Cambridge A Level Physics 9702 — 2025 Oct/Nov Paper 4 · Variant 1

1 (a) In terms of velocity and acceleration, describe uniform circular motion of an object. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Fig. 1.1 shows the view from above of a polystyrene ball undergoing horizontal circular motion of radius R. shadow of polystyrene ball screen P x polystyrene ball B θ O R path of ball light Fig. 1.1 The ball is illuminated by parallel light so that a shadow of the ball forms on a screen placed on the opposite side of the ball from the light source. The line joining points O and P is perpendicular to the screen. The angular speed of the circular motion is ω. (i) State an expression, in terms of R and ω, for the speed v of the ball. v = ......................................................... [1] (ii) Determine an expression, in terms of v and ω, for the centripetal acceleration of the ball. centripetal acceleration = ......................................................... [2] (c) The ball in (b) is in the position shown in Fig. 1.1, such that line OB is at an angle θ to the line OP. (i) Determine an expression, in terms of R and θ, for the displacement x of the shadow from P. x = ......................................................... [1] (ii) The value of θ is zero at time t = 0. State an expression for θ in terms of ω and t. θ = ......................................................... [1] (iii) Use your answers in (c)(i) and (c)(ii) to show that x is given by x = R sin ω t. [1] (iv) Explain, with reference to the equation in (c)(iii), why the motion of the shadow of the ball on the screen may be modelled as simple harmonic. ........................................................................................................................................... ..................................................................................................................................... [1] (d) The circular motion of the ball in Fig. 1.1 has a diameter of 0.46 m and an angular speed of 1.9 rad s–1. For the simple harmonic motion of the shadow of the ball in Fig. 1.1, calculate: (i) the amplitude amplitude = ......................................................m [1] (ii) the period period = ....................................................... s [2] (iii) the maximum acceleration. maximum acceleration = .................................................m s–2 [2] (e) On Fig. 1.1, draw, and label with the letter A, the position of the shadow on the screen when the shadow has its maximum positive acceleration. [1] [Total: 15]

Mark scheme: Question Answer Marks 1(a) velocity and acceleration both have constant magnitude B1 velocity is (always) perpendicular to acceleration B1 1(b)(i) v = R A1 1(b)(ii) a = R2 or a = v2 / R C1 a = v A1 1(c)(i) x = R sin  A1 1(c)(ii) = t A1 1(c)(iii) clear substitution of = t into x = R sin leading to x = R sin t A1 1(c)(iv) equation is of the form x = x0 sin t (so simple harmonic motion) B1 1(d)(i) amplitude = 0.46 / 2 A1 = 0.23 m 1(d)(ii) = 2 / T C1 period = 2 / 1.9 A1 = 3.3 s 1(d)(iii) a0 = 2x0 C1 = 1.92  0.23 A1 = 0.83 m s–2 1(e) shadow on screen, labelled A, above left-hand edge of the circular path B1

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2 (a) State two ways in which the first law of thermodynamics describes that the internal energy of a system may be changed. 1 ................................................................................................................................................ ................................................................................................................................................... 2 ................................................................................................................................................ ................................................................................................................................................... [2] (b) (i) Use the first law of thermodynamics to explain why a bicycle pump gets hot when it is used to pump up a tyre quickly. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) With reference to molecular energies, explain why the temperature of water remains at 100 °C when it vaporises in a kettle, even though it is being heated. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 8]

Mark scheme: 2(a) work done on / by system B1 thermal energy supplied to / removed from system B1 2(b)(i) no thermal energy transferred to / from system (due to lack of time) B1 work is done on the gas to compress it / to decrease its volume B1 internal energy increases so temperature increases B1 2(b)(ii) (during vaporisation) molecular separation increases B1 (heating causes) potential energy of molecules to increase B1 kinetic energy of molecules unchanged so temperature unchanged B1

