Cambridge A Level Physics 9702 — 2023 May/June Paper 4 · Variant 2

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A satellite is in a circular orbit around a planet. The radius of the orbit is R and the period of the orbit is T. The planet is a uniform sphere. Use Newton’s law of gravitation to show that R and T are related by 4π2R 3 = GMT 2 where M is the mass of the planet and G is the gravitational constant. [2] (c) The Earth may be considered to be a uniform sphere of mass 5.98 × 1024 kg and radius 6.37 × 106 m. A geostationary satellite is in orbit around the Earth. Use the expression in (b) to determine the height of the satellite above the Earth’s surface. height = ..................................................... m [3] (d) Another satellite is in a circular orbit around the Earth with the same orbital radius and period as the satellite in (c). (i) Calculate the angular speed of the satellite in this orbit. Give a unit with your answer. angular speed = .............................................. unit ................ [2] (ii) Despite having the same orbital period, the orbit of this satellite is not geostationary. Suggest two ways in which the orbit of this satellite could be different from the orbit of the satellite in (c). 1 ........................................................................................................................................ ........................................................................................................................................... 2 ........................................................................................................................................ ........................................................................................................................................... [2] [Total: 11]

Mark scheme: 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) GMm / R2 = mR2 M1  = 2 / T and algebra leading to 42R3 = GMT2 A1 or GMm / R2 = mv2 / R (M1) v = 2R / T and algebra leading to 42R3 = GMT2 (A1) 1(c) 42  R3 = 6.67  10–11  5.98  1024  (24  60  60)2 (R = 4.22  107 m) C1 h = R – (6.37  106) C1 h = (4.22  107) – (6.37  106) = 3.6  107 m A1 1(d)(i)  = 2 / T C1 = 2 / (24  60  60) = 7.3  10–5 rad s–1 A1 1(d)(ii) orbit is from east to west B1 orbit is not equatorial / orbit is polar B1

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2 (a) (i) State what is meant by an ideal gas. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) State the temperature, in degrees Celsius, of absolute zero. temperature = .................................................... °C [1] (b) A sealed vessel contains a mass of 0.0424 kg of an ideal gas at 227 °C. The pressure of the gas is 1.37 × 105 Pa and the volume of the gas is 0.640 m3. Calculate: (i) the number of molecules of the gas in the vessel number of molecules = ......................................................... [3] (ii) the mass of one molecule of the gas mass = .................................................... kg [1] (iii) the root-mean-square (r.m.s.) speed v of the molecules of the gas. v = ................................................ m s–1 [3] (c) The gas in (b) is now cooled gradually to absolute zero. On Fig. 2.1, sketch the variation with thermodynamic temperature T of the r.m.s. speed of the molecules of the gas. v r.m.s. speed 0 0 500 T / K Fig. 2.1 [2] [Total: 12]

Mark scheme: 2(a)(i) M1 where T is thermodynamic temperature A1 2(a)(ii) temperature = –273.15 °C A1 2(b)(i) pV = NkT C1 N = (1.37  105  0.640) / (1.38  10–23  (227 + 273)) C1 = 1.27  1025 A1 2(b)(ii) mass = 0.0424 / (1.27  1025) = 3.34  10–27 kg A1 2(b)(iii) ½m<c2> = (3 / 2)kT C1 3.34  10–27  v2 = 3  1.38  10–23  500 C1 v = 2490 m s–1 A1 or pV = ⅓(Nm) <c2> and Nm = mass of gas (C1) 0.0424  v2 = 3  1.37  105  0.640 (C1) v = 2490 m s–1 (A1) 2(c) sketch: line from (0, 0) to (500, v) B1 line with decreasing positive gradient throughout B1

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3 (a) State the first law of thermodynamics. Identify the meaning of any symbols that you use. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) The state of an ideal gas is continuously changed according to the cycle ABCDA shown in Fig. 3.1. C D pressure B A volume Fig. 3.1 (i) Complete Table 3.1 for the changes A to B and B to C by placing two ticks (3) in each row. Table 3.1 change in internal energy work done on gas change decrease no change increase negative zero positive A to B B to C [4] (ii) Use the first law of thermodynamics to describe and explain the energy transfers associated with one complete cycle ABCDA. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 9]

