » «J» ✓\n

\n

\n
\n energy transferred to air «=7.0 × 10\n \n −3\n \n − 1.4 × 10\n \n −3\n \n » = 5.6 × 10\n \n −3\n \n » «J» ✓\n

\n

\n

\n

\n any calculated answer to 2 sf ✓\n

\n

\n

\n

\n \n Allow\n \n [1]\n \n through the use of kinematics assuming constant acceleration.\n \n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n .\n \n

\n
\n
\n (c.ii)\n
\n

\n «gravitational» potential energy «of the raindrop» into thermal/internal energy «of the air» ✓\n

\n

\n

\n

\n \n Accept heat for thermal energy.\n \n

\n

\n \n Accept into kinetic energy of air particles.\n \n

\n

\n \n Ignore sound energy.\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n A nice introductory question answered correctly by most candidates. Most answers quoted the data booklet value, with a few 10's or 9.8's, or the answer in words. Very few lost the mark by just stating gravity, or zero.\n

\n
\n
\n (a)\n
\n

\n A nice introductory question answered correctly by most candidates. Most answers quoted the data booklet value, with a few 10's or 9.8's, or the answer in words. Very few lost the mark by just stating gravity, or zero.\n

\n
\n
\n (b)\n
\n

\n This was very well answered with most candidates scoring 3. The MP usually missed in candidates scoring 2 marks was MP2, to justify the variation of the magnitude of air resistance, although that rarely happened.\n

\n
\n
\n (c.i)\n
\n

\n Generally well answered, although several candidates lost a mark, usually as POT (power of ten) by quoting the value in kg leading to an answer of 5.3 J. Most candidates were able to score MP3 by rounding their calculation to two significant figures.\n

\n
\n
\n (c.ii)\n
\n

\n Of the wrong answers, the most common ones were gravitational potential to kinetic or the idea that because there was no change in velocity there was no energy transfer. A significant number, though, scored by identifying the change into thermal (most of them), kinetic of air particles (a few answers) or internal (very few).\n

\n
\n", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power" ] }, { "question_id": "22N.2.SL.TZ0.2", "Question": "
\n
\n (a.i)\n
\n
\n

\n Determine the minimum area of the solar heating panel required to increase the temperature of all the water in the tank to 30°C during a time of 1.0 hour.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (a.ii)\n
\n
\n

\n Estimate, in °C, the temperature of the roof tiles.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n State\n \n one\n \n way in which a real gas differs from an ideal gas.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n The water is heated. Explain why the quantity of air in the storage tank decreases.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n energy required = 250 × 4200 × (30 − 15) ✓\n

\n

\n energy available = 0.30 × 680 ×\n \n t\n \n ×\n \n A\n \n ✓\n

\n

\n \n A\n \n = «\n \n \n \n \n 250\n \n \n ×\n \n \n 4200\n \n \n ×\n \n \n 15\n \n \n \n \n 0\n \n \n .\n \n \n 30\n \n \n ×\n \n \n 680\n \n \n ×\n \n \n 60\n \n \n ×\n \n \n 60\n \n \n \n \n =\n \n \n » 21 «m\n \n 2\n \n »\n \n \n OR\n \n \n 22 «m\n \n 2\n \n » ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n and\n \n MP2\n \n .\n \n

\n

\n \n Accept the correct use of 0.30 in either\n \n MP1\n \n or\n \n MP2\n \n .\n \n

\n
\n
\n (a.ii)\n
\n

\n absorbed intensity = (1 − 0.2) × 680 «= 544» «W m\n \n −2\n \n »\n

\n

\n \n \n OR\n \n \n

\n

\n emitted intensity = 0.97 × 5.67 × 10\n \n −8\n \n × T\n \n 4\n \n ✓\n

\n

\n

\n

\n \n T\n \n \n \n =\n \n \n \n \n 544\n \n \n \n 0\n \n \n .\n \n \n 97\n \n \n ×\n \n \n 5\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 8\n \n \n \n \n \n \n 4\n \n \n \n =\n \n \n 315\n \n \n «K» ✓\n

\n

\n 42 «°C» ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n and\n \n MP2\n \n .\n \n

\n

\n \n Allow\n \n MP1\n \n if absorbed or emitted intensity is multiplied by area.\n \n

\n
\n
\n (b.i)\n
\n

\n can be liquefied ✓\n

\n

\n has intermolecular forces / potential energy ✓\n

\n

\n has atoms/molecules that are not point objects / take up volume ✓\n

\n

\n does not follow the ideal gas law «for all\n \n T\n \n and\n \n p\n \n » ✓\n

\n

\n collisions between particles are non-elastic ✓\n

\n

\n

\n

\n \n Accept the converse argument.\n \n

\n
\n
\n (b.ii)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n «constant\n \n p\n \n and\n \n V\n \n imply»\n \n nT\n \n = const ✓\n

\n

\n \n T\n \n increases hence\n \n n\n \n decreases ✓\n

\n

\n
\n \n \n ALTERNATIVE 2\n \n \n

\n

\n «constant\n \n p\n \n and\n \n n\n \n imply»\n \n V\n \n is proportional to\n \n T\n \n / air expands as it is heated ✓\n

\n

\n «original» air occupies a greater volume\n \n OR\n \n some air leaves through opening ✓\n

\n

\n

\n

\n \n \n MP2\n \n in\n \n ALT 2\n \n must come from expansion of air, not from expansion of water.\n \n

\n

\n \n Award\n \n [0]\n \n for an answer based on expansion of water.\n \n

\n

\n \n Award\n \n [1]\n \n \n max\n \n for an answer based on convection currents.\n \n

\n
\n", "Examiners report": "
\n (a.i)\n
\n

\n Most candidates had a good attempt at this but there were often slight slips. Some missed the efficiency of the process. Some included the albedo of the roof tiles. Some thought that the temperature rise needed to have 273 added to convert to kelvin. However, sometimes scoring through ECF (error carried forward), the average mark was around 2 marks.\n

\n
\n
\n (a.ii)\n
\n

\n This was a bit more hit and miss than the previous question part. One common mistake was not understanding what albedo meant. Some took it as the amount of energy absorbed rather than reflected. Emissivity was often missed. Several candidates, successfully answering the question or not, were able to score MP3 converting the final temperature into Celsius degrees.\n

\n
\n
\n (b.i)\n
\n

\n This was very well answered. Candidates showed an understanding of the differences between ideal and real gases.\n

\n
\n
\n (b.ii)\n
\n

\n It was surprising to see a large number of answers based on the expansion of water, as the stem of the question clearly states that the level of water remains constant. Most successful candidates scored by quoting\n \n pV\n \n constant so concluding with the inverse relationship of n and\n \n T\n \n , others also managed to score by explaining that the volume of air increases and therefore must go out through the opening. Answers based on convection currents were given partial credit.\n

\n
\n", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers", "b-2-greenhouse-effect", "b-3-gas-laws" ] }, { "question_id": "22N.2.SL.TZ0.3", "Question": "
\n
\n (a.i)\n
\n
\n

\n Draw, on the axes, a graph to show the variation with\n \n t\n \n of the displacement of particle Q.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (a.ii)\n
\n
\n

\n Calculate the speed of waves on the string.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Determine the fundamental SI unit for\n \n a\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n The tension force on the string is doubled. Describe the effect, if any, of this change on the frequency of the standing wave.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The standing wave on the string creates a travelling sound wave in the surrounding air.\n

\n

\n Outline\n \n two\n \n differences between a standing wave and a travelling wave.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c.i)\n
\n
\n

\n Outline\n \n one\n \n difference between a standing wave and a travelling wave.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c.ii)\n
\n
\n

\n The speed of sound in air is 340 m s\n \n −1\n \n and in water it is 1500 m s\n \n −1\n \n .\n

\n

\n Discuss whether the sound wave can enter the water.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n oscillation in antiphase ✓\n

\n

\n smaller amplitude than P ✓\n

\n

\n
\n \n

\n
\n
\n (a.i)\n
\n

\n oscillation in antiphase ✓\n

\n

\n smaller amplitude than P ✓\n

\n

\n
\n \n

\n
\n
\n (a.ii)\n
\n

\n wavelength\n \n \n =\n \n \n \n 2\n \n \n 3\n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 80\n \n \n =\n \n \n 0\n \n \n .\n \n \n 53\n \n \n «m» ✓\n

\n

\n speed\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 53\n \n \n \n \n 2\n \n \n .\n \n \n 8\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n =\n \n \n 190\n \n \n «m s\n \n −1\n \n » ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from incorrect wavelength.\n \n

\n
\n
\n (b.i)\n
\n

\n kg m s\n \n −2\n \n \n \n OR\n \n \n m\n \n 2\n \n s\n \n −2\n \n seen ✓\n

\n

\n kg m\n \n −1\n \n ✓\n

\n

\n

\n

\n \n Award\n \n [2]\n \n for a\n \n BCA\n \n .\n \n

\n
\n
\n (b.ii)\n
\n

\n speed increases hence frequency increases ✓\n

\n

\n by factor\n \n \n \n 2\n \n \n \n ✓\n

\n
\n
\n (c)\n
\n

\n travelling waves transfer energy\n \n \n OR\n \n \n standing waves don’t ✓\n

\n

\n amplitude of oscillation varies along a standing wave\n \n \n OR\n \n \n is constant along a travelling wave ✓\n

\n

\n standing waves have nodes and antinodes\n \n \n OR\n \n \n travelling waves don’t ✓\n

\n

\n points in an internodal region have same phase in standing waves\n \n \n OR\n \n \n different phase in travelling waves ✓\n

\n
\n
\n (c.i)\n
\n

\n travelling waves transfer energy\n \n \n OR\n \n \n standing waves don’t ✓\n

\n

\n amplitude of oscillation varies along a standing wave\n \n \n OR\n \n \n is constant along a travelling wave ✓\n

\n

\n standing waves have nodes / antinodes\n \n \n OR\n \n \n travelling waves don’t ✓\n

\n

\n points in an internodal region have same phase in standing waves\n \n \n OR\n \n \n different phase in travelling waves ✓\n

\n
\n
\n (c.ii)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n critical angle\n \n \n =\n \n \n 13\n \n \n °\n \n \n «from\n \n \n sin\n \n \n \n \n \n θ\n \n \n c\n \n \n \n =\n \n \n \n 340\n \n \n 1500\n \n \n \n » ✓\n

\n

\n the angle of incidence is greater than\n \n \n \n θ\n \n \n c\n \n \n \n hence the sound can’t enter water ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n \n sin\n \n \n \n \n \n θ\n \n \n r\n \n \n \n =\n \n \n \n 1500\n \n \n 340\n \n \n \n sin\n \n \n \n \n 30\n \n \n °\n \n \n ✓\n

\n

\n sine value greater than one hence the sound can’t enter water ✓\n

\n

\n

\n

\n \n Conclusion must be justified, award\n \n [0]\n \n for\n \n BCA\n \n .\n \n

\n
\n", "Examiners report": "
\n (a.i)\n
\n

\n Although there were good answers which scored full marks, there were a significant number of wrong answers where the amplitude was the same or not consistent throughout, or the wave drawn was not in antiphase of the original sketch.\n

\n
\n
\n (a.i)\n
\n

\n Although there were good answers which scored full marks, there were a significant number of wrong answers where the amplitude was the same or not consistent throughout, or the wave drawn was not in antiphase of the original sketch.\n

\n
\n
\n (a.ii)\n
\n

\n This was well answered, particularly MP1 to determine the wavelength, although several candidates misinterpreted the unit of time and obtained a very small value for the velocity of the wave.\n

\n
\n
\n (b.i)\n
\n

\n Students seem to be well prepared for this sort of question, as it was high-scoring.\n

\n
\n
\n (b.ii)\n
\n

\n This question was answered well, although the numerical aspect was often missing. It is worth highlighting that if there is a term like\n \n '\n \n doubled\n \n '\n \n in the question, it makes sense to expect a numerical answer.\n

\n
\n
\n (c)\n
\n

\n This question was answered well. Students showed to be familiar with the differences between standing and travelling waves. In SL they had to identify two differences, so that proved to be more challenging.\n

\n
\n
\n (c.i)\n
\n

\n This question was answered well. Students showed to be familiar with the differences between standing and travelling waves.\n

\n
\n
\n (c.ii)\n
\n

\n Surprisingly well answered as it was sound from air to water, rather than light from air to glass. A mixture of approaches but probably the most common was to calculate a sine value of over 1. Some went about calculating the critical angle but nowhere near as many.\n

\n
\n", "topics": [ "c-wave-behaviour", "tools" ], "subtopics": [ "c-2-wave-model", "c-3-wave-phenomena", "c-4-standing-waves-and-resonance", "tool-3-mathematics" ] }, { "question_id": "22N.2.SL.TZ0.4", "Question": "
\n
\n (a)\n
\n
\n

\n The scale diagram shows the weight\n \n W\n \n of the mass at an instant when the rod is horizontal.\n

\n

\n Draw, on the scale diagram, an arrow to represent the force exerted on the mass by the rod.\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Explain why the magnitude of the force exerted on the mass by the rod is not constant.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n horizontal component of any length to the left ✓\n

\n

\n vertical component two squares long upwards ✓\n

\n

\n E.g.\n

\n

\n \n

\n

\n

\n

\n \n Ignore point of application.\n \n

\n

\n \n Award\n \n [1]\n \n \n max\n \n if arrowhead not present.\n \n

\n
\n
\n (a)\n
\n

\n horizontal component of any length to the left ✓\n

\n

\n vertical component two squares long upwards ✓\n

\n

\n E.g.\n

\n

\n \n

\n

\n

\n

\n \n Ignore point of application.\n \n

\n

\n \n Award\n \n [1]\n \n \n max\n \n if arrowhead not present.\n \n

\n
\n
\n (b)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n the net/centripetal force has constant magnitude ✓\n

\n

\n the direction of the net/centripetal force constantly changes ✓\n

\n

\n this is achieved by vector-adding weight and the force from the rod\n

\n

\n \n \n OR\n \n \n

\n

\n the force from the rod is vector difference of the centripetal force and weight ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n at the top F\n \n rod\n \n = F\n \n c\n \n − W ✓\n

\n

\n at the bottom, F\n \n rod\n \n = F\n \n c\n \n + W ✓\n

\n

\n net F/F\n \n c\n \n is constant so the force from the rod is different «hence is changing» ✓\n

\n

\n

\n

\n \n Accept reference to centripetal or net force indistinctly.\n \n

\n

\n \n Allow reference to centripetal acceleration.\n \n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n Most just added the horizontal component. Not many centrifugal forces, but still a few. Very few were able to score both marks, so this question proved to be challenging for the candidates.\n

\n
\n
\n (a)\n
\n

\n Most just added the horizontal component. Not many centrifugal forces, but still a few. Very few were able to score both marks, so this question proved to be challenging for the candidates.\n

\n
\n
\n (b)\n
\n

\n Many got into a bit of a mess with this one and it was quite difficult to interpret some of the answers. If they started out with the net/centripetal force being constant, then it was often easy to follow the reasoning. Starting with force on the rod varying often led to confusion. Quite a few did not pick up on the constant speed vertical circle so there were complicated energy/speed arguments to pick through.\n

\n
\n", "topics": [ "a--space-time-and-motion", "tools" ], "subtopics": [ "a-2-forces-and-momentum", "tool-3-mathematics" ] }, { "question_id": "22N.2.SL.TZ0.5", "Question": "
\n
\n (a)\n
\n
\n

\n State what is meant by an ideal voltmeter.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Show that the internal resistance of the cell is about 0.7 Ω.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n Show that the internal resistance of the cell is about 0.7 Ω.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.iii)\n
\n
\n

\n Calculate the emf of the cell.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c.i)\n
\n
\n

\n Explain, by reference to charge carriers in the wire, how the magnetic force on the wire arises.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c.ii)\n
\n
\n

\n Identify the direction of the magnetic force on the wire.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n infinite resistance\n

\n

\n \n \n OR\n \n \n

\n

\n no current is flowing through it ✓\n

\n
\n
\n (b.i)\n
\n

\n current\n \n \n =\n \n \n \n \n 1\n \n \n .\n \n \n 47\n \n \n \n \n 50\n \n \n .\n \n \n 0\n \n \n \n \n =\n \n \n 2\n \n \n .\n \n \n 94\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n «A» ✓\n

\n

\n \n r\n \n \n \n =\n \n \n \n \n 1\n \n \n .\n \n \n 49\n \n \n \n \n 2\n \n \n .\n \n \n 94\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n -\n \n \n 50\n \n \n .\n \n \n 0\n \n \n \n \n OR\n \n \n \n \n 0\n \n \n .\n \n \n 68\n \n \n «Ω» ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , allow any other correctly substituted expression for r.\n \n

\n
\n
\n (b.ii)\n
\n

\n 29.4 (50.0 +\n \n r\n \n ) = 139 (10.0 +\n \n r\n \n ) ✓\n

\n

\n
\n attempt to solve for\n \n r\n \n , e.g. 29.4 × 50.0 − 139 × 10.0 =\n \n r\n \n (139 − 29.4)\n

\n

\n \n \n OR\n \n \n

\n

\n 0.73 «Ω» ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n allow working backwards from 0.7 Ω.\n \n

\n
\n
\n (b.iii)\n
\n

\n 139 × 10\n \n −3\n \n (10.0 + 0.73)\n

\n

\n \n \n OR\n \n \n

\n

\n 29.4 × 10\n \n −3\n \n (50.0 + 0.73)  ✓\n

\n

\n
\n 1.49 «V» ✓\n

\n

\n

\n

\n \n Watch for\n \n ECF\n \n from 5(b)(i).\n \n

\n
\n
\n (c.i)\n
\n

\n charge/carriers are moving in a magnetic field ✓\n

\n

\n

\n

\n there is a magnetic force on them / quote\n \n F = qvB\n \n

\n

\n \n \n OR\n \n \n

\n

\n this creates a magnetic field that interacts with the external magnetic field ✓\n

\n

\n

\n

\n \n Accept electrons.\n \n

\n

\n \n For\n \n MP2\n \n , the force must be identified as acting on charge / carriers.\n \n

\n
\n
\n (c.ii)\n
\n

\n into the plane «of the paper» ✓\n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n A majority of candidates scored a mark by simply stating infinite resistance. Several answers went the other way round, stating a resistance of zero.\n

\n
\n
\n (b.i)\n
\n

\n Many answers here produced a number that did not round to 0.7 but students claimed it did. The simultaneous equation approach was seen in the best candidates, getting the right answer. It is worthy of reminding about the need of showing one more decimal place when calculating a show that value type of question.\n

\n
\n
\n (b.ii)\n
\n

\n Many answers here produced a number that did not round to 0.7 but students claimed it did. The simultaneous equation approach was seen in the best candidates, getting the right answer. It is worthy of reminding about the need of showing one more decimal place when calculating a show that value type of question.\n

\n
\n
\n (b.iii)\n
\n

\n Usually well answered, regardless of b(ii), by utilising the show that value given.\n

\n
\n
\n (c.i)\n
\n

\n Many scored MP1 here but did not get MP2 as they jumped straight to the wire rather than continuing with the explanation of what was going on with the charge carriers.\n

\n
\n
\n (c.ii)\n
\n

\n Generally, a well answered question although there was some confusion on how to communicate it, with some contradictory answers indicating into or out, and also North or South at the same time.\n

\n
\n", "topics": [ "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "b-5-current-and-circuits", "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "22N.2.SL.TZ0.6", "Question": "
\n
\n (a)\n
\n
\n

\n Outline, by reference to nuclear binding energy, why the mass of a nucleus is less than the sum of the masses of its constituent nucleons.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.i)\n
\n
\n

\n Calculate, in MeV, the energy released in this decay.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b.ii)\n
\n
\n

\n The polonium nucleus was stationary before the decay.\n

\n

\n Show, by reference to the momentum of the particles, that the kinetic energy of the alpha particle is much greater than the kinetic energy of the lead nucleus.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b.iii)\n
\n
\n

\n In the decay of polonium-210, alpha emission can be followed by the emission of a gamma photon.\n

\n

\n State and explain whether the alpha particle or gamma photon will cause greater ionization in the surrounding material.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n according to Δ\n \n E\n \n = Δ\n \n mc\n \n \n 2\n \n / identifies mass energy equivalence ✓\n

\n

\n

\n

\n energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»\n

\n

\n \n \n OR\n \n \n

\n

\n energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓\n

\n

\n

\n

\n \n Accept protons and neutrons.\n \n

\n
\n
\n (a)\n
\n

\n according to Δ\n \n E\n \n = Δ\n \n mc\n \n \n 2\n \n / identifies mass energy equivalence ✓\n

\n

\n

\n

\n energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»\n

\n

\n \n \n OR\n \n \n

\n

\n energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓\n

\n

\n

\n

\n \n Accept protons and neutrons.\n \n

\n
\n
\n (b.i)\n
\n

\n (\n \n m\n \n \n polonium\n \n −\n \n m\n \n \n lead\n \n −\n \n m\n \n α\n \n \n )\n \n c\n \n \n 2\n \n \n \n OR\n \n \n (209.93676 − 205.92945 − 4.00151)\n

\n

\n \n \n OR\n \n \n

\n

\n mass difference = 5.8 × 10\n \n −3\n \n ✓\n

\n

\n

\n

\n conversion to MeV using 931.5 to give 5.4 «MeV» ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n .\n \n

\n

\n \n Award\n \n [2]\n \n for a\n \n BCA\n \n .\n \n

\n

\n \n Award\n \n [1]\n \n for 8.6 x 10\n \n −13\n \n J.\n \n

\n
\n
\n (b.ii)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n energy ratio expressed in terms of momentum, e.g.\n \n \n \n \n E\n \n \n α\n \n \n \n \n E\n \n \n lead\n \n \n \n \n =\n \n \n \n \n \n p\n \n \n α\n \n \n 2\n \n \n \n /\n \n \n 2\n \n \n \n m\n \n \n α\n \n \n \n \n \n \n p\n \n \n lead\n \n \n 2\n \n \n \n /\n \n \n 2\n \n \n \n m\n \n \n lead\n \n \n \n \n \n ✓\n

\n

\n \n \n \n p\n \n \n α\n \n \n \n =\n \n \n \n p\n \n \n lead\n \n \n \n hence\n \n \n \n \n E\n \n \n α\n \n \n \n \n E\n \n \n lead\n \n \n \n \n =\n \n \n \n \n m\n \n \n lead\n \n \n \n \n m\n \n \n α\n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n \n \n m\n \n \n lead\n \n \n \n \n m\n \n \n α\n \n \n \n \n ≃\n \n \n \n 206\n \n \n 4\n \n \n \n =\n \n \n 51\n \n \n .\n \n \n 5\n \n \n ⇒\n \n \n \n E\n \n \n α\n \n \n \n =\n \n \n 51\n \n \n .\n \n \n 5\n \n \n \n E\n \n \n lead\n \n \n \n «so\n \n \n α\n \n \n has a much greater KE»\n

\n

\n \n \n OR\n \n \n

\n

\n \n \n \n m\n \n \n lead\n \n \n \n «much» greater than\n \n \n \n m\n \n \n alpha\n \n \n \n «so\n \n \n α\n \n \n has a much greater KE» ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n alpha particle and lead particle have equal and opposite momenta ✓\n

\n

\n so their velocities are inversely proportional to mass ✓\n

\n

\n but KE ∝\n \n v\n \n \n 2\n \n «so\n \n \n α\n \n \n has a much greater KE» ✓\n

\n
\n
\n (b.iii)\n
\n

\n alpha particle ✓\n

\n

\n is electrically charged hence more likely to interact with electrons «in the surrounding material» ✓\n

\n
\n", "Examiners report": "
\n (a)\n
\n

\n Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.\n

\n
\n
\n (a)\n
\n

\n Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.\n

\n
\n
\n (b.i)\n
\n

\n Generally, well answered. There were quite a few who fell into the trap of multiplying by an unnecessary c\n \n 2\n \n as they were not sure of the significance of the unit of u.\n

\n
\n
\n (b.ii)\n
\n

\n Those who answered using the mass often did not get MP3 whereas those who converted to the number of particles or moles before the first calculation did, although that could be considered an unnecessary complication.\n

\n
\n
\n (b.iii)\n
\n

\n Many did not interpret the stem correctly and failed to compare the ionisation of alpha particles versus gamma rays.\n

\n
\n", "topics": [ "a--space-time-and-motion", "e-nuclear-and-quantum-physics" ], "subtopics": [ "a-2-forces-and-momentum", "a-3-work-energy-and-power", "e-3-radioactive-decay" ] }, { "question_id": "23M.2.HL.TZ1.3", "Question": "
\n
\n (a)\n
\n
\n

\n A transverse water wave travels to the right. The diagram shows the shape of the surface of the water at time\n \n t\n \n = 0. P and Q show two corks floating on the surface.\n

\n

\n \n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State what is meant by a transverse wave.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The frequency of the wave is 0.50 Hz. Calculate the speed of the wave.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Sketch on the diagram the position of P at time\n \n t\n \n = 0.50 s.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Show that the phase difference between the oscillations of the two corks is\n \n \n π\n \n \n radians.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Monochromatic light is incident on two very narrow slits. The light that passes through the slits is observed on a screen. M is directly opposite the midpoint of the slits.\n \n \n x\n \n \n represents the displacement from M in the direction shown.\n

\n

\n \n

\n

\n A student argues that what will be observed on the screen will be a total of two bright spots opposite the slits. Explain why the student’s argument is incorrect.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The graph shows the actual variation with displacement\n \n \n x\n \n \n from M of the intensity of the light on the screen.\n \n \n \n I\n \n \n 0\n \n \n \n is the intensity of light at the screen from one slit only.\n

\n

\n \n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Explain why the intensity of light at\n \n \n x\n \n \n = 0 is 4\n \n \n \n I\n \n \n 0\n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The slits are separated by a distance of 0.18 mm and the distance to the screen is 2.2 m. Determine, in m, the wavelength of light.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n The two slits are replaced by many slits of the same separation. State\n \n one\n \n feature of the intensity pattern that will remain the same and\n \n one\n \n that will change.\n

\n

\n Stays the same:\n

\n

\n

\n

\n Changes:\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n «A wave where the» displacement of particles/oscillations of particles/movement of particles/vibrations of particles is perpendicular/normal to the direction of energy transfer/wave travel/wave velocity/wave movement/wave propagation ✓\n

\n

\n \n Allow medium, material, water, molecules, or atoms for particles.\n \n

\n
\n
\n (i)\n
\n

\n «A wave where the» displacement of particles/oscillations of particles/movement of particles/vibrations of particles is perpendicular/normal to the direction of energy transfer/wave travel/wave velocity/wave movement/wave propagation ✓\n

\n

\n \n Allow medium, material, water, molecules, or atoms for particles.\n \n

\n
\n
\n (ii)\n
\n

\n \n v\n \n = «0.50 × 16 =» 8.0 «m s\n \n −1\n \n » ✓\n

\n
\n
\n (iii)\n
\n

\n P at (8, 1.2) ✓\n

\n
\n
\n (iv)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n
\n Phase difference is\n \n \n \n \n 2\n \n \n π\n \n \n \n λ\n \n \n \n ×\n \n \n \n λ\n \n \n 2\n \n \n \n ✓\n
\n «=\n \n \n π\n \n \n » ✓\n

\n

\n \n ALTERNATIVE 2\n \n

\n

\n One wavelength/period represents «phase difference» of 2\n \n \n π\n \n \n and «corks» are ½ wavelength/period apart so phase difference is\n \n \n π\n \n \n /\n \n \n OWTTE\n \n \n ✓\n

\n
\n
\n (b)\n
\n

\n light acts as a wave «and not a particle in this situation» ✓\n

\n

\n light at slits will diffract / create a diffraction pattern ✓\n

\n

\n light passing through slits will interfere / create an interference pattern «creating bright and dark spots» ✓\n

\n
\n
\n (c)\n
\n

\n The amplitude «at\n \n x\n \n = 0» will be doubled ✓\n

\n

\n intensity is proportional to amplitude squared /\n \n \n I\n \n \n ∝\n \n A\n \n \n 2\n \n ✓\n

\n
\n
\n (i)\n
\n

\n The amplitude «at\n \n x\n \n = 0» will be doubled ✓\n

\n

\n intensity is proportional to amplitude squared /\n \n \n I\n \n \n ∝\n \n A\n \n \n 2\n \n ✓\n

\n
\n
\n (ii)\n
\n

\n Use of\n \n s\n \n =\n \n \n \n \n λ\n \n \n D\n \n \n \n d\n \n \n \n ⇒\n \n \n λ\n \n \n =\n \n \n \n \n s\n \n \n d\n \n \n \n D\n \n \n \n \n \n OR\n \n \n \n s\n \n =\n \n \n \n \n n\n \n \n λ\n \n \n D\n \n \n \n d\n \n \n \n ⇒\n \n \n λ\n \n \n =\n \n \n \n \n s\n \n \n d\n \n \n \n \n n\n \n \n D\n \n \n \n \n ✓\n

\n

\n \n \n λ\n \n \n = «\n \n \n \n \n 0\n \n \n .\n \n \n 567\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 18\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n 2\n \n \n .\n \n \n 2\n \n \n \n \n =» 4.6 × 10\n \n −7\n \n «m» ✓\n

\n
\n
\n (iii)\n
\n

\n Stays the same: Position/location of maxima/distance/separation between maxima «will be the same» /\n \n \n OWTTE\n \n \n ✓\n

\n

\n Changes: Intensity/brightness/width/sharpness «of maxima will change»/\n \n \n OWTTE\n \n \n ✓\n

\n

\n \n Allow other phrasing for maxima (fringes, spots, etc).\n \n

\n
\n", "Examiners report": "None", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-3-wave-phenomena" ] }, { "question_id": "23M.2.HL.TZ1.5", "Question": "
\n
\n (a)\n
\n
\n

\n Identify with ticks [✓] in the table, the forces that can act on electrons and the forces that can act on quarks.\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Show that the energy released is about 18 MeV.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n \n two\n \n difficulties of energy production by nuclear fusion.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n \n one\n \n advantage of energy production by nuclear fusion compared to nuclear fission.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the nucleon number of the He isotope that\n \n \n \n H\n \n \n \n \n 1\n \n \n 3\n \n \n \n decays into.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Estimate the fraction of tritium remaining after one year.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Weak nuclear: 2 ticks ✓\n
\n Strong nuclear: quarks only ✓\n

\n
\n
\n (i)\n
\n

\n «𝜇» = 2.0141 + 3.0160 − (4.0026 + 1.008665) «= 0.0188 u»\n

\n

\n \n \n OR\n \n \n

\n

\n \n In\n \n MeV: 1876.13415 + 2809.404 − (3728.4219 + 939.5714475) ✓\n

\n

\n = 0.0188 × 931.5\n \n \n OR\n \n \n = 17.512 «MeV» ✓\n

\n

\n \n Must see either clear substitutions or answer to at least 3 s.f. for\n \n MP2\n \n .\n \n

