Calculate the electric potential at O.

Sketch, on the axes, the variation of the electric potential V with distance between X and Y.

Identify the direction of the resultant force acting on Z as it oscillates.

Deduce whether the motion of Z is simple harmonic.

use of $\frac{kQ}{r}$ ✓

$\frac{\left(8.99\times {10}^9\right)\left(4\times {10}^{-6}\right)}{0.15}$ OR 240 «kV» for one charge calculated ✓

480 «kV» for both ✓

MP1 can be seen or implied from calculation.

Allow ECF from MP2 for MP3.

symmetric curve around 0 with potential always positive, “bowl shape up” and curve not touching the horizontal axis. ✓

clear asymptotes at X and Y ✓

force is towards O ✓

always ✓

ALTERNATIVE 1

motion is not SHM ✓

«because SHM requires force proportional to r and» this force depends on $\frac{1}{r^2}$ ✓

ALTERNATIVE 2

motion is not SHM ✓

energy-distance «graph must be parabolic for SHM and this» graph is not parabolic ✓

This question was generally well approached. Two common errors were either starting with the wrong equation (electric potential energy or Coulomb's law) or subtracting the potentials rather than adding them.

Very few candidates drew a graph that was awarded two marks. Many had a generally correct shape, but common errors were drawing the graph touching the x-axis at O and drawing a general parabola with no clear asymptotes.

Many candidates were able to identify the direction of the force on the particle at position Z, but a common error was to miss that the question was about the direction as the particle was oscillating. Examiners were looking for a clear understanding that the force was always directed toward the equilibrium position, and not just at the moment shown in the diagram.

This was a challenging question for candidates. Most simply assumed that because the charge was oscillating that this meant the motion was simple harmonic. Some did recognize that it was not, and most of those candidates correctly identified that the relationship between force and displacement was an inverse square.

State what is meant by the Doppler effect.

A plate performs simple harmonic oscillations with a frequency of 39 Hz and an amplitude of 8.0 cm.

Show that the maximum speed of the oscillating plate is about 20 m s⁻¹.

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.

Determine the maximum frequency of the sound that is received back at the source after reflection at the plate.

State what will happen to the angular position of the primary maxima.

State what will happen to the width of the primary maxima.

State what will happen to the intensity of the secondary maxima.

the change in the observed frequency ✓

when there is relative motion between the source and the observer ✓

Do not award MP1 if they refer to wavelength.

use of $2\mathrm{\pi }fA$ ✓

maximum speed is $2\mathrm{\pi }\times 39\times 0.080=19.6$ «m s⁻¹» ✓

Award [2] for a bald correct answer.

frequency at plate $2400\times \frac{340+19.6}{340}$«$=2538$Hz»

at source $2538\times \frac{340}{340-19.6}=2694\approx 2700$ «Hz» ✓

Award [2] marks for a bald correct answer.

Award [1] mark when the effect is only applied once.

stays the same ✓

decreases ✓

decreases ✓

Police use radar to detect speeding cars. A police officer stands at the side of the road and points a radar device at an approaching car. The device emits microwaves which reflect off the car and return to the device. A change in frequency between the emitted and received microwaves is measured at the radar device.

The frequency change Δf is given by

$$\mathrm{\Delta }f=\frac{2fv}{c}$$

where f is the transmitter frequency, v is the speed of the car and c is the wave speed.

The following data are available.

(i) Explain the reason for the frequency change.

(ii) Suggest why there is a factor of 2 in the frequency-change equation.

(iii) Determine whether the speed of the car is below the maximum speed allowed.

Airports use radar to track the position of aircraft. The waves are reflected from the aircraft and detected by a large circular receiver. The receiver must be able to resolve the radar images of two aircraft flying close to each other.

The following data are available.

