Cambridge A Level Physics 9702 — 2020 May/June Paper 4 · Variant 1

1 (a) State what is meant by a gravitational force. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A binary star system consists of two stars S1 and S2, each in a circular orbit. The orbit of each star in the system has a period of rotation T. Observations of the binary star from Earth are represented in Fig. 1.1. S1 S1 S2 S2 T t = 0 t = — 4 S2 S1 S2 S1 T 3T t = — t = — 2 4 S1 S2 t = T Fig. 1.1 (not to scale) Observed from Earth, the angular separation of the centres of S1 and S2 is 1.2 × 10–5 rad. The distance of the binary star system from Earth is 1.5 × 1017 m. Show that the separation d of the centres of S1 and S2 is 1.8 × 1012 m. [1] (c) The stars S1 and S2 rotate with the same angular velocity ω about a point P, as illustrated in Fig. 1.2. d P S1 S2 x Fig. 1.2 (not to scale) Point P is at a distance x from the centre of star S1. The period of rotation of the stars is 44.2 years. (i) Calculate the angular velocity ω. ω = .............................................. rad s–1 [2] (ii) By considering the forces acting on the two stars, show that the ratio of the masses of the stars is given by mass of S1 d – x = . mass of S2 x [2] (iii) The mass M1 of star S1 is given by the expression GM1 = d 2 (d – x) ω 2 where G is the gravitational constant. The ratio in (ii) is found to be 1.5. Use data from (b) and your answer in (c)(i) to determine the mass M1. M1 = .................................................... kg [3] [Total: 9]

Mark scheme: 1(a) force acting between two masses or force on mass due to another mass or force on mass in a gravitational field B1 1(b) arc length = rθ d = 1.5 × 1017 × 1.2 × 10–5 = 1.8 × 1012 m A1 1(c)(i) ω = 2π / T C1 = 2π / (44.2 × 365 × 24 × 3600) = 4.5 × 10–9 rad s–1 A1 1(c)(ii) gravitational forces are equal or centripetal force about P is the same C1 M1xω2 = M2(d – x)ω2 so M1 / M2 = (d – x) / x A1 1(c)(iii) x = 0.4d C1 6.67 × 10–11 × M1 = (1.0 – 0.4) × (1.8 × 1012)3 × (4.5 × 10–9)2 C1 M1 = 1.1 × 1030 kg A1

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2 (a) State what is meant by the internal energy of a system. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) By reference to intermolecular forces, explain why the change in internal energy of an ideal gas is equal to the change in total kinetic energy of its molecules. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) State and explain the change, if any, in the internal energy of a solid metal ball as it falls under gravity in a vacuum. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 7]

Mark scheme: 2(a) total potential energy and kinetic energy (of molecules/atoms) M1 reference to random motion of molecules/atoms A1 2(b) (in ideal gas,) no intermolecular forces B1 no potential energy (so change in kinetic energy is change in internal energy) B1 2(c) (random) potential energy of molecules does not change M1 (random) kinetic energy of molecules does not change M1 so internal energy does not change A1 or decrease in total potential energy = gain in total kinetic energy (M1) no external energy supplied (M1) so internal energy does not change (A1) or no compression (of ball) so no work done on the ball (M1) no resistive forces so no heating of the ball (M1) so internal energy does not change (A1) Question Answer Marks 2(c) or no change of state so potential energy (of molecules) unchanged (M1) no temperature rise so kinetic energy (of molecules) unchanged (M1) so internal energy does not change (A1)

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3 The piston in the cylinder of a car engine moves in the cylinder with simple harmonic motion. The piston moves between a position of maximum height in the cylinder to a position of minimum height, as illustrated in Fig. 3.1. cylinder cylinder 9.8 cm piston piston maximum height minimum height Fig. 3.1 The distance moved by the piston between the positions shown in Fig. 3.1 is 9.8 cm. The mass of the piston is 640 g. At one particular speed of the engine, the piston completes 2700 oscillations in 1.0 minute. (a) For the oscillations of the piston in the cylinder, determine: (i) the amplitude amplitude = ................................................... cm [1] (ii) the frequency frequency = .................................................... Hz [1] (iii) the maximum speed maximum speed = ................................................ m s–1 [2] (iv) the speed when the top of the piston is 2.3 cm below its maximum height. speed = ............................................... m s–1 [2] (b) The acceleration of the piston varies. Determine the resultant force on the piston that gives rise to its maximum acceleration. force = ..................................................... N [3] [Total: 9]