\n
\n
\n (i)\n
\n

\n Requires high temp/pressure ✓\n
\n Must overcome Coulomb/intermolecular repulsion ✓\n
\n Difficult to contain / control «at high temp/pressure» ✓\n
\n Difficult to produce excess energy/often energy input greater than output /\n \n \n OWTTE\n \n \n ✓\n
\n Difficult to capture energy from fusion reactions ✓\n
\n Difficult to maintain/sustain a constant reaction rate ✓\n

\n
\n
\n (ii)\n
\n

\n Plentiful fuel supplies\n \n \n OR\n \n \n larger specific energy\n \n \n OR\n \n \n larger energy density\n \n \n OR\n \n \n little or no «major radioactive» waste products ✓\n

\n

\n \n Allow descriptions such as “more energy per unit mass” or “more energy per unit volume”\n \n

\n
\n
\n (i)\n
\n

\n 3 ✓\n

\n

\n \n Do\n \n not\n \n accept\n \n \n \n H\n \n \n \n \n 2\n \n \n 3\n \n \n \n e\n \n \n by itself.\n \n

\n
\n
\n (iii)\n
\n

\n \n \n λ\n \n \n = «\n \n \n \n \n ln\n \n \n \n \n 2\n \n \n \n \n 12\n \n \n .\n \n \n 3\n \n \n \n \n »0.056«y\n \n −1\n \n »\n \n \n OR\n \n \n \n \n 0\n \n \n .\n \n \n \n 5\n \n \n \n \n 1\n \n \n \n 12\n \n \n .\n \n \n 3\n \n \n \n \n \n \n \n \n OR\n \n \n \n \n \n e\n \n \n \n -\n \n \n 1\n \n \n ×\n \n \n \n \n ln\n \n \n \n \n 2\n \n \n \n \n 12\n \n \n .\n \n \n 3\n \n \n \n \n \n \n

\n

\n 0.945\n \n \n OR\n \n \n 94.5% ✓\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n \n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-3-radioactive-decay", "e-4-fission", "e-5-fusion-and-stars" ] }, { "question_id": "23M.2.HL.TZ1.6", "Question": "
\n
\n (a)\n
\n
\n

\n The centres of two identical fixed conducting spheres each of charge +\n \n Q\n \n are separated by a distance\n \n D\n \n . C is the midpoint of the line joining the centres of the spheres.\n

\n

\n \n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Sketch, on the axes, how the electric potential V due to the two charges varies with the distance r from the centre of the left charge. No numbers are required. Your graph should extend from r = 0 to r = D.\n

\n

\n \n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the work done to bring a small charge\n \n q\n \n from infinity to point C.\n

\n

\n Data given:\n

\n

\n \n Q\n \n = 2.0 × 10\n \n −3\n \n C,\n

\n

\n \n q\n \n = 4.0 × 10\n \n −9\n \n C\n

\n

\n \n D\n \n = 1.2 m\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n The magnitude of the net force on\n \n q\n \n is given by\n \n \n \n \n 32\n \n \n k\n \n \n Q\n \n \n q\n \n \n \n \n D\n \n \n 3\n \n \n \n \n x\n \n \n . Explain why the charge\n \n q\n \n will execute simple harmonic oscillations about C.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The mass of the charge\n \n q\n \n is 0.025 kg.\n

\n

\n Calculate the angular frequency of the oscillations using the data in (a)(ii) and the expression in (b)(i).\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The charges\n \n Q\n \n are replaced by neutral masses\n \n M\n \n and the charge\n \n q\n \n by a neutral mass\n \n m\n \n . The mass\n \n m\n \n is displaced away from C by a small distance\n \n \n x\n \n \n and released. Discuss whether the motion of\n \n m\n \n will be the same as that of\n \n q\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Constant, non-zero within spheres ✓\n

\n

\n A clear, non-zero positive minimum at C ✓\n

\n

\n Symmetric bowl shaped up curved shape in between ✓\n

\n

\n

\n

\n \n

\n

\n \n Do\n \n not\n \n allow a bowl shaped down curve for\n \n MP3\n \n .\n \n

\n
\n
\n (i)\n
\n

\n Constant, non-zero within spheres ✓\n

\n

\n A clear, non-zero positive minimum at C ✓\n

\n

\n Symmetric bowl shaped up curved shape in between ✓\n

\n

\n

\n

\n \n

\n

\n \n Do\n \n not\n \n allow a bowl shaped down curve for\n \n MP3\n \n .\n \n

\n
\n
\n (ii)\n
\n

\n \n V\n \n «= 2 ×\n \n \n \n \n 8\n \n \n .\n \n \n 99\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n 0\n \n \n .\n \n \n 60\n \n \n \n \n » = 6.0 × 10\n \n 7\n \n «V» ✓\n

\n

\n \n W\n \n = «qV = 6.0 × 10\n \n 7\n \n × 4.0 × 10\n \n −9\n \n =» 0.24 «J» ✓\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n \n

\n
\n
\n (i)\n
\n

\n The restoring force/acceleration is opposite to the displacement/towards equilibrium /\n \n \n OWTTE\n \n \n ✓\n

\n

\n and proportional to displacement from equilibrium /\n \n \n OWTTE✓\n \n \n

\n

\n \n Allow discussions based on the diagram (such as towards C for towards equilibrium).\n \n

\n

\n \n Accept F ∝ x\n \n OR\n \n a ∝ x for\n \n MP2\n \n \n

\n
\n
\n (ii)\n
\n

\n \n ω\n \n =\n \n \n \n \n \n 32\n \n \n k\n \n \n Q\n \n \n q\n \n \n \n \n m\n \n \n \n D\n \n \n 3\n \n \n \n \n \n \n \n \n OR\n \n \n use of\n \n F\n \n =\n \n mω\n \n \n 2\n \n \n r\n \n OR\n \n F\n \n = 1.33\n \n x\n \n \n \n OR\n \n \n \n a\n \n = 53.3\n \n x\n \n ✓\n

\n

\n «\n \n \n \n \n \n 32\n \n \n ×\n \n \n 8\n \n \n .\n \n \n 99\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n ×\n \n \n 4\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 9\n \n \n \n \n \n \n 0\n \n \n .\n \n \n 025\n \n \n ×\n \n \n 1\n \n \n .\n \n \n \n 2\n \n \n 3\n \n \n \n \n \n \n » = 7.299 «s\n \n −1\n \n »\n

\n
\n
\n (c)\n
\n

\n the net force will no longer be a restoring force/directed towards equilibrium\n

\n

\n \n \n OR\n \n \n

\n

\n the gravitational force is attractive/neutral mass would be pulled towards larger masses/\n \n \n OWTTE\n \n \n ✓\n

\n

\n «and so» no, motion will not be the same/no longer be SHM /\n \n \n OWTTE\n \n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "c-wave-behaviour", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "c-1-simple-harmonic-motion", "d-1-gravitational-fields", "d-2-electric-and-magnetic-fields" ] }, { "question_id": "23M.2.HL.TZ1.7", "Question": "
\n
\n (a)\n
\n
\n

\n Explain, by reference to Faraday’s law of electromagnetic induction, why there is an electromotive force (emf) induced in the loop as it leaves the region of magnetic field.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Just before the loop is about to completely exit the region of magnetic field, the loop moves with constant terminal speed\n \n v\n \n .\n

\n

\n The following data is available:\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
\n Mass of loop\n \n \n m\n \n = 4.0 g\n
\n Resistance of loop\n \n \n R\n \n = 25 mΩ\n
\n Width of loop\n \n \n L\n \n = 15 cm\n
\n Magnetic flux density\n \n \n B\n \n = 0.80 T\n
\n

\n Determine, in m s\n \n −1\n \n the terminal speed\n \n v\n \n .\n

\n
\n
\n

\n [4]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n The induced emf is equal/proportional/related to the «rate of» change of «magnetic» flux/flux linkage ✓\n

\n

\n Flux is changing because the area pierced/enclosed by magnetic field lines changes «decreases»\n

\n

\n \n \n OR\n \n \n

\n

\n Flux is changing because the loop is leaving/moving out of the «magnetic» field. ✓\n

\n

\n \n Need to see a connection between the EMF and change in flux for\n \n MP1\n \n .\n \n

\n

\n \n Need to see a connection between the area changing or leaving the field and the change in flux for\n \n MP2\n \n \n

\n
\n
\n (a)\n
\n

\n The induced emf is equal/proportional/related to the «rate of» change of «magnetic» flux/flux linkage ✓\n

\n

\n Flux is changing because the area pierced/enclosed by magnetic field lines changes «decreases»\n

\n

\n \n \n OR\n \n \n

\n

\n Flux is changing because the loop is leaving/moving out of the «magnetic» field. ✓\n

\n

\n \n Need to see a connection between the EMF and change in flux for\n \n MP1\n \n .\n \n

\n

\n \n Need to see a connection between the area changing or leaving the field and the change in flux for\n \n MP2\n \n \n

\n
\n
\n (b)\n
\n

\n \n mg\n \n =\n \n BIL\n \n \n \n OR\n \n \n I = 0.33 «A» ✓\n

\n

\n \n BvL\n \n =\n \n IR\n \n OR\n \n \n ℰ = 8.25 × 10\n \n −3\n \n «V»\n \n \n OR\n \n \n ℰ = 0.12\n \n v\n \n ✓\n

\n

\n Combining results to get v =\n \n \n \n \n m\n \n \n g\n \n \n R\n \n \n \n \n \n B\n \n \n 2\n \n \n \n \n L\n \n \n 2\n \n \n \n \n \n ✓\n

\n

\n \n v\n \n = «\n \n \n \n \n 0\n \n \n .\n \n \n 040\n \n \n ×\n \n \n 9\n \n \n .\n \n \n 81\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 025\n \n \n \n \n 0\n \n \n .\n \n \n \n 80\n \n \n 2\n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n \n 15\n \n \n 2\n \n \n \n \n \n =» 0.068 «m s\n \n −1\n \n »\n

\n

\n \n Allow\n \n ECF\n \n between steps if clear work is shown.\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "b-5-current-and-circuits", "d-3-motion-in-electromagnetic-fields", "d-4-induction" ] }, { "question_id": "23M.2.HL.TZ1.8", "Question": "
\n
\n (a)\n
\n
\n

\n Photons of wavelength 468 nm are incident on a metallic surface. The maximum kinetic energy of the emitted electrons is 1.8 eV.\n

\n

\n Calculate\n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the work function of the surface, in eV.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the longest wavelength of a photon that will eject an electron from this surface.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead (\n \n \n \n Pb\n \n \n \n \n 82\n \n \n 207\n \n \n \n ). Show that the distance of closest approach is about 4 × 10\n \n −14\n \n m.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The radius of a nucleus of\n \n \n \n Pb\n \n \n \n \n 82\n \n \n 207\n \n \n \n is 7.1 × 10\n \n −15\n \n m. Explain why there will be no deviations from Rutherford scattering in the experiment in (b)(i).\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Use of\n \n E\n \n \n max\n \n =\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n -\n \n \n ϕ\n \n \n ⇒\n \n \n ϕ\n \n \n =\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n -\n \n \n \n E\n \n \n max\n \n ✓\n

\n

\n \n \n ϕ\n \n \n =\n \n \n «\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n −\n \n E\n \n \n max\n \n =\n \n \n \n \n \n \n 6\n \n \n .\n \n \n 63\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 34\n \n \n \n \n \n \n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n 8\n \n \n \n \n \n \n \n \n \n 468\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 9\n \n \n \n \n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n \n \n \n \n − 1.81» = 0.85625\n \n \n ≈\n \n \n 0.86 «eV» ✓\n

\n

\n

\n
\n
\n (i)\n
\n

\n Use of\n \n E\n \n \n max\n \n =\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n -\n \n \n ϕ\n \n \n ⇒\n \n \n ϕ\n \n \n =\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n -\n \n \n \n E\n \n \n max\n \n ✓\n

\n

\n \n \n ϕ\n \n \n =\n \n \n «\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n −\n \n E\n \n \n max\n \n =\n \n \n \n \n \n \n 6\n \n \n .\n \n \n 63\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 34\n \n \n \n \n \n \n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n 8\n \n \n \n \n \n \n \n \n \n 468\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 9\n \n \n \n \n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n \n \n \n \n − 1.81» = 0.85625\n \n \n ≈\n \n \n 0.86 «eV» ✓\n

\n

\n

\n
\n
\n (ii)\n
\n

\n Use of\n \n \n \n \n h\n \n \n c\n \n \n \n λ\n \n \n \n =\n \n \n ϕ\n \n \n ⇒\n \n \n λ\n \n \n =\n \n \n \n \n h\n \n \n c\n \n \n \n ϕ\n \n \n \n ✓\n

\n

\n \n \n λ\n \n \n =\n \n \n «\n \n \n \n \n \n \n 6\n \n \n .\n \n \n 63\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 34\n \n \n \n \n \n \n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n 8\n \n \n \n \n \n \n \n \n \n 468\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 9\n \n \n \n \n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n \n \n \n \n =» 1.45 × 10\n \n −6\n \n «m»✓\n

\n

\n \n Allow\n \n ECF\n \n from\n \n a(i)\n \n \n

\n
\n
\n (i)\n
\n

\n 2e\n \n \n AND\n \n \n 82e seen\n

\n

\n \n \n OR\n \n \n

\n

\n 3.2 × 10\n \n −19\n \n «C»\n \n \n AND\n \n \n 1.312 × 10\n \n −17\n \n «C» seen ✓\n

\n

\n \n d\n \n =\n \n \n \n \n 8\n \n \n .\n \n \n 99\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n ×\n \n \n (\n \n \n 2\n \n \n e\n \n \n )\n \n \n (\n \n \n 82\n \n \n e\n \n \n )\n \n \n \n \n 5\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n ×\n \n \n e\n \n \n \n \n = 3.998 × 10\n \n −14\n \n ≈ 4 × 10\n \n −14\n \n «m» ✓\n

\n

\n \n Must see either clear substitutions or answer to at least 4 s.f. for\n \n MP2\n \n .\n \n

\n
\n
\n (ii)\n
\n

\n The closest approach is «significantly» larger than the radius of the nucleus / far away from the nucleus/\n \n \n OWTTE\n \n \n . ✓\n

\n

\n «Therefore» the strong nuclear force will not act on the alpha particle.✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-1-structure-of-the-atom", "e-2-quantum-physics" ] }, { "question_id": "23M.2.HL.TZ2.4", "Question": "
\n
\n (a)\n
\n
\n

\n State the value of the maximum distance between the stars that can be measured in any reference frame.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n State the speed of shuttle S relative to observer P using Galilean relativity.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate the distance between star A and star B relative to observer P.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Show that the speed of shuttle S relative to observer P is approximately 0.6\n \n c\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n Calculate the time, according to observer P, that the shuttle S takes to travel from star A to star B.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (f)\n
\n
\n

\n State and explain the reference frame in which the proper time for shuttle S to journey from star A to star B can be measured.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 4.8 «light years» ✓\n

\n
\n
\n (b)\n
\n

\n «−» 0.48\n \n c\n \n ✓\n

\n

\n

\n

\n \n Ignore sign\n \n

\n
\n
\n (c)\n
\n

\n \n \n y\n \n \n = 1.6 ✓\n

\n

\n \n D\n \n =\n \n \n \n \n 4\n \n \n .\n \n \n 8\n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n \n \n = 3 «ly» ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n \n

\n
\n
\n (d)\n
\n

\n \n \n =\n \n \n \n \n \n \n 0\n \n \n .\n \n \n 3\n \n \n c\n \n \n -\n \n \n 0\n \n \n .\n \n \n 78\n \n \n c\n \n \n \n \n 1\n \n \n -\n \n \n \n \n \n 0\n \n \n .\n \n \n 78\n \n \n c\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 3\n \n \n c\n \n \n \n \n c\n \n \n 2\n \n \n \n \n \n \n \n ✓\n

\n

\n «−» 0.63\n \n c\n \n ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n if\n \n MP1\n \n correct and correct answer given to 1 significant figure.\n \n

\n
\n
\n (e)\n
\n

\n \n \n Δ\n \n \n \n t\n \n \n p\n \n \n \n =\n \n \n \n \n 3\n \n \n \n \n ly\n \n \n \n \n (\n \n \n 0\n \n \n .\n \n \n 78\n \n \n -\n \n \n 0\n \n \n .\n \n \n 627\n \n \n )\n \n \n c\n \n \n \n \n

\n

\n \n \n OR\n \n \n

\n

\n \n \n Δ\n \n \n \n t\n \n \n p\n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 6\n \n \n \n \n \n \n 4\n \n \n .\n \n \n 8\n \n \n \n \n ly\n \n \n \n \n 0\n \n \n .\n \n \n 3\n \n \n c\n \n \n \n \n -\n \n \n \n \n 0\n \n \n .\n \n \n 78\n \n \n c\n \n \n ×\n \n \n 4\n \n \n .\n \n \n 8\n \n \n \n \n ly\n \n \n \n \n c\n \n \n 2\n \n \n \n \n \n \n ✓\n

\n

\n = 19\n \n \n OR\n \n \n 20 «years» ✓\n

\n

\n

\n

\n \n Award\n \n [2]\n \n for\n \n BCA\n \n \n

\n
\n
\n (f)\n
\n

\n shuttle measures proper time ✓\n

\n

\n as the events occur at the same place for the shuttle / shuttle is at both events ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-5-galilean-and-special-relativity" ] }, { "question_id": "23M.2.HL.TZ2.5", "Question": "
\n
\n (a)\n
\n
\n

\n Outline what is meant by an isotope.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (a)\n
\n
\n

\n Determine the speed of the spaceship relative to Earth.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Estimate, using the spacetime diagram, the time in seconds when the flash of light reaches the spaceship according to the Earth observer.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Determine the time coordinate\n \n ct′\n \n when the flash of light reaches the spaceship, according to an observer at rest in the spaceship.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n «An atom with» the same number of protons\n \n \n AND\n \n \n different numbers of neutrons\n
\n \n \n OR\n \n \n
\n Same chemical properties\n \n \n AND\n \n \n different physical properties ✓\n

\n

\n \n Do\n \n not\n \n allow just atomic number and mass number\n \n

\n
\n
\n (a)\n
\n

\n 0.6 c\n

\n

\n \n \n OR\n \n \n

\n

\n = 1.8 × 10\n \n 8\n \n «m s\n \n −1\n \n » ✓\n

\n
\n
\n (b)\n
\n

\n Line drawn at 45° from\n \n ct\n \n = 2 km to hit spaceship world line at\n \n ct\n \n = 5 km\n

\n

\n \n \n OR\n \n \n

\n

\n \n ct\n \n = 1.2/(\n \n c\n \n − 0.6\n \n c\n \n ) + 2 = 5 «km» ✓\n

\n

\n \n t\n \n =\n \n \n \n 5000\n \n \n c\n \n \n \n = 1.7 × 10\n \n −5\n \n «s» ✓\n

\n

\n

\n

\n \n Award\n \n [2]\n \n for\n \n BCA\n \n \n

\n
\n
\n (c)\n
\n

\n (\n \n ct\n \n ')\n \n 2\n \n − 0 = 5\n \n 2\n \n − 3\n \n 2\n \n

\n

\n \n \n OR\n \n \n

\n

\n \n \n γ\n \n \n = 1.25 ✓\n

\n

\n

\n

\n \n ct\n \n ' = 4 «km»\n

\n

\n \n \n OR\n \n \n

\n

\n \n t\n \n ' = 13 «\n \n µ\n \n s» ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n (b)\n \n \n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "e-nuclear-and-quantum-physics" ], "subtopics": [ "a-5-galilean-and-special-relativity", "e-3-radioactive-decay" ] }, { "question_id": "23M.2.HL.TZ2.6", "Question": "
\n
\n (a)\n
\n
\n

\n Determine\n \n \n \n \n g\n \n \n M\n \n \n \n \n g\n \n \n P\n \n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (a)\n
\n
\n

\n Show that the net torque on the system about the central axis is approximately 30 N m.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The system rotates from rest and reaches a maximum angular speed of 20 rad s\n \n −1\n \n in a time of 5.0 s. Calculate the angular acceleration of the system.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State and explain the magnitude of the resultant gravitational field strength at O.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Outline why the graph between P and O is negative.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Show that the gravitational potential\n \n V\n \n \n P\n \n at the surface of P due to the mass of P is given by\n \n V\n \n \n P\n \n = −\n \n g\n \n \n P\n \n \n R\n \n \n P\n \n where\n \n R\n \n \n P\n \n is the radius of the planet.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n The gravitational potential due to the mass of M at the surface of P can be assumed to be negligible.\n

\n

\n Estimate, using the graph, the gravitational potential at the surface of M due to the mass of M.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (v)\n
\n
\n

\n Draw on the axes the variation of gravitational potential between O and M.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Determine the moment of inertia of the system about the central axis.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Outline why the angular speed\n \n ω\n \n decreases when the spheres move outward.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Show that the rotational kinetic energy is\n \n \n \n 1\n \n \n 2\n \n \n \n \n Lω\n \n where\n \n L\n \n is the angular momentum of the system.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n When the spheres move outward, the angular speed decreases from 20 rad s\n \n −1\n \n to 12 rad s\n \n −1\n \n . Calculate the percentage change in rotational kinetic energy that occurs when the spheres move outward.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Work using\n \n g\n \n ∝\n \n \n \n m\n \n \n \n r\n \n \n 2\n \n \n \n \n ✓\n

\n

\n \n \n \n \n g\n \n \n M\n \n \n \n \n g\n \n \n P\n \n \n \n \n =\n \n \n \n \n m\n \n \n M\n \n \n \n \n m\n \n \n P\n \n \n \n \n \n \n \n \n r\n \n \n P\n \n \n \n \n r\n \n \n M\n \n \n \n \n \n 2\n \n \n \n = 0.75 ✓\n

\n
\n
\n (a)\n
\n

\n \n \n ΣΓ\n \n \n = 50 × 0.5 + 40 × 0.2\n

\n

\n \n \n OR\n \n \n

\n

\n 33 «Nm» ✓\n

\n

\n

\n

\n \n Accept opposite rotational sign convention\n \n

\n
\n
\n (b)\n
\n

\n «α =\n \n \n \n 20\n \n \n 5\n \n \n \n =» 4 «rad s\n \n −2\n \n » ✓\n

\n
\n
\n (i)\n
\n

\n g = 0 ✓\n

\n

\n As g «=\n \n \n -\n \n \n \n \n Δ\n \n \n \n V\n \n \n g\n \n \n \n \n \n Δ\n \n \n r\n \n \n \n \n which» is the gradient of the graph\n

\n

\n \n \n OR\n \n \n
\n As the force of attraction/field strength of P and M are equal ✓\n

\n
\n
\n (ii)\n
\n

\n The gravitational field is attractive so that energy is required «to move away from P» ✓\n

\n

\n the gravitational potential is defined as 0 at\n \n \n ∞\n \n \n , (the potential must be negative) ✓\n

\n
\n
\n (iii)\n
\n

\n \n V\n \n \n P\n \n =\n \n \n -\n \n \n \n \n G\n \n \n M\n \n \n \n \n R\n \n \n P\n \n \n \n \n \n \n AND\n \n g\n \n \n P\n \n =\n \n \n \n \n G\n \n \n M\n \n \n \n \n \n R\n \n \n P\n \n \n \n 2\n \n \n \n \n (at surface) ✓\n

\n

\n Suitable working and cancellation of\n \n G\n \n and\n \n M\n \n seen ✓\n

\n

\n \n V\n \n \n P\n \n = −\n \n g\n \n \n P\n \n \n R\n \n \n P\n \n

\n

\n \n Must see negative sign\n \n

\n

\n

\n
\n
\n (iv)\n
\n

\n «\n \n \n \n \n V\n \n \n M\n \n \n \n \n V\n \n \n P\n \n \n \n \n =\n \n \n \n \n \n g\n \n \n M\n \n \n \n \n R\n \n \n M\n \n \n \n \n \n \n g\n \n \n P\n \n \n \n \n R\n \n \n P\n \n \n \n \n \n = 0.75 × 0.27» = 0.20 ✓\n

\n

\n \n V\n \n \n M\n \n = «−6.4 × 10\n \n 7\n \n × 0.2 =» «−»1.3 × 10\n \n 7\n \n «J kg\n \n −1\n \n »✓\n

\n
\n
\n (v)\n
\n

\n Line always negative, of suitable shape and end point below −8 and above −20 unless awarding\n \n \n ECF\n \n \n from\n \n \n b(iv)\n \n \n ✓\n

\n

\n \n

\n
\n
\n (c)\n
\n

\n \n \n I\n \n \n =\n \n \n \n Γ\n \n \n α\n \n \n \n

\n

\n \n \n OR\n \n \n

\n

\n 33 =\n \n \n I\n \n \n × 4 ✓\n

\n

\n

\n

\n \n \n I\n \n \n = 8.25 «kg m\n \n 2\n \n » ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n (a)\n \n and\n \n (b)\n \n \n

\n

\n \n Award\n \n [2]\n \n for a\n \n BCA\n \n \n

\n
\n
\n (i)\n
\n

\n moment of inertia increases ✓\n

\n

\n Angular momentum is conserved ✓\n

\n

\n \n
\n Allow algebraic expressions e.g. ω =\n \n \n \n L\n \n \n I\n \n \n \n so ω decreases for\n \n MP2\n \n \n

\n
\n
\n (ii)\n
\n

\n \n E\n \n \n k\n \n «=\n \n \n \n 1\n \n \n 2\n \n \n \n I\n \n \n \n ω\n \n \n 2\n \n =»\n \n \n \n 1\n \n \n 2\n \n \n \n (\n \n \n I\n \n \n \n ω\n \n )\n \n ω\n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n \n Lω\n \n ✓\n

\n

\n

\n

\n \n Accept equivalent methods\n \n

\n
\n
\n (iii)\n
\n

\n \n «E\n \n \n k\n \n =»\n \n \n \n 1\n \n \n 2\n \n \n \n \n Lω\n \n \n 1\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n Lω\n \n 2\n \n \n ✓\n

\n

\n \n \n OR\n \n \n

\n

\n \n \n \n \n E\n \n \n \n k\n \n \n 1\n \n \n \n \n \n E\n \n \n \n k\n \n \n 2\n \n \n \n \n \n =\n \n \n \n \n ω\n \n \n 1\n \n \n \n \n ω\n \n \n 2\n \n \n \n \n

\n

\n \n \n OR\n \n \n

\n

\n «\n \n L\n \n is constant so»\n \n E\n \n \n k\n \n is proportional to\n \n ω\n \n ✓\n

\n

\n 40 % «energy loss» ✓\n

\n

\n

\n

\n \n \n MP1\n \n is for understanding that angular momentum is constant so change in rotational kinetic energy is proportional to change in angular velocity\n \n

\n

\n \n \n Award [0]\n \n if E = 0.5 I ω\n \n 2\n \n is used with the same I value for both values of E\n \n

\n

\n \n Award\n \n [2]\n \n for\n \n BCA\n \n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-4-rigid-body-mechanics", "d-1-gravitational-fields" ] }, { "question_id": "23M.2.HL.TZ2.7", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the pressure of the gas at B.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Sketch, on the\n \n pV\n \n diagram, the remaining two processes BC and CA that the gas undergoes.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Show that the temperature of the gas at C is approximately 350 °C.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Explain why the change of entropy for the gas during the process BC is equal to zero.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n Explain why the work done by the gas during the isothermal expansion AB is less than the work done on the gas during the adiabatic compression BC.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (f)\n
\n
\n

\n The quantity of trapped gas is 53.2 mol. Calculate the thermal energy removed from the gas during process CA.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n use of\n \n pV\n \n = constant ✓\n

\n

\n \n P\n \n \n B\n \n = 43 «Kpa» ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n \n

\n
\n
\n (b)\n
\n

\n concave curved line from B to locate C with a higher pressure than A ✓\n

\n

\n vertical line joining C to A ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n i.e.,\n \n award [1]\n \n for first process locating C at a lower pressure than A, then vertical line to A.\n \n

\n

\n \n Arrows on the processes are not needed.\n \n

\n

\n \n Point C need not be labelled.\n \n

\n

\n \n

\n
\n
\n (c)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n use of\n \n \n T\n \n \n \n V\n \n \n \n 2\n \n \n 3\n \n \n \n \n = constant «so\n \n \n 300\n \n \n \n \n (\n \n \n 3\n \n \n \n V\n \n \n A\n \n \n \n )\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n =\n \n \n \n T\n \n \n C\n \n \n \n \n \n (\n \n \n \n V\n \n \n A\n \n \n \n )\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n » ✓\n

\n

\n \n T\n \n \n C\n \n = 624 «K»\n \n \n OR\n \n \n \n T\n \n \n C\n \n = 351 «°C» ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n use of\n \n \n p\n \n \n \n V\n \n \n \n 5\n \n \n 3\n \n \n \n \n to get either\n \n p\n \n \n c\n \n =\n \n \n 43\n \n \n \n \n (\n \n \n 3\n \n \n )\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n \n OR\n \n \n \n p\n \n \n c\n \n = 268 «kPa» ✓\n

\n

\n «\n \n T\n \n \n C\n \n = 268 × 300/129 = so »\n

\n

\n \n T\n \n \n C\n \n = 624 «K»\n \n \n OR\n \n \n \n T\n \n \n C\n \n = 351 «°C» ✓\n

\n
\n
\n (d)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n «the process is adiabatic so» ΔQ = 0 ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n The compression is reversible «so ΔS = 0» ✓\n

\n

\n

\n

\n \n \n OWTTE\n \n \n

\n
\n
\n (e)\n
\n

\n area under curve AB is less than area under curve BC ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n allow\n \n ECF\n \n from part\n \n (b)\n \n \n

\n
\n
\n (f)\n
\n

\n «W = 0 so» Q = ΔU ✓\n

\n

\n «ΔU =\n \n \n \n 3\n \n \n 2\n \n \n \n × 53.2 × R × (351 – 27) so » ΔU = 2.15 × 10\n \n 5\n \n «J» ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "23M.2.HL.TZ2.8", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the net torque on the system about the central axis is approximately 30 N m.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The system rotates from rest and reaches a maximum angular speed of 20 rad s\n \n −1\n \n in a time of 5.0 s. Calculate the angular acceleration of the system.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Determine the moment of inertia of the system about the central axis.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Outline why the angular speed\n \n ω\n \n decreases when the spheres move outward.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Show that the rotational kinetic energy is\n \n \n \n 1\n \n \n 2\n \n \n \n \n Lω\n \n where\n \n L\n \n is the angular momentum of the system.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n When the spheres move outward, the angular speed decreases from 20 rad s\n \n −1\n \n to 12 rad s\n \n −1\n \n . Calculate the percentage change in rotational kinetic energy that occurs when the spheres move outward.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n Outline one reason why this model of a dancer is unrealistic.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n ΣΓ\n \n \n = 50 × 0.5 + 40 × 0.2\n