Mark scheme: 3(a) C1 increase in internal energy = work done on system + energy transferred to the system by heating A1 3(b)(i) AB change in internal energy: decrease B1 AB work done on gas: positive B1 BC change in internal energy: increase B1 BC work done on gas: zero B1 3(b)(ii) more work done by gas in CD than is done on gas in AB or (no work done on gas in BC and DA so) (overall) gas does work B1 (overall) change in internal energy is zero B1 (must be an overall) input of thermal energy B1

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4 A small steel sphere is oscillating vertically on the end of a spring, as shown in Fig. 4.1. spring steel sphere oscillations Fig. 4.1 The velocity v of the sphere varies with displacement x from its equilibrium position according to v = ± 9.7 (11 .6 - x 2) where v is in cm s–1 and x is in cm. (a) (i) Calculate the frequency of the oscillations. frequency = .................................................... Hz [2] (ii) Show that the amplitude of the oscillations is 3.4 cm. [1] (iii) Calculate the maximum acceleration a0 of the sphere. a0 = ................................................ m s–2 [2] (b) On Fig. 4.2, sketch the variation with x of the acceleration a of the sphere. 2 a0 a a0 0 – 4 – 2 0 2 4 x / cm – a0 – 2a0 Fig. 4.2 [3] (c) Describe, without calculation, the interchange between the potential energy and the kinetic energy of the oscillations. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 11]

Mark scheme: 4(a)(i) C1 f = 9.7 / 2 = 1.5 Hz A1 4(a)(ii) amplitude = √(11.6) = 3.4 cm A1 4(a)(iii) a0 = 2x0 C1 = 9.72  3.4  10–2 = 3.2 m s–2 A1 4(b) sketch: straight line through the origin with negative gradient B1 line with negative gradient passing through (+3.4, –a0) and (–3.4, +a0) B1 line with ends at x =  3.4 cm and a =  a0 B1 4(c) sum of potential energy and kinetic energy is constant B1 at maximum displacement, kinetic energy is zero or at maximum displacement, potential energy is maximum B1 at zero displacement, kinetic energy is maximum or at zero displacement, potential energy is minimum B1

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5 Two capacitors A and B are connected into the circuit shown in Fig. 5.1. X A S Y B Fig. 5.1 Capacitor A has capacitance C and capacitor B has capacitance 3C. The electromotive force (e.m.f.) of the cell is V. The two-way switch S is initially at position X, and capacitor B is initially uncharged. (a) State, in terms of V and C, expressions for: (i) the initial charge QA on the plates of capacitor A QA = ......................................................... [1] (ii) the initial energy EA stored in capacitor A. EA = .......................................................... [1] (b) The two-way switch S is now moved to position Y. (i) State and explain what happens to the charge that was initially on the plates of capacitor A. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Show that the final potential difference (p.d.) VB across capacitor B is given by V VB = . 4 Explain your reasoning. [3] (iii) Determine an expression, in terms of V and C, for the decrease ΔE in the total energy that is stored in the capacitors as a result of the change of the position of the switch. ΔE = ......................................................... [2] [Total: 9]

Mark scheme: 5(a)(i) QA = CV A1 5(a)(ii) EA = ½CV2 A1 5(b)(i) some of the charge transfers to (the plates of) capacitor B B1 transfer is because the p.d.s across the capacitors are not equal or transfer stops when the p.d.s across the capacitors become equal B1 5(b)(ii) VA = VB M1 charge on A + charge on B = CV M1 CVB + 3CVB = CV leading to VB = V / 4 A1 or CT = 4C (M1) QT = CV (M1) VB = CV / 4C = V / 4 (A1) 5(b)(iii) E = ½CV2 – nCV2, where n is a multiple that is less than ½ or total final energy = ½  4C  (V / 4)2 = ⅛CV2 C1 E = ½CV2 – ⅛CV2 = ⅜CV2 A1

