Cambridge A Level Physics 9702 — 2025 Oct/Nov Paper 4 · Variant 4

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A binary star consists of star A, of mass 4.0 × 1030 kg, and star B, of mass 2.0 × 1030 kg, separated by a distance of 3.3 × 1012 m. The stars are both in circular orbit around their common centre of gravity X, as shown in Fig. 1.1. orbit of A 3.3 × 1012 m star B star A X RA RB orbit of B Fig. 1.1 The radius RB of the orbit of star B is double the radius RA of the orbit of star A. (i) Use Newton’s law of gravitation to calculate the magnitude of the gravitational force exerted by each star on the other. force = ..................................................... N [2] (ii) Calculate the centripetal acceleration of star A. acceleration = ................................................ m s–2 [1] (iii) Use your answer in (b)(ii) to determine the period of the orbit of star A. period = ...................................................... s [3] (iv) By placing a tick (✓) in each row, complete Table 1.1 to show how the quantities indicated for star B compare with the same quantities for star A. Table 1.1 B less than A B equal to A B greater than A centripetal acceleration linear speed period [3] [Total: 11]

Mark scheme: Question Answer Marks 1(a) (gravitational) force is (directly) proportional to product of masses B1 force (between point masses) is inversely proportional to the square of their separation B1 1(b)(i) force = (6.67  10–11  4.0  1030  2.0  1030) / (3.3  1012)2 C1 = 4.9  1025 N A1 1(b)(ii) a = F / m = (4.9  1025) / (4.0  1030) A1 = 1.2  10–5 m s–2 1(b)(iii) a = r2 and = 2 / T C1 or a = v2 / r and v = (2r / T) r = (3.3  1012) / 3 C1 T = 2 [(3.3  1012) / (3  1.2  10–5)]½ A1 = 1.9  109 s 1(b)(iv) acceleration: B greater than A B1 linear speed: B greater than A B1 period: B equal to A B1

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2 (a) The equation of state for an ideal gas may be expressed as pV = NkT. (i) State the meaning of each of the symbols in this equation. p: ....................................................................................................................................... V: ....................................................................................................................................... N: ....................................................................................................................................... k: ....................................................................................................................................... T: ....................................................................................................................................... [3] (ii) Using the equation of state, derive an expression for the average translational kinetic energy EK of a particle in the gas in terms of some or all of N, k and T. EK = ......................................................... [2] (b) A molecule of hydrogen gas consists of two hydrogen atoms, each of nucleon number 1. A molecule of oxygen gas consists of two oxygen atoms, each of nucleon number 16. Assume that hydrogen and oxygen both behave as ideal gases. A sample of hydrogen gas is at the same temperature as a sample of oxygen gas. For the two samples, determine the ratio root-mean-square (r.m.s.) speed of hydrogen molecules . root-mean-square (r.m.s.) speed of oxygen molecules ratio = ......................................................... [2] [Total: 7]

Mark scheme: 2(a)(i) p = pressure (of gas), V = volume (of gas) and k = Boltzmann constant B1 N = number of molecules (in the gas) B1 T = thermodynamic temperature (of gas) B1 2(a)(ii) (pV =) NkT = ⅓Nm<c2> M1 EK = ½m<c2> = ½  3kT = (3 / 2)kT A1 2(b) (same temperature so) (½)m<c2> must be same for both gases C1 ratio = √(mO / mH) = √(32 / 2) A1 = 4.0

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3 (a) With reference to molecular kinetic energy and molecular potential energy, explain what is meant by the internal energy of an ideal gas. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A sample of an ideal gas is initially in state A, at a pressure of 2.0 × 105 Pa and with a volume of 0.016 m3, as shown in Fig. 3.1. 6 pressure / 105 Pa 4 2 A 0 0 0.01 0.02 0.03 0.04 volume / m3 Fig. 3.1 In state A, the temperature of the gas is 400 K. The gas undergoes two successive changes X and Y. In change X, it is heated at constant volume to a pressure of 4.0 × 105 Pa. At the end of change X, the gas is in state B. In change Y, it is then allowed to expand at constant temperature back to its original pressure. At the end of change Y, the gas is in state C. (i) Determine the internal energy of the gas in state A. internal energy = ...................................................... J [2] (ii) Determine the temperature of the gas in state B. temperature = ...................................................... K [1] (iii) Determine the volume of the gas in state C. volume = .................................................... m3 [1] (iv) On Fig. 3.1, draw two lines, one to represent change X and one to represent change Y. Label your lines X and Y respectively. [3] [Total: 9]