\n

\n \n \n OR\n \n \n

\n

\n 33 «Nm» ✓\n

\n

\n

\n

\n \n Accept opposite rotational sign convention\n \n

\n
\n
\n (b)\n
\n

\n «α =\n \n \n \n 20\n \n \n 5\n \n \n \n =» 4 «rad s\n \n −2\n \n » ✓\n

\n
\n
\n (c)\n
\n

\n \n \n I\n \n \n =\n \n \n \n Γ\n \n \n α\n \n \n \n

\n

\n \n \n OR\n \n \n

\n

\n 33 =\n \n \n I\n \n \n × 4 ✓\n

\n

\n

\n

\n \n \n I\n \n \n = 8.25 «kg m\n \n 2\n \n » ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n (a)\n \n and\n \n (b)\n \n \n

\n

\n \n Award\n \n [2]\n \n for a\n \n BCA\n \n \n

\n
\n
\n (i)\n
\n

\n moment of inertia increases ✓\n

\n

\n Angular momentum is conserved ✓\n

\n

\n \n
\n Allow algebraic expressions e.g. ω =\n \n \n \n L\n \n \n I\n \n \n \n so ω decreases for\n \n MP2\n \n \n

\n
\n
\n (ii)\n
\n

\n \n E\n \n \n k\n \n «=\n \n \n \n 1\n \n \n 2\n \n \n \n I\n \n \n \n ω\n \n \n 2\n \n =»\n \n \n \n 1\n \n \n 2\n \n \n \n (\n \n \n I\n \n \n \n ω\n \n )\n \n ω\n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n \n Lω\n \n ✓\n

\n

\n

\n

\n \n Accept equivalent methods\n \n

\n
\n
\n (iii)\n
\n

\n \n «E\n \n \n k\n \n =»\n \n \n \n 1\n \n \n 2\n \n \n \n \n Lω\n \n \n 1\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n Lω\n \n 2\n \n \n ✓\n

\n

\n \n \n OR\n \n \n

\n

\n \n \n \n \n E\n \n \n \n k\n \n \n 1\n \n \n \n \n \n E\n \n \n \n k\n \n \n 2\n \n \n \n \n \n =\n \n \n \n \n ω\n \n \n 1\n \n \n \n \n ω\n \n \n 2\n \n \n \n \n

\n

\n \n \n OR\n \n \n

\n

\n «\n \n L\n \n is constant so»\n \n E\n \n \n k\n \n is proportional to\n \n ω\n \n ✓\n

\n

\n 40 % «energy loss» ✓\n

\n

\n

\n

\n \n \n MP1\n \n is for understanding that angular momentum is constant so change in rotational kinetic energy is proportional to change in angular velocity\n \n

\n

\n \n \n Award [0]\n \n if E = 0.5 I ω\n \n 2\n \n is used with the same I value for both values of E\n \n

\n

\n \n Award\n \n [2]\n \n for\n \n BCA\n \n \n

\n
\n
\n (e)\n
\n

\n one example specified\n \n eg\n \n friction, air resistance, mass distribution not modelled ✓\n

\n

\n

\n

\n \n \n Award [1]\n \n for any reasonable physical parameter that is not consistent with the model\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "23M.2.HL.TZ2.9", "Question": "
\n
\n (a)\n
\n
\n

\n Show, using the data, that the energy released in the decay of one magnesium-27 nucleus is about 2.62 MeV.\n

\n

\n Mass of aluminium-27 atom = 26.98153 u\n
\n Mass of magnesium-27 atom = 26.98434 u\n
\n The unified atomic mass unit is 931.5 MeV c\n \n −2\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (a)\n
\n
\n

\n Calculate the pressure of the gas at B.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Sketch, on the\n \n pV\n \n diagram, the remaining two processes BC and CA that the gas undergoes.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the conclusion that can be drawn from the existence of these two routes.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the difference between the magnitudes of the total energy transfers in parts (a) and (b).\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Explain how the difference in part (b)(ii) arises.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Show that the temperature of the gas at C is approximately 350 °C.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n The smallest mass of magnesium that can be detected with this technique is 1.1 × 10\n \n −8\n \n kg.\n

\n

\n Show that the smallest number of magnesium atoms that can be detected with this technique is about 10\n \n 17\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n A sample of glass is irradiated with neutrons so that all the magnesium atoms become magnesium-27. The sample contains 9.50 × 10\n \n 15\n \n magnesium atoms.\n

\n

\n The decay constant of magnesium-27 is 1.22 × 10\n \n −3\n \n s\n \n −1\n \n .\n

\n

\n Determine the number of aluminium atoms that form in 10.0 minutes after the irradiation ends.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Estimate, in W, the average rate at which energy is transferred by the decay of magnesium-27 during the 10.0 minutes after the irradiation ends.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Explain why the change of entropy for the gas during the process BC is equal to zero.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n Explain why the work done by the gas during the isothermal expansion AB is less than the work done on the gas during the adiabatic compression BC.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (f)\n
\n
\n

\n The quantity of trapped gas is 53.2 mol. Calculate the thermal energy removed from the gas during process CA.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n (26.98434 - 26.98153) × 931.5\n

\n

\n \n \n OR\n \n \n

\n

\n 2.6175 «MeV» seen ✓\n

\n
\n
\n (a)\n
\n

\n use of\n \n pV\n \n = constant ✓\n

\n

\n \n P\n \n \n B\n \n = 43 «Kpa» ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n \n

\n
\n
\n (b)\n
\n

\n concave curved line from B to locate C with a higher pressure than A ✓\n

\n

\n vertical line joining C to A ✓\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n i.e.,\n \n award [1]\n \n for first process locating C at a lower pressure than A, then vertical line to A.\n \n

\n

\n \n Arrows on the processes are not needed.\n \n

\n

\n \n Point C need not be labelled.\n \n

\n

\n \n

\n
\n
\n (i)\n
\n

\n evidence for nuclear energy levels ✓\n

\n
\n
\n (ii)\n
\n

\n Difference = 2.6175 – (1.76656 +0.84376) = 2.6175 – 2.61032 = 0.007195 «MeV»\n

\n

\n \n \n OR\n \n \n

\n

\n Difference = 2.6175 – (1.59587 +1.01445) = 2.61032 = 0.007195 «MeV» ✓\n

\n
\n
\n (iii)\n
\n

\n Another particle/«anti» neutrino is emitted «that accounts for this mass / energy» ✓\n

\n
\n
\n (c)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n use of\n \n \n T\n \n \n \n V\n \n \n \n 2\n \n \n 3\n \n \n \n \n = constant «so\n \n \n 300\n \n \n \n \n (\n \n \n 3\n \n \n \n V\n \n \n A\n \n \n \n )\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n =\n \n \n \n T\n \n \n C\n \n \n \n \n \n (\n \n \n \n V\n \n \n A\n \n \n \n )\n \n \n \n \n 2\n \n \n 3\n \n \n \n \n » ✓\n

\n

\n \n T\n \n \n C\n \n = 624 «K»\n \n \n OR\n \n \n \n T\n \n \n C\n \n = 351 «°C» ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n use of\n \n \n p\n \n \n \n V\n \n \n \n 5\n \n \n 3\n \n \n \n \n to get either\n \n p\n \n \n c\n \n =\n \n \n 43\n \n \n \n \n (\n \n \n 3\n \n \n )\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n \n OR\n \n \n \n p\n \n \n c\n \n = 268 «kPa» ✓\n

\n

\n «\n \n T\n \n \n C\n \n = 268 × 300/129 = so »\n

\n

\n \n T\n \n \n C\n \n = 624 «K»\n \n \n OR\n \n \n \n T\n \n \n C\n \n = 351 «°C» ✓\n

\n
\n
\n (i)\n
\n

\n So 1.1 × 10\n \n −8\n \n kg ≡\n \n \n \n \n 1\n \n \n .\n \n \n 1\n \n \n \n \n 0\n \n \n .\n \n \n 027\n \n \n \n \n × 10\n \n −8\n \n «mol»\n

\n

\n \n \n OR\n \n \n

\n

\n Mass of atom = 27 × 1.66 × 10\n \n −27\n \n «kg» ✓\n

\n

\n 2.4 − 2.5 x 10\n \n 17\n \n atoms ✓\n

\n
\n
\n (ii)\n
\n

\n \n N\n \n \n 10\n \n = 9.50 × 10\n \n 15\n \n ×\n \n e\n \n \n −0.00122×60\n \n seen ✓\n

\n

\n \n N\n \n \n 10\n \n = 4.57 × 10\n \n 15\n \n ✓\n

\n

\n So number of aluminium-27 nuclei = (9.50 – 4.57) × 10\n \n 15\n \n = 4.9(3) × 10\n \n 15\n \n ✓\n

\n
\n
\n (iii)\n
\n

\n Total energy released = ans\n \n (c)(ii)\n \n × 2.62 × 10\n \n 6\n \n × 1.6 × 10\n \n −19\n \n «= 2100 J» ✓\n

\n

\n «\n \n \n \n 2100\n \n \n 600\n \n \n \n =» 3.4 −3.5 «W» ✓\n

\n
\n
\n (d)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n «the process is adiabatic so» ΔQ = 0 ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n The compression is reversible «so ΔS = 0» ✓\n

\n

\n

\n

\n \n \n OWTTE\n \n \n

\n
\n
\n (e)\n
\n

\n area under curve AB is less than area under curve BC ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n allow\n \n ECF\n \n from part\n \n (b)\n \n \n

\n
\n
\n (f)\n
\n

\n «W = 0 so» Q = ΔU ✓\n

\n

\n «ΔU =\n \n \n \n 3\n \n \n 2\n \n \n \n × 53.2 × R × (351 – 27) so » ΔU = 2.15 × 10\n \n 5\n \n «J» ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-1-thermal-energy-transfers", "b-4-thermodynamics", "e-3-radioactive-decay" ] }, { "question_id": "23M.2.SL.TZ1.1", "Question": "
\n
\n (a)\n
\n
\n

\n Deduce the unit of μ in terms of fundamental SI units.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n draw a free-body diagram for the ball.\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n show that the speed of the ball is about 4.3 m s\n \n −1\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n determine the tension in the string.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Show that the collision is elastic.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the maximum height risen by the centre of the ball.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Explain a possible reason for the systematic error.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Calculate the gradient of the graph.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The coefficient of dynamic friction between the block and the rough surface is 0.400.\n

\n

\n Estimate the distance travelled by the block on the rough surface until it stops.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n [μ] = «\n \n \n \n \n kg\n \n \n ×\n \n \n m\n \n \n \n \n \n s\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n \n s\n \n \n \n -\n \n \n 2\n \n \n \n \n ×\n \n \n \n m\n \n \n 2\n \n \n \n \n \n » kg × m\n \n −1\n \n

\n

\n

\n

\n \n Accept kg/m.\n \n

\n

\n \n Do\n \n not\n \n accept g m\n \n −1\n \n .\n \n

\n
\n
\n (i)\n
\n

\n Tension upwards, weight downwards ✓\n
\n Tension is clearly longer than weight ✓\n

\n

\n

\n

\n \n Look for:\n \n

\n

\n \n

\n
\n
\n (ii)\n
\n

\n \n v\n \n =\n \n \n \n 2\n \n \n ×\n \n \n 9\n \n \n .\n \n \n 81\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 95\n \n \n \n \n \n OR\n \n \n = 4.32 «m s\n \n −1\n \n » ✓\n

\n

\n

\n

\n \n Must see either full substitution or answer to at least 3 s.f.\n \n

\n
\n
\n (iii)\n
\n

\n \n T − mg = F\n \n net\n \n \n OR\n \n \n \n T − mg =\n \n \n \n \n \n m\n \n \n \n v\n \n \n 2\n \n \n \n \n r\n \n \n \n ✓\n

\n

\n \n T\n \n «= 0.800 × 9.81 +\n \n \n \n \n 0\n \n \n .\n \n \n 800\n \n \n ×\n \n \n 4\n \n \n .\n \n \n \n 317\n \n \n 2\n \n \n \n \n \n 0\n \n \n .\n \n \n 95\n \n \n \n \n » = 23.5 «N» ✓\n

\n
\n
\n (i)\n
\n

\n Use of conservation of momentum. ✓\n
\n Rebound speed = 2.16 «m s\n \n −1\n \n » ✓\n
\n Calculation of initial KE = «\n \n \n \n 1\n \n \n 2\n \n \n \n × 0.800 × 4.317\n \n 2\n \n » = 7.46 « J » ✓\n
\n Calculation of final KE = «\n \n \n \n 1\n \n \n 2\n \n \n \n × 0.800  × 2.16\n \n 2\n \n +\n \n \n \n 1\n \n \n 2\n \n \n \n × 2.40 × 2.16\n \n 2\n \n » = 7.46 «J» ✓\n
\n «hence elastic»\n

\n
\n
\n (ii)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n
\n Rebound speed is halved so energy less by a factor of 4 ✓\n
\n Hence height is\n \n \n \n 95\n \n \n 4\n \n \n \n =23.8 «cm» ✓\n

\n

\n \n \n ALTERNATIVE 2\n
\n \n \n Use of conservation of energy /\n \n \n \n 1\n \n \n 2\n \n \n \n × 0.800 × 2.16\n \n 2\n \n = 0.800 × 9.8 ×\n \n h\n
\n \n OR\n \n \n

\n

\n Use of proper kinematics equation (e.g. 0 = 2.16\n \n 2\n \n − 2 × 9.8 ×\n \n h\n \n ) ✓\n

\n

\n \n h\n \n = 23.8 «cm» ✓\n

\n

\n \n
\n Allow\n \n ECF\n \n from\n \n b(i)\n \n \n

\n
\n
\n (iii)\n
\n

\n mass of tray of weights neglected/friction at pulley/friction at slider/thickness of slider/zero off-set error ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n allow vague answers like friction neglected / error in length measurement.\n \n

\n
\n
\n (iv)\n
\n

\n large enough triangle Δ\n \n m\n \n ≥ 50 g✓\n

\n

\n answer in range 0.210 − 0.240 «kg m\n \n −1\n \n » ✓\n

\n

\n

\n

\n \n Accept answers in g m\n \n −2\n \n .\n \n

\n

\n \n Do\n \n not\n \n allow\n \n ECF\n \n from\n \n MP1\n \n .\n \n

\n
\n
\n (c)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n
\n Frictional force is\n \n f\n \n «= 0.400 × 2.40 × 9.81» = 9.42 «N» ✓\n
\n 9.42 ×\n \n d\n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n × 2.40 × 2.16\n \n 2\n \n \n \n OR\n \n \n \n d\n \n =\n \n \n \n \n 5\n \n \n .\n \n \n 5987\n \n \n \n \n 9\n \n \n .\n \n \n 42\n \n \n \n \n ✓\n
\n \n d\n \n = 0.594 «m» ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n
\n \n a\n \n = «\n \n \n \n f\n \n \n m\n \n \n \n =\n \n µg\n \n = 0.4 × 9.81 =» 3.924 «m s\n \n −2\n \n » ✓\n

\n

\n Proper use of kinematics equation(s) to determine ✓\n

\n

\n \n d\n \n = 0.594 «m» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "inquiry", "tools" ], "subtopics": [ "a-2-forces-and-momentum", "a-3-work-energy-and-power", "i-3-2-evaluating", "inquiry-3-concluding-and-evaluating", "tool-3-mathematics" ] }, { "question_id": "23M.2.SL.TZ1.2", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the average rate at which thermal energy is transferred into the chocolate is about 15 W.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n one variable that must be controlled,\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the main source of error in\n \n T\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n To determine\n \n T\n \n more precisely, the student measures the total time for 20 oscillations and divides by 20.\n

\n

\n Explain why this is preferable to measuring the time for just one oscillation.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Compare the internal energy of the chocolate at\n \n t\n \n = 2 minutes with that at\n \n t\n \n = 6 minutes.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n The student plots a graph with\n \n L\n \n on the horizontal axis. State the variable that must be plotted on the vertical axis in order to obtain a line of best fit that is straight.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculations using the data of the experiment show that\n \n g\n \n = 9.71622 m s\n \n −2\n \n with a percentage uncertainty of 8 %. Determine the value of\n \n g\n \n that can be obtained from this experiment. Include the absolute uncertainty in\n \n g\n \n to one significant figure.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Reads change in temperature to be 45 − 31\n \n \n OR\n \n \n 14 °C ✓\n

\n

\n
\n \n Q\n \n = 0.082 × 1.6 × 10\n \n 3\n \n × 14 = 1.84 × 10\n \n 3\n \n «J» ✓\n

\n

\n \n P\n \n =\n \n \n \n \n 1\n \n \n .\n \n \n 84\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n 60\n \n \n \n \n = 15.3\n \n \n ≈\n \n \n 15 «W»✓\n

\n

\n

\n

\n \n Must see either full substitution\n \n OR\n \n answer to at least 3 s.f. in\n \n MP3\n \n \n

\n
\n
\n (i)\n
\n

\n mass\n \n \n OR\n \n \n diameter\n \n \n OR\n \n \n material of bob\n \n \n OR\n \n \n « initial » amplitude/angle ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n allow statements about rulers, stopwatches, string, number of oscillations, constant gravity.\n \n

\n
\n
\n (ii)\n
\n

\n student’s reaction time «in starting and stopping stopwatch» / starting/stopping stopwatch ✓\n

\n
\n
\n (b)\n
\n

\n it reduces «the random» error/uncertainty ✓\n

\n

\n by a factor of 20 «compared to that in a single period measurement» ✓\n

\n

\n

\n

\n \n For\n \n MP1\n \n , allow increasing accuracy/precision.\n \n

\n

\n \n \n Award [0]\n \n for answers related to number of trials, 20 measurements of one period.\n \n

\n
\n
\n (c)\n
\n

\n Internal energy is greater at\n \n t\n \n = 6 min\n \n \n OR\n \n \n internal energy is lower at\n \n t\n \n = 2 min\n \n \n OR\n \n \n internal energy increases «as energy is added to the system» ✓\n

\n

\n Because kinetic energy «of the molecules» is the same\n \n \n AND\n \n \n potential energy «of the molecules» has increased /\n \n \n OWTTE\n \n \n ✓\n

\n
\n
\n (i)\n
\n

\n \n T\n \n \n 2\n \n ✓\n

\n
\n
\n (ii)\n
\n

\n \n g\n \n = 9.7 «m s\n \n −2\n \n » ✓\n

\n

\n Δ\n \n g\n \n = 0.8 «m s\n \n −2\n \n » ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "inquiry", "tools" ], "subtopics": [ "b-1-thermal-energy-transfers", "i-1-2-designing", "inquiry-1-exploring-and-designing", "tool-1-experimental-techniques", "tool-3-mathematics" ] }, { "question_id": "23M.2.SL.TZ1.3", "Question": "
\n
\n (a)\n
\n
\n

\n A transverse water wave travels to the right. The diagram shows the shape of the surface of the water at time\n \n t\n \n = 0. P and Q show two corks floating on the surface.\n

\n

\n \n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State what is meant by a transverse wave.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The frequency of the wave is 0.50 Hz. Calculate the speed of the wave.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Plot on the diagram the position of P at time\n \n t\n \n = 0.50 s.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Monochromatic light is incident on two very narrow slits. The light that passes through the slits is observed on a screen. M is directly opposite the midpoint of the slits.\n \n \n x\n \n \n represents the displacement from M in the direction shown.\n

\n

\n \n

\n

\n A student argues that what will be observed on the screen will be a total of two bright spots opposite the slits. Explain why the student’s argument is incorrect.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The graph shows the actual variation with displacement\n \n \n x\n \n \n from M of the intensity of the light on the screen.\n \n \n \n I\n \n \n 0\n \n \n \n is the intensity of light at the screen from one slit only.\n

\n

\n \n

\n

\n The slits are separated by a distance of 0.18 mm and the distance to the screen is 2.2 m. Determine, in m, the wavelength of light.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n «A wave where the» displacement of particles/oscillations of particles/movement of particles/vibrations of particles is perpendicular/normal to the direction of energy transfer/wave travel/wave velocity/wave movement/wave propagation ✓\n

\n

\n

\n

\n \n Allow medium, material, water, molecules, or atoms for particles.\n \n

\n
\n
\n (i)\n
\n

\n «A wave where the» displacement of particles/oscillations of particles/movement of particles/vibrations of particles is perpendicular/normal to the direction of energy transfer/wave travel/wave velocity/wave movement/wave propagation ✓\n

\n

\n

\n

\n \n Allow medium, material, water, molecules, or atoms for particles.\n \n

\n
\n
\n (ii)\n
\n

\n \n v\n \n = «0.50 × 16 =» 8.0 «m s\n \n −1\n \n » ✓\n

\n
\n
\n (iii)\n
\n

\n P at (8, 1.2) ✓\n

\n
\n
\n (b)\n
\n

\n light acts as a wave «and not a particle in this situation» ✓\n

\n

\n light at slits will diffract / create a diffraction pattern ✓\n

\n

\n light passing through slits will interfere / create an interference pattern «creating bright and dark spots» ✓\n

\n
\n
\n (c)\n
\n

\n Ue of\n \n s\n \n =\n \n \n \n \n λ\n \n \n D\n \n \n \n d\n \n \n \n ⇒\n \n \n λ\n \n \n =\n \n \n \n \n s\n \n \n d\n \n \n \n D\n \n \n \n \n \n OR\n \n \n \n s\n \n =\n \n \n \n \n n\n \n \n λ\n \n \n D\n \n \n \n d\n \n \n \n ⇒\n \n \n λ\n \n \n =\n \n \n \n \n s\n \n \n d\n \n \n \n \n n\n \n \n D\n \n \n \n \n ✓\n

\n

\n \n \n λ\n \n \n = «\n \n \n \n \n 0\n \n \n .\n \n \n 567\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 18\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n 2\n \n \n .\n \n \n 2\n \n \n \n \n =» 4.6 × 10\n \n −7\n \n «m» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-1-simple-harmonic-motion", "c-2-wave-model", "c-3-wave-phenomena" ] }, { "question_id": "23M.2.SL.TZ1.4", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the speed of the spacecraft is 0.80\n \n c\n \n as measured in S.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Show that the current in Q is 0.45 A.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the resistance of R.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Calculate the total power dissipated in the circuit.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Resistor P is removed. State and explain, without any calculations, the effect of this on the resistance of Q.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n An event has coordinates\n \n \n x\n \n \n = 0 and\n \n ct\n \n = 0.60 ly in S. Show, using a Lorentz transformation, that the time coordinate of this event in S′ is\n \n ct\n \n ′ = 1.00 ly .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Label, on the diagram with the letter P, the point on the\n \n ct\n \n ′ axis whose\n \n ct\n \n ′ coordinate is 1.00 ly.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Draw lines to indicate R on the diagram.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Determine, using the diagram or otherwise, the space coordinate\n \n \n x\n \n \n ′ of event R in S′.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n moves 4 ly in 5 years\n \n \n OR\n \n \n slope of angle with time axis is 0.8 ✓\n

\n

\n

\n

\n \n Allow evidence for mark on the graph.\n \n

\n
\n
\n (i)\n
\n

\n Voltage across P is 1.4 «V» ✓\n

\n

\n Voltage across Q is 4.6 «V» ✓\n

\n

\n And 6 – 1.4 = 4.6 «V» ✓\n

\n

\n

\n

\n \n Need to see a calculation involving the two voltages and the total voltage in the circuit for\n \n MP3\n \n (e.g. 1.4 + 4.6 = 6).\n \n

\n
\n
\n (ii)\n
\n

\n Current in R is «(0.45 − 0.4)=» 0.05 A ✓\n

\n

\n So resistance is «\n \n \n \n \n 1\n \n \n .\n \n \n 4\n \n \n \n \n 0\n \n \n .\n \n \n 05\n \n \n \n \n » = 28 «Ω»\n

\n

\n

\n

\n \n Allow\n \n ECF\n \n from\n \n a(i)\n \n \n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n \n

\n
\n
\n (iii)\n
\n

\n «0.45 × 6.0» = 2.7 «W»✓\n

\n
\n
\n (b)\n
\n

\n Q will have a smaller resistance ✓\n

\n

\n «Because total resistance in the circuit is now larger so» the current «through the\n
\n circuit/Q» is smaller /\n \n \n OWTTE\n \n \n ✓\n \n \n
\n \n \n

\n

\n \n Allow similar argument for\n \n MP2\n \n based on voltage across\n \n Q\n \n becoming smaller.\n \n

\n
\n
\n (i)\n
\n

\n \n γ\n \n = 1.67\n \n \n OR\n \n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n OR\n \n \n \n \n \n 1\n \n \n \n \n \n 1\n \n \n -\n \n \n 0\n \n \n .\n \n \n \n 8\n \n \n 2\n \n \n \n \n \n \n \n ✓\n

\n

\n \n ct\n \n ' = «\n \n γ\n \n (\n \n ct\n \n −\n \n \n \n \n v\n \n \n x\n \n \n \n c\n \n \n \n )» =\n \n \n \n 5\n \n \n 3\n \n \n \n × (0.60 +0) ✓\n

\n

\n «= 1.00 ly»\n

\n

\n

\n

\n \n For\n \n MP2\n \n , working should be seen.\n \n

\n
\n
\n (ii)\n
\n

\n identifies point with coordinates\n \n \n x\n \n \n = 0,\n \n ct\n \n = 0.60 on vertical axis ✓\n

\n

\n draws line parallel to the\n \n \n x\n \n \n prime axis until it intersects the prime\n \n ct\n \n axis ✓\n

\n

\n \n

\n

\n

\n

\n \n \n Award [2]\n \n for correct position of P without working shown.\n \n

\n
\n
\n (i)\n
\n

\n \n

\n

\n R located at (4,4) ✓\n

\n

\n «as intersection of» vertical line through 4 ly and photon worldline at 45 degrees✓\n

\n

\n

\n

\n \n Allow\n \n MP2\n \n even if one of the lines is not drawn.\n \n

\n
\n
\n (ii)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n \n Using diagram:\n \n

\n

\n line from R parallel to prime ct axis until it intersects space axis ✓\n

\n

\n use of scale from (b) to estimate coordinate to\n \n \n x\n \n \n ' = (1.3 ± 0.2) ly ✓\n

\n

\n \n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n Using Lorentz transformation:\n \n

\n

\n event R has coordinates\n \n \n x\n \n \n =\n \n ct\n \n = 4.00 ly in S ✓\n

\n

\n so\n \n \n x\n \n \n ' = «\n \n \n y\n \n \n (\n \n \n x\n \n \n −\n \n vt\n \n ) =\n \n \n \n 5\n \n \n 3\n \n \n \n × (4.00 − 0.80 × 4.00)» = 1.33 ly ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter" ], "subtopics": [ "a-5-galilean-and-special-relativity", "b-5-current-and-circuits" ] }, { "question_id": "23M.2.SL.TZ1.5", "Question": "
\n
\n (a)\n
\n
\n

\n According to laboratory observers\n \n \n \n \n number\n \n \n \n \n of\n \n \n \n \n muons\n \n \n \n \n detected\n \n \n \n \n number\n \n \n \n \n of\n \n \n \n \n muons\n \n \n \n \n produced\n \n \n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n .\n

\n

\n Calculate\n \n D\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Show that the energy released is about 18 MeV.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n determine the time taken for the detector to reach the muon source.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate, using the answers to (b)(i) and (b)(ii) the ratio\n \n \n \n \n number\n \n \n \n \n of\n \n \n \n \n muons\n \n \n \n \n detected\n \n \n \n \n number\n \n \n \n \n of\n \n \n \n \n muons\n \n \n \n \n produced\n \n \n \n \n in S.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n \n two\n \n difficulties of energy production by nuclear fusion.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n \n one\n \n advantage of energy production by nuclear fusion compared to nuclear fission.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Discuss the ratios in (a) and (c).\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the nucleon number of the He isotope that\n \n \n \n H\n \n \n \n \n 1\n \n \n 3\n \n \n \n decays into.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n T\n \n \n \n 1\n \n \n 2\n \n \n \n \n = 2.00 × 1.56 × 10\n \n −6\n \n or  3.12 × 10\n \n −6\n \n s ✓\n

\n

\n \n D\n \n = «3.12 × 10\n \n −6\n \n × 0.866 × 3 × 10\n \n 8\n \n =» 811 m\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n \n

\n
\n
\n (i)\n
\n

\n «𝜇» = 2.0141 + 3.0160 − (4.0026 + 1.008665) «= 0.0188 u»\n

\n

\n \n \n OR\n \n \n

\n

\n \n In\n \n MeV: 1876.13415 + 2809.404 − (3728.4219 + 939.5714475) ✓\n

\n

\n

\n

\n = 0.0188 × 931.5\n \n \n OR\n \n \n = 17.512 «MeV» ✓\n

\n

\n

\n

\n \n Must see either clear substitutions or answer to at least 3 s.f. for\n \n MP2\n \n .\n \n

\n
\n
\n (ii)\n
\n

\n transit time =\n \n \n \n 405\n \n \n \n 0\n \n \n .\n \n \n 866\n \n \n c\n \n \n \n \n = 1.56\n \n µs\n \n ✓\n

\n

\n

\n

\n \n \n Award [1]\n \n for\n \n BCA\n \n .\n \n

\n
\n
\n (c)\n
\n

\n transit time is one half life ✓\n

\n

\n so ratio has to be\n \n \n \n 1\n \n \n 2\n \n \n \n ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n .\n \n

\n
\n
\n (i)\n
\n

\n Requires high temp/pressure ✓\n
\n Must overcome Coulomb/intermolecular repulsion ✓\n
\n Difficult to contain / control «at high temp/pressure» ✓\n
\n Difficult to produce excess energy/often energy input greater than output /\n \n \n OWTTE\n \n \n ✓\n
\n Difficult to capture energy from fusion reactions ✓\n
\n Difficult to maintain/sustain a constant reaction rate ✓\n

\n
\n
\n (ii)\n
\n

\n Plentiful fuel supplies\n \n \n OR\n \n \n larger specific energy\n \n \n OR\n \n \n larger energy density\n \n \n OR\n \n \n little or no «major radioactive» waste products ✓\n

\n

\n

\n

\n \n Allow descriptions such as “more energy per unit mass” or “more energy per unit volume”\n \n

\n
\n
\n (d)\n
\n

\n the answers are the same ✓\n

\n

\n count rates cannot vary from frame to frame /\n \n \n OWTTE\n \n \n ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n allow\n \n ECF\n \n from\n \n (c)\n \n .\n \n

\n

\n \n \n Award [2]\n \n for “count rates cannot vary” if student made a mistake\n \n OR\n \n no answer in\n \n (c)\n \n and well discussed here.\n \n