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6 A heavy aluminium disc has a radius of 0.36 m. The disc rotates with the wheels of a vehicle and forms part of an electromagnetic braking system on the vehicle. In order to activate the braking system, a uniform magnetic field of flux density 0.17 T is switched on. This magnetic field is perpendicular to the plane of rotation of the disc, as shown in Fig. 6.1. aluminium disc, radius 0.36 m rim rotation of disc axle magnetic field, flux density 0.17 T Fig. 6.1 (a) (i) Define magnetic flux. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Calculate the magnetic flux through the disc. Give a unit with your answer. magnetic flux = ............................................... unit ................... [2] (b) The disc is rotating at a rate of 25 revolutions per second. Calculate the magnitude of the electromotive force (e.m.f.) induced between the axle and the rim of the disc. e.m.f. = ...................................................... V [3] (c) The axle and the rim are connected into an external circuit that enables the energy of the rotation of the disc to be stored for future use. The direction of rotation is shown in Fig. 6.1. Use Lenz’s law of electromagnetic induction to determine whether the current in the disc is from the rim to the axle or from the axle to the rim. Explain your reasoning. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 10]

Mark scheme: 6(a)(i) product of (magnetic) flux density and area M1 area perpendicular to the (magnetic) field A1 6(a)(ii) flux = B  r2 = 0.17    0.362 C1 = 6.9  10–2 Wb A1 6(b) time for one revolution = 1 / 25 s C1 e.m.f. = rate of cutting flux or  / t C1 = 0.069  25 = 1.7 V A1 6(c) current (in disc) is perpendicular to magnetic field or current causes force to act on disc B1 force opposes rotation of disc B1 left-hand rule indicates current is from rim to axle B1

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7 Four diodes are used in a bridge rectifier circuit to produce rectification of a sinusoidal a.c. input voltage VIN. Fig. 7.1 shows part of the circuit, but three of the diodes are missing. VIN VOUT R Fig. 7.1 The p.d. across the load resistor R is the output p.d. VOUT of the bridge rectifier. (a) (i) State the name of the type of rectification produced by a bridge rectifier. ..................................................................................................................................... [1] (ii) Complete Fig. 7.1 by drawing the three missing diodes, correctly connected. [2] (iii) On Fig. 7.1, draw an arrow to indicate the direction of the current in resistor R. [1] (b) VIN has amplitude V0 and period T. Fig. 7.2 shows the variation with time t of VIN. V0 VIN 0 0 0.5T 1.0T 1.5T 2.0T t –V0 Fig. 7.2 (i) On Fig. 7.3, sketch the variation of VOUT with t between t = 0 and t = 2.0T. V0 VOUT 0 0 0.5T 1.0T 1.5T 2.0T t –V0 Fig. 7.3 [3] (ii) The power dissipated in the resistor is P. On Fig. 7.4, sketch the variation of P with t between t = 0 and t = 2.0T. P 0 0 0.5T 1.0T 1.5T 2.0T t Fig. 7.4 [2] (iii) Suggest, with a reason, how the root-mean-square (r.m.s.) value of VOUT compares with the r.m.s. value of VIN. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 10]

Mark scheme: 7(a)(i) full-wave (rectification) B1 7(a)(ii) lower left diode shown pointing left B1 lower right and upper left diodes shown pointing left B1 7(a)(iii) arrow indicating current direction in resistor to the right B1 7(b)(i) sketch: periodic line showing minimum VOUT = 0 and maximum VOUT = +V0 B1 line showing peak VOUT at t = 0, 0.5T, 1.0T, 1.5T and 2.0T, with VOUT going to zero half-way in between each peak B1 line showing correct modulated sine shape B1 7(b)(ii) sketch: sinusoidal curve with troughs sitting on the time axis B1 peak power at t = 0, 0.5T, 1.0T, 1.5T and 2.0T and zero power half-way in between each peak B1 7(b)(iii) same power-time graph with or without rectification, so same Vrms or V2-time graph is same for both VOUT and VIN, so same Vrms or power does not depend on sign of V, so same Vrms B1

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8 Fig. 8.1 shows the lowest four energy levels of an electron in an isolated atom. n = 4 n = 3 n = 2 increasing energy n = 1 Fig. 8.1 Fig. 8.2 shows the lines in the emission spectrum of the atom that correspond to the transitions of the electron from n = 3 to n = 1 and from n = 4 to n = 1. increasing frequency Fig. 8.2 (a) Explain, with reference to photons, why there is a single frequency of electromagnetic radiation that corresponds to each of these transitions. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) (i) On Fig. 8.2, draw a line that corresponds to the transition of the electron from n = 2 to n = 1. Label this line A. [2] (ii) On Fig. 8.2, draw a line that corresponds to the transition of the electron from n = 3 to n = 2. Label this line B. [2] (c) The frequency of radiation represented by line A is fA. The frequency of radiation represented by line B is fB. The energy of the ground state (n = 1) is E1. Determine an expression, in terms of fA, fB, E1 and the Planck constant h, for the energy E3 of the energy level n = 3. E3 = ......................................................... [2] [Total: 8]