Calculate the minimum distance between the two aircraft when their images can just be resolved.

i
mention of Doppler effect
OR
«relative» motion between source and observer produces frequency/wavelength change
Accept answers which refer to a double frequency shift.
Award [0] if there is any suggestion that the wave speed is changed in the process.

the reflected waves come from an approaching “source”
OR
the incident waves strike an approaching “observer”

increased frequency received «by the device or by the car»

ii
the car is a moving “observer” and then a moving “source”, so the Doppler effect occurs twice
OR
the reflected radar appears to come from a “virtual image” of the device travelling at 2 v towards the device

iii
ALTERNATIVE 1
For both alternatives, allow ecf to conclusion if v OR Δf are incorrectly calculated.

v = «$\frac{\left(3\times {10}^8\right)\times \left(9.5\times {10}^3\right)}{\left(40\times {10}^9\right)\times 2}=$» 36 «ms^–1»

«36> 28» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.

ALTERNATIVE 2
reverse argument using speed limit.

$\mathrm{\Delta }f=$ «$\frac{2\times 40\times {10}^9\times 28}{3\times {10}^8}=$» 7500 «Hz»

« 9500> 7500» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.

$x=\frac{31\times {10}^3\times 1.22\times 2.5\times {10}^{-2}}{9.3}$
Award [2] for a bald correct answer.
Award [1 max] for POT error.

100 «m»
Award [1 max] for 83m (omits 1.22).

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.

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

Explain why zero intensity is observed at position A.

The distance from the centre of the pattern to A is 4.1 x 10^–2 m. The distance from the screen to the slits is 7.0 m.

Calculate the width of each slit.

Calculate the separation of the two slits.

State and explain the differences between the pattern on the screen due to the grating and the pattern due to the double slit.

The yellow light is made from two very similar wavelengths that produce two lines in the spectrum of sodium. The wavelengths are 588.995 nm and 589.592 nm. These two lines can just be resolved in the second-order spectrum of this diffraction grating. Determine the beam width of the light incident on the diffraction grating.

the diagram shows the combined effect of «single slit» diffraction and «double slit» interference

recognition that there is a minimum of the single slit pattern

OR

a missing maximum of the double slit pattern at A

waves «from the single slit» are in antiphase/cancel/have a path difference of (n + $\frac{1}{2}$)λ/destructive interference at A

θ = $\frac{4.1\times {10}^{-2}}{7.0}$ OR b = $\frac{\lambda }{\theta }$ «= $\frac{7.0\times 5.9\times {10}^{-7}}{4.1\times {10}^{-2}}$»

1.0 × 10^–4 «m»

Award [0] for use of double slit formula (which gives the correct answer so do not award BCA)

Allow use of sin or tan for small angles

use of s = $\frac{\lambda D}{d}$ with 3 fringes «$\frac{590\times {10}^{-9}\times 7.0}{4.1\times {10}^{-2}}$»

3.0 x 10^–4 «m»

Allow ECF.

fringes are further apart because the separation of slits is «much» less

intensity does not change «significantly» across the pattern or diffraction envelope is broader because slits are «much» narrower

the fringes are narrower/sharper because the region/area of constructive interference is smaller/there are more slits

intensity of peaks has increased because more light can pass through

Award [1 max] for stating one or more differences with no explanation

Award [2 max] for stating one difference with its explanation

Award [MP3] for a second difference with its explanation

Allow “peaks” for “fringes”

Δλ = 589.592 – 588.995

OR

Δλ = 0.597 «nm»

N = «$\frac{\lambda }{m\mathrm{\Delta }\lambda }$ =» $\frac{589}{2\times 0.597}$ «493»

beam width = «$\frac{493}{600}$ =» 8.2 x 10^–4 «m» or 0.82 «mm»

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.

Satellite X orbits 6600 km from the centre of the Earth.

Mass of the Earth = 6.0 x 10²⁴ kg

Show that the orbital speed of satellite X is about 8 km s^–1.

the orbital times for X and Y are different.

satellite Y requires a propulsion system.

The cable between the satellites cuts the magnetic field lines of the Earth at right angles.

Explain why satellite X becomes positively charged.

Satellite X must release ions into the space between the satellites. Explain why the current in the cable will become zero unless there is a method for transferring charge from X to Y.

