Calculate the wavelength of the wave.

Determine, for particle P, the magnitude and direction of the acceleration at t = 2.0 m s.

State the phase difference between the two waves.

Identify a time at which the displacement of P is zero.

Estimate the amplitude of the resultant wave.

Calculate the length of the tube.

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 $t=\frac{T}{8}$.

Draw on the diagram the standing wave at time $t=\frac{T}{4}$.

$T=4\times {10}^{-3}$«s» or $f=250$«Hz» ✓

$\lambda =340\times 4.0\times {10}^{-3}=1.36\approx 1.4$«m» ✓

Allow ECF from MP1.
Award [2] for a bald correct answer.

$\varpi =«\frac{2\mathrm{\pi }}{T}=»\frac{2\mathrm{\pi }}{4\times {10}^{-3}}$  OR  $1.57\times {10}^3$ «s⁻¹» ✓

$\text{a}=«{\varpi }^2{x}_0={\left(1.57\times {10}^3\right)}^2\times 6\times {10}^{-6}=14.8\approx »15$ «ms⁻²» ✓

«opposite to displacement so» to the right ✓

«±» $\frac{\mathrm{\pi }}{2}/90°$  OR  $\frac{3\mathrm{\pi }}{2}/270°$ ✓

1.5 «ms» ✓

8.0 OR 8.5 «μm» ✓

From the graph on the paper, value is 8.0. From the calculated correct trig functions, value is 8.49.

L = «$\frac{3}{4}\lambda =$» 0.90 «m» ✓

to the right ✓

displacement is getting less negative

OR

change of displacement is positive ✓

horizontal line drawn at the equilibrium position ✓

Sketch, on the diagram, the variation of displacement of the air molecules with distance along the pipe when t = $\frac{3}{4f}$.

An air molecule is situated at point X in the pipe at t = 0. Describe the motion of this air molecule during one complete cycle of the standing wave beginning from t = 0.

The speed of sound c for longitudinal waves in air is given by

$c=\sqrt{\frac{K}{\rho }}$

where ρ is the density of the air and K is a constant.

A student measures f to be 120 Hz when the length of the pipe is 1.4 m. The density of the air in the pipe is 1.3 kg m^–3. Determine the value of K for air. State your answer with the appropriate fundamental (SI) unit.

Demonstrate, using a second ray, that the image appears to come from the position indicated.

Outline why the observer detects a series of increases and decreases in the intensity of the received signal as the boat moves along the line XY.

horizontal line shown in centre of pipe ✔

«air molecule» moves to the right and then back to the left ✔

returns to X/original position ✔

wavelength = 2 × 1.4 «= 2.8 m» ✔

c = «f λ =» 120 × 2.8 «= 340 m s⁻¹» ✔

K = «ρc² = 1.3 × 340² =» 1.5 × 10⁵ ✔

kg m^–1s^–2 ✔

construction showing formation of image ✔

Another straight line/ray from image through the wall with line/ray from intersection at wall back to transmitter. Reflected ray must intersect boat.

interference pattern is observed

OR

interference/superposition mentioned ✔

maximum when two waves occur in phase/path difference is nλ

OR

minimum when two waves occur 180° out of phase/path difference is (n + ½)λ ✔

Describe the conditions required for an object to perform simple harmonic motion (SHM).

Calculate the mass of the wooden block.

In carrying out the experiment the student displaced the block horizontally by 4.8 cm from the equilibrium position. Determine the total energy in the oscillation of the wooden block.

A second identical spring is placed in parallel and the experiment in (b) is repeated. Suggest how this change affects the fractional uncertainty in the mass of the block.

State the direction of motion of P on the spring.

Explain whether P is at the centre of a compression or the centre of a rarefaction.

acceleration/restoring force is proportional to displacement
and in the opposite direction/directed towards equilibrium

ALTERNATIVE 1

$\frac{T_1^2}{T_2^2}=\frac{m_1}{m_2}$

mass = 0.38 / 0.39 «kg»

ALTERNATIVE 2

«use of T $=2\pi \sqrt{\frac{m}{k}}$» k = 28 «Nm^–1»

«use of T $=2\pi \sqrt{\frac{m}{k}}$» m = 0.38 / 0.39 «kg»

Allow ECF from MP1.