\n
\n
\n (i)\n
\n

\n 3 ✓\n

\n

\n

\n

\n \n Do\n \n not\n \n accept\n \n \n \n H\n \n \n \n \n 2\n \n \n 3\n \n \n \n e\n \n \n by itself.\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "e-nuclear-and-quantum-physics" ], "subtopics": [ "a-5-galilean-and-special-relativity", "e-3-radioactive-decay", "e-4-fission", "e-5-fusion-and-stars" ] }, { "question_id": "23M.2.SL.TZ1.6", "Question": "
\n
\n (a)\n
\n
\n

\n The moment of inertia of the rod about the axis is 0.180 kg m\n \n 2\n \n . Show that the moment of inertia of the rod–particle system is about 0.25 kg m\n \n 2\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Show that the angular speed of the system immediately after the collision is about 5.7 rad s\n \n −1\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate the energy lost during the collision.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the angular deceleration of the rod.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the number of revolutions made by the rod until it stops rotating.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n In another situation the rod rests on a horizontal frictionless surface with no pivot. Predict, without calculation, the motion of the rod–particle system after the collision.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 0.180 + 0.200 × 0.60\n \n 2\n \n «= 0.252 kg m\n \n 2\n \n » ✓\n

\n
\n
\n (b)\n
\n

\n angular speed of particle = «12/0.6 = » 20 «rad s\n \n −1\n \n »\n

\n

\n \n \n OR\n \n \n

\n

\n angular momentum of particle «0.200 × 12.0 × 0.60» = 1.44 «Js» ✓\n

\n

\n
\n «angular momentum of rod-particle system 0.252\n \n ω\n \n »\n

\n

\n equating\n \n ω\n \n = «\n \n \n \n \n 1\n \n \n .\n \n \n 44\n \n \n \n \n 0\n \n \n .\n \n \n 252\n \n \n \n \n » = 5.71 rad s\n \n −1\n \n ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , working or answer to at least 3 SF should be seen.\n \n

\n
\n
\n (c)\n
\n

\n \n \n \n 1\n \n \n 2\n \n \n \n × 0.200 × 12.0\n \n 2\n \n −\n \n \n \n 1\n \n \n 2\n \n \n \n (0.252) × 5.71\n \n 2\n \n ✓\n

\n

\n 10.3 J ✓\n

\n

\n

\n

\n \n \n Award [1]\n \n for answer 11.5\n \n J\n \n that neglects moment of inertia of particle but do not penalize this omission in\n \n (d)(i)\n \n .\n \n

\n
\n
\n (i)\n
\n

\n \n \n α\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 152\n \n \n \n \n 0\n \n \n .\n \n \n 252\n \n \n \n \n = 0.603 rads\n \n −2\n \n ✓\n

\n

\n

\n

\n \n Accept negative values.\n \n

\n
\n
\n (ii)\n
\n

\n \n θ\n \n =\n \n \n \n \n 5\n \n \n .\n \n \n \n 71\n \n \n 2\n \n \n \n \n \n 2\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 603\n \n \n \n \n = 27.0 rad ✓\n

\n

\n \n N\n \n =\n \n \n \n \n 27\n \n \n .\n \n \n 0\n \n \n \n \n 2\n \n \n ×\n \n \n π\n \n \n \n \n = 4.3 ✓\n

\n
\n
\n (e)\n
\n

\n the rod will rotate «about centre of mass» ✓\n

\n

\n «centre of mass» will move along straight line\n
\n «parallel to the particle’s initial velocity» ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , mention of translational motion is not enough.\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-3-work-energy-and-power", "a-4-rigid-body-mechanics" ] }, { "question_id": "23M.2.SL.TZ1.7", "Question": "
\n
\n (a)\n
\n
\n

\n Suggest why AC is the adiabatic part of the cycle.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Show that the volume at C is 3.33 × 10\n \n −2\n \n m\n \n 3\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Suggest, for the change A ⇒ B, whether the entropy of the gas is increasing, decreasing or constant.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Calculate the thermal energy (heat) taken out of the gas from B to C.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n The highest and lowest temperatures of the gas during the cycle are 602 K and 92 K.\n

\n

\n The efficiency of this engine is about 0.6. Outline how these data are consistent with the second law of thermodynamics.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n «considering expansions from A» an adiabatic process will reduce/change temperature ✓\n

\n

\n and so curve AC must be the steeper ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n temperature drop occurs for BC ✓\n

\n

\n therefore CA must increase temperature «via adiabatic process». ✓\n

\n
\n
\n (b)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n Use of adiabatic formula «\n \n \n \n p\n \n \n A\n \n \n \n \n \n V\n \n \n A\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n =\n \n \n \n p\n \n \n C\n \n \n \n \n \n V\n \n \n C\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n \n ⇒\n \n \n »\n \n \n \n V\n \n \n C\n \n \n \n =\n \n \n \n \n \n \n p\n \n \n A\n \n \n \n \n p\n \n \n C\n \n \n \n \n \n \n 3\n \n \n 5\n \n \n \n \n \n V\n \n \n A\n \n \n \n ✓\n

\n

\n \n \n \n V\n \n \n C\n \n \n \n =\n \n \n \n \n \n \n 5\n \n \n .\n \n \n 00\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n \n \n 4\n \n \n .\n \n \n 60\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n 3\n \n \n 5\n \n \n \n \n × 2.00 × 10\n \n −3\n \n «= 3.333 × 10\n \n −2\n \n m\n \n 3\n \n » ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , working or answer to at least 4 SF must be seen.\n \n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n V\n \n \n C\n \n =\n \n V\n \n \n B\n \n \n \n AND\n \n \n \n p\n \n \n A\n \n \n V\n \n \n A\n \n =\n \n p\n \n \n B\n \n \n V\n \n \n B\n \n ✓\n

\n

\n \n \n \n V\n \n \n C\n \n \n \n =\n \n \n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ×\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n 4\n \n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 3\n \n \n

\n

\n \n V\n \n \n C\n \n =\n \n V\n \n \n B\n \n \n \n AND\n \n \n \n n\n \n = 0.2 mol ✓\n

\n

\n \n V\n \n \n C\n \n = (0.2 × 8.31 × 602) / 4 × 10\n \n 4\n \n ✓\n

\n
\n
\n (c)\n
\n

\n Increasing ✓\n

\n

\n because thermal energy/heat is being provided to the gas « and temperature is constant,\n \n \n Δ\n \n \n S\n \n \n =\n \n \n \n \n Δ\n \n \n Q\n \n \n \n T\n \n \n \n ✓\n

\n
\n
\n (d)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n \n \n Q\n \n \n =\n \n \n Δ\n \n \n U\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n \n V\n \n \n C\n \n \n \n Δ\n \n \n P\n \n \n ✓\n

\n

\n \n \n Q\n \n \n =\n \n \n «\n \n \n \n 3\n \n \n 2\n \n \n \n × 3.33 × 10\n \n −2\n \n × (3.00 × 10\n \n 4\n \n − 4.60 × 10\n \n 3\n \n )» = 1268.7 ≈ 1270 «J» ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n .\n \n

\n

\n \n Accept negative values.\n \n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n \n R\n \n \n n\n \n \n =\n \n \n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ×\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n 602\n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 66\n \n \n \n \n OR\n \n \n \n T\n \n c\n \n \n = 4.6 × 10\n \n 3\n \n × 3.33 × 10\n \n −2\n \n × 1.66 = 92.2  ✓\n

\n

\n \n \n Δ\n \n \n U\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 661\n \n \n ×\n \n \n (\n \n \n 602\n \n \n -\n \n \n 92\n \n \n .\n \n \n 21\n \n \n )\n \n \n =\n \n \n 1270\n \n \n «J» ✓\n

\n

\n

\n

\n \n Award\n \n MP1\n \n if T\n \n c\n \n = 92 taken from\n \n (e)\n \n \n

\n
\n
\n (e)\n
\n

\n \n e\n \n \n c\n \n = 1 −\n \n \n \n 92\n \n \n 602\n \n \n \n = 0.847 ✓\n

\n

\n this engine has\n \n e\n \n <\n \n e\n \n \n c\n \n as it should ✓\n

\n

\n

\n

\n \n \n Award [0]\n \n if no calculation shown.\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "23M.2.SL.TZ1.8", "Question": "
\n
\n (a)\n
\n
\n

\n The moment of inertia of the rod about the axis is 0.180 kg m\n \n 2\n \n . Show that the moment of inertia of the rod–particle system is about 0.25 kg m\n \n 2\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Show that the angular speed of the system immediately after the collision is about 5.7 rad s\n \n −1\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate the energy lost during the collision.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the angular deceleration of the rod.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the number of revolutions made by the rod until it stops rotating.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n In another situation the rod rests on a horizontal frictionless surface with no pivot. Predict, without calculation, the motion of the rod–particle system after the collision.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 0.180 + 0.200 × 0.60\n \n 2\n \n «= 0.252 kg m\n \n 2\n \n » ✓\n

\n
\n
\n (b)\n
\n

\n angular speed of particle = «12/0.6 = » 20 «rad s\n \n −1\n \n »\n

\n

\n \n \n OR\n \n \n

\n

\n angular momentum of particle «0.200 × 12.0 × 0.60» = 1.44 «Js» ✓\n

\n

\n
\n «angular momentum of rod-particle system 0.252\n \n ω\n \n »\n

\n

\n equating\n \n ω\n \n = «\n \n \n \n \n 1\n \n \n .\n \n \n 44\n \n \n \n \n 0\n \n \n .\n \n \n 252\n \n \n \n \n » = 5.71 rad s\n \n −1\n \n ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , working or answer to at least 3 SF should be seen.\n \n

\n
\n
\n (c)\n
\n

\n \n \n \n 1\n \n \n 2\n \n \n \n × 0.200 × 12.0\n \n 2\n \n −\n \n \n \n 1\n \n \n 2\n \n \n \n (0.252) × 5.71\n \n 2\n \n ✓\n

\n

\n 10.3 J ✓\n

\n

\n

\n

\n \n \n Award [1]\n \n for answer 11.5\n \n J\n \n that neglects moment of inertia of particle but do not penalize this omission in\n \n (d)(i)\n \n .\n \n

\n
\n
\n (i)\n
\n

\n \n \n α\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 152\n \n \n \n \n 0\n \n \n .\n \n \n 252\n \n \n \n \n = 0.603 rads\n \n −2\n \n ✓\n

\n

\n

\n

\n \n Accept negative values.\n \n

\n
\n
\n (ii)\n
\n

\n \n θ\n \n =\n \n \n \n \n 5\n \n \n .\n \n \n \n 71\n \n \n 2\n \n \n \n \n \n 2\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 603\n \n \n \n \n = 27.0 rad ✓\n

\n

\n \n N\n \n =\n \n \n \n \n 27\n \n \n .\n \n \n 0\n \n \n \n \n 2\n \n \n ×\n \n \n π\n \n \n \n \n = 4.3 ✓\n

\n
\n
\n (e)\n
\n

\n the rod will rotate «about centre of mass» ✓\n

\n

\n «centre of mass» will move along straight line\n
\n «parallel to the particle’s initial velocity» ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , mention of translational motion is not enough.\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-3-work-energy-and-power", "a-4-rigid-body-mechanics" ] }, { "question_id": "23M.2.SL.TZ1.9", "Question": "
\n
\n (a)\n
\n
\n

\n Suggest why AC is the adiabatic part of the cycle.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Show that the volume at C is 3.33 × 10\n \n −2\n \n m\n \n 3\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Suggest, for the change A ⇒ B, whether the entropy of the gas is increasing, decreasing or constant.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Calculate the thermal energy (heat) taken out of the gas from B to C.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n The highest and lowest temperatures of the gas during the cycle are 602 K and 92 K.\n

\n

\n The efficiency of this engine is about 0.6. Outline how these data are consistent with the second law of thermodynamics.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n «considering expansions from A» an adiabatic process will reduce/change temperature ✓\n

\n

\n and so curve AC must be the steeper ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n temperature drop occurs for BC ✓\n

\n

\n therefore CA must increase temperature «via adiabatic process». ✓\n

\n
\n
\n (b)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n Use of adiabatic formula «\n \n \n \n p\n \n \n A\n \n \n \n \n \n V\n \n \n A\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n =\n \n \n \n p\n \n \n C\n \n \n \n \n \n V\n \n \n C\n \n \n \n \n 5\n \n \n 3\n \n \n \n \n \n \n ⇒\n \n \n »\n \n \n \n V\n \n \n C\n \n \n \n =\n \n \n \n \n \n \n p\n \n \n A\n \n \n \n \n p\n \n \n C\n \n \n \n \n \n \n 3\n \n \n 5\n \n \n \n \n \n V\n \n \n A\n \n \n \n ✓\n

\n

\n \n \n \n V\n \n \n C\n \n \n \n =\n \n \n \n \n \n \n 5\n \n \n .\n \n \n 00\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n \n \n 4\n \n \n .\n \n \n 60\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n 3\n \n \n 5\n \n \n \n \n × 2.00 × 10\n \n −3\n \n «= 3.333 × 10\n \n −2\n \n m\n \n 3\n \n » ✓\n

\n

\n

\n

\n \n For\n \n MP2\n \n , working or answer to at least 4 SF must be seen.\n \n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n V\n \n \n C\n \n =\n \n V\n \n \n B\n \n \n \n AND\n \n \n \n p\n \n \n A\n \n \n V\n \n \n A\n \n =\n \n p\n \n \n B\n \n \n V\n \n \n B\n \n ✓\n

\n

\n \n \n \n V\n \n \n C\n \n \n \n =\n \n \n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ×\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n 4\n \n \n \n \n \n ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 3\n \n \n

\n

\n \n V\n \n \n C\n \n =\n \n V\n \n \n B\n \n \n \n AND\n \n \n \n n\n \n = 0.2 mol ✓\n

\n

\n \n V\n \n \n C\n \n = (0.2 × 8.31 × 602) / 4 × 10\n \n 4\n \n ✓\n

\n
\n
\n (c)\n
\n

\n Increasing ✓\n

\n

\n because thermal energy/heat is being provided to the gas « and temperature is constant,\n \n \n Δ\n \n \n S\n \n \n =\n \n \n \n \n Δ\n \n \n Q\n \n \n \n T\n \n \n \n ✓\n

\n
\n
\n (d)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n \n \n Q\n \n \n =\n \n \n Δ\n \n \n U\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n \n V\n \n \n C\n \n \n \n Δ\n \n \n P\n \n \n ✓\n

\n

\n \n \n Q\n \n \n =\n \n \n «\n \n \n \n 3\n \n \n 2\n \n \n \n × 3.33 × 10\n \n −2\n \n × (3.00 × 10\n \n 4\n \n − 4.60 × 10\n \n 3\n \n )» = 1268.7 ≈ 1270 «J» ✓\n

\n

\n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n .\n \n

\n

\n \n Accept negative values.\n \n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n \n \n R\n \n \n n\n \n \n =\n \n \n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ×\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n 602\n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 66\n \n \n \n \n OR\n \n \n \n T\n \n c\n \n \n = 4.6 × 10\n \n 3\n \n × 3.33 × 10\n \n −2\n \n × 1.66 = 92.2  ✓\n

\n

\n \n \n Δ\n \n \n U\n \n \n =\n \n \n \n 3\n \n \n 2\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 661\n \n \n ×\n \n \n (\n \n \n 602\n \n \n -\n \n \n 92\n \n \n .\n \n \n 21\n \n \n )\n \n \n =\n \n \n 1270\n \n \n «J» ✓\n

\n

\n

\n

\n \n Award\n \n MP1\n \n if T\n \n c\n \n = 92 taken from\n \n (e)\n \n \n

\n
\n
\n (e)\n
\n

\n \n e\n \n \n c\n \n = 1 −\n \n \n \n 92\n \n \n 602\n \n \n \n = 0.847 ✓\n

\n

\n this engine has\n \n e\n \n <\n \n e\n \n \n c\n \n as it should ✓\n

\n

\n

\n

\n \n \n Award [0]\n \n if no calculation shown.\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "23M.2.SL.TZ2.1", "Question": "
\n
\n (a)\n
\n
\n

\n Estimate, using the graph, the maximum height of the bottle.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Estimate the acceleration of the bottle when it is at its maximum height.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate the fraction of the kinetic energy of the bottle that remains after the bounce.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The mass of the bottle is 27 g and it is in contact with the ground for 85 ms.\n

\n

\n Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n The maximum height reached by the bottle is greater with an air–water mixture than with only high-pressure air in the bottle.\n

\n

\n Assume that the speed at which the propellant leaves the bottle is the same in both cases.\n

\n

\n Explain why the bottle reaches a greater maximum height with an air–water mixture.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate the absolute uncertainty in\n \n \n \n F\n \n \n max\n \n \n 3\n \n \n \n for Δ\n \n p\n \n = 30 kPa. State an appropriate number of significant figures for your answer.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Draw the absolute uncertainty determined in part (d)(i) as an error bar on the graph.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Explain why the new hypothesis is supported.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n The maximum height reached by the bottle is greater with an air–water mixture than with only high-pressure air in the bottle.\n

\n

\n Assume that the speed at which the propellant leaves the bottle is the same in both cases.\n

\n

\n Explain why the bottle reaches a greater maximum height with an air–water mixture.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n Attempt to count squares ✓\n

\n

\n Area of one square found ✓\n

\n

\n 7.2 «m» (accept 6.4 – 7.4 m) ✓\n

\n

\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n Uses area equation for either triangle ✓\n

\n

\n Correct read offs for estimate of area of triangle ✓\n

\n

\n 7.2 «m» (accept 6.4 – 7.4) ✓\n

\n
\n
\n (b)\n
\n

\n Attempt to calculate gradient of line at\n \n t\n \n = 1.2 s ✓\n

\n

\n «−» 9.8 «m s\n \n −2\n \n » (accept 9.6 − 10.0) ✓\n

\n
\n
\n (i)\n
\n

\n Attempt to evaluate KE ratio as\n \n \n \n \n \n \n V\n \n \n final\n \n \n \n \n V\n \n \n initial\n \n \n \n \n \n 2\n \n \n \n ✓\n

\n

\n «\n \n \n \n \n \n \n 4\n \n \n .\n \n \n 5\n \n \n \n 10\n \n \n \n \n 2\n \n \n \n =» 0.20\n \n \n OR\n \n \n 20 %\n \n \n OR\n \n \n \n \n \n 1\n \n \n 5\n \n \n \n ✓\n

\n

\n

\n

\n \n Accept ± 0.5 velocity values from graph\n \n

\n
\n
\n (ii)\n
\n

\n Attempt to use force = momentum change ÷ time ✓\n

\n

\n «=\n \n \n \n \n (\n \n \n 4\n \n \n .\n \n \n 5\n \n \n +\n \n \n 10\n \n \n )\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 027\n \n \n \n \n 85\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n = 4.6»\n

\n

\n Force = «4.6 + 0.3» 4.9 «N» ✓\n

\n

\n Any answer to 2sf ✓\n

\n

\n \n Accept ± 0.5 velocity values from graph\n \n

\n
\n
\n (d)\n
\n

\n Mass «leaving the bottle per second» will be larger for air–water ✓\n

\n

\n the momentum change/force is greater ✓\n

\n

\n \n Allow opposite argument for air only\n \n

\n
\n
\n (i)\n
\n

\n 15 % seen anywhere ✓\n

\n

\n «Δ(\n \n F\n \n \n 3\n \n ) =» 39.4 × 10\n \n 5\n \n × 0.15 = 5.9 × 10\n \n 5\n \n ✓\n

\n

\n ±6 × 10\n \n 5\n \n ✓\n

\n

\n \n \n
\n MP1\n \n is for the propagation of 5 %. It can be shown differently, e.g. 3 × 5% Allow students to use 40 × 10\n \n 5\n \n (from the graph).\n \n

\n

\n \n Award\n \n MP3\n \n for\n \n any\n \n uncertainty rounded to 1 significant digit\n \n

\n

\n \n \n Award [3]\n \n for a\n \n BCA\n \n .\n \n

\n

\n \n Allow\n \n ECF\n \n from\n \n MP1\n \n and\n \n MP2\n \n \n

\n
\n
\n (ii)\n
\n

\n error bar drawn at 30 kPa from 34 × 10\n \n 5\n \n to 46 × 10\n \n 5\n \n N\n \n 3\n \n ✓\n

\n

\n \n
\n Allow ± half square on each side of the bar or one square overall (± 2 × 10\n \n 5\n \n )\n \n

\n

\n \n Allow\n \n ECF\n \n from\n \n d(i)\n \n .\n \n

\n
\n
\n (iii)\n
\n

\n a «straight» line can be drawn that passes through origin ✓\n

\n
\n
\n (e)\n
\n

\n Mass «leaving the bottle per second» will be larger for air–water ✓\n

\n

\n the momentum change/force is greater ✓\n

\n

\n

\n

\n \n Allow opposite argument for air only\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "inquiry", "tools" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power", "i-2-3-interpreting-results", "inquiry-2-collecting-and-processing-data", "tool-3-mathematics" ] }, { "question_id": "23M.2.SL.TZ2.11", "Question": "
\n
\n (i)\n
\n
\n

\n Determine which star will appear to move more.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate, in m, the distance to star X.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Determine the ratio\n \n \n \n \n luminosity\n \n \n \n \n of\n \n \n \n \n star\n \n \n \n \n X\n \n \n \n \n luminosity\n \n \n \n \n of\n \n \n \n \n star\n \n \n \n \n Y\n \n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (i)\n
\n

\n Star Y ✓\n

\n

\n because parallax angle is greater\n \n \n OR\n \n \n star Y is closer «and that means movement relative to distant stars is greater» ✓\n

\n

\n

\n

\n \n Allow reverse argument for star X\n \n

\n
\n
\n (ii)\n
\n

\n «distance =\n \n \n \n \n 1\n \n \n \n 0\n \n \n .\n \n \n 019\n \n \n \n \n \n × 3.26 × 9.46 × 10\n \n 15\n \n »\n

\n

\n 1.6 × 10\n \n 18\n \n «m» ✓\n

\n
\n
\n (iii)\n
\n

\n \n \n \n \n Luminosity\n \n \n \n \n of\n \n \n \n \n Star\n \n \n \n \n X\n \n \n \n \n Luminosity\n \n \n \n \n of\n \n \n \n \n Star\n \n \n \n \n Y\n \n \n \n \n =\n \n \n \n \n \n b\n \n \n x\n \n \n \n \n d\n \n \n x\n \n \n 2\n \n \n \n \n \n \n b\n \n \n y\n \n \n \n \n d\n \n \n y\n \n \n 2\n \n \n \n \n \n ✓\n

\n

\n = 10.8 ≈ 11 ✓\n

\n

\n

\n

\n \n Award\n \n MP1\n \n if ratio shown with distance or parallax angle.\n \n

\n

\n \n Award\n \n MP1\n \n for any correct substitution into ratio expression\n \n

\n

\n \n \n Award [2]\n \n for\n \n BCA\n \n \n

\n

\n \n Allow\n \n ECF\n \n for incorrect distances from\n \n b(i)\n \n or\n \n b(ii)\n \n .\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-1-thermal-energy-transfers", "e-5-fusion-and-stars" ] }, { "question_id": "23M.2.SL.TZ2.12", "Question": "
\n
\n (a)\n
\n
\n

\n State the main element that is undergoing nuclear fusion in star C.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Explain why star B has a greater surface area than star A.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n White dwarfs with similar volumes to each other are shown on the HR diagram.\n

\n

\n Sketch, on the HR diagram, to show the possible positions of other white dwarf stars with similar volumes to those marked on the HR diagram.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Hydrogen ✓\n

\n
\n
\n (b)\n
\n

\n stars have same/similar\n \n L\n \n \n \n AND\n \n \n star B has lower\n \n T\n \n ✓\n

\n

\n correct reference to luminosity formula (\n \n L\n \n α\n \n AT\n \n \n 4\n \n ) ✓\n

\n

\n

\n

\n \n \n MP1\n \n Allow reverse argument i.e., star A has higher T\n \n

\n
\n
\n (c)\n
\n

\n Any evidence of correct identification that three dots bottom left represent white dwarfs ✓\n

\n

\n line passing through all 3 white dwarfs\n \n \n OR\n \n \n line continuing from 3 white dwarfs with approximately same gradient, in either direction ✓\n

\n

\n

\n

\n \n Award\n \n MP2\n \n if no line drawn through the three dots but just beyond them in either direction\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-1-thermal-energy-transfers", "e-5-fusion-and-stars" ] }, { "question_id": "23M.2.SL.TZ2.17", "Question": "
\n
\n (a)\n
\n
\n

\n State the main element that is undergoing nuclear fusion in star C.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Explain why star B has a greater surface area than star A.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n White dwarfs with similar volumes to each other are shown on the HR diagram.\n

\n

\n Sketch, on the HR diagram, to show the possible positions of other white dwarf stars with similar volumes to those marked on the HR diagram.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Hydrogen ✓\n

\n
\n
\n (b)\n
\n

\n stars have same/similar\n \n L\n \n \n \n AND\n \n \n star B has lower\n \n T\n \n ✓\n

\n

\n correct reference to luminosity formula (\n \n L\n \n α\n \n AT\n \n \n 4\n \n ) ✓\n

\n

\n

\n

\n \n \n MP1\n \n Allow reverse argument i.e., star A has higher T\n \n

\n
\n
\n (c)\n
\n

\n Any evidence of correct identification that three dots bottom left represent white dwarfs ✓\n

\n

\n line passing through all 3 white dwarfs\n \n \n OR\n \n \n line continuing from 3 white dwarfs with approximately same gradient, in either direction ✓\n

\n

\n

\n

\n \n Award\n \n MP2\n \n if no line drawn through the three dots but just beyond them in either direction\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-1-thermal-energy-transfers", "e-5-fusion-and-stars" ] }, { "question_id": "23M.2.SL.TZ2.2", "Question": "
\n
\n (a)\n
\n
\n

\n State the unit for\n \n pV\n \n in fundamental SI units.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine, using the graph, whether the gas acts as an ideal gas.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate, in g, the mass of the gas.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n kg m\n \n 2\n \n s\n \n −2\n \n ✓\n

\n
\n
\n (b)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n Graph shown is a straight line/linear\n
\n \n \n OR\n \n \n
\n expected graph should be a straight line/linear ✓\n

\n

\n If ideal then\n \n T\n \n intercept must be at\n \n T\n \n = −273 °C ✓\n

\n

\n Use of\n \n y\n \n =\n \n mx\n \n +\n \n c\n \n to show that\n \n x\n \n = −273 °C when\n \n y\n \n = 0 ✓\n
\n (hence ideal)\n

\n

\n \n \n ALTERNATIVE 2\n \n \n

\n

\n Calculates\n \n \n \n \n p\n \n \n V\n \n \n \n T\n \n \n \n for two different points ✓\n
\n Obtains 1.50 «J K\n \n −1\n \n » for both ✓\n

\n

\n States that for ideal gas\n \n \n \n \n p\n \n \n V\n \n \n \n T\n \n \n \n =\n \n \n n\n \n \n R\n \n \n which is constant and concludes that gas is ideal ✓\n

\n
\n
\n (c)\n
\n

\n Use of\n \n \n n\n \n \n =\n \n \n \n \n p\n \n \n V\n \n \n \n \n R\n \n \n T\n \n \n \n \n \n \n OR\n \n \n \n \n N\n \n \n =\n \n \n \n \n p\n \n \n V\n \n \n \n \n k\n \n \n T\n \n \n \n \n ✓\n

\n

\n Mass of gas =\n \n n\n \n ×\n \n N\n \n \n A\n \n × mass of molecule\n
\n \n \n OR\n \n \n
\n Mass of gas = N\n \n \n × mass of molecule ✓\n

\n

\n 5.1 «g» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "tools" ], "subtopics": [ "b-3-gas-laws", "tool-3-mathematics" ] }, { "question_id": "23M.2.SL.TZ2.3", "Question": "
\n
\n (a)\n
\n
\n

\n Explain the pattern seen on the screen.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Calculate, in nm,\n \n \n λ\n \n \n .\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate, in nm,\n \n \n λ\n \n \n .\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The student moves the screen closer to the double slit and repeats the measurements. The instruments used to make the measurements are unchanged.\n

\n

\n Discuss the effect this movement has on the fractional uncertainty in the value of\n \n \n λ\n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The student changes the light source to one that emits two colours:\n
\n • blue light of wavelength\n \n \n λ\n \n \n , and\n
\n • red light of wavelength 1.5\n \n \n λ\n \n \n .\n

\n

\n Predict the pattern that the student will see on the screen.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Mention of interference / superposition ✓\n

\n

\n Bright fringe occurs when light from the slits arrives in phase ✓\n

\n

\n Dark fringe occurs when light from the slits arrives 180°/\n \n \n π\n \n \n out of phase ✓\n

\n
\n
\n (b)\n
\n

\n \n s\n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 15\n \n \n \n 8\n \n \n \n \n \n OR\n \n \n = 0.0188 «m» ✓\n

\n

\n use of\n \n \n λ\n \n \n =\n \n \n \n \n d\n \n \n s\n \n \n \n D\n \n \n \n ✓\n

\n

\n 450 «nm» ✓\n

\n
\n
\n (i)\n
\n

\n \n s\n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 15\n \n \n \n 8\n \n \n \n \n \n OR\n \n \n = 0.0188 «m» ✓\n

\n

\n use of\n \n \n λ\n \n \n =\n \n \n \n \n d\n \n \n s\n \n \n \n D\n \n \n \n ✓\n

\n

\n 450 «nm» ✓\n

\n
\n
\n (ii)\n
\n

\n «As the measurements decrease» the fractional uncertainty in D/s increases. ✓\n

\n

\n «Fractional uncertainties are additive here» so fractional uncertainty in\n \n \n λ\n \n \n increases ✓\n

\n

\n \n Answers can be described in symbols e.g. Δs/s\n \n

\n
\n
\n (c)\n
\n

\n Blue fringe is unchanged ✓\n

\n

\n Red fringes are farther apart than blue ✓\n

\n

\n By a factor of 1.5 ✓\n

\n

\n At some point/s the fringes coincide/are purple ✓\n

\n
\n", "Examiners report": "None", "topics": [ "c-wave-behaviour", "tools" ], "subtopics": [ "c-3-wave-phenomena", "tool-3-mathematics" ] }, { "question_id": "23M.2.SL.TZ2.4", "Question": "
\n
\n (a)\n
\n
\n

\n The designers state that the energy transferred by the resistor every second is 15 J.\n

\n

\n Calculate the current in the resistor.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n The resistor has a cross-sectional area of 9.6 × 10\n \n −6\n \n m\n \n 2\n \n .\n

\n

\n Show that a resistor made from carbon fibre will be suitable for the pad.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The power supply to the pad has a negligible internal resistance.\n

\n

\n State and explain the variation in current in the resistor as the temperature of the pad increases.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n outline the magnetic force acting on it due to the current in PQ.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n state and explain the net magnetic force acting on it due to the currents in PQ and TU.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n The design of the pad encloses the resistor in a material that traps air. The design also places the resistor close to the top surface of the pad.\n