Mark scheme: 8(a) transition (emits) (one) photon with energy equal to the difference in energy between the two levels B1 frequency of radiation corresponds to energy of photon B1 8(b)(i) line to the left of the pair in Fig. 8.2, labelled A B1 larger gap between line A and the nearest of the pair in Fig. 8.2 than between the lines in the pair B1 8(b)(ii) line to the left of both the pair in Fig. 8.2 and line A, labelled B B1 larger gap between line B and line A than between line A and the nearest one of the pair in Fig. 8.2 B1 8(c) E = hf C1 E3 = E1 + h(fA + fB) A1

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9 (a) Define mass defect. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Table 9.1 shows the mass defects of three nuclei. Table 9.1 nucleus mass defect / u 21H 0.002 388 31H 0.009 105 42He 0.030 377 The nuclear fusion process in a particular star is described by 21H + 31H 42He + X where X is a particle that has no mass defect. (i) State the name of particle X. ..................................................................................................................................... [1] (ii) Show that the energy released when one nucleus of 42He is formed in this fusion reaction is 2.8 × 10–12 J. [3] (c) The star in (b) has a radius of 2.3 × 109 m and a luminosity of 1.4 × 1028 W. All the energy released from the formation of 42He is radiated away from the star. All the energy that is radiated from the star has been released in the formation of 42He. Determine: (i) the mass of 42He produced per unit time by the fusion process mass per unit time = ............................................... kg s–1 [3] (ii) the surface temperature of the star. temperature = ...................................................... K [2] [Total: 11]

Mark scheme: 9(a) difference between mass of nucleus and (total) mass of nucleons M1 when infinitely separated A1 9(b)(i) neutron B1 9(b)(ii) E = m c2 C1 m = (0.030377 – 0.002388 – 0.009105)u ( = 0.018884u) C1 energy release = (0.030377 – 0.002388 – 0.009105)  1.66  10–27  (3.00  108)2 = 2.8  10–12 J A1 9(c)(i) number of atoms per unit time = (1.4  1028) / (2.8  10–12) ( = 5.0  1039 s–1) C1 mass of one atom = 4  1.66  10–27 or (4  10–3) / (6.02  1023) ( = 6.64  10–27 kg) C1 mass per unit time = 6.64  10–27  5.0  1039 = 3.3  1013 kg s–1 A1 9(c)(ii) L = 4σr2T4 1.4  1028 = 4  5.67  10–8  (2.3  109)2  T4 C1 T = 7800 K A1

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10 (a) X-rays for use in medical diagnosis are produced in an X-ray tube. In the X-ray tube, charged particles are accelerated towards a metal target by an applied potential difference (p.d.). (i) State the name of the charged particles that are accelerated by the applied p.d. ..................................................................................................................................... [1] (ii) Explain how X-rays are produced at the metal target. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (iii) Calculate the minimum wavelength of X-rays produced when the applied p.d. is 5.80 kV. wavelength = ..................................................... m [3] (b) X-rays pass through a medium that has an attenuation coefficient of 1.4 cm–1. Calculate the percentage of the X-ray energy that is absorbed by a 2.8 cm thickness of this medium. percentage absorbed = ..................................................... % [3] [Total: 9]

Mark scheme: 10(a)(i) electrons B1 10(a)(ii) electrons are decelerated / stopped on impact with the target B1 (kinetic) energy lost by electrons emitted as (X-ray) photons B1 10(a)(iii) eV = hc /  C1  = (6.63  10–34  3.00  108) / (1.60  10–19  5800) C1 = 2.14  10–10 m A1 10(b) I = I0 exp (–x) C1 IT / I0 = exp (–(1.4  2.8)) = 0.020 C1 % absorbed = (1.000 – 0.0198)  100 = 98% A1

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