ω = «$\frac{2\pi }{0.74}$» = 8.5 «rads^–1»

total energy = $\frac{1}{2}\times 0.39\times {8.5}^2\times (4.8\times {10}^{-2}{)}^2$

= 0.032 «J»

Allow ECF from (b) and incorrect ω.

Allow answer using k from part (b).

spring constant/k/stiffness would increase
T would be smaller
fractional uncertainty in T would be greater, so fractional uncertainty of mass of block would be greater

left

coils to the right of P move right and the coils to the left move left

hence P at centre of rarefaction

Do not allow a bald statement of rarefaction or answers that don’t include reference to the movement of coils.

Allow ECF from MP1 if the movement of the coils imply a compression.

Deduce that a minimum intensity of sound is heard at P.

A microphone moves along the line from P to Q. PQ is normal to the line midway between the loudspeakers.

The intensity of sound is detected by the microphone. Predict the variation of detected intensity as the microphone moves from P to Q.

When both loudspeakers are operating, the intensity of sound recorded at Q is $I_0$. Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to $I_{\mathrm{A}}$.

Estimate $\frac{I_{\mathrm{A}}}{I_0}$.

Explain why the frequency recorded by the microphone is lower than the frequency emitted by the loudspeaker.

Calculate $v$.

wavelength$=\frac{340}{850}=0.40 «\mathrm{m}»$ ✓

path difference $=1.8 «\mathrm{m}»$ ✓

$1.8 «\mathrm{m}»=4.5\lambda$  OR  $\frac{1.8}{0.20}=9$ «half-wavelengths» ✓

waves meet in antiphase «at P»
OR
destructive interference/superposition «at P» ✓

Allow approach where path length is calculated in terms of number of wavelengths; along path A ($\mathit{56}\mathit{.}\mathit{25}$) and
path B ($\mathit{60}\mathit{.}\mathit{75}$) for MP2, hence path difference $\mathit{4}\mathit{.}\mathit{5}$ wavelengths for MP3

«equally spaced» maxima and minima ✓

a maximum at Q ✓

four «additional» maxima «between P and Q» ✓

the amplitude of sound at Q is halved ✓
«intensity is proportional to amplitude squared hence» $\frac{I_{\mathrm{A}}}{I_0}=\frac{1}{4}$ ✓

speed of sound relative to the microphone is less ✓

wavelength unchanged «so frequency is lower»
OR
fewer waves recorded in unit time/per second «so frequency is lower» ✓

$845=850\times \frac{340-v}{340}$ ✓

$v=2.00 «\mathrm{m} {\mathrm{s}}^{-1}»$ ✓

This was answered very well, with those not scoring full marks able to, at least, calculate the wavelength.

Most candidates were able to score at least one mark by referring to a maximum at Q.

Most candidates earned 2 marks or nothing. A common answer was that intensity was 1/2 the original.

HL only. The majority of candidates answered this by describing the Doppler Effect for a moving source. Others reworded the question without adding any explanation. Correct explanations were rare.

HL only. This was answered well with the majority of candidates able to identify the correct formula and the correct values to substitute.

Outline what is meant by gravitational field strength at a point.

Newton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting Mars.

Mars has a mass of 6.4 × 10²³ kg. Show that, for Mars, k is about 9 × 10^–13s²m^–3.

The time taken for Mars to revolve on its axis is 8.9 × 10⁴ s. Calculate, in m s^–1, the orbital speed of the satellite.

Show that the intensity of solar radiation at the orbit of Mars is about 600 W m^–2.

Determine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.

The atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the mean temperature of Earth is strongly affected by gases in its atmosphere but that of Mars is not.

force per unit mass ✔

acting on a small/test/point mass «placed at the point in the field» ✔

Mars is spherical/a sphere «and of uniform density so behaves as a point mass» ✔

satellite has a much smaller mass/diameter/size than Mars «so approximates to a point mass» ✔

«$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$ hence» $v=\sqrt{\frac{GM}{R}}$. Also $v=\frac{2\pi R}{T}$

OR

$m{\omega }^{2}r=\frac{GMm}{r^2}$ hence ${\omega }^2=\frac{GM}{R^3}$ ✔

uses either of the above to get $T^2=\frac{4{\pi }^2}{GM}{R}^3$

OR

uses $k=\frac{4{\pi }^2}{GM}$ ✔

k = 9.2 × 10⁻¹³ / 9.3 × 10⁻¹³

Unit not required

$R^3=\frac{T^2}{k}=\frac{{\left(8.9\times {10}^4\right)}^2}{9.25\times {10}^{-13}}$  R = 2.04 × 10⁷ «m» ✔

v = «$\omega r=\frac{2\pi \times 2.04\times {10}^7}{89000}=$» 1.4 × 10³ «m s^–1»

OR

v = «$\sqrt{\frac{GM}{R}}=\sqrt{\frac{6.67\times {10}^{-11}\times 6.4\times {10}^{23}}{2.04\times {10}^7}}=$» 1.4 × 10³ «m s^–1» ✔

use of $I\propto \frac{1}{r^2}$ «1.36 × 10³ × $\frac{1}{{1.5}^2}$» ✔

604 «W m^–2» ✔

use of $\frac{600}{4}$ for mean intensity ✔

temperature/K = «$\sqrt[4]{\frac{600}{4\times 5.67\times {10}^{-8}}}=$» 230 ✔

reference to greenhouse gas/effect ✔

recognize the link between molecular density/concentration and pressure ✔

low pressure means too few molecules to produce a significant heating effect

OR

low pressure means too little radiation re-radiated back to Mars ✔

The greenhouse effect can be described, it doesn’t have to be named

Explain why intensity maxima are observed at X and Y.

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.

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

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.

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.

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

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

path difference = 0.062 «m»✔

so wavelength = 0.031 «m»✔

frequency = 9.7 × 10⁹ «Hz»✔

Award [2 max] for 4.8 x 10⁹ Hz.

intensity is modulated by a single slit diffraction envelope OR

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

cos² variation shown ✔

with zero at 90° (by eye) ✔

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

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.

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.

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.

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.

State the direction of the resultant force on the ball.

On the diagram, construct an arrow of the correct length to represent the weight of the ball.

Show that the magnitude of the net force F on the ball is given by the following equation.

                                          $$F=\frac{mg}{\tan \theta }$$

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

Outline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.

Outline why the ball will perform simple harmonic oscillations about the equilibrium position.

Show that the period of oscillation of the ball is about 6 s.

The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time t of the velocity v of the ball during one period.

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.

The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.

towards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

downward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required

eg: 

[2 marks]

ALTERNATIVE 1

F = N cos θ

mg = N sin θ

dividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

tan θ = $\frac{\text{O}}{A}=\frac{mg}{F}$

[3 marks]

$\frac{mg}{\tan \theta }$ = m$\frac{v^2}{r}$

r = R cos θ

v = $\sqrt{\frac{gR{\cos }^2\theta }{\sin \theta }}/\sqrt{\frac{gR\cos \theta }{\tan \theta }}/\sqrt{\frac{9.81\times 8.0\cos 22}{\tan 22}}$

v = 13.4/13 «ms ^–1»

Award [4] for a bald correct answer 

Award [3] for an answer of 13.9/14 «ms ^–1». MP2 omitted

[4 marks]

there is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

the «restoring» force/acceleration is proportional to displacement

Direction is not required

[1 mark]

ω = «$\sqrt{\frac{g}{R}}$» = $\sqrt{\frac{9.81}{8.0}}$ «= 1.107 s^–1»

T = «$\frac{2\pi }{\omega }$ = $\frac{2\pi }{1.107}$ =» 5.7 «s»

Allow use of or g = 9.8 or 10

Award [0] for a substitution into T = 2π$\sqrt{\frac{I}{g}}$

[2 marks]

sine graph

correct amplitude «0.13 m s^–1»

correct period and only 1 period shown

Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.

Amplitude should be correct to ±$\frac{1}{2}$ square for MP2

eg: v /m s^–1   

[3 marks]

speed before collision v = «$\sqrt{2gR}$ =» 12.5 «ms^–1»

«from conservation of momentum» common speed after collision is $\frac{1}{2}$ initial speed «v_c = $\frac{12.5}{2}$ = 6.25 ms^–1»

h = «$\frac{{v_c}^2}{2g}=\frac{{6.25}^2}{2\times 9.81}$» 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.