\n

\n \n

\n

\n Explain, with reference to thermal energy transfer, why the pad is designed in this way.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n I\n \n = «\n \n \n \n \n P\n \n \n R\n \n \n \n \n =» 1.9 «A» ✓\n

\n
\n
\n (i)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n (Calculation of length)\n
\n Read off from graph [2.8 − 3.2 × 10\n \n −5\n \n Ω m]✓\n
\n Use of\n \n \n I\n \n \n =\n \n \n \n \n R\n \n \n A\n \n \n \n ρ\n \n \n \n ✓\n
\n \n l\n \n = 1.3 − 1.4 «m» ✓\n

\n

\n \n \n ALTERNATIVE 2\n \n \n (Calculation of area)\n
\n Read off from graph [2.8 − 3.2 × 10\n \n −5\n \n Ω m]✓\n
\n Use of\n \n \n A\n \n \n =\n \n \n ρ\n \n \n \n I\n \n \n R\n \n \n \n ✓\n

\n

\n \n A\n \n = 8.3 − 9.5 × 10\n \n −6\n \n «m\n \n 2\n \n » ✓\n

\n

\n \n \n ALTERNATIVE 3\n \n \n (Calculation of resistance)\n
\n Read off from graph [2.8 − 3.2 × 10\n \n −5\n \n Ω m]✓\n
\n Use of\n \n \n R\n \n \n =\n \n \n ρ\n \n \n \n I\n \n \n A\n \n \n \n ✓\n

\n

\n \n R\n \n = 3.6 − 4.2 «Ω» ✓\n

\n

\n \n \n ALTERNATIVE 4\n \n \n (Calculation of resistivity)\n
\n Use of\n \n \n ρ\n \n \n =\n \n \n \n \n R\n \n \n A\n \n \n \n I\n \n \n \n ✓\n

\n

\n \n \n \n ρ\n \n \n \n = 3.2 × 10\n \n −5\n \n «Ω m» ✓\n

\n

\n Read off from graph 260 – 280 K ✓\n

\n
\n
\n (ii)\n
\n

\n «Resistivity and hence» resistance will decrease ✓\n

\n

\n «Pd across pad will not change because internal resistance is negligible»\n
\n Current will increase ✓\n

\n
\n
\n (i)\n
\n

\n «The force is» away from PQ/repulsive/to the right ✓\n

\n
\n
\n (ii)\n
\n

\n The magnetic fields «due to currents in PQ and TU» are in opposite directions\n
\n \n \n OR\n \n \n
\n There are two «repulsive» forces in opposite directions ✓\n

\n

\n Net force is zero ✓\n

\n
\n
\n (d)\n
\n

\n Air is a poor «thermal» conductor ✓\n

\n

\n Lack of convection due to air not being able to move in material ✓\n

\n

\n Appropriate statement about energy transfer between the pet, the resistor and surroundings ✓\n

\n

\n The rate of thermal energy transfer to the top surface is greater than the bottom «due to thinner material» ✓\n

\n

\n

\n

\n \n Accept air is a good insulator\n \n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "d-fields" ], "subtopics": [ "b-1-thermal-energy-transfers", "b-5-current-and-circuits", "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "23M.2.SL.TZ2.5", "Question": "
\n
\n (a)\n
\n
\n

\n Outline what is meant by an isotope.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n mass number.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n proton number.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n A beta-minus particle and an alpha particle have the same initial kinetic energy.\n

\n

\n Outline why the beta-minus particle can travel further in air than the alpha particle.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n «An atom with» the same number of protons\n \n \n AND\n \n \n different numbers of neutrons\n
\n \n \n OR\n \n \n
\n Same chemical properties\n \n \n AND\n \n \n different physical properties ✓\n

\n

\n \n Do\n \n not\n \n allow just atomic number and mass number\n \n

\n
\n
\n (i)\n
\n

\n 3 ✓\n

\n
\n
\n (ii)\n
\n

\n 2 ✓\n

\n
\n
\n (e)\n
\n

\n Alphas have double charge «and so are better ionisers »✓\n

\n

\n alphas have more mass and therefore slower «for same energy» ✓\n

\n

\n so longer time/more likely to interact with the «atomic» electrons/atoms «and therefore better ionisers» ✓\n

\n

\n \n Accept reverse argument in terms of betas travelling faster.\n \n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-3-radioactive-decay" ] }, { "question_id": "23M.2.SL.TZ2.6", "Question": "
\n
\n (a)\n
\n
\n

\n Determine\n \n \n \n \n g\n \n \n M\n \n \n \n \n g\n \n \n P\n \n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n Outline one reason why this model of a dancer is unrealistic.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Work using\n \n g\n \n ∝\n \n \n \n m\n \n \n \n r\n \n \n 2\n \n \n \n \n ✓\n

\n

\n \n \n \n \n g\n \n \n M\n \n \n \n \n g\n \n \n P\n \n \n \n \n =\n \n \n \n \n m\n \n \n M\n \n \n \n \n m\n \n \n P\n \n \n \n \n \n \n \n \n r\n \n \n P\n \n \n \n \n r\n \n \n M\n \n \n \n \n \n 2\n \n \n \n = 0.75 ✓\n

\n
\n
\n (e)\n
\n

\n one example specified\n \n eg\n \n friction, air resistance, mass distribution not modelled ✓\n

\n

\n

\n

\n \n \n Award [1]\n \n for any reasonable physical parameter that is not consistent with the model\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-4-rigid-body-mechanics", "d-1-gravitational-fields" ] }, { "question_id": "EXE.2.HL.TZ0.1", "Question": "
\n
\n (a.i)\n
\n
\n

\n State the principal energy change in nuclear fission.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the principal energy change in nuclear fission.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The energy released in the reaction is about 180 MeV. Estimate, in J, the energy released when 1 kg of\n \n \n \n U\n \n \n \n \n 92\n \n \n 235\n \n \n \n undergoes fission.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State and explain the decay mode of\n \n \n \n Te\n \n \n \n \n 52\n \n \n 132\n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate, in s\n \n −1\n \n , the initial activity of the sample.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Show that the decay constant of a nuclide is given by −\n \n m\n \n , where\n \n m\n \n is the slope of the graph of ln\n \n A\n \n against\n \n t\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Determine, in days, the half-life of\n \n \n \n Te\n \n \n \n \n 52\n \n \n 132\n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Outline the role of the heat exchanger in a nuclear power station.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the maximum temperature of the gas during the cycle.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Outline why the entropy of the gas remains constant during changes BC and DA.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Determine the efficiency of the cycle.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Show that the rotational kinetic energy of the turbine decreases at a constant rate.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n Mass-energy «of uranium» into kinetic energy of fission products ✓\n

\n
\n
\n (i)\n
\n

\n Mass-energy «of uranium» into kinetic energy of fission products ✓\n

\n
\n
\n (ii)\n
\n

\n Mass of uranium nucleus\n \n \n ≈\n \n \n 235\n \n \n u\n \n \n ✓\n

\n

\n \n \n \n energy\n \n \n mass\n \n \n \n =\n \n \n \n \n 180\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 60\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n \n \n 235\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 661\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 27\n \n \n \n \n \n \n ✓\n

\n

\n \n \n 7\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 13\n \n \n \n «J» ✓\n

\n
\n
\n (i)\n
\n

\n beta minus decay ✓\n

\n

\n \n \n \n Te\n \n \n \n \n 52\n \n \n 132\n \n \n \n has more neutrons / higher\n \n \n \n N\n \n \n Z\n \n \n \n ratio than stable nuclides of similar\n \n A\n \n «and beta minus reduces\n \n \n \n N\n \n \n Z\n \n \n \n » ✓\n

\n
\n
\n (ii)\n
\n

\n \n \n \n e\n \n \n 25\n \n \n \n =\n \n \n 7\n \n \n .\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n 10\n \n \n \n «s\n \n −1\n \n » ✓\n

\n
\n
\n (iii)\n
\n

\n Takes ln of both sides of\n \n \n A\n \n \n =\n \n \n \n A\n \n \n 0\n \n \n \n \n e\n \n \n \n -\n \n \n λ\n \n \n t\n \n \n \n \n , leading to\n \n \n ln\n \n \n \n \n A\n \n \n =\n \n \n ln\n \n \n \n \n \n A\n \n \n 0\n \n \n \n -\n \n \n λ\n \n \n t\n \n \n ✓\n

\n

\n «hence slope\n \n \n =\n \n \n -\n \n \n λ\n \n \n » ✓\n

\n
\n
\n (iv)\n
\n

\n Slope =«−»\n \n \n 2\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n «s\n \n −1\n \n » ✓\n

\n

\n \n \n \n T\n \n \n \n 1\n \n \n 2\n \n \n \n \n =\n \n \n «\n \n \n \n \n ln\n \n \n \n \n 2\n \n \n \n \n 2\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n ×\n \n \n 60\n \n \n ×\n \n \n 60\n \n \n ×\n \n \n 24\n \n \n \n \n =\n \n \n » 3.2 «days» ✓\n

\n
\n
\n (i)\n
\n

\n Collects thermal energy from the coolant and delivers it to the gas ✓\n

\n

\n Prevents the «irradiated» coolant from leaving the reactor vessel ✓\n

\n
\n
\n (ii)\n
\n

\n Correct read offs of pressure and volume at B ✓\n

\n

\n \n \n 8\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 0\n \n \n ×\n \n \n 8\n \n \n .\n \n \n 31\n \n \n ×\n \n \n T\n \n \n ✓\n

\n

\n \n \n T\n \n \n =\n \n \n 1500\n \n \n «K»  ✓\n

\n
\n
\n (iii)\n
\n

\n From\n \n \n Δ\n \n \n S\n \n \n =\n \n \n \n \n Δ\n \n \n Q\n \n \n \n T\n \n \n \n , the change in entropy is zero when\n \n \n Δ\n \n \n Q\n \n \n =\n \n \n 0\n \n \n ✓\n

\n
\n
\n (iv)\n
\n

\n Net work done  = «\n \n \n 8\n \n \n .\n \n \n 23\n \n \n +\n \n \n 9\n \n \n .\n \n \n 11\n \n \n -\n \n \n 4\n \n \n .\n \n \n 32\n \n \n -\n \n \n 3\n \n \n .\n \n \n 25\n \n \n » 9.77 «kJ» ✓\n

\n

\n Efficiency =«\n \n \n \n \n 9\n \n \n .\n \n \n 77\n \n \n \n \n 20\n \n \n .\n \n \n 58\n \n \n \n \n =\n \n \n » 0.47 ✓\n

\n
\n
\n (d)\n
\n

\n Rotational KE is proportional to\n \n \n \n ω\n \n \n 2\n \n \n \n ✓\n

\n

\n Calculation of\n \n \n \n ω\n \n \n 2\n \n \n \n for at least four points, e.g. {96.1, 76.7, 57.6, 38.4, 19.3}×10\n \n 3\n \n

\n

\n Shows that the differences in equal time intervals are approximately the same, e.g. {19.4, 19.1, 19.2, 19.1, 19.3}×10\n \n 3\n \n ✓\n

\n

\n

\n

\n \n Allow a tolerance of ±1×10\n \n 3\n \n s\n \n −2\n \n from the values stated in MP2.\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "a-4-rigid-body-mechanics", "b-3-gas-laws", "b-4-thermodynamics", "e-3-radioactive-decay", "e-4-fission" ] }, { "question_id": "EXE.2.HL.TZ0.10", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the angular impulse applied to the flywheel.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The angular speed of the flywheel increased by 280 rad s\n \n −1\n \n during the application of the angular impulse.\n

\n

\n Determine the moment of inertia of the flywheel.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The flywheel was rotating at 150 rev per minute before the application of the angular impulse. Determine the change in angular rotational energy of the flywheel during the application of the flywheel.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Attempt to find area of triangle ✓\n

\n

\n \n \n \n 1\n \n \n 2\n \n \n \n ×\n \n \n 8\n \n \n .\n \n \n 0\n \n \n ×\n \n \n 180\n \n \n =\n \n \n 720\n \n \n «kg m\n \n 2\n \n s\n \n −1\n \n » ✓\n

\n
\n
\n (a)\n
\n

\n Attempt to find area of triangle ✓\n

\n

\n \n \n \n 1\n \n \n 2\n \n \n \n ×\n \n \n 8\n \n \n .\n \n \n 0\n \n \n ×\n \n \n 180\n \n \n =\n \n \n 720\n \n \n «kg m\n \n 2\n \n s\n \n −1\n \n » ✓\n

\n
\n
\n (b)\n
\n

\n Use of\n \n \n Δ\n \n \n L\n \n \n =\n \n \n Δ\n \n \n \n \n I\n \n \n ω\n \n \n \n \n ✓\n

\n

\n 2.6 kg m\n \n 2\n \n ✓\n

\n
\n
\n (c)\n
\n

\n Correct conversions to a consistent set of\n \n \n ω\n \n \n units ✓\n

\n

\n \n \n \n 1\n \n \n 2\n \n \n \n ×\n \n \n L\n \n \n ×\n \n \n \n \n \n ω\n \n \n f\n \n \n 2\n \n \n \n -\n \n \n \n ω\n \n \n i\n \n \n 2\n \n \n \n \n \n or correct substitution seen ✓\n

\n

\n 113 kJ ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "EXE.2.HL.TZ0.11", "Question": "
\n
\n (a)\n
\n
\n

\n Explain why the gas in configuration B has a greater number of microstates than in A.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Deduce, with reference to entropy, that the expansion of the gas from the initial configuration A is irreversible.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Configuration A has only one microstate ✓\n

\n

\n In configuration B, pairs of particles can be swapped between the halves ✓\n

\n

\n Every such change gives rise to a new microstate «so there is a large number of microstates in B» ✓\n

\n
\n
\n (a)\n
\n

\n Configuration A has only one microstate ✓\n

\n

\n In configuration B, pairs of particles can be swapped between the halves ✓\n

\n

\n Every such change gives rise to a new microstate «so there is a large number of microstates in B» ✓\n

\n
\n
\n (b)\n
\n

\n The entropy of the gas is related to the number of microstates\n

\n

\n \n OR\n \n

\n

\n \n \n \n S\n \n \n A\n \n \n \n =\n \n \n \n k\n \n \n B\n \n \n \n \n \n ln\n \n \n \n \n \n Ω\n \n \n A\n \n \n \n and\n \n \n \n S\n \n \n A\n \n \n \n =\n \n \n \n k\n \n \n B\n \n \n \n \n \n ln\n \n \n \n \n \n Ω\n \n \n B\n \n \n \n ✓\n

\n

\n
\n Since\n \n \n \n \n \n Ω\n \n \n B\n \n \n \n >\n \n \n \n Ω\n \n \n A\n \n \n \n , the entropy in configuration B is greater ✓\n

\n

\n A process that results in an increase of entropy in an isolated system is irreversible ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "EXE.2.HL.TZ0.12", "Question": "
\n
\n (a)\n
\n
\n

\n State what is meant by an isolated system.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n State and explain the number of microstates of the system in configuration A.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Configuration B has 120 microstates. Calculate the entropy difference between configurations B and A. State the answer in terms of\n \n \n \n k\n \n \n B\n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n The system is initially in configuration A. Comment, with reference to the second law of thermodynamics and your answer in (c), on the likely evolution of the system.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Neither mass nor energy is exchanged with the surroundings ✓\n

\n
\n
\n (a)\n
\n

\n Neither mass nor energy is exchanged with the surroundings ✓\n

\n
\n
\n (b)\n
\n

\n 6 microstates ✓\n

\n

\n Any of the six particles can be the one of the highest energy ✓\n

\n
\n
\n (c)\n
\n

\n \n \n \n k\n \n \n B\n \n \n \n \n \n ln\n \n \n \n \n 120\n \n \n -\n \n \n \n k\n \n \n B\n \n \n \n \n \n ln\n \n \n \n \n 6\n \n \n ✓\n

\n

\n \n \n 3\n \n \n .\n \n \n 0\n \n \n \n k\n \n \n B\n \n \n \n ✓\n

\n
\n
\n (d)\n
\n

\n The second law predicts that isolated systems spontaneously evolve towards high-entropy states ✓\n

\n

\n From (c), the entropy of B is greater than that of A ✓\n

\n

\n The final state will likely be similar to B / contain relatively many low-energy particles «of different energies» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "EXE.2.HL.TZ0.14", "Question": "
\n
\n (a)\n
\n
\n

\n Determine the fractional number of throws for which the three most likely macrostates occur.\n

\n

\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n A throw is made once every minute. Estimate the average time required before a throw occurs where all coins are heads or all coins are tails.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n In one throw the coins all land heads upwards. The following throw results in 7 heads and 3 tails. Calculate, in terms of\n \n \n \n k\n \n \n B\n \n \n \n , the change in entropy between the two throws.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Recognises that 4 heads and 6 tails also required ✓\n

\n

\n Total number of microstates = 672 ✓\n

\n

\n Fractional number = 672/1024 = 0.66 ✓\n

\n

\n

\n

\n \n Allow ecf for MP3\n \n

\n
\n
\n (a)\n
\n

\n Recognises that 4 heads and 6 tails also required ✓\n

\n

\n Total number of microstates = 672 ✓\n

\n

\n Fractional number = 672/1024 = 0.66 ✓\n

\n

\n

\n

\n \n Allow ecf for MP3\n \n

\n
\n
\n (b)\n
\n

\n Two chances in 1024 so once every 512 throws ✓\n

\n

\n 512 throws take 8.5 h so (a reasonable estimate is half way through) on average 4.3 h ✓\n

\n
\n
\n (c)\n
\n

\n \n \n Δ\n \n \n S\n \n \n =\n \n \n \n k\n \n \n B\n \n \n \n \n \n \n \n ln\n \n \n \n \n 120\n \n \n -\n \n \n ln\n \n \n \n \n 1\n \n \n \n \n ✓\n

\n

\n \n \n =\n \n \n 4\n \n \n .\n \n \n 8\n \n \n \n k\n \n \n B\n \n \n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-4-thermodynamics" ] }, { "question_id": "EXE.2.HL.TZ0.15", "Question": "
\n
\n (i)\n
\n
\n

\n orbital speed;\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n escape speed from its orbit.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n in its initial circular orbit;\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n in its final orbit.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (i)\n
\n

\n «\n \n \n \n \n \n 6\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n ×\n \n \n 5\n \n \n .\n \n \n 97\n \n \n ×\n \n \n \n 10\n \n \n 24\n \n \n \n \n \n 6\n \n \n .\n \n \n 70\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n \n \n =\n \n \n »\n \n \n 7\n \n \n .\n \n \n 71\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n «m s\n \n −1\n \n » ✓\n

\n
\n
\n (ii)\n
\n

\n «\n \n \n \n \n \n 2\n \n \n ×\n \n \n 6\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n ×\n \n \n 5\n \n \n .\n \n \n 97\n \n \n ×\n \n \n \n 10\n \n \n 24\n \n \n \n \n \n 6\n \n \n .\n \n \n 70\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n \n \n = »\n \n \n 1\n \n \n .\n \n \n 09\n \n \n ×\n \n \n \n 10\n \n \n 4\n \n \n \n \n \n «\n \n \n m\n \n \n \n \n \n s\n \n \n \n -\n \n \n 1\n \n \n \n \n »\n \n \n ✓\n

\n
\n
\n (i)\n
\n

\n Negative ✓\n

\n
\n
\n (ii)\n
\n

\n Positive ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields" ] }, { "question_id": "EXE.2.HL.TZ0.16", "Question": "
\n
\n (a)\n
\n
\n

\n The radius of the dwarf planet Pluto is 1.19 x 10\n \n 6\n \n m. The acceleration due to gravity at its surface is 0.617 m s\n \n −2\n \n .\n

\n

\n Determine the escape speed for an object at the surface of Pluto.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Pluto rotates about an axis through its centre. Its rotation is in the opposite sense to that of the Earth, i.e. from east to west.\n

\n

\n Explain the advantage of an object launching from the equator of Pluto and travelling to the west.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n v\n \n \n esc\n \n \n \n =\n \n \n \n \n \n 2\n \n \n G\n \n \n M\n \n \n \n r\n \n \n \n \n AND\n \n \n g\n \n \n =\n \n \n \n \n G\n \n \n M\n \n \n \n \n r\n \n \n 2\n \n \n \n \n seen ✓\n

\n

\n \n \n \n v\n \n \n esc\n \n \n \n =\n \n \n \n \n \n 2\n \n \n g\n \n \n \n r\n \n \n 2\n \n \n \n \n r\n \n \n \n \n ✓\n

\n

\n Leading to\n \n \n \n v\n \n \n esc\n \n \n \n =\n \n \n \n 2\n \n \n g\n \n \n r\n \n \n \n ✓\n

\n

\n 1.2 km s\n \n −1\n \n ✓\n \n \n

\n
\n
\n (a)\n
\n

\n \n \n \n v\n \n \n esc\n \n \n \n =\n \n \n \n \n \n 2\n \n \n G\n \n \n M\n \n \n \n r\n \n \n \n \n AND\n \n \n g\n \n \n =\n \n \n \n \n G\n \n \n M\n \n \n \n \n r\n \n \n 2\n \n \n \n \n seen ✓\n

\n

\n \n \n \n v\n \n \n esc\n \n \n \n =\n \n \n \n \n \n 2\n \n \n g\n \n \n \n r\n \n \n 2\n \n \n \n \n r\n \n \n \n \n ✓\n

\n

\n Leading to\n \n \n \n v\n \n \n esc\n \n \n \n =\n \n \n \n 2\n \n \n g\n \n \n r\n \n \n \n ✓\n

\n

\n 1.2 km s\n \n −1\n \n ✓\n \n \n

\n
\n
\n (b)\n
\n

\n Object at equator has the maximum linear/tangential speed possible ✓\n

\n

\n It therefore has maximum kinetic energy before takeoff (and this is not required from the fuel) ✓\n

\n

\n Idea that the object is already moving in direction of planet before takeoff ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields" ] }, { "question_id": "EXE.2.HL.TZ0.17", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the energy of the scattered photon is about 16 keV.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine the wavelength of the incident photon.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Outline why the results of the experiment are inconsistent with the wave model of electromagnetic radiation.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Calculate the scattering angle of the photon.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n \n 1\n \n \n .\n \n \n 24\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n 7\n \n \n .\n \n \n 80\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n \n \n «= 15.9 keV» ✓\n

\n
\n
\n (a)\n
\n

\n \n \n \n \n 1\n \n \n .\n \n \n 24\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n 7\n \n \n .\n \n \n 80\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n \n \n «= 15.9 keV» ✓\n

\n
\n
\n (b)\n
\n

\n Energy of incident photon =«\n \n \n 15\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n +\n \n \n 500\n \n \n =\n \n \n »\n \n \n 16\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n «eV» ✓\n

\n

\n Wavelength of incident photon =«\n \n \n \n \n 1\n \n \n .\n \n \n 24\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n 16\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n =\n \n \n »\n \n \n 7\n \n \n .\n \n \n 56\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n «m» ✓\n

\n
\n
\n (c)\n
\n

\n The wavelength of the X-rays changes ✓\n

\n

\n According to the wave model, the wavelength of the incident and scattered X-rays should be the same ✓\n

\n
\n
\n (d)\n
\n

\n \n \n \n \n 7\n \n \n .\n \n \n 80\n \n \n -\n \n \n 7\n \n \n .\n \n \n 56\n \n \n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 11\n \n \n \n \n =\n \n \n \n \n 1\n \n \n .\n \n \n 24\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n 0\n \n \n .\n \n \n 511\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n \n \n \n 1\n \n \n -\n \n \n cos\n \n \n \n \n θ\n \n \n \n \n ✓\n

\n

\n \n \n θ\n \n \n =\n \n \n 89\n \n \n °\n \n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-2-quantum-physics" ] }, { "question_id": "EXE.2.HL.TZ0.18", "Question": "
\n
\n (a)\n
\n
\n

\n Outline why the pattern observed on the screen is an evidence for matter waves.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n A typical interatomic distance in the graphite crystal is of the order of\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 10\n \n \n \n \n m. Estimate the minimum value of\n \n U\n \n for the pattern in (a) to be formed on the screen.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Protons can also be accelerated by the same potential difference\n \n U\n \n . Compare, without calculation, the de Broglie wavelength of the protons to that of the electrons.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n The pattern is formed when the electrons scattered from adjacent planes in the graphite crystal undergo interference / diffract ✓\n

\n

\n Interference / diffraction is a property of waves only ✓\n

\n
\n
\n (a)\n
\n

\n The pattern is formed when the electrons scattered from adjacent planes in the graphite crystal undergo interference / diffract ✓\n

\n

\n Interference / diffraction is a property of waves only ✓\n

\n
\n
\n (b)\n
\n

\n The de Broglie wavelength of the electrons should be comparable to or shorter than the interatomic distance /\n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 10\n \n \n \n \n m ✓\n

\n

\n Momentum of electrons\n \n \n ≈\n \n \n \n \n 6\n \n \n .\n \n \n 63\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 34\n \n \n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 10\n \n \n \n \n \n \n «\n \n \n =\n \n \n 3\n \n \n .\n \n \n 32\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n N s» ✓\n

\n

\n Kinetic energy of electrons = «\n \n \n \n \n p\n \n \n 2\n \n \n \n \n 2\n \n \n m\n \n \n \n \n =\n \n \n \n \n \n \n 3\n \n \n .\n \n \n 32\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 24\n \n \n \n \n \n \n 2\n \n \n \n \n 2\n \n \n ×\n \n \n 9\n \n \n .\n \n \n 11\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 31\n \n \n \n \n \n \n =\n \n \n »\n \n \n 6\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 18\n \n \n \n \n «J»  ✓\n

\n

\n \n U\n \n = «\n \n \n \n E\n \n \n e\n \n \n \n =\n \n \n \n \n 6\n \n \n .\n \n \n 10\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 18\n \n \n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n \n \n =\n \n \n »40 «V» ✓\n

\n
\n
\n (c)\n
\n

\n The protons have the same energy but greater mass hence a greater momentum than the electrons ✓\n

\n

\n From\n \n \n λ\n \n \n =\n \n \n \n h\n \n \n p\n \n \n \n , the protons will have a shorter wavelength ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-2-quantum-physics" ] }, { "question_id": "EXE.2.HL.TZ0.19", "Question": "
\n
\n (a)\n
\n
\n

\n State the de Broglie hypothesis.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Calculate the maximum speed of the electrons in the beam.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n After passing through the circular hole the electrons strike a fluorescent screen.\n

\n

\n Predict whether an apparatus such as this can demonstrate that moving electrons have wave properties.\n

\n
\n
\n

\n [4]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n a moving particle has wave properties ✓\n

\n

\n de Broglie wavelength = Planck constant÷momentum (must define p if quoted as equation) ✓\n

\n
\n
\n (a)\n
\n

\n a moving particle has wave properties ✓\n

\n

\n de Broglie wavelength = Planck constant÷momentum (must define p if quoted as equation) ✓\n

\n
\n
\n (b)\n
\n

\n use of ½\n \n mv\n \n \n 2\n \n =\n \n eV ✓\n \n

\n

\n 1.33 × 10\n \n 7\n \n m s\n \n −1\n \n ✓\n

\n
\n
\n (c)\n
\n

\n idea that de Broglie wavelength and hole size must be similar ✓\n

\n

\n Use of\n \n \n λ\n \n \n =\n \n \n \n h\n \n \n \n 2\n \n \n m\n \n \n e\n \n \n V\n \n \n \n \n ✓\n

\n

\n Leading to\n \n \n λ\n \n \n around 5 x 10\n \n −11\n \n m (which is unrealistic for a practical situation) ✓\n

\n

\n It would not be possible to construct this hole\n
\n \n OR\n \n
\n hole must be smaller than an atom so impossible ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-2-quantum-physics" ] }, { "question_id": "EXE.2.HL.TZ0.2", "Question": "
\n
\n (a.i)\n
\n
\n

\n State the movement direction for which the geophone has its greatest sensitivity.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the movement direction for which the geophone has its greatest sensitivity.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Outline how an emf is generated in the coil.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Explain why the magnitude of the emf is related to the amplitude of the ground movement.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n In one particular event, a maximum emf of 65 mV is generated in the geophone. The geophone coil has 150 turns.\n

\n

\n Calculate the rate of flux change that leads to this emf.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Suggest\n \n two\n \n changes to the system that will make the geophone more sensitive.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Show that, when sound travels from clay to sandstone, the critical angle is approximately 40°.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The angle between the clay–air surface and\n \n path 1\n \n is 80°.\n

\n

\n Draw, on the diagram, the subsequent path of a sound wave that travels initially in the clay along\n \n path 1\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Calculate\n \n d\n \n .\n

\n
\n
\n

\n [4]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n Vertical direction / parallel to springs ✓\n

\n
\n
\n (i)\n
\n

\n Vertical direction / parallel to springs ✓\n

\n
\n
\n (ii)\n
\n

\n The magnetic field moves relative to the coil ✓\n

\n

\n As field lines cut the coil, forces act on (initially stationary) electrons in the wire (and these move producing an emf) ✓\n

\n
\n
\n (iii)\n
\n

\n The springs have a natural time period for the oscillation ✓\n

\n

\n A greater amplitude of movement leads to higher magnet speed (with constant time period) ✓\n

\n

\n So field lines cut coil more quickly leading to greater emf ✓\n

\n
\n
\n (iv)\n
\n

\n Use of\n \n \n ε\n \n \n =\n \n \n -\n \n \n \n \n Δ\n \n \n \n \n N\n \n \n Φ\n \n \n \n \n \n \n Δ\n \n \n t\n \n \n \n \n ✓\n

\n

\n \n \n \n ΔΦ\n \n \n \n Δ\n \n \n t\n \n \n \n \n =\n \n \n \n -\n \n \n \n \n \n 65\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n 150\n \n \n \n

\n

\n \n \n =\n \n \n 0\n \n \n .\n \n \n 43\n \n \n mWb s\n \n −1\n \n ✓\n

\n
\n
\n (b)\n
\n

\n Any two suggestions from:\n

\n

\n Increase number of turns in coil ✓\n
\n Because more flux cutting per cycle ✓\n

\n

\n Increase field strength of magnet  ✓\n
\n So that there are more field lines ✓\n

\n

\n Change mass-spring system so that time period decreases ✓\n
\n So magnet will be moving faster for given amplitude of movement ✓\n

\n
\n
\n (i)\n
\n

\n \n \n c\n \n n\n \n s\n \n \n \n \n =\n \n \n \n \n 3\n \n \n .\n \n \n 0\n \n \n \n \n 4\n \n \n .\n \n \n 7\n \n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 57\n \n \n ✓\n

\n

\n Critical angle\n \n \n =\n \n \n \n sin\n \n \n \n -\n \n \n 1\n \n \n \n \n \n \n 1\n \n \n \n 1\n \n \n .\n \n \n 57\n \n \n \n \n \n =\n \n \n 39\n \n \n .\n \n \n 6\n \n \n °\n \n \n ✓\n