Mark scheme: 3(a) total kinetic energy associated with random motion of molecules B1 potential energy (of molecules) is zero B1 3(b)(i) pV = NkT and U = (3 / 2)NkT C1 U = (3 / 2)pV A1 = (3 / 2)  2.0  105  0.016 = 4800 J 3(b)(ii) temperature = 800 K A1 3(b)(iii) volume = 0.032 m3 A1 3(b)(iv) straight vertical line labelled X between A and (0.016, 4.0) B1 curve from B with continuously decreasing negative gradient, labelled Y B1 line labelled Y between (0.016, 4.0) and (0.032, 2.0) B1

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4 (a) State what is meant by the frequency of the oscillations of an oscillating object. ................................................................................................................................................... ............................................................................................................................................. [1] (b) An object is oscillating. Fig. 4.1 shows the variation of the acceleration a of the object with its displacement x from the equilibrium position. Fig. 4.2 shows the variation of the kinetic energy EK of the object with time t. 2 8 a / m s–2 EK / 10– 4 J 0 4 x / m –0.02 0 0.02 –2 0 0 0.2 0.4 0.6 0.8 t / s Fig. 4.1 Fig. 4.2 (i) Explain how Fig. 4.2 shows that the period of the oscillations is 0.80 s. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Calculate the angular frequency ω of the oscillations. ω = .............................................. rad s–1 [2] (iii) Apart from the period, frequency and angular frequency of the oscillations, determine three other conclusions about the object and its oscillations that may be drawn from Fig. 4.1 and Fig. 4.2. The conclusions may be qualitative or quantitative. Use the space below for any working. 1 ........................................................................................................................................ ........................................................................................................................................... 2 ........................................................................................................................................ ........................................................................................................................................... 3 ........................................................................................................................................ ........................................................................................................................................... [3] (iv) Describe the interchange between kinetic energy and potential energy during the oscillations. Numerical values are not required. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 10]

Mark scheme: 4(a) number of oscillations per unit time B1 4(b)(i) kinetic energy (of object) reaches maximum / minimum / zero twice in a cycle A1 4(b)(ii)  = 2 / T C1 = 2 / 0.80 or a0 = 2x0 = √(1.0 / 0.016) = 7.9 rad s–1 A1 4(b)(iii) Any three points from: B3 • oscillations are simple harmonic • amplitude = 0.016 m • maximum speed = 0.13 m s–1 • total energy of oscillations = 7.0  10–4 J • mass of object = 0.087 kg • maximum momentum = 0.011 kg m s–1 or 0.011 N s 4(b)(iv) Any two points from: B2 • kinetic energy is a maximum at zero displacement or kinetic energy is zero at maximum displacement • potential energy is zero at zero displacement or potential energy is a maximum at maximum displacement • kinetic energy is maximum when the potential energy is zero or potential energy is a maximum when the kinetic energy is zero kinetic energy + potential energy is constant B1

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5 (a) Explain why the electric potential near an isolated proton is positive. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) An isolated metal sphere is positively charged and has radius R, as shown in Fig. 5.1. sphere + + + + R + + P X Y + + Q + + + + x Fig. 5.1 Line XY passes through the centre of the sphere. Point P lies on line XY at a variable displacement x from the centre of the sphere. Point Q is at a fixed position that is not on line XY. The electric field strength at the surface of the sphere is E0. (i) On Fig. 5.1, draw an arrow at point Q to show the direction of the electric field at that point. [1] (ii) On Fig. 5.2, sketch the variation of the electric field E at point P with x for values of x between x = –3R and x = 3R. Do not include the region inside the sphere between x = –R and x = R. E0 E ½E0 0 –3R –2R –R 0 R 2R 3R x –½E0 –E0 Fig. 5.2 [3] (c) The proton and the electron in a hydrogen atom are separated by a distance of 5.3 × 10–11 m. Calculate the electric potential energy of the proton and the electron. electric potential energy = ...................................................... J [2] [Total: 9]