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3 (a) Define gravitational field at a point. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Fig. 3.1 shows an isolated point mass of mass M. mass M P x Fig. 3.1 Point P is at distance x from the point mass. (i) By considering the force exerted by the point mass on a test mass of mass m placed at P, derive an equation for the gravitational field strength g at P, in terms of M and x. Identify any other symbols you use. [2] (ii) On Fig. 3.1, draw an arrow to indicate the direction of the gravitational field at P. [1] x (iii) Point Q is at distance from the point mass, on the opposite side of the mass from P, as 2 shown in Fig. 3.2. Q mass M P x 2 x Fig. 3.2 Compare the gravitational field at Q with that at P. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) Two identical isolated uniform spheres X and Y each have radius R. The centres of the spheres are separated by distance L, as shown in Fig. 3.3. X Y P x L Fig. 3.3 Point P lies on the line joining the centres of X and Y, and is at a variable displacement x from the centre of sphere X. The gravitational field strength at the surface of each sphere is g0. On Fig. 3.4, sketch the variation with x of the gravitational field g at point P between x = R and x = L – R. g0 g 1 2 g0 0 R L / 2 L – R x – 1 2 g0 –g0 Fig. 3.4 [3] [Total: 9]

Mark scheme: 3(a) force per unit mass B1 3(b)(i) F = GMm / x2 C1 g = F / m A1 g = [GMm / x2] / m = GM / x2 and G = gravitational constant 3(b)(ii) arrow drawn at P pointing directly towards the point mass B1 3(b)(iii) fields are in opposite directions B1 field strength at Q is four times the field strength at P B1 3(c) line starting at (R, –g0) and ending at (L – R, +g0) B1 line passing through (L / 2, 0) B1 curve becoming shallower from R to (L / 2) and then steeper from (L / 2) to (L – R) B1

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4 (a) State the value of absolute zero on: (i) the Celsius temperature scale temperature = ..................................................... °C [1] (ii) the thermodynamic temperature scale. Give a unit with your answer. temperature = ....................................... unit .......... [1] (b) A sample contains a fixed amount of gas. The gas has pressure p, volume V and thermodynamic temperature T. Fig. 4.1 shows the variation of pV with kT for the sample, where k is the Boltzmann constant. 300 pV / J 200 100 0 0 2 4 6 8 kT / 10–21 J Fig. 4.1 (i) State what is indicated about the nature of the gas from the variation shown in Fig. 4.1. ..................................................................................................................................... [1] (ii) Determine the number N of molecules of the gas in the sample. N = ......................................................... [2] (iii) Use your answer in (b)(ii) to determine the amount n of gas in the sample. n = ................................................... mol [1] (c) The root-mean-square (r.m.s.) speed of the molecules of the gas is 1900 m s–1 when pV is equal to 270 J. Determine the mass, in u, of one molecule of the gas, where u is the unified atomic mass unit. mass = ...................................................... u [4] [Total: 10]

Mark scheme: 4(a)(i) temperature = –273.15 °C A1 4(a)(ii) temperature = 0 K A1 4(b)(i) gas is ideal B1 4(b)(ii) pV = NkT C1 N = 270 / (8.0  10–21) A1 = 3.4  1022 4(b)(iii) n = (3.4  1022) / (6.02  1023) A1 = 0.056 mol 4(c) ½ m<c2> = (3 / 2) kT C1 ½  m  19002 = 1.5  8.0  10–21 C1 (m = 6.65  10–27 kg) m = (6.65  10–27) / (1.66  10–27) C1 = 4.0 u A1

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5 (a) Define electric potential at a point. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A hydrogen atom may be considered to consist of a proton and an electron separated by a distance of 120 pm, as shown in Fig. 5.1. proton P electron x 120 pm Fig. 5.1 The two particles may be considered as point charges. Point P lies on the line joining the electron and the proton and is at a variable distance x from the proton. (i) Show that the electric potential V at point P when x = 10 pm is equal to 130 V. [2] (ii) Calculate, to two significant figures, V when x = 30 pm. V = .......................................................V [2] (iii) On Fig. 5.1, draw a cross (×) at one position, other than infinity, where the electric potential is zero. [1] (iv) On Fig. 5.2, sketch the variation of V with x between x = 10 pm and x = 110 pm. 160 V / V 80 0 10 30 50 70 90 110 x / pm –80 –160 Fig. 5.2 [3] [Total: 10]

Mark scheme: 5(a) work done per unit charge B1 work (done) moving positive charge from infinity (to the point) B1 5(b)(i) potential (due to proton) = (1.60  10–19) / (4  8.85  10–12  10  10–12) C1 or potential (due to electron) = (–1.60  10–19) / (4  8.85  10–12  110  10–12) V = [(1.60  10–19) / (4  8.85  10–12)]  [(10–1 – 110–1)  1012] = 130 V A1 5(b)(ii) V = [(1.60  10–19) / (4  8.85  10–12)]  [(30–1 – 90–1)  1012] C1 = (+) 32 V A1 5(b)(iii) cross drawn midway between the electron and the proton B1 5(b)(iv) line from (10, +130) to (110, –130) B1 curve getting shallower until x = 60 pm, crossing V = 0 at (60, 0) and then getting steeper after x = 60 pm B1 curve passing through (30, ±32) and (90, ±32) B1