\n
\n
\n (ii)\n
\n

\n ray shown reflected back into the clay (and then to Z) at (by eye) the incidence angle ✓\n

\n

\n ray shown refracted into the sandstone with angle of refraction greater than angle of incidence (by eye) ✓\n

\n
\n
\n (d)\n
\n

\n distance difference\n \n \n =\n \n \n \n \n 3000\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 00667\n \n \n =\n \n \n \n \n 19\n \n \n .\n \n \n 8\n \n \n m ✓\n

\n

\n ½ distance difference\n \n \n =\n \n \n 9\n \n \n .\n \n \n 9\n \n \n m so YZ\n \n \n =\n \n \n 49\n \n \n .\n \n \n 9\n \n \n m ✓\n

\n

\n \n \n d\n \n \n =\n \n \n \n \n \n YZ\n \n \n 2\n \n \n \n -\n \n \n \n \n \n XZ\n \n \n 2\n \n \n \n \n 2\n \n \n \n \n \n =\n \n \n \n 49\n \n \n .\n \n \n \n 9\n \n \n 2\n \n \n \n -\n \n \n \n 40\n \n \n 2\n \n \n \n \n ✓\n

\n

\n
\n 29.8 m ✓\n

\n

\n \n OR\n \n

\n

\n Recognises situation as (almost) 3:4:5 triangle ✓\n

\n

\n
\n 30 m (\n \n 1 sf answer only accepted in this route\n \n ) ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "c-wave-behaviour", "d-fields" ], "subtopics": [ "a-1-kinematics", "c-1-simple-harmonic-motion", "c-3-wave-phenomena", "d-4-induction" ] }, { "question_id": "EXE.2.HL.TZ0.20", "Question": "
\n
\n (a)\n
\n
\n

\n The quantity\n \n \n \n h\n \n \n \n \n m\n \n \n e\n \n \n \n c\n \n \n \n \n is known as the Compton wavelength.\n

\n

\n Show that the Compton wavelength is about 2.4 pm.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the wavelength of the photon after the interaction.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Outline why the wavelength of the photon has changed.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Deduce the scattering angle for the photon.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Determine, in J, the kinetic energy of the electron after the interaction.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 2.43 pm ✓\n

\n
\n
\n (a)\n
\n

\n 2.43 pm ✓\n

\n
\n
\n (i)\n
\n

\n 8.43 pm ✓\n

\n
\n
\n (ii)\n
\n

\n (Energy of photon inversely prop to wavelength)\n

\n

\n

\n

\n photon transfers some of its energy to the electron. ✓\n

\n

\n If its energy decreases so its wavelength increases. ✓\n

\n
\n
\n (iii)\n
\n

\n (for this interaction)\n

\n

\n \n \n Δ\n \n \n λ\n \n \n =\n \n \n \n h\n \n \n \n \n m\n \n \n e\n \n \n \n c\n \n \n \n \n \n \n 1\n \n \n -\n \n \n cos\n \n \n \n \n θ\n \n \n \n \n =\n \n \n \n h\n \n \n \n \n m\n \n \n e\n \n \n \n c\n \n \n \n \n and therefore\n \n \n \n \n 1\n \n \n -\n \n \n cos\n \n \n \n \n θ\n \n \n \n \n must equal 1 ✓\n

\n

\n so cos theta = 0 and theta = 90 deg ✓\n

\n
\n
\n (iv)\n
\n

\n (Because energy is conserved)\n

\n

\n
\n \n \n \n E\n \n \n k\n \n \n \n =\n \n \n h\n \n \n c\n \n \n \n \n \n 1\n \n \n \n λ\n \n \n i\n \n \n \n \n -\n \n \n \n 1\n \n \n \n λ\n \n \n f\n \n \n \n \n \n \n

\n

\n \n \n =\n \n \n 6\n \n \n .\n \n \n 63\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 34\n \n \n \n \n ×\n \n \n 3\n \n \n .\n \n \n 00\n \n \n ×\n \n \n \n 10\n \n \n 8\n \n \n \n \n \n \n 1\n \n \n \n 6\n \n \n .\n \n \n 00\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 12\n \n \n \n \n \n \n -\n \n \n \n 1\n \n \n \n 8\n \n \n .\n \n \n 43\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 12\n \n \n \n \n \n \n \n \n ✓\n

\n

\n 9.6 fJ ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-2-quantum-physics" ] }, { "question_id": "EXE.2.HL.TZ0.21", "Question": "
\n
\n (a)\n
\n
\n

\n Suggest one problem that is faced in dealing with the waste from nuclear fission reactors. Go on to outline how this problem is overcome.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Strontium-90 is a waste product from nuclear reactors that has a decay constant of 7.63 x 10\n \n −10\n \n s\n \n −1\n \n . Determine, in s, the time that it takes for the activity of strontium-90 to decay to 2% of its original activity.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate the energy released when one mole of strontium-90 decays to 2% of its original activity forming the stable daughter product.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Strontium-90 decays to Zirconium-90 via two successive beta emissions. Discuss whether all the energy released when strontium-90 decays to Zirconium-90 can be transferred to a thermal form.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Waste is very hot …\n

\n

\n … So has to be placed in cooling ponds to transfer the (thermal) energy away ✓\n

\n

\n

\n

\n \n OR\n \n

\n

\n Waste is very radioactive … ✓\n

\n

\n … So has to be placed in cooling ponds to absorb this radiation\n

\n

\n \n OR\n \n

\n

\n … So has to be handled remotely\n

\n

\n \n OR\n \n

\n

\n … So has to be transported in crash resistant casings / stored on site ✓\n

\n

\n

\n

\n \n OR\n \n

\n

\n Waste will be radioactive for thousands of years … ✓\n

\n

\n … So storage needs to be (eventually) in geologically stable areas ✓\n

\n
\n
\n (a)\n
\n

\n Waste is very hot …\n

\n

\n … So has to be placed in cooling ponds to transfer the (thermal) energy away ✓\n

\n

\n

\n

\n \n OR\n \n

\n

\n Waste is very radioactive … ✓\n

\n

\n … So has to be placed in cooling ponds to absorb this radiation\n

\n

\n \n OR\n \n

\n

\n … So has to be handled remotely\n

\n

\n \n OR\n \n

\n

\n … So has to be transported in crash resistant casings / stored on site ✓\n

\n

\n

\n

\n \n OR\n \n

\n

\n Waste will be radioactive for thousands of years … ✓\n

\n

\n … So storage needs to be (eventually) in geologically stable areas ✓\n

\n
\n
\n (b)\n
\n

\n \n \n ln\n \n \n \n \n \n \n 0\n \n \n .\n \n \n 02\n \n \n \n \n =\n \n \n -\n \n \n \n \n 7\n \n \n .\n \n \n 63\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 10\n \n \n \n \n \n \n t\n \n \n or equivalent seen ✓\n

\n

\n \n \n t\n \n \n =\n \n \n 5\n \n \n .\n \n \n 1\n \n \n Gs ✓\n

\n
\n
\n (i)\n
\n

\n Idea that the Yttrium half life is much less than Strontium so can assume all Yttrium energy is included. ✓\n

\n

\n \n \n 0\n \n \n .\n \n \n 98\n \n \n ×\n \n \n 6\n \n \n .\n \n \n 02\n \n \n ×\n \n \n \n 10\n \n \n 23\n \n \n \n ×\n \n \n \n \n 0\n \n \n .\n \n \n 52\n \n \n +\n \n \n 2\n \n \n .\n \n \n 3\n \n \n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n seen ✓\n

\n

\n \n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 67\n \n \n ×\n \n \n 1030\n \n \n \n \n eV\n \n \n \n \n

\n

\n Answer\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n =\n \n \n 270\n \n \n GJ ✓\n

\n
\n
\n (ii)\n
\n

\n (No)\n
\n (anti-)neutrinos are released in (both) decays ✓\n

\n

\n Carrying away energy because they interact poorly with matter ✓\n

\n

\n

\n

\n \n Ignore arguments relating to energy transferred to nucleus as this appears eventually as thermal energy.\n \n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-4-fission" ] }, { "question_id": "EXE.2.HL.TZ0.3", "Question": "
\n
\n (a.i)\n
\n
\n

\n State the vertical component of the total momentum of the balls after the collision.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the vertical component of the total momentum of the balls after the collision.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Hence, calculate the vertical component of the velocity of ball B after the collision.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine the angle\n \n θ\n \n that the velocity of ball B makes with the initial direction of motion of ball A.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Predict whether the collision is elastic.\n

\n
\n
\n

\n [4]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n Zero ✓\n

\n
\n
\n (i)\n
\n

\n Zero ✓\n

\n
\n
\n (ii)\n
\n

\n \n \n m\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 0\n \n \n ×\n \n \n sin\n \n \n \n \n 45\n \n \n °\n \n \n +\n \n \n m\n \n \n ×\n \n \n \n v\n \n \n y\n \n \n \n =\n \n \n 0\n \n \n ✓\n

\n

\n \n \n \n v\n \n \n y\n \n \n \n = «−»0.71 «m s\n \n −1\n \n » ✓\n

\n
\n
\n (b)\n
\n

\n The use of conservation of momentum in the horizontal direction, e.g.\n \n \n m\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 0\n \n \n ×\n \n \n cos\n \n \n \n \n 45\n \n \n °\n \n \n +\n \n \n m\n \n \n ×\n \n \n \n v\n \n \n x\n \n \n \n =\n \n \n m\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ✓\n

\n

\n \n \n \n v\n \n \n x\n \n \n \n =\n \n \n «\n \n \n 2\n \n \n .\n \n \n 0\n \n \n -\n \n \n 1\n \n \n .\n \n \n 0\n \n \n ×\n \n \n cos\n \n \n \n \n 45\n \n \n °\n \n \n =\n \n \n »\n \n \n 1\n \n \n .\n \n \n 3\n \n \n «m s\n \n −1\n \n » ✓\n

\n

\n \n \n θ\n \n \n =\n \n \n \n tan\n \n \n \n -\n \n \n 1\n \n \n \n \n \n \n \n v\n \n \n y\n \n \n \n \n v\n \n \n x\n \n \n \n \n \n =\n \n \n 29\n \n \n °\n \n \n ✓\n

\n
\n
\n (c)\n
\n

\n Initial kinetic energy\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n \n \n 2\n \n \n .\n \n \n 0\n \n \n \n \n 2\n \n \n \n =\n \n \n 2\n \n \n .\n \n \n 0\n \n \n \n \n \n m\n \n \n \n

\n

\n Final kinetic energy\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n \n \n \n \n 1\n \n \n .\n \n \n 0\n \n \n \n \n 2\n \n \n \n +\n \n \n \n \n \n 0\n \n \n .\n \n \n 71\n \n \n \n \n 2\n \n \n \n +\n \n \n \n \n \n 1\n \n \n .\n \n \n 3\n \n \n \n \n 2\n \n \n \n \n \n ✓\n

\n

\n \n \n 1\n \n \n .\n \n \n 6\n \n \n \n \n (\n \n \n m\n \n \n )\n \n \n ✓\n

\n

\n Final energy is less than the initial energy hence inelastic  ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "EXE.2.HL.TZ0.4", "Question": "
\n
\n (a.i)\n
\n
\n

\n State what is meant by an elastic collision.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State what is meant by an elastic collision.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n No unbalanced external forces act on the system of the curling stones. Outline why the momentum of the system does not change during the collision.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Show that\n \n \n \n v\n \n \n B\n \n \n \n =\n \n \n \n \n sin\n \n \n \n \n 70\n \n \n °\n \n \n \n \n sin\n \n \n \n \n 20\n \n \n °\n \n \n \n \n \n v\n \n \n A\n \n \n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Determine\n \n v\n \n \n A\n \n . State the answer in terms of\n \n v\n \n .\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n No change in the kinetic energy of the system  ✓\n

\n
\n
\n (i)\n
\n

\n No change in the kinetic energy of the system  ✓\n

\n
\n
\n (ii)\n
\n

\n From\n \n \n F\n \n \n =\n \n \n \n \n Δ\n \n \n p\n \n \n \n \n Δ\n \n \n t\n \n \n \n \n , zero net force on the system implies that\n \n \n Δ\n \n \n p\n \n \n =\n \n \n 0\n \n \n

\n

\n \n OR\n \n

\n

\n From Newton’s third law, the impulse delivered to A is equal but opposite to the impulse delivered to B, hence\n \n \n Δ\n \n \n p\n \n \n =\n \n \n 0\n \n \n for the system  ✓\n

\n
\n
\n (b)\n
\n

\n The vertical momentum is zero hence\n \n \n m\n \n \n \n v\n \n \n A\n \n \n \n sin\n \n \n \n \n 70\n \n \n °\n \n \n -\n \n \n m\n \n \n \n v\n \n \n B\n \n \n \n sin\n \n \n \n \n 20\n \n \n °\n \n \n =\n \n \n 0\n \n \n ✓\n

\n

\n «leading to the expected result»  ✓\n

\n
\n
\n (c)\n
\n

\n Energy is conserved hence\n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n \n v\n \n \n A\n \n \n \n 2\n \n \n \n +\n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n \n v\n \n \n B\n \n \n \n 2\n \n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n v\n \n \n 2\n \n \n \n ✓\n

\n

\n Eliminate\n \n \n \n v\n \n \n B\n \n \n \n using the result of part (a), e.g.,\n \n \n \n \n v\n \n \n A\n \n \n \n 2\n \n \n \n +\n \n \n \n \n \n \n sin\n \n \n \n \n 70\n \n \n °\n \n \n \n \n sin\n \n \n \n \n 20\n \n \n °\n \n \n \n \n \n 2\n \n \n \n \n \n v\n \n \n A\n \n \n \n 2\n \n \n \n =\n \n \n \n v\n \n \n 2\n \n \n \n ✓\n

\n

\n \n \n \n v\n \n \n A\n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 342\n \n \n v\n \n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "EXE.2.HL.TZ0.5", "Question": "
\n
\n (a.i)\n
\n
\n

\n Calculate the component of momentum of the first curling stone perpendicular to the initial direction.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate the component of momentum of the first curling stone perpendicular to the initial direction.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the velocity component of the first curling stone in the initial direction.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Determine the velocity of the first curling stone.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Deduce whether this collision is elastic.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n \n \n \n v\n \n \n 17\n \n \n \n \n \n sin\n \n \n \n \n θ\n \n \n =\n \n \n \n \n 19\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 5\n \n \n ×\n \n \n sin\n \n \n \n 30\n \n \n \n \n 17\n \n \n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 279\n \n \n \n \n m\n \n \n \n \n \n s\n \n \n \n -\n \n \n 1\n \n \n \n \n \n \n ✓\n

\n
\n
\n (i)\n
\n

\n \n \n \n v\n \n \n 17\n \n \n \n \n \n sin\n \n \n \n \n θ\n \n \n =\n \n \n \n \n 19\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 5\n \n \n ×\n \n \n sin\n \n \n \n 30\n \n \n \n \n 17\n \n \n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 279\n \n \n \n \n m\n \n \n \n \n \n s\n \n \n \n -\n \n \n 1\n \n \n \n \n \n \n ✓\n

\n
\n
\n (ii)\n
\n

\n \n \n \n v\n \n \n 17\n \n \n \n \n \n cos\n \n \n \n \n θ\n \n \n =\n \n \n \n \n 17\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 5\n \n \n -\n \n \n 19\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 50\n \n \n ×\n \n \n cos\n \n \n \n 30\n \n \n \n \n 17\n \n \n \n ✓\n

\n

\n \n \n =\n \n \n 2\n \n \n .\n \n \n 02\n \n \n \n \n \n m s\n \n \n \n -\n \n \n 1\n \n \n \n \n ✓\n

\n
\n
\n (iii)\n
\n

\n 2.04 m s\n \n −1\n \n ✓\n

\n

\n At 7.9° to initial direction  ✓\n

\n
\n
\n (b)\n
\n

\n Total angle between stones is 38°, angle will be 90° when elastic\n

\n

\n \n OR\n \n

\n

\n Compares kinetic energies in a correct calculation (initial ke = 53 J, final ke = 34 J +2.4 J) ✓\n

\n

\n
\n Collision is not elastic ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "EXE.2.HL.TZ0.6", "Question": "
\n
\n (a)\n
\n
\n

\n Determine the recoil velocity of the cannon.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Calculate the initial kinetic energy of the cannon.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Suggest what happens to the vertical component of momentum of the cannon when the shell is fired.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Momentum must be conserved in initial direction of shell (20° above horizontal)  ✓\n

\n

\n Recoil velocity is 4.2 m s\n \n −1\n \n at 20° below horizontal  ✓\n

\n

\n 3.95 m s\n \n −1\n \n ✓\n

\n
\n
\n (a)\n
\n

\n Momentum must be conserved in initial direction of shell (20° above horizontal)  ✓\n

\n

\n Recoil velocity is 4.2 m s\n \n −1\n \n at 20° below horizontal  ✓\n

\n

\n 3.95 m s\n \n −1\n \n ✓\n

\n
\n
\n (b)\n
\n

\n \n \n \n E\n \n \n k\n \n \n \n =\n \n \n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n v\n \n \n 2\n \n \n \n \n \n =\n \n \n 11\n \n \n .\n \n \n 7\n \n \n \n \n kJ\n \n \n ✓\n

\n
\n
\n (c)\n
\n

\n Must be transferred into the ground beneath the cannon\n \n OR\n \n into the suspension system  ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "EXE.2.HL.TZ0.7", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the angular impulse delivered to the flywheel during the acceleration.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine the average magnitude of\n \n \n \n F\n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n State\n \n two\n \n assumptions of your calculation in part (b).\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n Δ\n \n \n L\n \n \n =\n \n \n l\n \n \n Δ\n \n \n ω\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 8\n \n \n ×\n \n \n 0\n \n \n .\n \n \n \n 15\n \n \n 2\n \n \n \n \n \n \n \n 9\n \n \n .\n \n \n 0\n \n \n -\n \n \n 4\n \n \n .\n \n \n 0\n \n \n \n \n ✓\n

\n

\n \n \n 0\n \n \n .\n \n \n 10\n \n \n «N m s»  ✓\n

\n
\n
\n (a)\n
\n

\n \n \n Δ\n \n \n L\n \n \n =\n \n \n l\n \n \n Δ\n \n \n ω\n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 8\n \n \n ×\n \n \n 0\n \n \n .\n \n \n \n 15\n \n \n 2\n \n \n \n \n \n \n \n 9\n \n \n .\n \n \n 0\n \n \n -\n \n \n 4\n \n \n .\n \n \n 0\n \n \n \n \n ✓\n

\n

\n \n \n 0\n \n \n .\n \n \n 10\n \n \n «N m s»  ✓\n

\n
\n
\n (b)\n
\n

\n Average torque\n \n \n =\n \n \n \n \n Δ\n \n \n L\n \n \n \n \n Δ\n \n \n t\n \n \n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 10\n \n \n \n \n 0\n \n \n .\n \n \n 24\n \n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 42\n \n \n «N m» ✓\n

\n

\n Average force\n \n \n =\n \n \n \n τ\n \n \n R\n \n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 42\n \n \n \n \n 0\n \n \n .\n \n \n 15\n \n \n \n \n =\n \n \n 2\n \n \n .\n \n \n 8\n \n \n «N» ✓\n

\n
\n
\n (c)\n
\n

\n No other forces than\n \n F\n \n provide the torque ✓\n

\n

\n The thread unwinds without slipping ✓\n

\n

\n The thread is weightless ✓\n

\n

\n \n F\n \n is always tangent to the flywheel ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "EXE.2.HL.TZ0.8", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate:\n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the angular acceleration of the ring;\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the angular velocity of the ring after a time of 5.0 s.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the angular impulse delivered to the disc and to the ring during the first 5.0 s.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the final kinetic energy of the disc and the ring.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n α\n \n \n =\n \n \n \n τ\n \n \n l\n \n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 20\n \n \n \n \n 0\n \n \n .\n \n \n 32\n \n \n ×\n \n \n 0\n \n \n .\n \n \n \n 25\n \n \n 2\n \n \n \n \n \n ✓\n

\n

\n \n \n 10\n \n \n «rad s\n \n −2\n \n »  ✓\n

\n
\n
\n (i)\n
\n

\n \n \n α\n \n \n =\n \n \n \n τ\n \n \n l\n \n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 20\n \n \n \n \n 0\n \n \n .\n \n \n 32\n \n \n ×\n \n \n 0\n \n \n .\n \n \n \n 25\n \n \n 2\n \n \n \n \n \n ✓\n

\n

\n \n \n 10\n \n \n «rad s\n \n −2\n \n »  ✓\n

\n
\n
\n (ii)\n
\n

\n \n \n 10\n \n \n ×\n \n \n 5\n \n \n .\n \n \n 0\n \n \n =\n \n \n 50\n \n \n «rad s\n \n −1\n \n »  ✓\n

\n
\n
\n (i)\n
\n

\n The angular impulse is the product of torque and time ✓\n

\n

\n Both factors are the same so the angular impulse is the same ✓\n

\n
\n
\n (ii)\n
\n

\n The disc has a smaller moment of inertia «because its mass is distributed closer to the axis of rotation» ✓\n

\n

\n From\n \n \n \n E\n \n \n k\n \n \n \n =\n \n \n \n \n L\n \n \n 2\n \n \n \n \n 2\n \n \n l\n \n \n \n \n , the disc will achieve a greater kinetic energy «because\n \n \n \n L\n \n \n \n is the same for both» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "EXE.2.HL.TZ0.9", "Question": "
\n
\n (a)\n
\n
\n

\n For the propellor,\n \n \n L\n \n \n =\n \n \n 5\n \n \n .\n \n \n 0\n \n \n cm\n \n \n and\n \n \n M\n \n \n =\n \n \n 0\n \n \n .\n \n \n 035\n \n \n \n \n kg\n \n \n .\n

\n

\n Calculate the moment of inertia of the propellor.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate the angular impulse that acts on the propellor.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate, using your answer to (b)(i), the time taken by the propellor to attain this rotational speed.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n State and explain the effect of the angular impulse on the body of the aeroplane.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n 7\n \n \n .\n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n \n kg m\n \n \n 2\n \n \n \n ✓\n

\n
\n
\n (a)\n
\n

\n \n \n 7\n \n \n .\n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n \n \n \n kg m\n \n \n 2\n \n \n \n ✓\n

\n
\n
\n (i)\n
\n

\n 190 rev s\n \n −1\n \n = 1190 rad s\n \n −1\n \n ✓\n

\n

\n \n \n Δ\n \n \n L\n \n \n =\n \n \n Δ\n \n \n \n \n I\n \n \n ω\n \n \n \n \n =\n \n \n 7\n \n \n .\n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n ×\n \n \n 1190\n \n \n =\n \n \n 8\n \n \n .\n \n \n 7\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n \n \n kg m\n \n \n 2\n \n \n \n \n \n \n s\n \n \n \n -\n \n \n 1\n \n \n \n \n ✓\n

\n
\n
\n (ii)\n
\n

\n \n \n Δ\n \n \n t\n \n \n =\n \n \n \n \n Δ\n \n \n L\n \n \n \n τ\n \n \n \n used ✓\n

\n

\n 0.25 s ✓\n

\n
\n
\n (iii)\n
\n

\n As the motor is internal, angular momentum is conserved (ignoring the torques due to resistive forces) ✓\n

\n

\n The body of the plane will (try to) rotate in the opposite direction to the propellor ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-4-rigid-body-mechanics" ] }, { "question_id": "EXE.2.SL.TZ0.10", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the pressure of the gas in the container.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine the mass of the gas in the container.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate the average translational speed of the gas particles.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n The temperature of the gas in the container is increased.\n

\n

\n Explain, using the kinetic theory, how this change leads to a change in pressure in the container.\n

\n
\n
\n

\n [4]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n p\n \n \n =\n \n \n \n \n n\n \n \n R\n \n \n T\n \n \n \n V\n \n \n \n seen\n \n \n =\n \n \n 12\n \n \n \n \n kPa\n \n \n ✓\n

\n
\n
\n (a)\n
\n

\n \n \n p\n \n \n =\n \n \n \n \n n\n \n \n R\n \n \n T\n \n \n \n V\n \n \n \n seen\n \n \n =\n \n \n 12\n \n \n \n \n kPa\n \n \n ✓\n

\n
\n
\n (b)\n
\n

\n \n \n 6\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n kg\n \n \n ✓\n

\n
\n
\n (c)\n
\n

\n \n \n v\n \n \n =\n \n \n \n \n \n 3\n \n \n P\n \n \n V\n \n \n \n M\n \n \n \n \n =\n \n \n 450\n \n \n \n \n \n m s\n \n \n \n -\n \n \n 1\n \n \n \n \n ✓\n

\n
\n
\n (d)\n
\n

\n increased temperature means increased average KE and hence increased average translational speed ✓\n

\n

\n This increases the momentum transfer at the walls for each collision /\n \n mv\n \n is greater per collision ✓\n

\n

\n This increases the frequency of collisions at the walls / particles cover the distance between walls more quickly ✓\n

\n

\n Ideas that\n \n \n \n F\n \n \n wall\n \n \n \n =\n \n \n \n \n Δ\n \n \n p\n \n \n \n \n Δ\n \n \n t\n \n \n \n \n AND\n \n \n P\n \n \n =\n \n \n \n F\n \n \n A\n \n \n \n (can be in words) so that force increases and pressure\n \n increases\n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-3-gas-laws" ] }, { "question_id": "EXE.2.SL.TZ0.11", "Question": "
\n
\n (a)\n
\n
\n

\n A comet orbits the Sun in an elliptical orbit. A and B are two positions of the comet.\n

\n

\n \n

\n

\n Explain, with reference to Kepler’s second law of planetary motion, the change in the kinetic energy of the comet as it moves from A to B.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n An asteroid (minor planet) orbits the Sun in a circular orbit of radius 4.5 × 10\n \n 8\n \n km. The radius of Earth’s orbit is 1.5 × 10\n \n 8\n \n km. Calculate, in years, the orbital period of the asteroid.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n The areas swept out in unit time by the Sun-comet line are the same at A and B ✓\n

\n

\n At B, the distance is greater hence the orbital speed/distance moved in unit time is lower «so that the area remains the same» ✓\n

\n

\n A decrease in speed means that the kinetic energy also decreases ✓\n

\n
\n
\n (a)\n
\n

\n The areas swept out in unit time by the Sun-comet line are the same at A and B ✓\n

\n

\n At B, the distance is greater hence the orbital speed/distance moved in unit time is lower «so that the area remains the same» ✓\n

\n

\n A decrease in speed means that the kinetic energy also decreases ✓\n

\n
\n
\n (b)\n
\n

\n An attempt to use Kepler’s 3\n \n rd\n \n law, e.g.,\n \n \n \n \n \n T\n \n \n 1\n \n \n \n \n 2\n \n \n \n =\n \n \n \n \n \n \n 4\n \n \n .\n \n \n 5\n \n \n \n \n 1\n \n \n .\n \n \n 5\n \n \n \n \n \n 3\n \n \n \n ✓\n

\n

\n \n \n T\n \n \n =\n \n \n «\n \n \n \n \n \n \n 4\n \n \n .\n \n \n 5\n \n \n \n \n 1\n \n \n .\n \n \n 5\n \n \n \n \n \n \n 1\n \n \n .\n \n \n 5\n \n \n \n \n =\n \n \n » 5.2 «years» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields" ] }, { "question_id": "EXE.2.SL.TZ0.12", "Question": "
\n
\n (a)\n
\n
\n

\n Show that\n \n \n k\n \n \n =\n \n \n \n 1\n \n \n \n G\n \n \n M\n \n \n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The table gives data relating to the two moons of Mars.\n

\n

\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
\n Moon\n \n \n T\n \n / hour\n \n \n r\n \n / Mm\n
\n Phobos\n \n 7.66\n \n 9.38\n
\n Deimos\n \n 30.4\n \n -\n
\n

\n

\n

\n Determine\n \n r\n \n for Deimos.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Determine the mass of Mars.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Equates centripetal force (with Newton’s law of gravitation\n \n \n m\n \n \n r\n \n \n \n \n \n ω\n \n \n 2\n \n \n \n =\n \n \n \n \n G\n \n \n M\n \n \n m\n \n \n \n \n r\n \n \n 2\n \n \n \n \n )\n

\n

\n \n \n OR\n \n \n

\n

\n \n \n T\n \n \n =\n \n \n \n \n 2\n \n \n π\n \n \n \n ω\n \n \n \n ✓\n

\n

\n
\n Uses both equation correctly with clear re-arrangement ✓\n

\n
\n
\n (a)\n
\n

\n Equates centripetal force (with Newton’s law of gravitation\n \n \n m\n \n \n r\n \n \n \n \n \n ω\n \n \n 2\n \n \n \n =\n \n \n \n \n G\n \n \n M\n \n \n m\n \n \n \n \n r\n \n \n 2\n \n \n \n \n )\n

\n

\n \n \n OR\n \n \n

\n

\n \n \n T\n \n \n =\n \n \n \n \n 2\n \n \n π\n \n \n \n ω\n \n \n \n ✓\n

\n

\n
\n Uses both equation correctly with clear re-arrangement ✓\n

\n
\n
\n (b)\n
\n

\n \n \n \n r\n \n \n De\n \n \n 3\n \n \n \n =\n \n \n \n \n \n T\n \n \n De\n \n \n 2\n \n \n \n \n r\n \n \n Ph\n \n \n 3\n \n \n \n \n \n T\n \n \n Ph\n \n \n 2\n \n \n \n \n seen or correct substitution ✓\n

\n

\n 23.5 Mm ✓\n

\n
\n
\n (c)\n
\n

\n Converts\n \n T\n \n to 27.6 ks\n \n and\n \n converts to m from Mm ✓\n

\n

\n \n \n k\n \n \n =\n \n \n 7\n \n \n .\n \n \n 33\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 14\n \n \n \n \n «s\n \n 2\n \n m\n \n −3\n \n » ✓\n

\n

\n «\n \n \n M\n \n \n =\n \n \n \n 1\n \n \n \n k\n \n \n G\n \n \n \n \n »\n \n \n =\n \n \n 2\n \n \n .\n \n \n 04\n \n \n ×\n \n \n \n 10\n \n \n 23\n \n \n \n «kg» ✓\n

\n

\n

\n

\n \n MP1 can be implicit\n \n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields" ] }, { "question_id": "EXE.2.SL.TZ0.13", "Question": "
\n
\n (a)\n
\n
\n

