A Level Physics 9702 Paper 43, May/June 2025

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Question paper

9702/43 · May/June 2025

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1
a

Define gravitational potential at a point.

[2]
b(i)

Mars is a planet that may be considered to be an isolated uniform sphere of radius 3.4×106 m. A satellite of mass 122 kg is in orbit around Mars at a constant height of 1.7×106 m above the surface of the planet. The height of the orbit is increased to 6.8×106 m above the surface. This increases the gravitational potential energy of the satellite by 5.1×108 J. Show that the mass of Mars is 6.4×1023 kg.

[3]
b(ii)

Calculate the gravitational potential ϕ at the surface of Mars. Give a unit with your answer.

[2]
c(i)

The satellite in (b) is moved to an orbit in which the satellite remains at the same point above the surface of Mars. The orbit has a period of 25 hours. State what can be deduced from this about the rotation of Mars on its axis.

[1]
c(ii)

State one other feature of this orbit.

[1]
2

A helium atom may be modelled as a nucleus surrounded by two electrons in diametrically opposite circular orbits, each of radius 170 pm, as shown in Fig. 2.1.

Figure 1 of 1 for question 2
a

State Coulomb’s law.

[2]
b(i)

State the charge on the nucleus, in terms of the elementary charge e.

[1]
b(ii)

Show that the electric force between the nucleus and one of the electrons is 1.6×10−8 N.

[1]
c(i)

Assume that the force in (b)(ii) is the only force on the electrons. Calculate the speed of the orbiting electrons.

[2]
c(ii)

Calculate the period of the orbit of the electrons.

[2]
d(i)

In practice, the orbit of each electron is affected by the presence of the other electron. For the position of one of the electrons, determine the ratio electric field strength due to the other electronelectric field strength due to the nucleus.

[2]
d(ii)

Use your answer in (d)(i) to suggest and explain how the orbit of the electron is affected by the presence of the other electron.

[1]
3
a

Define specific latent heat.

[2]
b

Explain why, for a substance, the specific latent heat of vaporisation is usually greater than the specific latent heat of fusion.

[3]
c

An ice cube of mass 37.0 g at temperature 0.0 ∘C is placed in a beaker containing water of mass 208 g at temperature 26.4 ∘C. When all the ice has melted, and all the water in the beaker has reached thermal equilibrium, the final temperature of all the water is 10.3 ∘C. The specific heat capacity of water is 4.18 J g−1 ∘C−1. The beaker has negligible specific heat capacity and is perfectly insulated from the surroundings. Determine a value, to three significant figures, for the specific latent heat of fusion of water.

[4]
4
a(i)

State what is meant by the internal energy of a system.

[2]
a(ii)

Explain why the internal energy of an ideal gas is directly proportional to the thermodynamic temperature of the gas.

[2]
b

A sample of an ideal gas at thermodynamic temperature T has internal energy U. The gas is compressed so that its temperature increases to 3T. During this compression, work W is done on the gas. The gas is then cooled at constant volume so that its temperature decreases to 2T. Complete Table 4.1 to show, in terms of some or all of W, T and U, the work done on the gas, the thermal energy supplied to the gas and the increase in internal energy of the gas for each of the two processes.

[4]
5

A cuboidal block floats in a liquid with its base horizontal, as shown in Fig. 5.1. The base of the block is at a depth h below the surface of the liquid. The block is displaced downwards by a small distance and then released so that it oscillates. Fig. 5.2 shows the variation with h of the acceleration a of the block. Fig. 5.3 shows the variation with h of the kinetic energy EK of the block.

Figure 1 of 2 for question 5
Figure 2 of 2 for question 5
a(i)

Determine the amplitude of the oscillations.

[1]
a(ii)

State what the line in Fig. 5.2 shows about the nature of the oscillations.

[1]
b

State three other quantitative conclusions that can be drawn from Fig. 5.2 and Fig. 5.3 about the block and its oscillations. Use the space for any working.

[3]
c

On Fig. 5.4, sketch the variation with h of the potential energy EP of the oscillations.

[3]
6

Fig. 6.1 shows a circuit that rectifies an alternating input voltage VIN and produces an output voltage VOUT across a resistor R. The four terminals of the rectification circuit are labelled W, X, Y and Z. A capacitor C is connected in parallel with resistor R.

Figure 1 of 3 for question 6
Figure 2 of 3 for question 6
Figure 3 of 3 for question 6
a(i)

State what is meant by rectification.

[1]
a(ii)

State the purpose of capacitor C.