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6 (a) Two parallel plate capacitors C1 and C2 are connected to a supply that has a potential difference (p.d.) VS. The capacitors may be connected in series or in parallel. The supply provides charge QS and the plates of the two capacitors acquire charges Q1 and Q2 respectively. The p.d.s across the plates of the capacitors are V1 and V2 respectively. Complete Table 6.1 to indicate how QS, Q1 and Q2 relate to each other, and how VS, V1 and V2 relate to each other, for series and parallel connections of the capacitors to the supply. Table 6.1 relationship between charges relationship between p.d.s series parallel [4] (b) An isolated capacitor of capacitance 470 μF stores 19 mJ of energy. (i) Calculate the p.d. across the capacitor. p.d. = .......................................................V [2] (ii) Calculate the charge on the capacitor. charge = ...................................................... C [2] (iii) The capacitor is now connected in parallel with a capacitor of capacitance 180 μF that is initially uncharged. Determine the total energy, in mJ, now stored in the two capacitors. energy = .................................................... mJ [3] [Total: 11]

Mark scheme: 6(a) series charges: QS = Q1 = Q2 B1 series p.d.s: VS = V1 + V2 B1 parallel charges: QS = Q1 + Q2 B1 parallel p.d.s: VS = V1 = V2 B1 6(b)(i) E = ½ CV2 C1 p.d. = [(2  19  10–3) / (470  10–6)]½ A1 = 9.0 V 6(b)(ii) E = Q2 / 2C or C = Q / V C1 Q = (19 × 10–3 × 2 × 470 × 10–6)½ A1 or Q = 470 × 10–6 × 9.0 Q = 4.2  10–3 C 6(b)(iii) total charge unchanged C1 total capacitance = (470 + 180)  10–6 (F) C1 E = Q2 / 2C = (4.23  10–3)2 / (2  650  10–6) (= 0.014 J) A1 E = 14 mJ

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7 (a) State Faraday’s law of electromagnetic induction. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) An aircraft is flying horizontally at constant speed v through the Earth’s magnetic field, as shown in Fig. 7.1. vertical component of Earth’s aircraft magnetic field, 38 μT P v v 68 m Q VIEW FROM SIDE VIEW FROM ABOVE Fig. 7.1 At the location of the aircraft, the vertical component of the Earth’s magnetic field is 38 μT towards the ground. The distance between the wingtips P and Q of the aircraft is 68 m. As the aircraft moves through the magnetic field, an electromotive force (e.m.f.) of 0.54 V is induced between the wingtips P and Q. (i) Calculate the magnetic flux cut by the wings of the aircraft in a time of 15 s. Give a unit with your answer. magnetic flux = …………………………………… unit ….……. [2] (ii) Determine the area of flux cut by the wings in a time of 15 s. area = .....................................................m2 [2] (iii) Use your answer in (b)(ii) to determine the speed v of the aircraft. v = .................................................m s–1 [2] (iv) Use Lenz’s law of electromagnetic induction to explain which of the wingtips P and Q is at the higher induced potential. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 11]

Mark scheme: 7(a) (induced) e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b)(i) flux = e.m.f.  time C1 flux = 0.54  15 A1 = 8.1 Wb 7(b)(ii) = BA C1 area = 8.1 / (38  10–6) A1 = 2.1  105 m2 7(b)(iii) area = speed  time  width C1 v = (2.1  105) / (15  68) A1 = 210 m s–1 7(b)(iv) opposing force (due to current in wings) must be backwards B1 from Fleming’s left-hand rule, current (in wings) must be from Q to P B1 current is from – to + inside an e.m.f. source so P is at higher potential B1