The magnetic field strength of the Earth is 31 μT at the orbital radius of the satellites. The cable is 15 km in length. Calculate the emf induced in the cable.

Estimate the value of k in the following expression.

T = $2\pi \sqrt{\frac{m}{k}}$

Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation of satellite X.

Describe the energy changes in the satellite Y-cable system during one cycle of the oscillation.

«$v=\sqrt{\frac{G{M}_E}{r}}$» = $\sqrt{\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}}{6600\times {10}^3}}$

7800 «m s^–1»

Full substitution required

Must see 2+ significant figures.

Y has smaller orbit/orbital speed is greater so time period is less

Allow answer from appropriate equation

Allow converse argument for X

to stop Y from getting ahead

to remain stationary with respect to X

otherwise will add tension to cable/damage satellite/pull X out of its orbit

cable is a conductor and contains electrons

electrons/charges experience a force when moving in a magnetic field

use of a suitable hand rule to show that satellite Y becomes negative «so X becomes positive»

Alternative 2

cable is a conductor

so current will flow by induction flow when it moves through a B field

use of a suitable hand rule to show current to right so «X becomes positive»

Marks should be awarded from either one alternative or the other.

Do not allow discussion of positive charges moving towards X

electrons would build up at satellite Y/positive charge at X

preventing further charge flow

by electrostatic repulsion

unless a complete circuit exists

«ε = Blv =» 31 x 10^–6 x 7990 x 15000

3600 «V»

Allow 3700 «V» from v = 8000 m s^–1.

use of k = «$\frac{4{\pi }^{2}m}{T^2}=$» $\frac{4\times {\pi }^2\times 350}{{5.2}^2}$

510

N m^–1 or kg s^–2

Allow MP1 and MP2 for a bald correct answer

Allow 500

Allow N/m etc.

E_p in the cable/system transfers to E_k of Y

and back again twice in each cycle

Exclusive use of gravitational potential energy negates MP1

State the phase change when a ray is reflected at B.

Explain the condition for w that eliminates reflection for a particular light wavelength in air ${\lambda }_{\text{air}}$.

State the Rayleigh criterion for resolution.

The painting contains a pattern of red dots with a spacing of 3 mm. Assume the wavelength of red light is 700 nm. The average diameter of the pupil of a human eye is 4 mm. Calculate the maximum possible distance at which these red dots are distinguished.

$\text{«change is» }\pi /180°$ ✓

«to eliminate reflection» destructive interference is required ✓

phase change is the same at both boundaries / no relative phase change due to reflections ✓

therefore $2w{n}_{\text{coating}}=\left(m+\frac{1}{2}\right){\lambda }_{\text{air}}$

OR

$w=\frac{{\lambda }_{\text{coating}}}{4}$

 OR

$w=\frac{{\lambda }_{\text{air}}}{4n_{\text{coating}}}$ ✓

central maximum of one diffraction pattern lies over the central/first minimum of the other diffraction pattern ✓

$\theta =«1.22\frac{\lambda }{b}=1.22\frac{700\times {10}^{-9}}{4\times {10}^{-3}}=»2.14\times {10}^{-4}«\text{rad}»$ ✓

$D=14«0.05 \text{m}»$ ✓ 

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

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

$\frac{62}{3}=21 «\text{cm}»$ ✓

State the phase difference between the waves at V and Y.

State, in terms of λ, the path length between points X and Z.

The separation of adjacent slits is d. Show that for the second-order diffraction maximum $2\lambda =d\sin \theta$.

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.

Determine a mean value for the number of slits per millimetre of the grating.

using a light source with a smaller wavelength.

increasing the distance between the diffraction grating and the screen.

