Cambridge A Level Physics 9702 — 2025 May/June Paper 4 · Variant 1

1 (a) Define gravitational potential at a point. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) 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. (i) Show that the mass of Mars is 6.4 × 1023 kg. [3] (ii) Calculate the gravitational potential φ at the surface of Mars. Give a unit with your answer. φ = .................................. unit ............... [2] (c) The satellite in (b) is moved to an orbit in which the satellite remains at the same point above the surface of Mars. (i) The orbit has a period of 25 hours. State what can be deduced from this about the rotation of Mars on its axis. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State one other feature of this orbit. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 9]

Mark scheme: Question Answer Marks 1(a) work done per unit mass B1 work (done in) moving mass from infinity (to the point) B1 1(b)(i) 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) C1 6.67 × 10–11 × M × 122 × [(5.1 × 106)–1 – (10.2 × 106)–1] = 5.1 × 108 A1 leading to M = 6.4 × 1023 kg 1(b)(ii)  = (–) (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106) C1 = –1.3 × 107 J kg–1 A1 1(c)(i) Mars takes (just under) 25 hours to rotate once on its axis B1 1(c)(ii) orbit is equatorial B1 or orbit is in same direction as direction of rotation of Mars

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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. orbit of electrons 170 pm electron electron nucleus Fig. 2.1 (a) State Coulomb’s law. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) (i) State the charge on the nucleus, in terms of the elementary charge e. charge = ....................................................... e [1] (ii) Show that the electric force between the nucleus and one of the electrons is 1.6 × 10–8 N. [1] (c) Assume that the force in (b)(ii) is the only force on the electrons. (i) Calculate the speed of the orbiting electrons. speed = ................................................ m s–1 [2] (ii) Calculate the period of the orbit of the electrons. period = ....................................................... s [2] (d) In practice, the orbit of each electron is affected by the presence of the other electron. (i) For the position of one of the electrons, determine the ratio electric field strength due to the other electron . electric field strength due to the nucleus ratio = ......................................................... [2] (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] [Total: 11]

Mark scheme: 2(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 2(b)(i) charge = (+)2e A1 2(b)(ii) F = 2 × (1.60 × 10–19)2 / [4 × 8.85 × 10–12 × (170 × 10–12)2] = 1.6 × 10–8 N A1 2(c)(i) F = mv2 / r C1 v = [ (1.6 × 10–8 × 170 × 10–12) / (9.11 × 10–31) ]½ A1 = 1.7 × 106 m s–1 2(c)(ii) F = mr2 and = 2 / T C1 F = 42mr / T2 T = [ (42 × 9.11 × 10–31 × 170 × 10–12) / (1.6 × 10–8) ]½ A1 = 6.2 × 10–16 s or v = 2r / T (C1) T = (2 × 170 × 10–12) / (1.73 × 106) (A1) = 6.2 × 10–16 s 2(d)(i) E  Q / r2 C1 ratio = [1.60 × 10–19 × (170 × 10–12)2] / [3.2 × 10–19 × (340 × 10–12)2] A1 = 0.13 2(d)(ii) resultant force slightly less (than 1.6 × 10–8 N) so speed lower B1 or resultant force slightly less (than 1.6 × 10–8 N) so period greater

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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. specific latent heat of fusion = ................................................. J g–1 [4] [Total: 9]

Mark scheme: 3(a) (thermal) energy per unit mass (to cause state change) B1 (thermal) energy to change state at constant temperature B1 3(b) (for vaporisation): B1 involves greater change in volume (of substance) or involves greater increase in separation of molecules more work has to be done by molecules (to separate) M1 or greater increase in potential energy of molecules kinetic energy of molecules unchanged, so more thermal energy needed A1 3(c) Q = mc and Q = mL C1 for the water = 26.4 – 10.3 C1 (37.0 × L) + (37.0 × 4.18 × 10.3) = (208 × 4.18 × 16.1) C1 L = 335 J g–1 A1

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4 (a) (i) State what is meant by the internal energy of a system. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (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. Table 4.1 thermal energy increase in internal work done on gas supplied to gas energy of gas compression +W cooling [4] [Total: 8]