[1]
b(i)

Fig. 6.2 shows the variations with time t of the potential differences (p.d.s) VIN and VOUT. The variation of VIN with t can be represented by VIN=Acos⁡Bt where A and B are constants. Determine the values of A and B. Give a unit with your answer for A.

[2]
b(ii)

Determine the type of rectification produced by the circuit in Fig. 6.1.

[1]
b(iii)

On Fig. 6.3, draw the circuit diagram for the components inside the rectification circuit.

[2]
b(iv)

Determine a value for the time constant for the discharge of the capacitor C through the resistor R in Fig. 6.1.

[3]
c

The capacitor C has a capacitance of 570 μF. Use your answer in (b)(iv) to determine the resistance of resistor R.

[2]
7
Figure 1 of 1 for question 7
a

Define magnetic flux density.

[2]
b(i)

A particle of mass m and charge +Q moves at speed v into a region where there is a uniform magnetic field, as shown in Fig. 7.1. The uniform magnetic field is into the page and has flux density B. The particle enters the region of the field at point Y. State an expression, in terms of some or all of m, Q, B and v, for the magnetic force F that acts on the particle when it is at point Y.

[1]
b(ii)

On Fig. 7.1, draw an arrow at point Y to indicate the direction of the force in (b)(i).

[1]
b(iii)

On Fig. 7.1, draw a line to show a possible path for the particle through the region of the magnetic field.

[1]
c(i)

Explain how an electric field can be used with the magnetic field to ensure that the particle in (b) now passes through point Z.

[3]
c(ii)

Derive an expression for v in terms of B and the electric field strength E.

[2]
8
a

State what is meant by the de Broglie wavelength.

[1]
b

Calculate the de Broglie wavelength of an electron moving at a speed of 4.9×107 m s−1.

[2]
c

State one similarity and one difference between an electron and a positron.

[2]
d(i)

An electron moving at a speed of 4.9×107 m s−1 collides with a positron that is travelling at the same speed in the opposite direction. As a result of the collision, two gamma-ray photons are produced. State the name of this type of reaction.

[1]
d(ii)

State what happens to the electron and to the positron.

[2]
d(iii)

Explain why two gamma-ray photons are produced, rather than just one.

[1]
d(iv)

Show that the kinetic energy of the electron before the collision is 1.1×10−15 J.

[1]
d(v)

Use the information in (d)(iv) to determine, to three significant figures, the wavelength associated with the gamma radiation emitted in the collision.

[3]
9
a

Define activity of a radioactive sample.

[1]
b

Explain why the variation with time of the activity of a radioactive sample is exponential in nature.

[3]
c(i)

A sample contains a single radioactive isotope that decays to form a stable isotope. The sample has an activity of 180 Bq at time t=0. At a time 8.4 minutes later, the activity is 120 Bq. Determine the decay constant, in min−1, of the radioactive isotope.

[2]
c(ii)

Use your answer in (c)(i) to determine the half-life, in min, of the radioactive isotope.

[1]
c(iii)

On Fig. 9.1, sketch the variation of the activity A of the sample with t for values of t between t=0 and t=24 min.

[3]
10
a

State Hubble’s law.

[2]
b(i)

A star in a distant galaxy emits radiation that has a maximum intensity of emission at a wavelength of 4.62×10−7 m. Observations of the galaxy made on the Earth detect the maximum intensity of emission from the star at a wavelength of 4.91×10−7 m. Explain why the observed wavelength and the emitted wavelength have different values.

[2]
b(ii)

Calculate the speed of the star relative to the Earth.

[2]
b(iii)

The wavelength of maximum intensity of emission is used to determine a value for the surface temperature of the star. Explain how the temperature determined using the observed wavelength compares with the true value of temperature determined using the emitted wavelength.

[2]
c

A value for the Hubble constant is 2.3×10−18 s−1. Use your answer in (b)(ii) to determine the distance of the star in (b) from the Earth.

[2]

Mark scheme

9702/43 · May/June 2025

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1a · 2 marks

  • B1

    work done per unit mass

  • B1

    work (done in) moving mass from infinity (to the point)

1b(i) · 3 marks

  • C1

    evidence of addition of 3.4 × 10⁶ to 1.7 × 10⁶ or 6.8 × 10⁶

  • C1

    GM × 122 / (5.1 × 10⁶) or GM × 122 / (10.2 × 10⁶)

  • A1

    6.67 × 10⁻¹¹ × M × 122 × [(5.1 × 10⁶)⁻¹ – (10.2 × 10⁶)⁻¹] = 5.1 × 10⁸ leading to M = 6.4 × 10²³ kg