0 OR 2π OR 360° ✓

4λ ✓

$\sin \theta «=\frac{\text{XZ}}{\text{VX}}»=\frac{4\lambda }{2d}$✓

Do not award ECF from(a)(ii).

identifies gradient with $\frac{\lambda }{d}$ OR use of $d \sin \theta =n\lambda$ ✓

gradient = 0.08 OR correct replacement in equation with coordinates of a point ✓

$d=\frac{633\times {10}^{-9}}{0.080}=«7.91\times {10}^{-6} \text{m}»$ ✓

$1.26\times {10}^2 \mathit{\text{OR}} 1.27\times 102«{\text{mm}}^{-1}»$ ✓

Allow ECF from MP3

gradient smaller ✓

no change ✓

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]

At position B the rope starts to extend. Calculate the speed of the block at position B.

Determine the magnitude of the average resultant force acting on the block between B and C.

Sketch on the diagram the average resultant force acting on the block between B and C. The arrow on the diagram represents the weight of the block.

Calculate the magnitude of the average force exerted by the rope on the block between B and C.

between A and B.

between B and C.

The length reached by the rope at C is 77.4 m. Suggest how energy considerations could be used to determine the elastic constant of the rope.

Calculate the time taken for the block to return to the equilibrium position for the first time. 

Calculate the speed of the block as it passes the equilibrium position. 

use of conservation of energy

OR

v² = u² + 2as

v = «$\sqrt{2\times 60.0\times 9.81}$» = 34.3 «ms^–1»

[2 marks]

use of impulse F_ave × Δt = Δp

OR

use of F = ma with average acceleration

OR

F = $\frac{80.0\times 34.3}{0.759}$

3620«N»

Allow ECF from (a).

[2 marks]

upwards

clearly longer than weight

For second marking point allow ECF from (b)(i) providing line is upwards.

[2 marks]

3620 + 80.0 × 9.81

4400 «N»

Allow ECF from (b)(i).

[2 marks]

(loss in) gravitational potential energy (of block) into kinetic energy (of block)

Must see names of energy (gravitational potential energy and kinetic energy) – Allow for reasonable variations of terminology (eg energy of motion for KE).

[1 mark]

(loss in) gravitational potential and kinetic energy of block into elastic potential energy of rope

See note for 1(c)(i) for naming convention.

Must see either the block or the rope (or both) mentioned in connection with the appropriate energies.

[1 mark]

k can be determined using EPE = $\frac{1}{2}$kx²

correct statement or equation showing

GPE at A = EPE at C

OR

(GPE + KE) at B = EPE at C

Candidate must clearly indicate the energy associated with either position A or B for MP2.

[2 marks]

T = 2π$\sqrt{\frac{80.0}{400}}$ = 2.81 «s»

time = $\frac{T}{4}$ = 0.702 «s»

Award [0] for kinematic solutions that assume a constant acceleration.

[2 marks]

ALTERNATIVE 1

ω = $\frac{2\pi }{2.81}$ = 2.24 «rad s^–1»

v = 2.24 × 3.50 = 7.84 «ms^–1»

ALTERNATIVE 2

$\frac{1}{2}$kx² = $\frac{1}{2}$mv² OR $\frac{1}{2}$400 × 3.5² = $\frac{1}{2}$80v²

v = 7.84 «ms^–1»

Award [0] for kinematic solutions that assume a constant acceleration.

Allow ECF for T from (e)(i).

[2 marks]

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]

Monochromatic light from two identical lamps arrives on a screen.

The intensity of light on the screen from each lamp separately is I₀.

On the axes, sketch a graph to show the variation with distance x on the screen of the intensity I of light on the screen.

Monochromatic light from a single source is incident on two thin, parallel slits.

The following data are available.

The intensity I of light at the screen from each slit separately is I₀. Sketch, on the axes, a graph to show the variation with distance x on the screen of the intensity of light on the screen for this arrangement.

The slit separation is increased. Outline one change observed on the screen.

horizontal straight line through I = 2

Accept a curve that falls from I = 2 as distance increases from centre but not if it falls to zero.