Mark scheme: 5(a) potential is (defined as) zero at infinity B1 proton has a positive charge and so repels another positive charge B1 work is done on two (positive) charges to move them towards each other B1 or work is done by two (positive) charges as they move apart from each other 5(b)(i) arrow drawn through Q in a WSW direction directly away from the centre of the sphere B1 5(b)(ii) curve in at least one quadrant passing through (R, E0) and (2R, ¼E0) B1 curve between –3R and –R of increasing magnitude of gradient B1 and curve between R and 3R of decreasing magnitude of gradient two lines drawn, one in the top right quadrant, the other in the bottom left quadrant B1 5(c) EP = – (1.60  10–19)2 / [4  8.85  10–12  (5.3  10–11)] C1 = –4.3  10–18 J A1

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6 Fig. 6.1 shows part of a bridge rectifier circuit that can be used for rectification of an alternating input voltage VIN. VIN VOUT R Fig. 6.1 The circuit contains four diodes, one of which is shown. The rectified output voltage VOUT is applied across load resistor R. (a) (i) State what is meant by rectification. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State the name of the type of rectification produced by a bridge rectifier circuit. ..................................................................................................................................... [1] (iii) Complete the circuit in Fig. 6.1 by drawing the three missing diodes inside the dashed circles. [2] (b) The input voltage varies with time t according to the equation VIN = 34 sin 18t where VIN is in V and t is in s. (i) Show that the period of the input voltage is 0.35 s. [2] (ii) Calculate the root-mean-square (r.m.s.) input voltage. r.m.s. voltage = ...................................................... V [1] (iii) On Fig. 6.2, sketch the variation of VOUT with t from t = 0 to t = 0.35 s. 40 VOUT / V 0 0 0.1 0.2 0.3 0.4 t / s – 40 Fig. 6.2 [3] (c) Resistor R has a resistance of 56 kΩ. A capacitor of capacitance 12 μF is connected into the circuit of Fig. 6.1 in order to smooth the output voltage. (i) On Fig. 6.1, draw the capacitor correctly connected into the circuit. [1] (ii) Calculate the time constant of the smoothing circuit. time constant = ...................................................... s [2] (iii) During each discharge cycle, the time for which the capacitor is discharging is 0.14 s. Determine the minimum value of the smoothed output voltage. minimum voltage = ...................................................... V [2] [Total: 15]

Mark scheme: 6(a)(i) conversion of a.c. to d.c. B1 6(a)(ii) full-wave (rectification) B1 6(a)(iii) three diodes with correct symbols connected into the circuit B1 lower-left diode shown pointing towards the right and upper-left and lower-right diodes both shown pointing towards the left B1 6(b)(i) = 18 (rad s–1) C1 T = 2 / 18 = 0.35 s A1 6(b)(ii) Vr.m.s.= 34 / √2 A1 = 24 V 6(b)(iii) sinusoidal ‘humps’ with minimum VOUT = 0 and non-zero VOUT all same sign B1 peak VOUT shown as 34 V or –34 V B1 two ‘humps’ shown, with VOUT always with same sign and VOUT = 0 at t = 0, t = 0.175 s and t = 0.350 s and non-zero in B1 between 6(c)(i) correct circuit symbol for capacitor, connected in parallel with resistor B1 6(c)(ii) time constant = RC C1 = 56  103  12  10–6 A1 = 0.67 s 6(c)(iii) Vmin = 34 exp (–0.14 / 0.67) C1 = 28 V A1

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7 (a) State Lenz’s law of electromagnetic induction. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A helicopter hovering in stationary equilibrium has four rotors, each of length 12 m, as shown in the view from above in Fig. 7.1. direction of rotation rotors X 12 m O Fig. 7.1 The vertical component of the Earth’s magnetic field at the helicopter is downwards with a flux density of 0.047 mT. The rotors each rotate in a horizontal plane in the direction shown with a frequency of 85 Hz. (i) Calculate the magnetic flux Φ cut by rotor OX during one complete rotation. Give a unit with your answer. Φ = .................................... unit ............ [3] (ii) Determine the magnitude of the electromotive force (e.m.f.) induced across the length of rotor OX. e.m.f. = ...................................................... V [2] (iii) Use Lenz’s law to explain whether end O or end X of the rotor is at the higher potential. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 9]

Mark scheme: 7(a) direction of induced e.m.f. M1 such as to (produce effects that) oppose the change that caused it A1 7(b)(i)  = BA C1 = 0.047  10–3    122 C1 = 0.021 Wb A1 7(b)(ii) e.m.f. = flux cut / time C1 = 0.021 / (1 / 85) A1 = 1.8 V 7(b)(iii) force (due to current) cause anticlockwise moment (to oppose rotation) B1 (from the LH rule) current is from O to X, so X is at the higher potential B1