\n Show that\n \n \n T\n \n \n ∝\n \n \n \n r\n \n \n \n 3\n \n \n 2\n \n \n \n \n for the planets in a solar system where\n \n \n \n T\n \n \n \n is the orbital period of a planet and\n \n \n \n r\n \n \n \n is the radius of circular orbit of planet about its sun.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Outline what is meant by one astronomical unit (1 AU)\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Pluto is a dwarf planet of the Sun that orbits at a distance of 5.9 × 10\n \n 9\n \n km from the Sun. Determine, in years, the orbital period of Pluto.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Equates centripetal force (with Newton’s law gravitation\n \n \n m\n \n \n r\n \n \n \n \n \n ω\n \n \n 2\n \n \n \n =\n \n \n \n \n G\n \n \n M\n \n \n m\n \n \n \n \n r\n \n \n 2\n \n \n \n \n )\n

\n

\n AND\n

\n

\n \n \n T\n \n \n =\n \n \n \n \n 2\n \n \n π\n \n \n \n ω\n \n \n \n ✓\n

\n

\n leads to\n \n \n \n T\n \n \n 2\n \n \n \n =\n \n \n \n r\n \n \n 3\n \n \n \n ×\n \n \n \n \n \n 4\n \n \n \n π\n \n \n 2\n \n \n \n \n \n G\n \n \n M\n \n \n \n \n \n hence result ✓\n

\n
\n
\n (a)\n
\n

\n Equates centripetal force (with Newton’s law gravitation\n \n \n m\n \n \n r\n \n \n \n \n \n ω\n \n \n 2\n \n \n \n =\n \n \n \n \n G\n \n \n M\n \n \n m\n \n \n \n \n r\n \n \n 2\n \n \n \n \n )\n

\n

\n AND\n

\n

\n \n \n T\n \n \n =\n \n \n \n \n 2\n \n \n π\n \n \n \n ω\n \n \n \n ✓\n

\n

\n leads to\n \n \n \n T\n \n \n 2\n \n \n \n =\n \n \n \n r\n \n \n 3\n \n \n \n ×\n \n \n \n \n \n 4\n \n \n \n π\n \n \n 2\n \n \n \n \n \n G\n \n \n M\n \n \n \n \n \n hence result ✓\n

\n
\n
\n (i)\n
\n

\n «mean» Distance from centre of Sun to centre of Earth ✓\n

\n

\n \n \n OR\n \n \n

\n

\n Suitable ratio in terms of parsec and arcsecond ✓\n

\n
\n
\n (ii)\n
\n

\n \n \n \n T\n \n \n Pluto\n \n \n 2\n \n \n \n =\n \n \n \n T\n \n \n Earth\n \n \n 2\n \n \n \n \n \n r\n \n \n Pluto\n \n \n 3\n \n \n \n \n r\n \n \n Earth\n \n \n 3\n \n \n \n \n used ✓\n

\n

\n Earth orbital radius = 1.5 × 10\n \n 11\n \n m (from AU)\n \n AND\n \n uses 1 earth year (in any units) ✓\n

\n

\n 247 years ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-1-gravitational-fields" ] }, { "question_id": "EXE.2.SL.TZ0.14", "Question": "
\n
\n (a)\n
\n
\n

\n State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Calculate the current in wire Q.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n State the direction of the current in R, relative to the current in P.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Deduce the current in R.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓\n

\n

\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n \n \n N\n \n \n ✓\n

\n
\n
\n (a)\n
\n

\n From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓\n

\n

\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n \n \n N\n \n \n ✓\n

\n
\n
\n (b)\n
\n

\n \n \n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n \n \n 0\n \n \n .\n \n \n 50\n \n \n \n \n =\n \n \n \n \n 4\n \n \n π\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n ×\n \n \n 5\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n I\n \n \n Q\n \n \n \n \n \n 2\n \n \n π\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 10\n \n \n \n \n ✓\n

\n

\n \n \n \n I\n \n \n Q\n \n \n \n =\n \n \n 4\n \n \n .\n \n \n 0\n \n \n «A» ✓\n

\n
\n
\n (i)\n
\n

\n Opposite ✓\n

\n
\n
\n (ii)\n
\n

\n The force on Q due to R must have the same magnitude «but opposite direction» as the force on Q due to P ✓\n

\n

\n The distance is halved therefore one half of the current is needed to produce the same force, so 2.5 A ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "EXE.2.SL.TZ0.15", "Question": "
\n
\n (a)\n
\n
\n

\n State the fundamental SI units for permeability of free space,\n \n \n \n μ\n \n \n 0\n \n \n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n magnetic field at A;\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n magnetic force on section AB of the loop.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n magnitude of the net force acting on the loop;\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n direction of the net force acting on the loop.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n kg m s\n \n −2\n \n A\n \n −2\n \n ✓\n

\n
\n
\n (a)\n
\n

\n kg m s\n \n −2\n \n A\n \n −2\n \n ✓\n

\n
\n
\n (i)\n
\n

\n Into the page ✓\n

\n
\n
\n (ii)\n
\n

\n Repulsive / to the right ✓\n

\n
\n
\n (i)\n
\n

\n \n \n \n \n \n \n 4\n \n \n π\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 0\n \n \n \n \n 2\n \n \n π\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 30\n \n \n \n \n -\n \n \n \n \n 4\n \n \n π\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n 1\n \n \n .\n \n \n 0\n \n \n \n \n 2\n \n \n π\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 50\n \n \n \n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 20\n \n \n ✓\n

\n

\n \n \n 1\n \n \n .\n \n \n 1\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n «N»  ✓\n

\n
\n
\n (ii)\n
\n

\n Repulsive / to the right ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "EXE.2.SL.TZ0.16", "Question": "
\n
\n (a)\n
\n
\n

\n Determine the magnetic force acting on the 15 Ω wire due to the current in the 30 Ω wire.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The magnetic field strength of Earth’s field at the location of the wires is 45 μT.\n

\n

\n Discuss the assumption made in this question.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Use of combination of resistors OR\n \n \n V\n \n \n =\n \n \n I\n \n \n R\n \n \n ✓\n

\n

\n To show that current in 30 Ω wire is 5.0 A ✓\n

\n

\n \n \n \n F\n \n \n I\n \n \n \n =\n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n \n I\n \n \n 15\n \n \n \n \n I\n \n \n 30\n \n \n \n \n r\n \n \n \n =\n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n ×\n \n \n 10\n \n \n ×\n \n \n 5\n \n \n \n \n 0\n \n \n .\n \n \n 20\n \n \n \n \n ✓\n

\n

\n \n \n \n \n F\n \n \n =\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 25\n \n \n =\n \n \n \n \n \n \n 1\n \n \n .\n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n N ✓\n

\n
\n
\n (a)\n
\n

\n Use of combination of resistors OR\n \n \n V\n \n \n =\n \n \n I\n \n \n R\n \n \n ✓\n

\n

\n To show that current in 30 Ω wire is 5.0 A ✓\n

\n

\n \n \n \n F\n \n \n I\n \n \n \n =\n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n \n I\n \n \n 15\n \n \n \n \n I\n \n \n 30\n \n \n \n \n r\n \n \n \n =\n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n ×\n \n \n 10\n \n \n ×\n \n \n 5\n \n \n \n \n 0\n \n \n .\n \n \n 20\n \n \n \n \n ✓\n

\n

\n \n \n \n \n F\n \n \n =\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 25\n \n \n =\n \n \n \n \n \n \n 1\n \n \n .\n \n \n 3\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 5\n \n \n \n \n N ✓\n

\n
\n
\n (b)\n
\n

\n Use of\n \n \n F\n \n \n =\n \n \n B\n \n \n I\n \n \n l\n \n \n ✓\n

\n

\n \n \n 45\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 6\n \n \n \n \n ×\n \n \n 10\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 25\n \n \n =\n \n \n 1\n \n \n .\n \n \n 15\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 4\n \n \n \n \n \n \n N\n \n \n ✓\n

\n

\n Concludes that the assumption is not valid ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "EXE.2.SL.TZ0.17", "Question": "
\n
\n (a.i)\n
\n
\n

\n Draw the magnetic field lines due to A.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Draw the magnetic field lines due to A.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n State and explain, using your diagram, why a force acts on B due to A in the plane of the paper.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Both wires are 7.5 m long and are 0.25 m apart. The current in both wires is 12 A. Determine the force that acts on one wire due to the other.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n At least one circle centred on centre of wire A\n
\n \n \n AND\n
\n \n \n indication of clockwise direction ✓\n

\n

\n More than 2 circles with increasing separation between circles from centre outwards (by eye) ✓\n

\n
\n
\n (i)\n
\n

\n At least one circle centred on centre of wire A\n
\n \n \n AND\n
\n \n \n indication of clockwise direction ✓\n

\n

\n More than 2 circles with increasing separation between circles from centre outwards (by eye) ✓\n

\n
\n
\n (ii)\n
\n

\n B lies in magnetic field of A OWTTE ✓\n

\n

\n Explained use of appropriate rule together with drawn indication of rule operating in this case ✓\n

\n

\n To show that force on B is to left and in plane of paper ✓\n
\n \n \n OR\n
\n \n \n Magnetic field lines of B merge with those of A to give combined field line pattern ✓\n

\n

\n Sketch of combined pattern to show null point somewhere on line between wires. ✓\n

\n

\n Wires will move to reduce stored energy and this is achieved by moving together so force on B is to left ✓\n

\n
\n
\n (b)\n
\n

\n \n \n \n F\n \n \n l\n \n \n \n =\n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n \n I\n \n \n A\n \n \n \n \n I\n \n \n B\n \n \n \n \n r\n \n \n \n =\n \n \n \n \n 2\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 7\n \n \n \n \n ×\n \n \n \n 12\n \n \n 2\n \n \n \n \n \n 0\n \n \n .\n \n \n 25\n \n \n \n \n ✓\n

\n

\n \n \n 8\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 4\n \n \n \n \n \n \n N\n \n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-3-motion-in-electromagnetic-fields" ] }, { "question_id": "EXE.2.SL.TZ0.18", "Question": "
\n
\n (a)\n
\n
\n

\n Outline, with reference to the decay equation above, the role of chain reactions in the operation of a nuclear power station.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Calculate, in MeV, the energy released in the reaction.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Two nuclides present in spent nuclear fuel\n \n \n \n Cs\n \n \n \n \n 55\n \n \n 137\n \n \n \n are  and cerium-144 (\n \n \n \n Ce\n \n \n \n \n 58\n \n \n 144\n \n \n \n ). The initial activity of a sample of pure\n \n \n \n Ce\n \n \n \n \n 58\n \n \n 144\n \n \n \n is about 40 times greater than the activity of the same amount of pure\n \n \n \n Cs\n \n \n \n \n 55\n \n \n 137\n \n \n \n .\n

\n

\n Discuss which of the two nuclides is more likely to require long-term storage once removed from the reactor.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n The four neutrons released in the reaction may initiate further fissions ✓\n

\n

\n «If sufficient U-235 is available,» the reaction is self-sustained ✓\n

\n

\n Allowing for the continuous production of energy ✓\n

\n

\n The number of neutrons available is controlled with control rods «to maintain the desired reaction rate» ✓\n

\n
\n
\n (a)\n
\n

\n The four neutrons released in the reaction may initiate further fissions ✓\n

\n

\n «If sufficient U-235 is available,» the reaction is self-sustained ✓\n

\n

\n Allowing for the continuous production of energy ✓\n

\n

\n The number of neutrons available is controlled with control rods «to maintain the desired reaction rate» ✓\n

\n
\n
\n (b)\n
\n

\n \n \n 137\n \n \n ×\n \n \n 8\n \n \n .\n \n \n 389\n \n \n +\n \n \n 95\n \n \n ×\n \n \n 8\n \n \n .\n \n \n 460\n \n \n -\n \n \n 235\n \n \n ×\n \n \n 7\n \n \n .\n \n \n 591\n \n \n ✓\n

\n

\n 169 «MeV» ✓\n

\n
\n
\n (c)\n
\n

\n «For the same number of nuclei,» the activity is inversely related to half-life ✓\n

\n

\n Thus\n \n \n \n Cs\n \n \n \n \n 55\n \n \n 137\n \n \n \n has a longer half-life and will likely require longer storage ✓\n

\n

\n Half-lives of their decay products need also be considered when planning storage ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-3-radioactive-decay", "e-4-fission" ] }, { "question_id": "EXE.2.SL.TZ0.19", "Question": "
\n
\n (a)\n
\n
\n

\n Compare and contrast spontaneous and neutron-induced nuclear fission.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Every neutron-induced fission reaction of uranium-235 releases an energy of about 200 MeV. A nuclear power station transfers an energy of about 2.4 GJ per second.\n

\n

\n Determine the mass of uranium-235 that undergoes fission in one day in this power station.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n State\n \n two\n \n properties of the products of nuclear fission due to which the spent nuclear fuel needs to be kept safe.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Spontaneous fission occurs with no external influence, neutron-induced fission requires an interaction with a neutron «of appropriate energy» ✓\n

\n

\n Both result in the release of energy\n
\n \n \n OR\n \n \n
\n both have a large number of possible pairs of products ✓\n

\n
\n
\n (a)\n
\n

\n Spontaneous fission occurs with no external influence, neutron-induced fission requires an interaction with a neutron «of appropriate energy» ✓\n

\n

\n Both result in the release of energy\n
\n \n \n OR\n \n \n
\n both have a large number of possible pairs of products ✓\n

\n
\n
\n (b)\n
\n

\n Fissions per day\n \n \n =\n \n \n \n \n 2\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n ×\n \n \n 60\n \n \n ×\n \n \n 60\n \n \n ×\n \n \n 24\n \n \n \n \n 200\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n \n \n «\n \n \n =\n \n \n 6\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 24\n \n \n \n » ✓\n

\n

\n Mass of uranium\n \n \n =\n \n \n \n \n 6\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 24\n \n \n \n \n \n 6\n \n \n .\n \n \n 05\n \n \n ×\n \n \n \n 10\n \n \n 23\n \n \n \n \n \n ×\n \n \n 0\n \n \n .\n \n \n 235\n \n \n ✓\n

\n

\n 2.5 «kg» ✓\n

\n
\n
\n (c)\n
\n

\n Have relatively short half-lives / high activity ✓\n

\n

\n Their decay products are «usually» also radioactive ✓\n

\n

\n Volatile / chemically active ✓\n

\n

\n Biologically active / easily absorbed by living matter ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-3-radioactive-decay", "e-4-fission" ] }, { "question_id": "EXE.2.SL.TZ0.2", "Question": "
\n
\n (a)\n
\n
\n

\n A tram is just leaving the lower railway station.\n

\n

\n Determine, as the train leaves the lower station,\n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the pd across the motor of the tram,\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the mechanical power output of the motor.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Discuss the variation in the power output of the motor with distance from the lower station.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The total friction in the system acting on the tram is equivalent to an opposing force of 750 N.\n

\n

\n For one particular journey, the tram is full of passengers.\n

\n

\n Estimate the maximum speed\n \n v\n \n of the tram as it leaves the lower station.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n The tram travels at\n \n v\n \n throughout the journey. Two trams are available so that one is returning to the lower station on another line while the other is travelling to the village. The journeys take the same time.\n

\n

\n It takes 1.5 minutes to unload and 1.5 minutes to load each tram. Ignore the time taken to accelerate the tram at the beginning and end of the journey.\n

\n

\n Estimate the maximum number of passengers that can be carried up to the village in one hour.\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (e)\n
\n
\n

\n There are eight wheels on each tram with a brake system for each wheel. A pair of brake pads clamp firmly onto an annulus made of steel.\n

\n

\n The train comes to rest from speed\n \n v\n \n . Ignore the energy transferred to the brake pads and the change in the gravitational potential energy of the tram during the braking.\n

\n

\n Calculate the temperature change in each steel annulus as the tram comes to rest.\n

\n

\n Data for this question\n

\n

\n The inner radius of the annulus is 0.40 m and the outer radius is 0.50 m.\n

\n

\n The thickness of the annulus is 25 mm.\n

\n

\n The density of the steel is 7860 kg m\n \n −3\n \n

\n

\n The specific heat capacity of the steel is 420 J kg\n \n −1\n \n K\n \n −1\n \n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (f)\n
\n
\n

\n The speed of the tram is measured by detecting a beam of microwaves of wavelength 2.8 cm reflected from the rear of the tram as it moves away from the station. Predict the change in wavelength of the microwaves at the stationary microwave detector in the station.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Resistance of cable = 0.072 Ω ✓\n

\n

\n Pd is (500 − 0.072 × 600) = 457 V ✓\n

\n
\n
\n (i)\n
\n

\n Resistance of cable = 0.072 Ω ✓\n

\n

\n Pd is (500 − 0.072 × 600) = 457 V ✓\n

\n
\n
\n (ii)\n
\n

\n Power input = 457 × 600 = 274 kW ✓\n

\n

\n Power output = 0.9 × 274 = 247 kW ✓\n

\n
\n
\n (b)\n
\n

\n The pd across the motor increases as the tram travels up the track ✓\n

\n

\n (As the current is constant), the power output also rises ✓\n

\n
\n
\n (c)\n
\n

\n Total weight of tram = 75 × 710 + 5 × 10\n \n 4\n \n = 1.03 × 10\n \n 5\n \n N ✓\n

\n

\n Total force down track = 750 + 1.03 × 10\n \n 5\n \n sin (10) = 1.87 × 10\n \n 4\n \n N ✓\n

\n

\n Use of\n \n P\n \n =\n \n F ×\n \n \n v ✓\n \n

\n

\n (\n \n v\n \n = 247 000 ÷ 1.87 × 10\n \n 4\n \n )= 13 m s\n \n −1\n \n ✓\n

\n
\n
\n (d)\n
\n

\n Time for run =\n \n s/v\n \n = 3000 ÷ 13.2 = 227 s ✓\n

\n

\n 3 minutes loading = 180 s\n

\n

\n So one trip = 407 s ✓\n

\n

\n And there are 3600/407 trips per hour = 8.84 ✓\n

\n

\n So 8\n \n complete\n \n trips with 75 = 600 passengers ✓\n

\n
\n
\n (e)\n
\n

\n Work leading to volume = 7.1 x 10\n \n −3\n \n m\n \n 3\n \n ✓\n

\n

\n Work leading to mass of steel = 55 .8 kg ✓\n

\n

\n Kinetic Energy transferred per annulus =\n \n \n \n \n \n \n 1\n \n \n 2\n \n \n \n \n m\n \n \n \n v\n \n \n 2\n \n \n \n \n 8\n \n \n \n =\n \n \n \n 1\n \n \n 16\n \n \n \n ×\n \n \n \n \n 1\n \n \n .\n \n \n 03\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n \n \n 9\n \n \n .\n \n \n 81\n \n \n \n \n ×\n \n \n \n 13\n \n \n 2\n \n \n \n

\n

\n = 110 kJ ✓\n

\n

\n \n \n Δ\n \n \n T\n \n \n =\n \n \n \n \n E\n \n \n k\n \n \n \n \n m\n \n \n c\n \n \n \n \n =\n \n \n \n \n 1\n \n \n .\n \n \n 1\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n \n \n 55\n \n \n .\n \n \n 8\n \n \n ×\n \n \n 420\n \n \n \n \n =\n \n \n 4\n \n \n .\n \n \n 7\n \n \n K ✓\n

\n
\n
\n (f)\n
\n

\n Use of\n \n \n \n \n Δ\n \n \n λ\n \n \n \n λ\n \n \n \n ≈\n \n \n \n v\n \n \n c\n \n \n \n ✓\n

\n

\n 1.2 nm ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "b-the-particulate-nature-of-matter", "c-wave-behaviour" ], "subtopics": [ "a-1-kinematics", "a-2-forces-and-momentum", "a-3-work-energy-and-power", "b-1-thermal-energy-transfers", "b-5-current-and-circuits", "c-5-doppler-effect" ] }, { "question_id": "EXE.2.SL.TZ0.20", "Question": "
\n
\n (a)\n
\n
\n

\n State\n \n one\n \n source of the radioactive waste products from nuclear fission reactions.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Outline how this waste is treated after it has been removed from the fission reactor.\n

\n
\n
\n

\n [4]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Fission fragments from the fuel rods\n

\n

\n \n OR\n \n activated materials in (e.g.) fuel rod casings\n

\n

\n \n OR\n \n nuclei formed by neutron activation from U-235\n

\n

\n \n OR\n \n stated products, e.g. Pu, U-236 etc. ✓\n

\n
\n
\n (a)\n
\n

\n Fission fragments from the fuel rods\n

\n

\n \n OR\n \n activated materials in (e.g.) fuel rod casings\n

\n

\n \n OR\n \n nuclei formed by neutron activation from U-235\n

\n

\n \n OR\n \n stated products, e.g. Pu, U-236 etc. ✓\n

\n
\n
\n (b)\n
\n

\n Waste (fuel rod) is placed in cooling ponds for a number of years ✓\n

\n

\n After most active products have decayed the uranium is separated to be recycled/reprocessed ✓\n

\n

\n The remaining highly active waste is vitrified / made into a solid form ✓\n

\n

\n And stored (deep) underground ✓\n

\n
\n", "Examiners report": "None", "topics": [ "e-nuclear-and-quantum-physics" ], "subtopics": [ "e-4-fission" ] }, { "question_id": "EXE.2.SL.TZ0.3", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate:\n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the initial temperature gradient through the base of the pot. State an appropriate unit for your answer.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the initial rate, in kW, of thermal energy transfer by conduction through the base of the pot.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The electrical power rating of the hot plate is 1 kW. Comment, with reference to this value, on your answer in (a)(ii).\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Describe how thermal energy is distributed throughout the volume of the water in the pot.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n \n 180\n \n \n -\n \n \n 10\n \n \n \n \n 0\n \n \n .\n \n \n 005\n \n \n \n \n =\n \n \n 3\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 4\n \n \n \n ✓\n

\n

\n K m\n \n −1\n \n ✓\n

\n
\n
\n (i)\n
\n

\n \n \n \n \n 180\n \n \n -\n \n \n 10\n \n \n \n \n 0\n \n \n .\n \n \n 005\n \n \n \n \n =\n \n \n 3\n \n \n .\n \n \n 4\n \n \n ×\n \n \n \n 10\n \n \n 4\n \n \n \n ✓\n

\n

\n K m\n \n −1\n \n ✓\n

\n
\n
\n (ii)\n
\n

\n 45 × 0.15 × 3.4 × 10\n \n 4\n \n = 230 «kW» ✓\n

\n
\n
\n (b)\n
\n

\n The answer is unrealistically large / impossible to sustain ✓\n

\n

\n Due to much lower actual power, a lower temperature gradient through the base of the pot is quickly established ✓\n

\n

\n The surface of the hot plate becomes colder from contact with pot\n
\n \n \n OR\n \n \n
\n there is a temperature gradient also through the hot plate ✓\n

\n
\n
\n (c)\n
\n

\n By means of convection currents ✓\n

\n

\n That arise due to density difference between hot and cold water ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "EXE.2.SL.TZ0.4", "Question": "
\n
\n (a)\n
\n
\n

\n The temperature of the air outside of the bottle is 20 °C. The surface area of the bottle is 4.0 × 10\n \n −2\n \n m\n \n 2\n \n . Calculate the initial rate of thermal energy transfer by conduction through the bottle.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Explain why the rate calculated in part (a) is decreasing.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Estimate the initial rate of the change of the temperature of the water in the bottle. State your answer in K s\n \n −1\n \n . The specific heat capacity of water is 4200 J kg\n \n −1\n \n K\n \n −1\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n 0\n \n \n .\n \n \n 90\n \n \n ×\n \n \n 4\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n ×\n \n \n \n \n 60\n \n \n -\n \n \n 20\n \n \n \n \n 0\n \n \n .\n \n \n 003\n \n \n \n \n ✓\n

\n

\n 480 «W» ✓\n

\n
\n
\n (a)\n
\n

\n \n \n 0\n \n \n .\n \n \n 90\n \n \n ×\n \n \n 4\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n ×\n \n \n \n \n 60\n \n \n -\n \n \n 20\n \n \n \n \n 0\n \n \n .\n \n \n 003\n \n \n \n \n ✓\n

\n

\n 480 «W» ✓\n

\n
\n
\n (b)\n
\n

\n The temperature gradient decreases as the water cools down ✓\n

\n

\n The rate of energy transfer is proportional to the temperature gradient ✓\n

\n
\n
\n (c)\n
\n

\n \n \n 480\n \n \n =\n \n \n 4200\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 50\n \n \n ×\n \n \n \n \n Δ\n \n \n T\n \n \n \n \n Δ\n \n \n t\n \n \n \n \n ✓\n

\n

\n \n \n \n \n Δ\n \n \n T\n \n \n \n \n Δ\n \n \n t\n \n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 23\n \n \n «K s\n \n −1\n \n » ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "EXE.2.SL.TZ0.5", "Question": "
\n
\n (a)\n
\n
\n

\n Discuss the mechanism that accounts for the greatest rate of energy transfer when:\n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n \n T\n \n \n t\n \n >\n \n T\n \n \n b\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n \n T\n \n \n b\n \n >\n \n T\n \n \n t\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The liquid now freezes so that the vertical column is entirely of ice. Suggest how your answer to (a)(ii) will change.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n Conduction identified ✓\n

\n

\n energy transfer through interaction of particles in liquid at atomic scale ✓\n

\n
\n
\n (i)\n
\n

\n Conduction identified ✓\n

\n

\n energy transfer through interaction of particles in liquid at atomic scale ✓\n

\n
\n
\n (ii)\n
\n

\n Convection identified ✓\n

\n

\n energy transfer through movement of bodies of liquid at different densities ✓\n

\n
\n
\n (b)\n
\n

\n the solid cannot now move relative to material above it ✓\n

\n

\n so conduction only ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "EXE.2.SL.TZ0.6", "Question": "
\n
\n (a)\n
\n
\n

\n The side of the rod can be unlagged or ideally lagged. Explain the difference in energy transfer for these two cases.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Calculate the temperature at Y.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The temperatures are now reversed so that X is at 45 °C and Z is at 90 °C. Show that the rate of energy transfer is unchanged.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n When ideally lagged, no energy transfer can occur through the sides of the bar. ✓\n

\n

\n All the power input/ energy input per second at one end will emerge at the other end. ✓\n

\n

\n When unlagged, energy transfer occurs from the sides of the bar and the power /energy input per second at input > the energy output per second at the other end. ✓\n

\n

\n

\n

\n \n Max 1 if answer does not refer to rate of energy transfer in MP2 and MP3.\n \n

\n
\n
\n (a)\n
\n

\n When ideally lagged, no energy transfer can occur through the sides of the bar. ✓\n

\n

\n All the power input/ energy input per second at one end will emerge at the other end. ✓\n

\n

\n When unlagged, energy transfer occurs from the sides of the bar and the power /energy input per second at input > the energy output per second at the other end. ✓\n

\n

\n

\n

\n \n Max 1 if answer does not refer to rate of energy transfer in MP2 and MP3.\n \n

\n
\n
\n (i)\n
\n

\n idea that\n \n \n \n \n d\n \n \n Q\n \n \n \n \n d\n \n \n t\n \n \n \n \n is same in both bars because lagged ✓\n

\n

\n work to show that\n \n \n 90\n \n \n -\n \n \n θ\n \n \n =\n \n \n 2\n \n \n \n \n θ\n \n \n -\n \n \n 45\n \n \n \n \n
\n \n \n OR\n \n \n
\n Temperature difference across XY is twice temperature difference across YZ ✓\n

\n

\n solves to show that\n \n \n θ\n \n \n =\n \n \n 60\n \n \n \n \n °\n \n \n C\n \n \n ✓\n

\n
\n
\n (ii)\n
\n

\n repeats calculation to show that\n \n θ\n \n = 75 °C ✓\n

\n

\n temperature difference across YZ is still 15 K which gives the same rate of energy transfer (but in opposite direction) ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "EXE.2.SL.TZ0.7", "Question": "
\n
\n (a)\n
\n
\n

\n State\n \n two\n \n assumptions of the kinetic model of an ideal gas that refer to intermolecular collisions.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Discuss how the motion of the molecules of a gas gives rise to pressure in the gas.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The average speed of the molecules of a gas is 500 m s\n \n −1\n \n . The density of the gas is 1.2 kg m\n \n −3\n \n . Calculate, in kPa, the pressure of the gas.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n The collisions are elastic ✓\n

\n

\n The time for a collision is much shorter than the time between collisions ✓\n

\n

\n The intermolecular forces are only present during collisions ✓\n

\n
\n
\n (a)\n
\n

\n The collisions are elastic ✓\n

\n

\n The time for a collision is much shorter than the time between collisions ✓\n

\n

\n The intermolecular forces are only present during collisions ✓\n

\n
\n
\n (b)\n
\n

\n The momentum of a molecule changes when it collides with a container wall ✓\n

\n

\n From\n \n \n F\n \n \n =\n \n \n \n \n Δ\n \n \n p\n \n \n \n \n Δ\n \n \n t\n \n \n \n \n and Newton’s third law, this leads to a force exerted on the wall by the molecule ✓\n

\n

\n The average force exerted by all the molecules on a unit area of the wall is equivalent to pressure ✓\n

\n

\n

\n
\n
\n (c)\n
\n

\n \n \n \n 1\n \n \n 3\n \n \n \n ×\n \n \n 1\n \n \n .\n \n \n 2\n \n \n ×\n \n \n \n 500\n \n \n 2\n \n \n \n ✓\n

\n

\n 100 «kPa» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-3-gas-laws" ] }, { "question_id": "EXE.2.SL.TZ0.8", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the average translational speed of air molecules.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The air is a mixture of nitrogen, oxygen and other gases. Explain why the component gases of air in the container have different average translational speeds.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n The temperature of the sample is increased without a change in pressure. Outline the effect it has on the density of the gas.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n 1\n \n \n .\n \n \n 8\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n =\n \n \n \n 1\n \n \n 3\n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n v\n \n \n 2\n \n \n \n ✓\n

\n

\n \n \n v\n \n \n =\n \n \n 520\n \n \n «m s\n \n −1\n \n »  ✓\n

\n
\n
\n (a)\n
\n

\n \n \n 1\n \n \n .\n \n \n 8\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n =\n \n \n \n 1\n \n \n 3\n \n \n \n ×\n \n \n 2\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n v\n \n \n 2\n \n \n \n ✓\n

\n

\n \n \n v\n \n \n =\n \n \n 520\n \n \n «m s\n \n −1\n \n »  ✓\n

\n
\n
\n (b)\n
\n

\n Average kinetic energy of the molecules is determined by the temperature only ✓\n

\n

\n The mass of a molecule is different for each component gas ✓\n

\n

\n From\n \n \n \n E\n \n \n k\n \n \n \n =\n \n \n \n 1\n \n \n 2\n \n \n \n m\n \n \n \n v\n \n \n 2\n \n \n \n , the same\n \n \n \n E\n \n \n k\n \n \n \n and different mass implies a different average velocity  ✓\n

\n
\n
\n (c)\n
\n

\n \n \n ALTERNATIVE 1\n \n \n

\n

\n The average translational speed increases «because\n \n T\n \n increases» ✓\n

\n

\n From\n \n \n P\n \n \n =\n \n \n \n 1\n \n \n 3\n \n \n \n ρ\n \n \n \n v\n \n \n 2\n \n \n \n , the density decreases «to keep\n \n \n \n P\n \n \n \n constant»  ✓\n