[1 mark]

«standard two slit pattern»

general shape with a maximum at x = 0

maxima at 4I₀

maxima separated by «$\frac{D\lambda }{s}$ =» 2.0 cm

Accept single slit modulated pattern provided central maximum is at 4. ie height of peaks decrease as they go away from central maximum. Peaks must be of the same width

[3 marks]

fringe width/separation decreases

OR

more maxima seen

[1 mark]

Show that the magnitude of the resultant electric field at P is 3 MN C⁻¹

State the direction of the resultant electric field at P.

Explain why q will perform simple harmonic oscillations when it is released.

Calculate the period of oscillations of q.

At t = 0, the switch is connected to X. On the axes, draw a sketch graph to show the variation with time of the voltage V_R across R.

The switch is then connected to Y and C discharges through the 20 MΩ resistor. The voltage V_c drops to 50 % of its initial value in 5.0 s. Determine the capacitance of C.

«electric field at P from one charge is $\frac{kQ}{r^2}=$» $\frac{8.99\times {10}^9\times 44\times {10}^{-6}}{0.{48}^2}$

OR

$1.7168\times {10}^6$ «NC⁻¹» ✓

« net field is » $2\times 1.7168\times {10}^6\times \cos 30°=2.97\times {10}^6$ «NC⁻¹» ✓

directed vertically up «on plane of the page» ✓

Allow an arrow pointing up on the diagram.

force «on q» is proportional to the displacement ✓

and opposite to the displacement / directed towards equilibrium ✓

$«a=\frac{F}{m}=»{\omega }^{2}x=\frac{115x}{0.25}$ ✓

$T=«\frac{2\mathrm{\pi }}{\omega }=» 0.29 «\text{s}»$ ✓

Award [2] marks for a bald correct answer.

Allow ECF for MP2.

decreasing from 12 ✓

correct shape as shown ✓

Do not penalize if the graph does not touch the t axis.

$\frac{1}{2}=e^{-\frac{5.0}{20\times {10}^6 C}}$ ✓

$C=3.6\times {10}^{-7}$ «F» ✓

Award [2] for a bald correct answer.

Determine the energy of a photon of blue light (435nm) emitted in the hydrogen spectrum.

Identify, with an arrow labelled B on the diagram, the transition in the hydrogen spectrum that gives rise to the photon with the energy in (a)(i).

Explain your answer to (a)(ii).

The light illuminates a width of 3.5 mm of the grating. The deuterium and hydrogen red lines can just be resolved in the second-order spectrum of the diffraction grating. Show that the grating spacing of the diffraction grating is about 2 × 10^–6m.

Calculate the angle between the first-order line of the red light in the hydrogen spectrum and the second-order line of the violet light in the hydrogen spectrum.

The light source is changed so that white light is incident on the diffraction grating. Outline the appearance of the diffraction pattern formed with white light.

identifies λ = 435 nm ✔

E = «$\frac{hc}{\lambda }$ =» $\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{4.35\times {10}^{-7}}$ ✔

4.6 ×10⁻¹⁹ «J» ✔

–0.605 OR –0.870 OR –1.36 to –5.44 AND arrow pointing downwards ✔

Arrow MUST match calculation in (a)(i)

Allow ECF from (a)(i)

Difference in energy levels is equal to the energy of the photon ✔

Downward arrow as energy is lost by hydrogen/energy is given out in the photon/the electron falls from a higher energy level to a lower one ✔

$\frac{\lambda }{2\mathrm{\Delta }\lambda }=\frac{656.20}{0.181\times 2}=1813$ «lines» ✔

so spacing is $\frac{3.5\times {10}^{-3}}{1813}$ «= 1.9 × 10⁻⁶ m» ✔

Allow use of either wavelength or the mean value

Must see at least 2 SF for a bald correct answer

2 × 4.1 × 10⁻⁷ = 1.9 × 10⁻⁶ sin θ_v seen

OR

6.6 × 10⁻⁷ = 1.9 × 10⁻⁶ sin θ_r seen ✔

θ_v = 24 − 26 «°»

OR

θ_r = 19 − 20 «°» ✔

Δθ = 5 − 6 «°» ✔

For MP3 answer must follow from answers in MP2

For MP3 do not allow ECF from incorrect angles

centre of pattern is white ✔

coloured fringes are formed ✔

blue/violet edge of order is closer to centre of pattern

OR

red edge of order is furthest from centre of pattern ✔

the greater the order the wider the pattern ✔

there are gaps between «first and second order» spectra ✔

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.