Mark scheme: 3(a)(i) amplitude = 4.9 cm A1 3(a)(ii) frequency = 2700 / 60 = 45 Hz A1 3(a)(iii) v0 = x0ω and ω = 2πf C1 v0 = 4.9 × 10–2 × 2π × 45 = 14 m s–1 A1 3(a)(iv) v = ω (x02 – x2)½ = 2π × 45 × [(4.9 × 10–2)2 – (2.6 × 10–2)2]½ C1 = 12 m s–1 A1 Question Answer Marks 3(b) F = ma and a0 = v0ω or a0 = x0ω2 C1 F = 0.64 × 13.9 × 2π × 45 or 0.64 × 4.9 × (2π × 45)2 C1 = 2500 N A1

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4 (a) (i) By reference to an ultrasound wave, explain what is meant by specific acoustic impedance. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) An ultrasound wave is incident normally on the boundary between two media. The media have specific acoustic impedances Z1 and Z2. State how the ratio intensity of ultrasound reflected from boundary intensity of ultrasound incident on boundary depends on the relative magnitudes of Z1 and Z2. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) (i) State what is meant by the attenuation of an ultrasound wave. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) A parallel beam of ultrasound is passing through a medium. The incident intensity I0 is reduced to 0.35 I0 on passing through a thickness of 0.046 m of the medium. Calculate the linear attenuation coefficient μ of the ultrasound beam in the medium. μ = .................................................. m–1 [2] [Total: 7]

Mark scheme: 4(a)(i) product of density and speed M1 speed of ultrasound in medium A1 4(a)(ii) the greater the difference between Z1 and Z2, the closer the ratio is to 1 or if difference between Z1 and Z2 large, ratio is close to 1 B1 the closer together Z1 and Z2, the closer the ratio is to 0 or if difference between Z1 and Z2 small, ratio close to 0 B1 4(b)(i) loss of intensity/amplitude/power (of the wave) B1 4(b)(ii) I = I0 e–μx C1 0.35 = e–0.046μ μ = 23 m–1 A1

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5 (a) State one similarity and one difference between the fields of force produced by an isolated point charge and by an isolated point mass. similarity: ................................................................................................................................... ................................................................................................................................................... difference: ................................................................................................................................. ................................................................................................................................................... [2] (b) An isolated solid metal sphere A of radius R has charge +Q, as illustrated in Fig. 5.1. R P 2R sphere A charge +Q Fig. 5.1 A point P is distance 2R from the surface of the sphere. Determine an expression that includes the terms R and Q for the electric field strength E at point P. E = ......................................................... [2] (c) A second identical solid metal sphere B is now placed near sphere A. The centres of the spheres are separated by a distance 6R, as shown in Fig. 5.2. R R P sphere A sphere B charge +Q charge –Q 6R Fig. 5.2 Point P lies midway between spheres A and B. Sphere B has charge –Q. Explain why: (i) the magnitude of the electric field strength at P is given by the sum of the magnitudes of the field strengths due to each sphere ........................................................................................................................................... ..................................................................................................................................... [1] (ii) the electric field strength at point P due to the charged metal spheres is not, in practice, equal to 2E, where E is the electric field strength determined in (b). ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 7]

Mark scheme: 5(a) similarity: both are radial or both have inverse square (variations) B1 difference: direction is always/only towards the mass or direction can be towards or away from charge B1 5(b) field strength = Q / 4πε0x2 C1 E = Q / 36πε0R 2 A1 5(c)(i) fields (due to each sphere) are in same direction B1 5(c)(ii) charges on spheres attract/affect each other or charge distribution on each sphere distorted by the other sphere or charges on the surface of the spheres move B1 spheres are not point charges (at their centres) B1

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6 (a) The transmission of signals using optic fibres has, to a great extent, replaced the use of coaxial cables. Advantages of optic fibres include greater bandwidth and very little crosslinking. (i) Suggest an advantage of greater bandwidth. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State what is meant by crosslinking. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) In telecommunications, a signal power of 1.0 mW is used as a reference power. Signal powers relative to this reference power and expressed in dB are said to be measured in ‘dBm’. Show that a signal power of 13 dBm is equivalent to 20 mW. [2] (c) A signal of input power 20 mW is transmitted along an optic fibre for an uninterrupted distance of 45 km. The optic fibre has an attenuation per unit length of 0.18 dB km–1. Calculate the output power P from the optic fibre. P = .................................................. mW [2] [Total: 7]

Mark scheme: 6(a)(i) greater information carrying capacity B1 6(a)(ii) power/energy is radiated B1 signal picked up by adjacent fibre/wire B1 6(b) ratio / dB = 10 lg(P2 / P1) C1 13 = 10 lg [P / (1.0 × 10–3)] and so P = 20 mW A1 6(c) 45 × 0.18 = 10 lg (20 / P) C1 P = 3.1 mW A1