\n

\n
\n \n \n ALTERNATIVE 2\n \n \n

\n

\n From the ideal gas law, the volume of the gas increases ✓\n

\n

\n Since\n \n \n ρ\n \n \n =\n \n \n \n m\n \n \n v\n \n \n \n and\n \n \n \n m\n \n \n \n is constant, the density decreases  ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-3-gas-laws" ] }, { "question_id": "EXE.2.SL.TZ0.9", "Question": "
\n
\n (a)\n
\n
\n

\n Outline how the concept of absolute zero of temperature is interpreted in terms of:\n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n the ideal gas law,\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n the kinetic energy of particles in an ideal gas.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n A container holds a mixture of argon and helium atoms at a temperature of 37 °C.\n

\n

\n Calculate the average translational speed of the argon atoms.\n

\n

\n The molar mass of argon is 4.0 × 10\n \n −2\n \n kg mol\n \n −1\n \n .\n

\n
\n
\n

\n [4]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Discuss how the mean kinetic energy of the argon atoms in the mixture compares with that of the helium atoms.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n it is the temperature at which the volume\n

\n

\n \n \n OR\n \n \n

\n

\n the pressure extrapolates to zero (can be shown by sketch)  ✓\n

\n
\n
\n (i)\n
\n

\n it is the temperature at which the volume\n

\n

\n \n \n OR\n \n \n

\n

\n the pressure extrapolates to zero (can be shown by sketch)  ✓\n

\n
\n
\n (ii)\n
\n

\n it is the temperature at which all the (random) motion stops\n

\n

\n \n \n OR\n \n \n

\n

\n at which all the motion can be extrapolated to stop\n

\n

\n \n \n OR\n \n \n

\n

\n at which the kinetic energy of all particles is zero ✓\n

\n
\n
\n (b)\n
\n

\n Use of\n \n \n \n \n 4\n \n \n .\n \n \n 0\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 2\n \n \n \n \n \n \n N\n \n \n A\n \n \n \n \n \n \n =\n \n \n 6\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 26\n \n \n \n \n \n \n kg\n \n \n \n \n ✓\n

\n

\n Work showing that\n \n \n v\n \n \n \n \n =\n \n \n \n \n \n 3\n \n \n p\n \n \n \n ρ\n \n \n \n =\n \n \n \n \n \n 3\n \n \n p\n \n \n V\n \n \n \n m\n \n \n \n \n \n \n \n =\n \n \n \n \n \n 3\n \n \n k\n \n \n T\n \n \n \n m\n \n \n \n \n ✓\n

\n

\n Correct substitution AND conversion to K (310 K)  ✓\n

\n

\n 430/440 «m s\n \n −1\n \n » ✓\n

\n
\n
\n (c)\n
\n

\n the gases are in the same container at the same temperature so are\n \n in equilibrium\n \n ✓\n

\n

\n they must have the same mean/average kinetic energy ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-3-gas-laws" ] }, { "question_id": "SPM.1B.HL.TZ0.1", "Question": "
\n
\n (a)\n
\n
\n

\n The students vary\n \n V\n \n and measure the time\n \n T\n \n for the ball to move\n \n once\n \n from one plate to the other. The table shows some of the data.\n

\n

\n \n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n \n V\n \n is provided by two identical power supplies connected in series. The potential difference of each of the power supplies is known with an uncertainty of 0.01 kV.\n

\n

\n State the uncertainty in the potential difference\n \n V\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n \n T\n \n is measured with an electronic stopwatch that measures to the nearest 0.1 s.\n

\n

\n Describe how an uncertainty in\n \n T\n \n of less than 0.1 s can be achieved using this stopwatch.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Outline why it is unlikely that the relationship between\n \n T\n \n and\n \n V\n \n is linear.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Calculate the largest fractional uncertainty in\n \n T\n \n for these data.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Determine\n \n A\n \n by drawing the line of best fit.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n State the units of\n \n A\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n The theoretical relationship assumes that the ball is only affected by the electric force.\n

\n

\n Suggest why, in order to test the relationship, the length of the string should be much greater than the distance between the plates.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 0.02 «kV» ✓\n

\n
\n
\n (i)\n
\n

\n 0.02 «kV» ✓\n

\n
\n
\n (ii)\n
\n

\n by measuring the time for many bounces ✓\n

\n

\n and dividing the result by the number of bounces ✓\n

\n
\n
\n (iii)\n
\n

\n it is not possible to draw a straight line through all the error bars ✓\n

\n
\n
\n (iv)\n
\n

\n \n T\n \n = 0.5 s ✓\n

\n

\n «\n \n \n \n \n 0\n \n \n .\n \n \n 1\n \n \n \n \n 0\n \n \n .\n \n \n 5\n \n \n \n \n =\n \n \n » 0.2 ✓\n

\n
\n
\n (i)\n
\n

\n a best-fit line drawn through the entire range of the data ✓\n

\n

\n large triangle greater than half a line or two data points on the line greater than half a line apart ✓\n

\n

\n correct read offs consistent with the line,\n \n eg\n \n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n -\n \n \n 0\n \n \n \n \n 0\n \n \n .\n \n \n 40\n \n \n -\n \n \n 0\n \n \n \n \n =\n \n \n 4\n \n \n .\n \n \n 0\n \n \n ✓\n

\n

\n \n Accept answer in the range 3.8–4.2\n \n

\n
\n
\n (ii)\n
\n

\n kV s ✓\n

\n
\n
\n (iii)\n
\n

\n the angle between the string and the vertical should be very small «for any position of the ball» ✓\n

\n

\n

\n

\n so that the tension in the string is «almost» balanced by the ball’s weight\n

\n

\n \n \n OR\n \n \n

\n

\n restoring force from the string / horizontal component of tension negligibly small «compared with electric force» ✓\n

\n

\n

\n

\n \n OWTTE\n \n

\n
\n", "Examiners report": "None", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-1-2-designing", "inquiry-1-exploring-and-designing", "tool-1-experimental-techniques", "tool-3-mathematics" ] }, { "question_id": "SPM.1B.HL.TZ0.2", "Question": "
\n
\n (a)\n
\n
\n

\n The group obtains the following repeated readings for\n \n d\n \n for\n \n one\n \n value of\n \n W\n \n .\n

\n

\n \n

\n

\n The group divides into two subgroups, A and B, to analyse the data.\n

\n

\n Group A quotes the mean value of\n \n d\n \n as 2.93 cm.\n

\n

\n Group B quotes the mean value of\n \n d\n \n as 2.8 cm.\n

\n

\n Discuss the values that the groups have quoted.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The variation of\n \n d\n \n with\n \n W\n \n is shown.\n

\n

\n \n

\n

\n Outline\n \n one\n \n experimental reason why the graph does not go through the origin.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Theory predicts that\n

\n

\n \n \n d\n \n \n ∝\n \n \n \n \n \n W\n \n \n x\n \n \n \n \n L\n \n \n y\n \n \n \n \n \n E\n \n \n I\n \n \n \n \n

\n

\n where\n \n \n \n E\n \n \n \n and\n \n \n I\n \n \n are constants. The fundamental units of\n \n \n I\n \n \n are m\n \n 4\n \n and those of\n \n E\n \n are kg m\n \n −1\n \n s\n \n −2\n \n .\n

\n

\n Calculate\n \n \n x\n \n \n and\n \n \n y\n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Suggest an appropriate measuring instrument for determining\n \n b\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the percentage uncertainty in the value of\n \n A\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 3 sf is inappropriate for\n \n A\n \n ✓\n

\n

\n rejects trial 3 as outlier for\n \n B\n \n ✓\n

\n
\n
\n (a)\n
\n

\n 3 sf is inappropriate for\n \n A\n \n ✓\n

\n

\n rejects trial 3 as outlier for\n \n B\n \n ✓\n

\n
\n
\n (b)\n
\n

\n beam bends under its own weight / weight of pan\n

\n

\n \n \n OR\n \n \n

\n

\n specified systematic error in\n \n d\n \n ✓\n

\n
\n
\n (c)\n
\n

\n units of\n \n \n \n W\n \n \n \n : kg m s\n \n −2\n \n ✓\n

\n

\n work leading to\n \n \n x\n \n \n =\n \n \n 1\n \n \n and\n \n \n y\n \n \n =\n \n \n 3\n \n \n ✓\n

\n
\n
\n (i)\n
\n

\n instrument (capable of reading to 0.05 mm) with reason related to resolution of instrument ✓\n

\n

\n \n eg micrometer screw gauge, Vernier caliper, travelling microscope\n \n

\n
\n
\n (ii)\n
\n

\n attempt to calculate fractional uncertainty in either\n \n a\n \n or\n \n b\n \n [0.0357, 0.0167] ✓\n

\n

\n 0.0357 + 0.0167 = 0.05 = 5% ✓\n

\n
\n", "Examiners report": "None", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-2-3-interpreting-results", "inquiry-2-collecting-and-processing-data", "tool-1-experimental-techniques", "tool-3-mathematics" ] }, { "question_id": "SPM.1B.SL.TZ0.1", "Question": "
\n
\n (a)\n
\n
\n

\n The students vary\n \n V\n \n and measure the time\n \n T\n \n for the ball to move\n \n once\n \n from one plate to the other. The table shows some of the data.\n

\n

\n \n

\n
\n
\n

\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n \n V\n \n is provided by two identical power supplies connected in series. The potential difference of each of the power supplies is known with an uncertainty of 0.01 kV.\n

\n

\n State the uncertainty in the potential difference\n \n V\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n \n T\n \n is measured with an electronic stopwatch that measures to the nearest 0.1 s.\n

\n

\n Describe how an uncertainty in\n \n T\n \n of less than 0.1 s can be achieved using this stopwatch.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Outline why it is unlikely that the relationship between\n \n T\n \n and\n \n V\n \n is linear.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iv)\n
\n
\n

\n Calculate the largest fractional uncertainty in\n \n T\n \n for these data.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Determine\n \n A\n \n by drawing the line of best fit.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n State the units of\n \n A\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n The theoretical relationship assumes that the ball is only affected by the electric force.\n

\n

\n Suggest why, in order to test the relationship, the length of the string should be much greater than the distance between the plates.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 0.02 «kV» ✓\n

\n
\n
\n (i)\n
\n

\n 0.02 «kV» ✓\n

\n
\n
\n (ii)\n
\n

\n by measuring the time for many bounces ✓\n

\n

\n and dividing the result by the number of bounces ✓\n

\n
\n
\n (iii)\n
\n

\n it is not possible to draw a straight line through all the error bars ✓\n

\n
\n
\n (iv)\n
\n

\n \n T\n \n = 0.5 s ✓\n

\n

\n «\n \n \n \n \n 0\n \n \n .\n \n \n 1\n \n \n \n \n 0\n \n \n .\n \n \n 5\n \n \n \n \n =\n \n \n » 0.2 ✓\n

\n
\n
\n (i)\n
\n

\n a best-fit line drawn through the entire range of the data ✓\n

\n

\n large triangle greater than half a line or two data points on the line greater than half a line apart ✓\n

\n

\n correct read offs consistent with the line,\n \n eg\n \n \n \n \n \n 1\n \n \n .\n \n \n 6\n \n \n -\n \n \n 0\n \n \n \n \n 0\n \n \n .\n \n \n 40\n \n \n -\n \n \n 0\n \n \n \n \n =\n \n \n 4\n \n \n .\n \n \n 0\n \n \n ✓\n

\n

\n \n Accept answer in the range 3.8–4.2\n \n

\n
\n
\n (ii)\n
\n

\n kV s ✓\n

\n
\n
\n (iii)\n
\n

\n the angle between the string and the vertical should be very small «for any position of the ball» ✓\n

\n

\n so that the tension in the string is «almost» balanced by the ball’s weight\n

\n

\n \n \n OR\n \n \n

\n

\n restoring force from the string / horizontal component of tension negligibly small «compared with electric force» ✓\n

\n

\n \n OWTTE\n \n

\n
\n", "Examiners report": "None", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-1-1-exploring", "i-1-2-designing", "i-1-3-controlling-variables", "inquiry-1-exploring-and-designing", "tool-1-experimental-techniques", "tool-3-mathematics" ] }, { "question_id": "SPM.1B.SL.TZ0.2", "Question": "
\n
\n (a)\n
\n
\n

\n The group obtains the following repeated readings for\n \n d\n \n for\n \n one\n \n value of\n \n W\n \n .\n

\n

\n \n

\n

\n The group divides into two subgroups, A and B, to analyse the data.\n

\n

\n Group A quotes the mean value of\n \n d\n \n as 2.93 cm.\n

\n

\n Group B quotes the mean value of\n \n d\n \n as 2.8 cm.\n

\n

\n Discuss the values that the groups have quoted.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n The variation of\n \n d\n \n with\n \n W\n \n is shown.\n

\n

\n \n

\n

\n Outline\n \n one\n \n experimental reason why the graph does not go through the origin.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Theory predicts that\n

\n

\n \n \n d\n \n \n ∝\n \n \n \n \n \n W\n \n \n x\n \n \n \n \n L\n \n \n y\n \n \n \n \n \n E\n \n \n I\n \n \n \n \n

\n

\n where\n \n \n E\n \n \n and\n \n \n I\n \n \n are constants. The fundamental units of\n \n \n I\n \n \n are m\n \n 4\n \n and those of\n \n \n E\n \n \n are kg m\n \n −1\n \n s\n \n −2\n \n .\n

\n

\n Calculate\n \n \n x\n \n \n and\n \n \n y\n \n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Suggest an appropriate measuring instrument for determining\n \n b\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Calculate the percentage uncertainty in the value of\n \n A\n \n .\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n 3 sf is inappropriate for\n \n A\n \n ✓\n

\n

\n rejects trial 3 as outlier for\n \n B\n \n ✓\n

\n
\n
\n (a)\n
\n

\n 3 sf is inappropriate for\n \n A\n \n ✓\n

\n

\n rejects trial 3 as outlier for\n \n B\n \n ✓\n

\n
\n
\n (b)\n
\n

\n beam bends under its own weight / weight of pan\n

\n

\n \n \n OR\n \n \n

\n

\n specified systematic error in\n \n d\n \n ✓\n

\n
\n
\n (c)\n
\n

\n units of\n \n W\n \n : kg m s\n \n −2\n \n ✓\n

\n

\n work leading to\n \n \n x\n \n \n =\n \n \n 1\n \n \n and\n \n \n y\n \n \n =\n \n \n 3\n \n \n ✓\n

\n
\n
\n (i)\n
\n

\n instrument (capable of reading to 0.05 mm) with reason related to resolution of instrument ✓\n

\n

\n \n eg micrometer screw gauge, Vernier caliper, travelling microscope\n \n

\n
\n
\n (ii)\n
\n

\n attempt to calculate fractional uncertainty in either\n \n a\n \n or\n \n b\n \n [0.0357, 0.0167] ✓\n

\n

\n 0.0357 + 0.0167 = 0.05 = 5 % ✓\n

\n
\n", "Examiners report": "None", "topics": [ "inquiry", "tools" ], "subtopics": [ "i-2-3-interpreting-results", "inquiry-2-collecting-and-processing-data", "tool-1-experimental-techniques", "tool-3-mathematics" ] }, { "question_id": "SPM.2.HL.TZ0.1", "Question": "
\n
\n (a)\n
\n
\n

\n State the nature and direction of the force that accelerates the 15 kg object.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (b)\n
\n
\n

\n Determine the largest magnitude of\n \n F\n \n for which the block and the object do not move relative to each other.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n static friction force «between blocks»\n

\n

\n \n \n AND\n \n \n

\n

\n directed to the right ✓\n

\n
\n
\n (a)\n
\n

\n static friction force «between blocks»\n

\n

\n \n \n AND\n \n \n

\n

\n directed to the right ✓\n

\n
\n
\n (b)\n
\n

\n \n F\n \n = 60\n \n a ✓\n \n

\n

\n \n F\n \n \n f\n \n = 0.6 × 15 × 9.8 «= 88.2 N» ✓\n

\n

\n \n \n 88\n \n \n .\n \n \n 2\n \n \n =\n \n \n 15\n \n \n ×\n \n \n \n F\n \n \n 60\n \n \n \n ⇒\n \n \n F\n \n \n =\n \n \n 350\n \n \n «N» ✓\n

\n

\n

\n

\n \n Allow use of a =\n \n 0.6\n \n g leading to 353 N.\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "SPM.2.HL.TZ0.2", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate\n \n \n γ\n \n \n for a speed of 0.80\n \n c\n \n .\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n γ\n \n \n =\n \n \n \n 1\n \n \n \n 1\n \n \n -\n \n \n 0\n \n \n .\n \n \n \n 80\n \n \n 2\n \n \n \n \n \n =\n \n \n \n 5\n \n \n 3\n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 67\n \n \n ✓\n

\n
\n
\n (a)\n
\n

\n \n \n γ\n \n \n =\n \n \n \n 1\n \n \n \n 1\n \n \n -\n \n \n 0\n \n \n .\n \n \n \n 80\n \n \n 2\n \n \n \n \n \n =\n \n \n \n 5\n \n \n 3\n \n \n \n =\n \n \n 1\n \n \n .\n \n \n 67\n \n \n ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-5-galilean-and-special-relativity" ] }, { "question_id": "SPM.2.HL.TZ0.3", "Question": "
\n
\n (a)\n
\n
\n

\n Outline how this standing wave pattern of melted spots is formed.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n standing waves form «in the oven» by superposition / constructive interference ✓\n

\n

\n energy transfer is greatest at the antinodes «of the standing wave pattern» ✓\n

\n
\n
\n (a)\n
\n

\n standing waves form «in the oven» by superposition / constructive interference ✓\n

\n

\n energy transfer is greatest at the antinodes «of the standing wave pattern» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-4-standing-waves-and-resonance" ] }, { "question_id": "SPM.2.HL.TZ0.4", "Question": "
\n
\n (a)\n
\n
\n

\n Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n arrow normal to the orbit towards the Earth ✓\n

\n
\n
\n (a)\n
\n

\n arrow normal to the orbit towards the Earth ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "d-1-gravitational-fields" ] }, { "question_id": "SPM.2.HL.TZ0.5", "Question": "
\n
\n (a)\n
\n
\n

\n Outline why the magnetic flux in ring B increases.\n

\n
\n
\n

\n [1]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Outline why work must be done on ring B as it moves towards ring A at a constant speed.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n ring B cuts an increasing number of magnetic field lines ✓\n

\n

\n \n OR\n \n

\n

\n magnetic field from current in A increases at the position of B ✓\n

\n
\n
\n (a)\n
\n

\n ring B cuts an increasing number of magnetic field lines ✓\n

\n

\n \n OR\n \n

\n

\n magnetic field from current in A increases at the position of B ✓\n

\n
\n
\n (d)\n
\n

\n the current induced in B gives rise to a magnetic field opposing that of A\n

\n

\n \n OR\n \n

\n

\n there will be a magnetic force opposing the motion ✓\n

\n

\n work must be done to move B in the opposite direction to this force ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-3-work-energy-and-power", "d-4-induction" ] }, { "question_id": "SPM.2.HL.TZ0.6", "Question": "
\n
\n (a)\n
\n
\n

\n A nucleus of americium-241 has 146 neutrons. This nuclide decays to neptunium through alpha emission.\n

\n

\n Complete the nuclear equation for this decay.\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (d)\n
\n
\n

\n Calculate the maximum current in the chamber due to the electrons when there is no smoke in the chamber.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n Am\n \n \n \n \n 95\n \n \n 241\n \n \n \n ✓\n

\n

\n \n \n \n Np\n \n \n \n \n 93\n \n \n 237\n \n \n \n +\n \n \n \n α\n \n \n \n \n 2\n \n \n 4\n \n \n \n ✓\n

\n
\n
\n (d)\n
\n

\n Each alpha gives rise to\n \n \n \n \n 5\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n 15\n \n \n \n =\n \n \n 3\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ion pairs ✓\n

\n

\n So\n \n \n \n \n 3\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ×\n \n \n 42000\n \n \n \n 3\n \n \n \n =\n \n \n 5\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n ion pairs per second ✓\n

\n

\n current\n \n \n =\n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n ×\n \n \n 5\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 82\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 9\n \n \n \n \n «A» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-5-current-and-circuits", "e-3-radioactive-decay" ] }, { "question_id": "SPM.2.HL.TZ0.7", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the surface temperature of\n \n δ\n \n Vel A is about 9000 K.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n correct substitution into\n \n \n \n λ\n \n \n max\n \n \n \n =\n \n \n \n \n 2\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n T\n \n \n \n OR 9350 K ✓\n

\n
\n
\n (a)\n
\n

\n correct substitution into\n \n \n \n λ\n \n \n max\n \n \n \n =\n \n \n \n \n 2\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n T\n \n \n \n OR 9350 K ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "SPM.2.HL.TZ0.8", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the theoretical equilibrium temperature of the mixture.\n

\n
\n
\n

\n [3]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n The mixture was held in a large metal container during the mixing.\n

\n

\n Explain\n \n one\n \n change to the procedure that will reduce the difference in (b)(i).\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 7\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 5\n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n T\n \n \n -\n \n \n 47\n \n \n \n \n

\n

\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 150\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n 240\n \n \n -\n \n \n T\n \n \n \n \n

\n

\n One heat capacity term correctly substituted ✓\n

\n

\n latent heat correctly substituted\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n ✓\n

\n

\n \n \n T\n \n \n =\n \n \n 190\n \n \n «°C» ✓\n

\n
\n
\n (a)\n
\n

\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 7\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 5\n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n T\n \n \n -\n \n \n 47\n \n \n \n \n

\n

\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 150\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n 240\n \n \n -\n \n \n T\n \n \n \n \n

\n

\n One heat capacity term correctly substituted ✓\n

\n

\n latent heat correctly substituted\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n ✓\n

\n

\n \n \n T\n \n \n =\n \n \n 190\n \n \n «°C» ✓\n

\n
\n
\n (ii)\n
\n

\n Insulate the container\n

\n

\n \n OR\n \n

\n

\n Carry out experiment quicker\n

\n

\n \n OR\n \n

\n

\n Use larger volumes of substances ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "inquiry" ], "subtopics": [ "b-1-thermal-energy-transfers", "i-1-3-controlling-variables", "inquiry-1-exploring-and-designing" ] }, { "question_id": "SPM.2.HL.TZ0.9", "Question": "
\n
\n (a.i)\n
\n
\n

\n Draw the electric field lines due to the charged plates.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Draw the electric field lines due to the charged plates.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (ii)\n
\n
\n

\n Draw the forces acting on the oil drop, ignoring the buoyancy force.\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (iii)\n
\n
\n

\n Show that the electric charge on the oil drop is given by\n

\n

\n \n \n q\n \n \n =\n \n \n \n \n \n ρ\n \n \n o\n \n \n \n g\n \n \n V\n \n \n \n E\n \n \n \n

\n

\n where\n \n \n \n ρ\n \n \n o\n \n \n \n is the density of oil and\n \n \n V\n \n \n is the volume of the oil drop.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n equally spaced arrows «by eye» all pointing down ✓\n

\n

\n edge effects also shown with arrows ✓\n

\n
\n
\n (i)\n
\n

\n equally spaced arrows «by eye» all pointing down ✓\n

\n

\n edge effects also shown with arrows ✓\n

\n
\n
\n (ii)\n
\n

\n Weight vertically down\n \n \n AND\n \n \n electric force vertically up ✓\n

\n

\n Of equal length «by eye» ✓\n

\n
\n
\n (iii)\n
\n

\n Mass of drop is\n \n \n \n ρ\n \n \n o\n \n \n \n V\n \n \n ✓\n

\n

\n \n \n q\n \n \n E\n \n \n =\n \n \n \n \n \n ρ\n \n \n o\n \n \n \n V\n \n \n \n \n g\n \n \n ✓\n

\n

\n «hence answer»\n

\n

\n
\n \n MP1 must be shown implicitly for credit.\n \n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "d-fields", "tools" ], "subtopics": [ "a-2-forces-and-momentum", "d-2-electric-and-magnetic-fields", "tool-3-mathematics" ] }, { "question_id": "SPM.2.SL.TZ0.1", "Question": "
\n
\n (a)\n
\n
\n

\n State the nature and direction of the force that accelerates the 15 kg object.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n static friction force «between blocks»\n

\n

\n \n AND\n \n

\n

\n directed to the right ✓\n

\n
\n
\n (a)\n
\n

\n static friction force «between blocks»\n

\n

\n \n AND\n \n

\n

\n directed to the right ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion" ], "subtopics": [ "a-2-forces-and-momentum" ] }, { "question_id": "SPM.2.SL.TZ0.2", "Question": "
\n
\n (a)\n
\n
\n

\n Outline how this standing wave pattern of melted spots is formed.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n standing waves form «in the oven» by superposition / constructive interference ✓\n

\n

\n energy transfer is greatest at the antinodes «of the standing wave pattern» ✓\n

\n
\n
\n (a)\n
\n

\n standing waves form «in the oven» by superposition / constructive interference ✓\n

\n

\n energy transfer is greatest at the antinodes «of the standing wave pattern» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "c-wave-behaviour" ], "subtopics": [ "c-4-standing-waves-and-resonance" ] }, { "question_id": "SPM.2.SL.TZ0.3", "Question": "
\n
\n (a)\n
\n
\n

\n Draw an arrow on the diagram to represent the direction of the acceleration of the satellite.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n arrow normal to the orbit towards the Earth ✓\n

\n
\n
\n (a)\n
\n

\n arrow normal to the orbit towards the Earth ✓\n

\n
\n", "Examiners report": "None", "topics": [ "a--space-time-and-motion", "d-fields" ], "subtopics": [ "a-2-forces-and-momentum", "d-1-gravitational-fields" ] }, { "question_id": "SPM.2.SL.TZ0.4", "Question": "
\n
\n (a)\n
\n
\n

\n A nucleus of americium-241 has 146 neutrons. This nuclide decays to neptunium through alpha emission.\n

\n

\n Complete the nuclear equation for this decay.\n

\n

\n \n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (c)\n
\n
\n

\n Calculate the maximum current in the chamber due to the electrons when there is no smoke in the chamber.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n Am\n \n \n \n \n 95\n \n \n 241\n \n \n \n ✓\n

\n

\n \n \n \n Np\n \n \n \n \n 93\n \n \n 237\n \n \n \n +\n \n \n \n α\n \n \n \n \n 2\n \n \n 4\n \n \n \n ✓\n

\n
\n
\n (c)\n
\n

\n Each alpha gives rise to\n \n \n \n \n 5\n \n \n .\n \n \n 5\n \n \n ×\n \n \n \n 10\n \n \n 6\n \n \n \n \n 15\n \n \n \n =\n \n \n 3\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ion pairs ✓\n

\n

\n So\n \n \n \n \n 3\n \n \n .\n \n \n 67\n \n \n ×\n \n \n \n 10\n \n \n 5\n \n \n \n ×\n \n \n 42000\n \n \n \n 3\n \n \n \n =\n \n \n 5\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n ion pairs per second ✓\n

\n

\n current\n \n \n =\n \n \n 1\n \n \n .\n \n \n 6\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 19\n \n \n \n \n ×\n \n \n 5\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 9\n \n \n \n =\n \n \n 0\n \n \n .\n \n \n 82\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 9\n \n \n \n \n «A» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter", "e-nuclear-and-quantum-physics" ], "subtopics": [ "b-5-current-and-circuits", "e-3-radioactive-decay" ] }, { "question_id": "SPM.2.SL.TZ0.5", "Question": "
\n
\n (a)\n
\n
\n

\n Show that the surface temperature of\n \n δ\n \n Vel A is about 9000 K.\n

\n
\n
\n

\n [1]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n correct substitution into\n \n \n \n λ\n \n \n max\n \n \n \n =\n \n \n \n \n 2\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n T\n \n \n \n OR 9350 K ✓\n

\n
\n
\n (a)\n
\n

\n correct substitution into\n \n \n \n λ\n \n \n max\n \n \n \n =\n \n \n \n \n 2\n \n \n .\n \n \n 9\n \n \n ×\n \n \n \n 10\n \n \n \n -\n \n \n 3\n \n \n \n \n \n T\n \n \n \n OR 9350 K ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "SPM.2.SL.TZ0.6", "Question": "
\n
\n (a)\n
\n
\n

\n Calculate the theoretical equilibrium temperature of the mixture.\n

\n
\n
\n

\n [3]\n

\n
\n
\n", "Markscheme": "
\n (a)\n
\n

\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 7\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 5\n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n T\n \n \n -\n \n \n 47\n \n \n \n \n

\n

\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 150\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n 240\n \n \n -\n \n \n T\n \n \n \n \n

\n

\n One heat capacity term correctly substituted ✓\n

\n

\n latent heat correctly substituted\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n ✓\n

\n

\n \n \n T\n \n \n =\n \n \n 190\n \n \n «°C» ✓\n

\n
\n
\n (a)\n
\n

\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 0\n \n \n .\n \n \n 7\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n ×\n \n \n 5\n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n +\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n T\n \n \n -\n \n \n 47\n \n \n \n \n

\n

\n \n \n =\n \n \n \n \n 0\n \n \n .\n \n \n 150\n \n \n ×\n \n \n 2\n \n \n .\n \n \n 13\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n \n \n 240\n \n \n -\n \n \n T\n \n \n \n \n

\n

\n One heat capacity term correctly substituted ✓\n

\n

\n latent heat correctly substituted\n \n \n \n \n 0\n \n \n .\n \n \n 030\n \n \n ×\n \n \n 220\n \n \n ×\n \n \n \n 10\n \n \n 3\n \n \n \n \n \n ✓\n

\n

\n \n \n T\n \n \n =\n \n \n 190\n \n \n «°C» ✓\n

\n
\n", "Examiners report": "None", "topics": [ "b-the-particulate-nature-of-matter" ], "subtopics": [ "b-1-thermal-energy-transfers" ] }, { "question_id": "SPM.2.SL.TZ0.7", "Question": "
\n
\n (a.i)\n
\n
\n

\n Draw the electric field lines due to the charged plates.\n

\n
\n
\n

\n [2]\n

\n
\n
\n
\n
\n (i)\n
\n
\n

\n Draw the electric field lines due to the charged plates.\n

\n
\n
\n

\n [2]\n

\n
\n
\n", "Markscheme": "
\n (a.i)\n
\n

\n equally spaced arrows «by eye» all pointing down ✓\n

\n

\n edge effects also shown with arrows ✓\n

\n
\n
\n (i)\n
\n

\n equally spaced arrows «by eye» all pointing down ✓\n

\n

\n edge effects also shown with arrows ✓\n

\n
\n", "Examiners report": "None", "topics": [ "d-fields" ], "subtopics": [ "d-2-electric-and-magnetic-fields" ] } ]