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.

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.

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

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

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

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

Show that the collision is inelastic.

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.

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′.

Explain why f and f′ are different.

identifies T as 2.25 s ✔

L = 1.26 m ✔

1.3 / 1.26 «m» ✔

Accept any number of s.f. for MP2.

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

X labels any point on the curve where E_K   $\frac{1}{4}$ of maximum/5 mJ ✔

$\frac{1}{2}$ mv² = 20 × 10⁻³ seen OR $\frac{1}{2}$ × 7.5 × 10^-2 × v² = 20 × 10^-3 ✔

0.73 «m s⁻¹» ✔

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

0.40 «m s^-1» ✔

initial energy 24 mJ and final energy 12 mJ ✔

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

inelastic collisions occur when energy is lost ✔

graph with same period but inverted ✔

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

mention of Doppler effect ✔

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

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

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.

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.

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.

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.

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.

Predict whether reflected ray X undergoes a phase change.

state, in terms of d, the path difference between the reflected rays X and Y.

calculate the smallest value of d that will result in destructive interference between ray X and ray Y.

discuss a practical advantage of this arrangement.

Draw, on the axes, the variation with diffraction angle of the intensity of light incident on the retina of the observer.

Estimate, in rad, the smallest angular separation of two distinct point sources of light of wavelength 656 nm that can be resolved by the eye of this observer.

there is a phase change ✔
of $\mathrm{\pi }$ OR as it is reflected off a medium of higher refractive index ✔

2d ✔
NOTE: Accept 2dn

$2dn=\frac{\lambda }{2}$ ✔

$d=«\frac{\lambda }{4n}=\frac{656}{4\times 1.38}=» 119 «\mathrm{nm}»$ ✔

NOTE: Award [2] for bald correct answer

reflection from «front surface of» lens eliminated/reduced
OR
energy reaching sensor increased ✔

at one wavelength ✔

NOTE: Accept reference to reduction of glare for MP1

standard single slit diffraction pattern with correct overall shape ✔

secondary maxima of right size ✔

NOTE: Secondary maximum not to exceed 1/5^th of maximum intensity

Ignore width of maxima

use of $\theta =\frac{1.22\lambda }{b}$✔

$\theta =2.9\times {10}^{-4}$ «rad» ✔

NOTE: Award [2] for bald correct answer

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.

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 two reasons why both models predict that the motion is simple harmonic when $a$ is small.

Determine the time period of the system when $a$ is small.

Outline, without calculation, the change to the time period of the system for the model represented by graph B when $a$ is large.

The graph shows for model A the variation with $x$ of elastic potential energy E_p stored in the spring.

Describe the graph for model B.

For both models:
displacement is ∝ to acceleration/force «because graph is straight and through origin» ✓

displacement and acceleration / force in opposite directions «because gradient is negative»
OR
acceleration/«restoring» force is always directed to equilibrium ✓

attempted use of ${\omega }^2=\left(-\right)\frac{a}{x}$ ✓

suitable read-offs leading to gradient of line = 28 « s^-2» ✓

$T=\frac{2\pi }{\omega }$«$\frac{2\pi }{\sqrt{28}}$» ✓

$T=1.2$ s ✓

time period increases ✓

because average ω «for whole cycle» is smaller

OR

slope / acceleration / force at large x is smaller

OR

area under graph B is smaller so average speed is smaller. ✓

same curve OR shape for small amplitudes «to about 0.05 m» ✓

for large amplitudes «outside of 0.05 m» E_p smaller for model B / values are lower than original / spread will be wider ✓  OWTTE

Accept answers drawn on graph – e.g.

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.

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.

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.

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.

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.