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7 The output of a microphone is processed using a non-inverting amplifier. The amplifier incorporates an operational amplifier (op-amp). (a) State, by reference to the input and output signals, the function of a non-inverting amplifier. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) The circuit for the microphone and amplifier is shown in Fig. 7.1. 15 kΩ +5 V – P + –5 V VOUT 2.0 kΩ R Fig. 7.1 The output potential difference VOUT is 2.6 V when the potential at point P is 84 mV. Determine: (i) the gain of the amplifier circuit gain = ......................................................... [1] (ii) the resistance of resistor R. resistance = ..................................................... Ω [2] (c) For the circuit of Fig. 7.1: (i) suggest a suitable device to connect to the output such that the shape of the waveform of the sound received by the microphone may be examined ..................................................................................................................................... [1] (ii) state and explain the effect on the output potential difference VOUT of increasing the resistance of resistor R. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: 7(a) output signal proportional to input signal B1 output signal has same sign/polarity as input signal B1 7(b)(i) gain = VOUT / VIN = 2.6 / 0.084 = 31 A1 7(b)(ii) 31 = 1 + (15 × 103) / R C1 R = 500 Ω A1 7(c)(i) e.g. cathode-ray oscilloscope/CRO B1 7(c)(ii) gain is reduced B1 (so) VOUT is smaller B1

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8 (a) Define the tesla. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) A magnet produces a uniform magnetic field of flux density B in the space between its poles. A rigid copper wire carrying a current is balanced on a pivot. Part PQLM of the wire is between the poles of the magnet, as illustrated in Fig. 8.1. 5.6 cm M P L weight W N S rigid copper Q wire pivot magnet Fig. 8.1 (not to scale) The wire is balanced horizontally by means of a small weight W. The section of the wire between the poles of the magnet is shown in Fig. 8.2. rigid copper wire M P L N S Q pole of magnet pole of magnet Fig. 8.2 (not to scale) Explain why: (i) section QL of the wire gives rise to a moment about the pivot ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) sections PQ and LM of the wire do not affect the equilibrium of the wire. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) Section QL of the wire has length 0.85 cm. The perpendicular distance of QL from the pivot is 5.6 cm. When the current in the wire is changed by 1.2 A, W is moved a distance of 2.6 cm along the wire in order to restore equilibrium. The mass of W is 1.3 × 10–4 kg. (i) Show that the change in moment of W about the pivot is 3.3 × 10–5 N m. [2] (ii) Use the information in (i) to determine the magnetic flux density B between the poles of the magnet. B = ...................................................... T [3] [Total: 13]

Mark scheme: 8(a) magnetic field normal to current B1 newton per ampere B1 newton per metre B1 8(b)(i) current in wire QL gives rise to a force or wire QL is perpendicular to the magnetic field B1 force on wire QL is vertical B1 force does not act through the pivot B1 8(b)(ii) forces act through the same line or forces are horizontal B1 forces are equal (in magnitude) and opposite (in direction) B1 8(c)(i) change = mg × (Δ)L C1 = 1.3 × 10–4 × 9.81 × 2.6 × 10–2 = 3.3 × 10–5 N m–1 A1 8(c)(ii) change = B × (Δ)I × L × x C1 3.3 × 10–5 = B × 1.2 × 0.85 × 10–2 × 5.6 × 10–2 C1 B = 0.058 T A1

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9 (a) A coil of wire is situated in a uniform magnetic field of flux density B. The coil has diameter 3.6 cm and consists of 350 turns of wire, as illustrated in Fig. 9.1. uniform magnetic field 3.6 cm flux density B coil, 350 turns Fig. 9.1 The variation with time t of B is shown in Fig. 9.2. 50 40 B / mT 30 20 10 0 0 0.2 0.4 0.6 0.8 t / s Fig. 9.2 (i) Show that, for the time t = 0 to time t = 0.20 s, the electromotive force (e.m.f.) induced in the coil is 0.080 V. [2] (ii) On the axes of Fig. 9.3, show the variation with time t of the induced e.m.f. E for time t = 0 to time t = 0.80 s. 0.2 E / V 0.1 0 0 0.2 0.4 0.6 0.8 t / s –0.1 –0.2 Fig. 9.3 (b) A bar magnet is held a small distance above the surface of an aluminium disc by means of a rod, as illustrated in Fig. 9.4. rotating magnet fixed aluminium disc Fig. 9.4 The aluminium disc is supported horizontally and held stationary. The magnet is rotated about a vertical axis at constant speed. Use laws of electromagnetic induction to explain why there is a torque acting on the aluminium disc. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] [Total: 10]