The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10^–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.

Estimate the speed of the train.

Determine the width of one of the slits.

Suggest the variation in the output voltage from the light sensor that will be observed as the train moves beyond the first diffraction minimum.

In another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier.

The graph shows the variation with time of the output voltage from the sounds sensor.

Explain how this effect arises.

«light» superposes/interferes

pattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»

voltage peaks correspond to interference maxima

«$s=\frac{\lambda D}{d}=\frac{6.3\times {10}^{-7}\times 5.0}{1.5\times {10}^{-3}}=$» 2.1 x 10^–3 «m» 

If no unit assume m.
Correct answer only.

correct read-off from graph of 25 m s

v = «$\frac{x}{t}=\frac{2.1\times {10}^{-3}}{25\times {10}^{-3}}=$» 8.4 x 10^–2 «m s^–1»

Allow ECF from (b)(i)

angular width of diffraction minimum = $\frac{0.13}{5.0}$ «= 0.026 rad»

slit width = «$\frac{\lambda }{d}=\frac{6.3\times {10}^{-7}}{0.026}=$» 2.4 x 10^–5 «m»

Award [1 max] for solution using 1.22 factor.

«beyond the first diffraction minimum» average voltage is smaller

«voltage minimum» spacing is «approximately» same
OR
rate of variation of voltage is unchanged

OWTTE

«reflection at barrier» leads to two waves travelling in opposite directions 

mention of formation of standing wave

maximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position

Outline how a standing wave is produced on the string.

Show that the speed of the wave on the string is about 240 m s⁻¹.

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

Calculate, in m s⁻¹, the maximum velocity of vibration of point P when it is vibrating with a frequency of 195 Hz.

Calculate, in terms of g, the maximum acceleration of P.

Estimate the displacement needed to double the energy of the string.

The string is made to vibrate in its third harmonic. State the distance between consecutive nodes. 

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

superposition/interference of incident and reflected waves ✓

the superposition of the reflections is reinforced only for certain wavelengths ✓ 

$\lambda =2l=2\times 0.62=«1.24 \text{m}»$ ✓

$v=f\lambda =195\times 1.24=242 «{\text{m s}}^{-1}»$ ✓

Answer must be to 3 or more sf or working shown for MP2.

straight line through origin with negative gradient ✓

max velocity occurs at x = 0 ✓

$v=«\left(2\pi \right)\left(195\right)\sqrt{0.{004}^2}»=4.9 «{\text{m s}}^{-1}»$ ✓

$a={\left(2_{\mathrm{\pi }} 195\right)}^2\times 0.004=6005 «{\text{m\hspace{0.05em}s}}^{-2}»$ ✓

$=600 \text{g}$ ✓

use of $E\propto A^2\mathit{\text{ OR }}{x_{\text{o}}}^2$ ✓

$A=0.4\sqrt{2}=0.57 «\text{cm}» \cong 0.6 «\text{cm}»$ ✓

$\frac{62}{3}=21 «\text{cm}»$ ✓

Outline how the standing wave is formed.

Draw an arrow on the diagram to represent the direction of motion of the molecule at X.

Label a position N that is a node of the standing wave.

The speed of sound is 340 m s^–1 and the length of the pipe is 0.30 m. Calculate, in Hz, the frequency of the sound.

The speed of sound in air is 340 m s^–1 and in water it is 1500 m s^–1.

The wavefronts make an angle θ with the surface of the water. Determine the maximum angle, θ_max, at which the sound can enter water. Give your answer to the correct number of significant figures.