1b(ii) · 2 marks

  • C1

    ϕ = (–) (6.67 × 10⁻¹¹ × 6.4 × 10²³) / (3.4 × 10⁶)

  • A1

    = –1.3 × 10⁷ J kg⁻¹

1c(i) · 1 mark

  • B1

    Mars takes (just under) 25 hours to rotate once on its axis

1c(ii) · 1 mark

  • B1

    orbit is equatorial or orbit is in same direction as direction of rotation of Mars

2a · 2 marks

  • B1

    (electric) force is (directly) proportional to product of charges

  • B1

    force (between point charges) is inversely proportional to the square of their separation

2b(i) · 1 mark

  • A1

    charge = (+)2e

2b(ii) · 1 mark

  • A1

    F = 2 × (1.60 × 10⁻¹⁹)² / [4π × 8.85 × 10⁻¹² × (170 × 10⁻¹²)²] = 1.6 × 10⁻⁸ N

2c(i) · 2 marks

  • C1

    F = mv² / r

  • A1

    v = [ (1.6 × 10⁻⁸ × 170 × 10⁻¹²) / (9.11 × 10⁻³¹) ]½ = 1.7 × 10⁶ m s⁻¹

2c(ii) · 2 marks

  • C1

    F = mrω² and ω = 2π / T F = 4π²mr / T²

  • A1

    T = [ (4π² × 9.11 × 10⁻³¹ × 170 × 10⁻¹²) / (1.6 × 10⁻⁸) ]½ = 6.2 × 10⁻¹⁶ s

2d(i) · 2 marks

  • C1

    E ∝ Q / r²

  • A1

    ratio = [1.60 × 10⁻¹⁹ × (170 × 10⁻¹²)²] / [3.2 × 10⁻¹⁹ × (340 × 10⁻¹²)²] = 0.13

2d(ii) · 1 mark

  • B1

    resultant force slightly less (than 1.6 × 10⁻⁸ N) so speed lower or resultant force slightly less (than 1.6 × 10⁻⁸ N) so period greater

3a · 2 marks

  • B1

    (thermal) energy per unit mass (to cause state change)

  • B1

    (thermal) energy to change state at constant temperature

3b · 3 marks

  • B1

    (for vaporisation): involves greater change in volume (of substance) or involves greater increase in separation of molecules

  • M1

    more work has to be done by molecules (to separate) or greater increase in potential energy of molecules

  • A1

    kinetic energy of molecules unchanged, so more thermal energy needed

3c · 4 marks

  • C1

    Q = mcΔθ and Q = mL

  • C1

    Δθ for the water = 26.4 – 10.3

  • C1

    (37.0 × L) + (37.0 × 4.18 × 10.3) = (208 × 4.18 × 16.1)

  • A1

    L = 335 J g⁻¹

4a(i) · 2 marks

  • B1

    sum of potential energy and kinetic energy

  • B1

    (total) energy of random motion of particles

4a(ii) · 2 marks

  • B1

    potential energy (of molecules) (in an ideal gas) is zero, so the internal energy of the gas is equal to the total kinetic energy (of molecules)

  • B1

    kinetic energy of molecules is proportional to (thermodynamic) temperature (so internal energy is proportional to (thermodynamic) temperature))

4b · 4 marks

  • B1

    cooling work done = 0

  • B1

    compression increase in internal energy = +2U

  • B1

    cooling change in internal energy = –U

  • B1

    both rows: thermal energy adds to work to give increase in internal energy in terms of U and/or W (if fully correct, thermal energy for compression = 2U – W and thermal energy for cooling = –U:

    compression+W2U – W+2U
    cooling0–U–U

    )

5a(i) · 1 mark

  • A1

    amplitude = 0.60 m

5a(ii) · 1 mark

  • B1

    oscillations are simple harmonic

5b · 3 marks

  • B3

    mean / equilibrium position is at h = 1.4 m

  • 1 mark

    total energy of oscillations = 9.0 J

  • 1 mark

    angular frequency of oscillations = 1.2 rad s⁻¹ or period of oscillations = 5.1 s or frequency of oscillation = 0.19 Hz

  • 0 marks

    maximum speed of block = 0.73 m s⁻¹

  • 0 marks

    mass of block = 33 kg

5c · 3 marks

  • B1

    U-shaped curve resting on h axis (with minimum at E_P = 0)

  • B1

    curve from h = 0.8 m to h = 2.0 m, with minimum E_P shown at h = 1.4 m

  • B1

    both end-points of curve shown at E_P = 9.0 J

6a(i) · 1 mark

  • B1

    conversion from a.c. to d.c.