Mark scheme: 9(a)(i) e.m.f. = (Δ)B × AN / t C1 = 45 × 10–3 × π × (1.8 × 10–2)2 × 350 / 0.20 = 0.080 V A1 9(a)(ii) 0 to 0.2 s: straight horizontal line at 0.080 V or –0.080 V B1 0.2 s to 0.4 s: zero B1 0.4 s to 0.8 s: straight horizontal line at 0.040 V or –0.040 V B1 opposite polarity to 0 to 0.2 s line B1 9(b) either disc cuts flux lines (of the magnet) or there is a changing flux in the disc B1 (by Faraday’s law) e.m.f. is induced in the disc B1 e.m.f. causes (eddy) currents in the disc B1 current in the magnetic field (of the magnet) causes force on disc B1

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10 (a) White light passes through a cloud of cool low-pressure gas, as illustrated in Fig. 10.1. cool gas white emergent light light Fig. 10.1 For light that has passed through the gas, its continuous spectrum is seen to contain a number of darker lines. Use the concept of discrete electron energy levels to explain the existence of these darker lines. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (b) The uppermost electron energy bands in a solid are illustrated in Fig. 10.2. conduction band (CB) forbidden band (FB) valence band (VB) Fig. 10.2 Use band theory to explain the dependence on light intensity of the resistance of a light-dependent resistor (LDR). ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [5] [Total: 9]

Mark scheme: 10(a) • photon gives energy to electron (in an inner shell) or electron (in an inner shell) absorbs a photon • electron moves (from lower) to higher energy level • energy (of photon) is equal to difference in energy levels • electron de-excites giving off photon (of same energy) • photons emitted in all directions Any four points, 1 mark each B4 10(b) (in light) photons gives energy to electrons in VB or (in light) electrons in VB absorb photons B1 electron crosses FB/jumps to CB B1 (positive) holes left/created in VB B1 low intensity: few electrons in CB/most electrons in VB or high intensity: more photons so more electrons in CB or electron-hole pairs are charge carriers B1 more charge carriers results in lower resistance B1

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11 An electron, at rest, has mass me and charge –q. A positron is a particle that, at rest, has mass me and charge +q. A positron interacts with an electron. The electron and the positron may be considered to be at rest. The outcome of this interaction is that the electron and the positron become two gamma-ray (γ-ray) photons, each having the same energy. (a) Calculate, for one of the γ-ray photons: (i) the photon energy, in J energy = ...................................................... J [2] (ii) its momentum. momentum = ................................................... N s [2] (b) State and explain the direction, relative to each other, in which the γ-ray photons are emitted. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 6]

Mark scheme: 11(a)(i) E = mc2 C1 = 9.11 × 10–31 × (3.0 × 108)2 = 8.2 × 10–14 J A1 11(a)(ii) p = h / λ and E = hc / λ or E = pc C1 p = (8.2 × 10–14) / (3.0 × 108) = 2.7 × 10–22 N s A1 11(b) total momentum (before and after interaction) is zero or momentum must be conserved (in the interaction) or momentum of the photons must be equal and opposite B1 (photons emitted in) opposite directions B1

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12 (a) The decay of a sample of a radioactive isotope is said to be random and spontaneous. Explain what is meant by the decay being: (i) random ........................................................................................................................................... ..................................................................................................................................... [1] (ii) spontaneous. ........................................................................................................................................... ..................................................................................................................................... [1] (b) A radioactive isotope X has a half-life of 1.4 hours. Initially, a pure sample of this isotope X has an activity of 3.6 × 105 Bq. Determine the activity of the isotope X in the sample after a time of 2.0 hours. activity = .................................................... Bq [3] (c) The variation with time t of the actual activity A of the sample in (b) is shown in Fig. 12.1. 4 A / 105 Bq 3 2 1 0 0 1 2 3 4 5 6 t / hours Fig. 12.1 (i) The initial activity of isotope X in the sample is 3.6 × 105 Bq. Use information from (b) to sketch, on the axes of Fig. 12.1, the variation with time t of the activity of a pure sample of isotope X. [1] (ii) Suggest an explanation for any difference between the actual activity of the sample shown in Fig. 12.1 and the curve you have drawn for the activity of isotope X. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: 12(a)(i) time at which a nucleus will decay cannot be predicted or constant probability of decay of a nucleus B1 12(a)(ii) decay (of a nucleus) not affected by environmental factors B1 12(b) A = A0e–λt and λ = ln 2 / t½ C1 = 3.6 × 105 × exp [–(2 × ln 2) / 1.4] C1 or A = A0 × 0.5N (C1) = 3.6 × 105 × 0.5N where N = 2 / 1.4 (C1) A = 1.3 × 105 Bq A1 12(c)(i) smooth curve, starting at (0, 3.6 × 105) and passing through (1.4, 1.8 × 105) and (2.0, 1.3 × 105) B1 12(c)(ii) (activity of sample is greater than activity of X so) there must be an additional source of activity C1 the decay product (of isotope X) is radioactive A1

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