Draw lines on the diagram to complete wavefronts A and B in water for θ < θ_max.

the incident wave «from the speaker» and the reflected wave «from the closed end»

superpose/combine/interfere

Allow superimpose/add up

Do not allow meet/interact

[1 mark]

Horizontal arrow from X to the right

MP2 is dependent on MP1

Ignore length of arrow

[1 mark]

P at a node

[1 mark]

wavelength is λ = «$\frac{4\times 0.30}{3}$ =» 0.40 «m»

f = «$\frac{340}{0.40}$» 850 «Hz»

Award [2] for a bald correct answer

Allow ECF from MP1

[2 marks]

$\frac{\sin {\theta }_c}{340}=\frac{1}{1500}$

θ_c = 13«°»

Award [2] for a bald correct answer

Award [2] for a bald answer of 13.1

Answer must be to 2/3 significant figures to award MP2

Allow 0.23 radians

[2 marks]

correct orientation

greater separation

Do not penalize the lengths of A and B in the water

Do not penalize a wavefront for C if it is consistent with A and B

MP1 must be awarded for MP2 to be awarded

[2 marks]

A series of dark and bright fringes appears on the screen. Explain how a dark fringe is formed.

Outline why the beam has to be coherent in order for the fringes to be visible.

The wavelength of the beam as observed on Earth is 633.0 nm. The separation between a dark and a bright fringe on the screen is 4.50 mm. Calculate D.

Calculate the angular separation between the central peak and the missing peak in the double-slit interference intensity pattern. State your answer to an appropriate number of significant figures.

Deduce, in mm, the width of one slit.

The wavelength of the light in the beam when emitted by the galaxy was 621.4 nm.

Explain, without further calculation, what can be deduced about the relative motion of the galaxy and the Earth.

superposition of light from each slit / interference of light from both slits

with path/phase difference of any half-odd multiple of wavelength/any odd multiple of $\pi$ (in words or symbols)

producing destructive interference

Ignore any reference to crests and troughs.

[3 marks]

light waves (from slits) must have constant phase difference / no phase difference / be in phase

OWTTE

[1 mark]

evidence of solving for D «D $=\frac{sd}{\lambda }$» ✔

«$\frac{4.50\times {10}^{-3}\times 0.300\times {10}^{-3}}{633.0\times {10}^{-9}}\times 2$» = 4.27 «m» ✔

Award [1] max for 2.13 m.

sin θ = $\frac{4\times 633.0\times {10}^{-9}}{0.300\times {10}^{-3}}$

sin θ = 0.0084401…

final answer to three sig figs (eg 0.00844 or 8.44 × 10^–3)

Allow ECF from (a)(iii).

Award [1] for 0.121 rad (can award MP3 in addition for proper sig fig)

Accept calculation in degrees leading to 0.481 degrees.

Award MP3 for any answer expressed to 3sf.

[3 marks]

use of diffraction formula «b = $\frac{\lambda }{\theta }$»

OR

$\frac{633.0\times {10}^{-9}}{0.00844}$

«=» 7.5«00» × 10^–2 «mm»

Allow ECF from (b)(i).

[2 marks]

wavelength increases (so frequency decreases) / light is redshifted

galaxy is moving away from Earth

Allow ECF for MP2 (ie wavelength decreases so moving towards).

[2 marks]

Particle P in the metal sheet performs simple harmonic oscillations. When the displacement of P is 3.2 μm the magnitude of its acceleration is 7.9 m s^-2. Calculate the magnitude of the acceleration of P when its displacement is 2.3 μm.

The wave is incident at point Q on the metal–air boundary. The wave makes an angle of 54° with the normal at Q. The speed of sound in the metal is 6010 m s^–1 and the speed of sound in air is 340 m s^–1. Calculate the angle between the normal at Q and the direction of the wave in air.

The frequency of the sound wave in the metal is 250 Hz. Determine the wavelength of the wave in air.

On the diagram, at time T, draw an arrow to indicate the acceleration of this molecule.

On the diagram, at time T, label with the letter C a point in the pipe that is at the centre of a compression.

Calculate the frequency heard by the observer.

Calculate the wavelength measured by the observer.

Expression or statement showing acceleration is proportional to displacement ✔

so «$7.9\times \frac{2.3}{3.2}$» = 5.7«ms^–2» ✔

$\sin \theta =\frac{340}{6010}\times \sin {54}^0$   ✔

$\theta ={2.6}^0$   ✔

$\lambda =«\frac{340}{250}=»1.36\approx 1.4«\text{m}»$  ✔

horizontal arrow «at M» pointing left ✔

any point labelled C on the vertical line shown below ✔

eg:

$f^{\prime }=2500\times \frac{340}{340+280}$   ✔

$f^{\prime }=1371\approx 1400$«Hz»   ✔

${\lambda }^{\prime }=\frac{340}{1371}\approx 0.24/0.25$«m»  ✔

This was well answered at both levels.