6a(ii) · 1 mark

  • B1

    smoothing

6b(i) · 2 marks

  • A1

    A = 12 V

  • A1

    B = 2π / (20 × 10⁻³) = 310 rad s⁻¹

6b(ii) · 1 mark

  • B1

    full-wave (rectification)

6b(iii) · 2 marks

  • B1

    four diodes shown, with correct circuit symbols

  • B1

    four diodes correctly connected to form a bridge rectifier

6b(iv) · 3 marks

  • C1

    V = V₀ exp (–t / τ) or V = V₀ exp (–t / RC) and τ = RC

  • C1

    8.0 = 12 exp (– 7.3 × 10⁻³ / τ)

  • A1

    τ = 0.018 s

6c · 2 marks

  • C1

    time constant = RC

  • A1

    R = (0.018 / 570 × 10⁻⁶) = 32 Ω

7a · 2 marks

  • B2

    force per unit length

  • 1 mark

    force per unit current

  • 0 marks

    length / current perpendicular to field

7b(i) · 1 mark

  • B1

    F = BQv

7b(ii) · 1 mark

  • B1

    arrow at Y pointing vertically upwards

7b(iii) · 1 mark

  • B1

    upwards deflection showing circular path

7c(i) · 3 marks

  • B1

    electric field applied vertically downwards (may be shown on a labelled diagram)

  • B1

    electric force on particle in opposite direction to magnetic force (may be shown on a labelled diagram)

  • B1

    particle undeflected when magnitudes of electric and magnetic forces are equal

7c(ii) · 2 marks

  • B1

    EQ = BQv

  • A1

    v = E / B

8a · 1 mark

  • B1

    wavelength associated with a moving particle

8b · 2 marks

  • C1

    λ = h / p

  • A1

    = (6.63 × 10⁻³⁴) / (9.11 × 10⁻³¹ × 4.9 × 10⁷) = 1.5 × 10⁻¹¹ m

8c · 2 marks

  • B1

    similarity: same mass / same magnitude of charge / both leptons

  • B1

    difference: electron has negative charge, positron has positive charge / positron is anti-particle of electron / electron is a particle, positron is an anti-particle

8d(i) · 1 mark

  • B1

    (pair) annihilation

8d(ii) · 2 marks

  • B1

    their mass gets converted into energy

  • B1

    (their mass–energy) becomes the energy of the gamma photons

8d(iii) · 1 mark

  • B1

    they travel in opposite directions to conserve momentum

8d(iv) · 1 mark

  • A1

    kinetic energy = ½ × 9.11 × 10⁻³¹ × (4.9 × 10⁷)² = 1.1 × 10⁻¹⁵ J

8d(v) · 3 marks

  • C1

    E = mc²

  • C1

    E = hc / λ or E = hf and c = fλ

  • A1

    (1.1 × 10⁻¹⁵) + (9.11 × 10⁻³¹ × (3.00 × 10⁸)²) = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / λ λ = 2.39 × 10⁻¹² m

9a · 1 mark

  • B1

    number of nuclear disintegrations per unit time

9b · 3 marks

  • B1

    activity is proportional to the number of undecayed nuclei

  • B1

    activity = (–) rate of change of number of undecayed nuclei

  • B1

    N is proportional to the rate of change of N (so exponential variation)

9c(i) · 2 marks

  • C1

    120 = 180 exp (– λ × 8.4)

  • A1

    λ = 0.048 min⁻¹

9c(ii) · 1 mark

  • A1

    half-life = ln 2 / 0.048 = 14 min

9c(iii) · 3 marks

  • B1

    line with negative gradient throughout, starting at (0, 180)

  • B1

    curve with negative gradient passing through (8.4, 120)

  • B1

    curve with decreasing negative gradient, from t = 0 to t = 24 min, passing through (14, 90)

10a · 2 marks

  • M1

    speed is (directly) proportional to distance

  • A1

    speed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer

10b(i) · 2 marks

  • B1

    galaxy is receding from the Earth

  • B1

    observed wavelength is redshifted from emitted wavelength

10b(ii) · 2 marks

  • C1

    Δλ / λ = v / c (4.91 – 4.62) / 4.62 = v / (3.00 × 10⁸)

  • A1

    v = 1.9 × 10⁷ m s⁻¹

10b(iii) · 2 marks

  • B1

    wavelength (of maximum intensity) is inversely proportional to temperature

  • B1

    observed wavelength too high, so determined temperature too low

10c · 2 marks

  • C1

    v = H₀d

  • A1

    d = (1.9 × 10⁷) / (2.3 × 10⁻¹⁸) = 8.3 × 10²⁴ m

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