Mark scheme: 4(a)(i) sum of potential energy and kinetic energy B1 (total) energy of random motion of particles B1 4(a)(ii) potential energy (of molecules) (in an ideal gas) is zero, so the internal energy of the gas is equal to the total kinetic energy B1 (of molecules) kinetic energy of molecules is proportional to (thermodynamic) temperature (so internal energy is proportional to B1 (thermodynamic) temperature)) 4(b) 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 B1 (if fully correct, thermal energy for compression = 2U – W and thermal energy for cooling = –U: compression +W 2U – W +2U cooling 0 –U –U )

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5 A cuboidal block floats in a liquid with its base horizontal, as shown in Fig. 5.1. block liquid surface h 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. 1.0 a / m s–2 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m –1.0 Fig. 5.2 Fig. 5.3 shows the variation with h of the kinetic energy EK of the block. 10 EK / J 5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m Fig. 5.3 (a) (i) Determine the amplitude of the oscillations. amplitude = ...................................................... m [1] (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. 1 ................................................................................................................................................ ................................................................................................................................................... 2 ................................................................................................................................................ ................................................................................................................................................... 3 ................................................................................................................................................ ................................................................................................................................................... [3] (c) On Fig. 5.4, sketch the variation with h of the potential energy EP of the oscillations. 10 EP / J 5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m Fig. 5.4 [3] [Total: 8]

Mark scheme: 5(a)(i) amplitude = 0.60 m A1 5(a)(ii) oscillations are simple harmonic B1 5(b) Any three points from: B3 • mean / equilibrium position is at h = 1.4 m • total energy of oscillations = 9.0 J • angular frequency of oscillations = 1.2 rad s–1 or period of oscillations = 5.1 s or frequency of oscillation = 0.19 Hz • maximum speed of block = 0.73 m s–1 • mass of block = 33 kg 5(c) U-shaped curve resting on h axis (with minimum at EP = 0) B1 curve from h = 0.8 m to h = 2.0 m, with minimum EP shown at h = 1.4 m B1 both end-points of curve shown at EP = 9.0 J B1

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6 Fig. 6.1 shows a circuit that rectifies an alternating input voltage VIN and produces an output voltage VOUT across a resistor R. W Y rectification VIN C R VOUT circuit X Z Fig. 6.1 The four terminals of the rectification circuit are labelled W, X, Y and Z. A capacitor C is connected in parallel with resistor R. (a) (i) State what is meant by rectification. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State the purpose of capacitor C. ........................................................................................................................................... ..................................................................................................................................... [1] (b) Fig. 6.2 shows the variations with time t of the potential differences (p.d.s) VIN and VOUT. 12 8 VOUT p.d. / V 4 0 0 10 20 30 40 t / ms –4 –8 VIN –12 Fig. 6.2 (i) The variation of VIN with t can be represented by VIN = A cos Bt where A and B are constants. Determine the values of A and B. Give a unit with your answer for A. A = ........................................ unit ............... B = .................................................... rad s–1 [2] (ii) Determine the type of rectification produced by the circuit in Fig. 6.1. ..................................................................................................................................... [1] (iii) On Fig. 6.3, draw the circuit diagram for the components inside the rectification circuit. W Y X Z Fig. 6.3 [2] (iv) Determine a value for the time constant for the discharge of the capacitor C through the resistor R in Fig. 6.1. time constant = ....................................................... s [3] (c) The capacitor C has a capacitance of 570 μF. Use your answer in (b)(iv) to determine the resistance of resistor R. resistance = ...................................................... Ω [2] [Total: 12]

Mark scheme: 6(a)(i) conversion from a.c. to d.c. B1 6(a)(ii) smoothing B1 6(b)(i) A = 12 V A1 B = 2 / (20 × 10–3) A1 = 310 rad s–1 6(b)(ii) full-wave (rectification) B1 6(b)(iii) four diodes shown, with correct circuit symbols B1 four diodes correctly connected to form a bridge rectifier B1 6(b)(iv) V = V0 exp (–t / ) C1 or V = V0 exp (–t / RC) and  = RC 8.0 = 12 exp (– 7.3 × 10–3 / ) C1  = 0.018 s A1 6(c) time constant = RC C1 R = (0.018 / 570 × 10–6) A1 = 32 