Many scored full marks on this question. Common errors were using the calculator in radian mode or getting the equation upside down.

This was very well answered.

Very few candidates could interpret this situation and most arrows were shown in a vertical plane.

This was answered well at both levels.

This was answered well with the most common mistake being to swap the speed of sound and the speed of the aircraft.

Answered well with ECF often being awarded to those who answered the previous part incorrectly.

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻².

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.

Show that the equilibrium surface temperature of Titan is about 90 K.

The mass of Titan is 0.025 times the mass of the Earth and its radius is 0.404 times the radius of the Earth. The escape speed from Earth is 11.2 km s⁻¹. Show that the escape speed from Titan is 2.8 km s⁻¹.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

Show that the mass of a nitrogen molecule is 4.7 × 10⁻²⁶ kg.

Estimate the root mean square speed of nitrogen molecules in the Titan atmosphere. Assume an atmosphere temperature of 90 K.

Discuss, by reference to the answer in (b), whether it is likely that Titan will lose its atmosphere of nitrogen.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

$v=«\sqrt{\frac{2GM}{R}}=»\sqrt{\frac{0.025}{0.404}}\times 11.2$

OR

$2.79$ «km s⁻¹» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

$m=\frac{28\times {10}^{-3}}{6.02\times {10}^{23}}$

OR

$4.65\times {10}^{-26}$ «kg» ✓

$«\frac{1}{2}m{v}^2=\frac{3}{2}kT\Rightarrow »v=\sqrt{\frac{3kT}{m}}$ ✓

$v=«\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 90}{4.651\times {10}^{-26}}}=»283\approx 300$ «ms⁻¹» ✓

Award [2] marks for a bald correct answer.

Allow 282 from a rounded mass.

no, molecular speeds much less than escape speed ✓

Allow ECF from incorrect (d)(ii).

Outline why the cylinder performs simple harmonic motion when released.

The mass of the cylinder is $118 \mathrm{kg}$ and the cross-sectional area of the cylinder is $2.29\times {10}^{-1} {\mathrm{m}}^2$. The density of water is $1.03\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$. Show that the angular frequency of oscillation of the cylinder is about $4.4 \mathrm{rad} {\mathrm{s}}^{-1}$.

Determine the maximum kinetic energy $E_{\mathrm{kmax}}$ of the cylinder.

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

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

Allow use of symbols i.e. $F\propto -x$ or $a\propto -x$

Evidence of equating $m{\omega }^{2}x=\rho Agx$ «to obtain $\frac{\rho Ag}{m}={\omega }^2$» ✓

$\omega =\sqrt{\frac{1.03\times {10}^3\times 2.29\times {10}^{-1}\times 9.81}{118}}$ OR $4.43«\mathrm{rad} {\mathrm{s}}^{-1}»$ ✓

Answer to at least $\mathit{3}$ s.f.

«$E_{\mathrm{K}}$ is a maximum when $x=0$ hence» $E_{\mathrm{K}, \max }=\frac{1}{2}\times 118\times 4.4^2\left(0.{250}^2-0^2\right)$ ✓

$71.4 «\mathrm{J}»$ ✓

energy never negative ✓

correct shape with two maxima ✓

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

Again, very well answered.

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

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

Explain why the received intensity varies between maximum and minimum values.

State and explain the wavelength of the sound measured at M.

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

Speed of sound = 340 m s⁻¹

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.

Determine the difference between the frequency detected at M and that emitted by B.

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» ✓

interference
OR
superposition «of waves» ✓

maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or $\pi$ » out of phase / path difference = (n+½) x lambda ✓

wavelength = 26 cm ✓

peak to peak distance is the path difference which is one wavelength

OR

this is the distance B moves to be back in phase «with A» ✓

Allow 25 – 27 cm for MP1.

«$\frac{\lambda }{2}$» = 13 cm ✓

$f=$«$\frac{c}{\lambda }=\frac{340}{0.13}=$» 2.6 «kHz» ✓

Allow ½ of wavelength from (b) or data from graph for MP1.

Allow ECF from MP1.