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8 Oxygen-15 (158O) is radioactive and has a half-life of 2.04 minutes. The decay of oxygen-15 produces positrons. For this reason, oxygen-15 is sometimes used as a tracer in positron emission tomography (PET scanning). (a) State what is meant by a tracer. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) The equation for the decay of oxygen-15 is 158O Q XP + RS β+ + Z where X is the nucleus formed during the decay and Z is another particle. (i) State the values of the integers P, Q, R and S. P = ............................................... R = ............................................. Q = .............................................. S = .............................................. [2] (ii) State the name of particle Z. ..................................................................................................................................... [1] (c) (i) Define the activity of a sample. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Calculate the decay constant of oxygen-15. Give a unit with your answer. decay constant = .................................... unit ............ [2] (iii) Determine the rate at which positrons are produced in a sample of oxygen-15 that has a mass of 2.85 × 10–6 kg. rate = ................................................... s–1 [4] (d) The particles that are emitted from the body and detected outside it during PET scanning are not positrons but another type of particle. (i) State the name of the particles that are detected. ..................................................................................................................................... [1] (ii) Explain how these particles are formed inside the body. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 15]

Mark scheme: 8(a) (radioactive) substance introduced into the body B1 substance absorbed by the tissues being studied B1 8(b)(i) P = 15 and R = 0 A1 Q = 7 and S = (+)1 A1 8(b)(ii) (electron) neutrino B1 8(c)(i) number of nuclear disintegrations per unit time B1 8(c)(ii) decay constant = ln 2 / (2.04  60) C1 = 5.66  10–3 s–1 A1 8(c)(iii) N = (2.85  10–6) / (15  1.66  10–27) C1 A = N C1 rate = (5.66  10–3)  (2.85  10–6) / (15  1.66  10–27) C1 = 6.48  1017 s–1 A1 8(d)(i) gamma photons B1 8(d)(ii) positron collides with electron (in body) B1 annihilation results in their masses becoming photon energy B1

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9 (a) State what is meant by the photoelectric effect. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) The photoelectric effect is investigated using two clean metal plates. One plate is made from metal X and the other is made from metal Y. Metal X has work function energy Φ. Metal Y has work function energy 2Φ. Metal X has threshold frequency F. State expressions, in terms of either or both of Φ and F, for (i) the threshold frequency of metal Y threshold frequency = ......................................................... [1] (ii) the Planck constant. Planck constant = ......................................................... [1] (c) The maximum kinetic energy EK of photoelectrons is determined for each of the plates in (b) for different frequencies f of incident radiation. On Fig. 9.1, sketch the variation of EK with f for each plate. Label your lines X and Y to identify which line relates to which plate. 2Φ EK Φ 0 0 F 2F 3F f –Φ –2Φ Fig. 9.1 [4] [Total: 8]

Mark scheme: 9(a) emission of electrons (from a metal surface) M1 when electromagnetic radiation is incident (on surface) A1 9(b)(i) threshold frequency = 2F A1 9(b)(ii) Planck constant =  / F A1 9(c) two diagonal straight lines with positive gradient in the positive EK region only, starting at non-zero values of f B1 line labelled X passing through (F, 0) and line labelled Y passing through (2F,0) and extending to 3F or +2 B1 two diagonal straight lines with equal gradients B1 X line extrapolates back to (0, –) and Y line extrapolates back to (0, –2) B1

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10 (a) State what is meant by redshift. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Explain how observations of redshift lead to the idea that the universe is expanding. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) Explain how Hubble’s law leads to the Big Bang theory of the origin of the universe. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 7]

Mark scheme: 10(a) recession of galaxy (from observer) causes emitted light to have B1 increase in observed wavelength / decrease in observed frequency B1 10(b) (distant) galaxies show redshift so galaxies are moving apart B1 galaxies moving apart means universe must be expanding B1 10(c) the speed of recession is proportional to the distance of the galaxies from each other B1 or more distant galaxies are receding faster Any two points from: B2 • more distant galaxies represent further back in time • a long time ago, all matter in the universe must have been very close together • a long time ago, all matter in the universe must have been moving apart very fast

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