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7 (a) Define magnetic flux density. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) 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. path of region of particle magnetic field particle Y Z 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. (i) 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. F = ......................................................... [1] (ii) On Fig. 7.1, draw an arrow at point Y to indicate the direction of the force in (b)(i). [1] (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] (ii) Derive an expression for v in terms of B and the electric field strength E. v = ......................................................... [2] [Total: 10]

Mark scheme: 7(a) • force per unit length B2 • force per unit current • length / current perpendicular to field 1 mark for any two points, 2 marks for all three points. 7(b)(i) F = BQv B1 7(b)(ii) arrow at Y pointing vertically upwards B1 7(b)(iii) upwards deflection showing circular path B1 7(c)(i) 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 B1 7(c)(ii) EQ = BQv B1 v = E / B A1

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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. wavelength = ...................................................... m [2] (c) State one similarity and one difference between an electron and a positron. similarity: ................................................................................................................................... ................................................................................................................................................... difference: ................................................................................................................................. ................................................................................................................................................... [2] (d) 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. (i) State the name of this type of reaction. ..................................................................................................................................... [1] (ii) State what happens to the electron and to the positron. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (iii) Explain why two gamma-ray photons are produced, rather than just one. ........................................................................................................................................... ..................................................................................................................................... [1] (iv) Show that the kinetic energy of the electron before the collision is 1.1 × 10–15 J. [1] (v) Use the information in (d)(iv) to determine, to three significant figures, the wavelength associated with the gamma radiation emitted in the collision. wavelength = ...................................................... m [3] [Total: 13]

Mark scheme: 8(a) wavelength associated with a moving particle B1 8(b)  = h / p C1 = (6.63 × 10–34) / (9.11 × 10–31 × 4.9 × 107) A1 = 1.5 × 10–11 m 8(c) similarity: any one point from: B1 • same mass • same magnitude of charge • both leptons difference: any one point from: B1 • electron has negative charge, positron has positive charge • positron is anti-particle of electron • electron is a particle, positron is an anti-particle 8(d)(i) (pair) annihilation B1 8(d)(ii) their mass gets converted into energy B1 (their mass–energy) becomes the energy of the gamma photons B1 8(d)(iii) they travel in opposite directions to conserve momentum B1 8(d)(iv) kinetic energy = ½ × 9.11 × 10–31 × (4.9 × 107)2 = 1.1 × 10–15 J A1 8(d)(v) E = mc2 C1 E = hc /  C1 or E = hf and c = f (1.1 × 10–15) + (9.11 × 10–31 × (3.00 × 108)2) = (6.63 × 10–34 × 3.00 × 108) /  A1  = 2.39 × 10–12 m

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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) 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. (i) Determine the decay constant, in min–1, of the radioactive isotope. decay constant = ................................................ min–1 [2] (ii) Use your answer in (c)(i) to determine the half-life, in min, of the radioactive isotope. half-life = ................................................... min [1] (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. 200 150 A / Bq 100 50 0 0 4 8 12 16 20 24 t / min Fig. 9.1 [3] [Total: 10]

Mark scheme: 9(a) number of nuclear disintegrations per unit time B1 9(b) 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) B1 9(c)(i) 120 = 180 exp (–  × 8.4) C1  = 0.048 min–1 A1 9(c)(ii) half-life = ln 2 / 0.048 A1 = 14 min 9(c)(iii) 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) B1

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10 (a) State Hubble’s law. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) 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. (i) Explain why the observed wavelength and the emitted wavelength have different values. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Calculate the speed of the star relative to the Earth. speed = ................................................ m s–1 [2] (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. distance = ...................................................... m [2] [Total: 10]

Mark scheme: 10(a) speed is (directly) proportional to distance M1 speed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1 10(b)(i) galaxy is receding from the Earth B1 observed wavelength is redshifted from emitted wavelength B1 10(b)(ii)  /  = v / c C1 (4.91 – 4.62) / 4.62 = v / (3.00 × 108) v = 1.9 × 107 m s–1 A1 10(b)(iii) wavelength (of maximum intensity) is inversely proportional to temperature B1 observed wavelength too high, so determined temperature too low B1 10(c) v = H0d C1 d = (1.9 × 107) / (2.3 × 10–18) A1 = 8.3 × 1024 m

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