ALTERNATIVE 1
use of $f\text{'}=f\frac{v}{v+u_0}$ (+ sign must be seen) OR $f\text{'}$= 2987 «Hz» ✓
« $\Delta f$» = 13 «Hz» ✓

ALTERNATIVE 2
Attempted use of $\frac{\Delta f}{f}$≈ $\frac{v}{c}$

« Δf » = 13 «Hz» ✓

This was an "explain" questions, so examiners were looking for a clear discussion of the movement of speaker B creating a changing path difference between B and the microphone and A and the microphone. This path difference would lead to interference, and the examiners were looking for a connection between specific phase differences or path differences for maxima or minima. Some candidates were able to discuss basic concepts of interference (e.g. "there is constructive and destructive interference"), but failed to make clear connections between the physical situation and the given graph. A very common mistake candidates made was to think the question was about intensity and to therefore describe the decrease in peak height of the maxima on the graph. Another common mistake was to approach this as a Doppler question and to attempt to answer it based on the frequency difference of B.

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.

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.

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.

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

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

d = 0.12 mm

D = 1.5 m

Distance MP = 7.0 mm

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

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

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

there is constructive interference at M

OR

the amplitude doubles at M ✔

intensity is «proportional to» amplitude² ✔

88 «W m⁻²» ✔

«$s=\frac{\lambda D}{d}\Rightarrow »\lambda =\frac{sd}{D}/\frac{0.12\times {10}^{-3}\times 7.0\times {10}^{-3}}{1.5}$    ✔

$\lambda =560«\text{nm}$»  ✔

«the interference pattern will be modulated by»

single slit diffraction ✔

«envelope and so it will be less»

ALTERNATIVE 1

the angular position of this point is $\theta =\frac{28\times {10}^{-3}}{1.5}=0.01867$«rad»  ✔

which coincides with the first minimum of the diffraction envelope

$\theta =\frac{\lambda }{b}=\frac{560\times {10}^{-9}}{0.030\times {10}^{-3}}=0.01867$ «rad» ✔

«so intensity will be zero»

ALTERNATIVE 2

the first minimum of the diffraction envelope is at $\theta =\frac{\lambda }{b}=\frac{560\times {10}^{-9}}{0.030\times {10}^{-3}}=0.01867$«rad»   ✔

distance on screen is $y=1.50\times 0.01867=28$«mm»  ✔

«so intensity will be zero»

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.

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.

Many recognised the significance of the single slit diffraction envelope.

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.

Outline the conditions necessary for simple harmonic motion (SHM) to occur.

A wave of amplitude 4.3 m and wavelength 35 m, moves with a speed of 3.4 m s^–1. Calculate the maximum vertical speed of the buoy.

Sketch a graph to show the variation with time of the generator output power. Label the time axis with a suitable scale.

Outline, with reference to energy changes, the operation of a pumped storage hydroelectric system.

The water in a particular pumped storage hydroelectric system falls a vertical distance of 270 m to the turbines. Calculate the speed at which water arrives at the turbines. Assume that there is no energy loss in the system.

The hydroelectric system has four 250 MW generators. Determine the maximum time for which the hydroelectric system can maintain full output when a mass of 1.5 x 10¹⁰ kg of water passes through the turbines.

Not all the stored energy can be retrieved because of energy losses in the system. Explain two such losses.

force/acceleration proportional to displacement «from equilibrium position»

and directed towards equilibrium position/point
OR
and directed in opposite direction to the displacement from equilibrium position/point

Do not award marks for stating the defining equation for SHM.
Award [1 max] for a ω–=²x with a and x defined.

frequency of buoy movement $=\frac{3.4}{35}$ or 0.097 «Hz»

OR

time period of buoy $=\frac{35}{3.4}$ or 10.3 «s» or 10 «s»

v = «$\frac{2\pi x_0}{T}$ or $2\pi f{x}_0$» $=\frac{2\times \pi \times 4.3}{10.3}$ or $2\times \pi \times 0.097\times 4.3$

2.6 «m s^–1»

peaks separated by gaps equal to width of each pulse «shape of peak roughly as shown»

one cycle taking 10 s shown on graph

Judge by eye.
Do not accept cos₂ or sin₂ graph
At least two peaks needed.
Do not allow square waves or asymmetrical shapes.
Allow ECF from (b)(i) value of period if calculated.