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8 (a) State what is meant by a photon. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] 238 (b) A stationary nucleus of uranium-238 ( 92U) undergoes alpha decay to produce a nucleus 234 of thorium-234 ( 90Th). The kinetic energy of the emitted alpha particle is 4.200 MeV. A gamma-ray photon is also emitted during the decay. Assume that the rebound kinetic energy of the thorium nucleus is negligible. Table 8.1 shows the masses of the nuclides involved in the decay reaction. The mass of the uranium-238 nuclide is missing. Table 8.1 nuclide nuclide mass / u 4 4.000 407 2α 234 233.915 174 90Th 238 92U The total energy released in the decay of the nucleus of uranium-238 is 4.274 MeV. (i) Calculate the mass, in u, of the uranium-238 nuclide. Give your answer to five decimal places. mass = ...................................................... u [3] (ii) Determine a value for the wavelength of the gamma radiation emitted during the decay of the uranium-238 nucleus. wavelength = ......................................................m [3] (iii) In practice, the rebound kinetic energy of the thorium nucleus is not negligible. Explain, without further calculation, how your answer in (b)(ii) compares with the true wavelength of gamma radiation emitted during the decay of the uranium-238 nucleus. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (c) Gamma radiation emitted during the decay of a sample of uranium-238 has a single wavelength. decay by beta emission, and also emit gamma radiation in the Nuclei of cobalt-60 (6027Co) process. Suggest why there is not a single wavelength for the gamma radiation emitted during the decay of a sample of cobalt-60. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 11]

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) E = c2m C1 m = (4.274  106  1.60  10–19) / (1.66  10–27  (3.00  108)2) C1 ( = 0.00458 u) m = 233.915174 + 4.000407 + 0.00458 A1 = 237.92016 u 8(b)(ii) E = hc /  C1 or E = hf and c = f (4.274 – 4.200)  1.60  10–13 = (6.63  10–34  3.00  108) /  C1 = 1.7  10–11 m A1 8(b)(iii) (true) energy of gamma photon is smaller so (true) wavelength is larger B1 8(c) (anti)neutrinos are emitted during beta decay B1 particles emitted during beta decay carry varying amounts of energy, so energy of gamma photon is also variable (between B1 decays)

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9 (a) State Wien’s displacement law. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Fig. 9.1 shows the variation with d –2 of the radiant flux intensity F observed from a star X, where d is the distance of the observer from the star. Fig. 9.2 shows the variation with wavelength λ of the rates of emission P of radiation by star X and the Sun. 8 star X F / 103 W m–2 P 4 Sun 0 0 1 2 3 0 5 10 15 d–2 / 10–23 m–2 λ / 10–7 m Fig. 9.1 Fig. 9.2 The surface temperature of the Sun is 5770 K. State three conclusions about star X that can be drawn from this data. The conclusions may be qualitative or quantitative. Use the space for any working. 1 ................................................................................................................................................ ................................................................................................................................................... 2 ................................................................................................................................................ ................................................................................................................................................... 3 ................................................................................................................................................ ................................................................................................................................................... [3] (c) Star X is in a galaxy that is moving away from the Earth. Suggest, with a reason, how the line for star X in Fig. 9.2 would appear differently if it had been obtained from data measured on the Earth. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 7]

Mark scheme: 9(a) temperature inversely proportional to wavelength M1 temperature is thermodynamic temperature of surface of star and wavelength is the wavelength at which maximum A1 emission rate from star occurs 9(b) Any three points from: B3 • (surface) temperature of star X = 7000 K or star X has a higher temperature than the Sun • star X has a higher luminosity than the Sun • luminosity of star X = 2.7  1027 W • radius of star X = 1.3  109 m 9(c) light (from star X) is redshifted B1 wavelength of peak emission rate would be greater (using observed data) B1

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10 (a) Define specific acoustic impedance. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Explain how ultrasound waves are detected by a piezoelectric crystal. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) Table 10.1 shows the specific acoustic impedance Z for body tissue, water and steel. Table 10.1 material Z / kg m–2 s–1 body tissue 1.38 × 106 water 1.48 × 106 steel 4.04 × 107 (i) Calculate the intensity reflection coefficient for ultrasound incident on a water–steel boundary. intensity reflection coefficient = ......................................................... [2] (ii) Explain, without calculation, what is likely to happen when ultrasound is incident on a body tissue–water boundary. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: 10(a) product of density and speed M1 speed of sound in medium (and density of the medium) A1 10(b) ultrasound waves cause crystal to vibrate B1 vibrations (of crystal) cause induced e.m.f. (across crystal) B1 10(c)(i) intensity reflection coefficient= (40.4 – 1.48)2 / (40.4 + 1.48)2 C1 = 0.86 A1 10(c)(ii) Z values are very similar B1 (almost) all the ultrasound will be transmitted B1 or (almost) none of the ultrasound will be reflected

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