PE of water is converted to KE of moving water/turbine to electrical energy «in generator/turbine/dynamo»

idea of pumped storage, ie: pump water back during night/when energy cheap to buy/when energy not in demand/when there is a surplus of energy

specific energy available = «gh =» 9.81 x 270 «= 2650J kg^–1»

OR

mgh $=\frac{1}{2}$mv²

OR

v² = 2gh

v = 73 «ms^–1»

Do not allow 72 as round from 72.8

total energy = «mgh = 1.5 x 10¹⁰ x 9.81 x 270=» 4.0 x 10¹³ «J»

OR

total energy = «$\frac{1}{2}m{v}^2=\frac{1}{2}\times 1.5\times {10}^{10}\times$ (answer (c)(ii))² =» 4.0 x 10¹³ «J»

time = «$\frac{4.0\times {10}^{13}}{4\times 2.5\times {10}^8}$» 11.1h or 4.0 x 10⁴ s

Use of 3.97 x 10¹³ «J» gives 11 h.

For MP2 the unit must be present.

friction/resistive losses in pipe/fluid resistance/turbulence/turbine or generator «bearings»
OR
sound energy losses from turbine/water in pipe 

thermal energy/heat losses in wires/components

water requires kinetic energy to leave system so not all can be transferred

Must see “seat of friction” to award the mark.

Do not allow “friction” bald.

Calculate the frequency of the oscillation for both tests.

Deduce $\frac{k_1}{k_2}$.

Determine the amplitude of oscillation for test 1.

In test 2, the maximum elastic potential energy E_p stored in the spring is 44 J.

When t = 0 the value of E_p for test 2 is zero.

Sketch, on the axes, the variation with time of E_p for test 2.

The motion sensor operates by detecting the sound waves reflected from the base of the mass. The sensor compares the frequency detected with the frequency emitted when the signal returns.

The sound frequency emitted by the sensor is 35 kHz. The speed of sound is 340 m s⁻¹.

Determine the maximum frequency change detected by the sensor for test 2.

1.3 «Hz» ✓

$k\propto m$  OR  $\frac{m_1}{k_1}=\frac{m_2}{k_2}$ ✓

0.25  OR  $\frac{1}{4}$ ✓

v_max = 4.8 «m s⁻¹» ✓

$x_0=$«$\frac{v}{\omega }=\frac{vT}{2\pi }=\frac{4.8\times 0.80}{2\pi }$» = 0.61 «m» ✓

Allow a range of 4.7 to 4.9 for MP1

Allow a range of 0.58 to 0.62 for MP2

Allow ECF from (a)(i)

Allow ECF from MP1.

all energy shown positive ✓

curve starting and finishing at E = 0 with two peaks with at least one at 44 J
OR
curve starting and finishing at E = 0 with one peak at 44 J ✓

Do not accept straight lines or discontinuous curves for MP2

read off of 9.4 «m s⁻¹» ✓

use of $f\text{'}=f\left(\frac{v}{v\pm u_s}\right)$  OR  $f\text{'}=f\left(\frac{v\pm u_o}{v}\right)$ ✓

f = 36 «kHz» OR 34 «kHz» ✓

«recognition that there are two shifts so» change in f = 2 «kHz» OR f = 37 «kHz» OR 33 «kHz» ✓

Allow a range of 9.3 to 9.5 for MP1

Allow ECF from MP1.

MP4 can also be found by applying the Doppler effect twice.

ai) The majority managed to answer this question correctly.

aii) A very well answered question where most worked correctly from the formula for the period of oscillation of a spring.

aiii) Quite a few answers had v_max from the wrong test.

aiv) Most common answers were a correct 2 peak curve, a correct 1 peak curve and a sine curve. Several alternatives were included in the MS as the original data provided in the question was inconsistent, i.e. 44 J is not the maximum kinetic energy available for the second test, and that was taken into account not to disadvantage any candidate´s interpretation.

b) Many got the first three marks for a correct Doppler shift calculation from the correct speed. . There were very few good correct full answers, might be a question to look at for 6/7 during grading.