Cambridge A Level Physics 9702 — 2017 May/June Paper 4 · Variant 1

1 (a) Explain how a satellite may be in a circular orbit around a planet. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) The Earth and the Moon may be considered to be uniform spheres that are isolated in space. The Earth has radius R and mean density ρ. The Moon, mass m, is in a circular orbit about the Earth with radius nR, as illustrated in Fig. 1.1. Earth radius R Moon nR Fig. 1.1 The Moon makes one complete orbit of the Earth in time T. Show that the mean density ρ of the Earth is given by the expression 3πn3 ρ = 2 . GT [4] (c) The radius R of the Earth is 6.38 × 103 km and the distance between the centre of the Earth and the centre of the Moon is 3.84 × 105 km. The period T of the orbit of the Moon about the Earth is 27.3 days. Use the expression in (b) to calculate ρ. ρ = ............................................... kg m–3 [3] [Total: 9]

Mark scheme: 1(a) gravitational force (of attraction between satellite and planet) B1 provides / is centripetal force (on satellite about the planet) B1 1(b) M = (4/3) × πR3ρ B1 ω = 2π / T or v = 2πnR / T B1 GM / (nR)2 = nRω2 or v 2 / nR M1 substitution clear to give ρ = 3πn3 / GT 2 A1 1(c) n = (3.84 × 105) / (6.38 × 103) = 60.19 or 60.2 C1 ρ = 3π × 60.193 / [(6.67 × 10–11) × (27.3 × 24 × 3600)2] C1 ρ = 5.54 × 103 kg m–3 A1

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2 A bar magnet of mass 180 g is suspended from the free end of a spring, as illustrated in Fig. 2.1. spring magnet coil Fig. 2.1 The magnet hangs so that one pole is near the centre of a coil of wire. The coil is connected in series with a resistor and a switch. The switch is open. The magnet is displaced vertically and then allowed to oscillate with one pole remaining inside the coil. The other pole remains outside the coil. At time t = 0, the magnet is oscillating freely as it passes through its equilibrium position. At time t = 3.0 s, the switch in the circuit is closed. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 2.2. 2.0 1.5 y / cm 1.0 0.5 0 0 1 2 3 4 5 6 7 8 9 t / s –0.5 –1.0 –1.5 –2.0 Fig. 2.2 (a) Determine, to two significant figures, the frequency of oscillation of the magnet. frequency = .................................................... Hz [2] (b) State whether the closing of the switch gives rise to light, heavy or critical damping. ...............................................................................................................................................[1] (c) Calculate the change in the energy ΔE of oscillation of the magnet between time t = 2.7 s and time t = 7.5 s. Explain your working. ΔE = ....................................................... J [6] [Total: 9]

Mark scheme: 2(a) e.g. period = 3 / 2.5 C1 frequency = 0.83 Hz A1 2(b) light (damping) B1 2(c) at 2.7 s, A0 = 1.5 (cm) B1 energy = ½ m × 4π2f 2A0 2 B1 = ½ × 0.18 × 4π2 × 0.832 × (1.5 × 10–2)2 = 5.51 × 10–4 (J) C1 at 7.5 s, A0 = 0.75 (cm) B1 energy = ¼ × 5.51 × 10–4 or energy = ½ × 0.18 × 4π2 × 0.832 × (0.75 × 10–2)2 C1 energy = 1.38 × 10–4 (J) change = (5.51 × 10–4 – 1.38 × 10–4) = 4.13 J A1

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3 The digital transmission of speech may be illustrated using the block diagram of Fig. 3.1. serial -to - ADC X parallel Y optic fibre converter Fig. 3.1 (a) (i) State what is meant by a digital signal. ........................................................................................................................................... .......................................................................................................................................[1] (ii) State the names of the components labelled X and Y on Fig. 3.1. X: ...................................................................................................................................... Y: ...................................................................................................................................... [2] (iii) Describe the function of the ADC. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) The optic fibre has length 84 km and the attenuation per unit length in the fibre is 0.19 dB km–1. The input power to the optic fibre is 9.7 mW. At the output from the optic fibre, the signal-to- noise ratio is 28 dB. Calculate (i) in dB, the ratio input power to optic fibre noise power at output of optic fibre, ratio = .................................................... dB [2] (ii) the noise power at the output of the optic fibre. noise power = ..................................................... W [3] [Total: 10]

Mark scheme: 3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1 3(a)(ii) component X: parallel-to-serial converter B1 component Y: DAC/digital-to-analogue converter B1 3(a)(iii) sample the (analogue) signal M1 at regular intervals and converts the analogue number to a digital number A1 3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1 ratio = 16 + 28 = 44 dB A1 3(b)(ii) ratio / dB = 10 lg (P2 / P1) C1 44 = 10 lg ({9.7 × 10–3} / P) or –44 = 10 lg (P / {9.7 × 10–3}) C1 power = 3.9 × 10–7 W A1

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4 (a) Describe the motion of molecules in a gas, according to the kinetic theory of gases. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Describe what is observed when viewing Brownian motion that provides evidence for your answer in (a). ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) At a pressure of 1.05 × 105 Pa and a temperature of 27 °C, 1.00 mol of helium gas has a volume of 0.0240 m3. The mass of 1.00 mol of helium gas, assumed to be an ideal gas, is 4.00 g. (i) Calculate the root-mean-square (r.m.s.) speed of an atom of helium gas for a temperature of 27 °C. r.m.s. speed = ................................................. m s–1 [3] (ii) Using your answer in (i), calculate the r.m.s. speed of the atoms at 177 °C. r.m.s. speed = ................................................. m s–1 [3] [Total: 10]

Mark scheme: 4(a) random/haphazard B1 constant velocity or speed in a straight line between collisions or distribution of speeds/different directions B1 4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1 moving haphazardly/randomly/jerky/in a zigzag fashion A1 4(c)(i) pV = ⅓ Nm〈c2〉 1.05 × 105 × 0.0240 = ⅓ × 4.00 × 10–3 × 〈c2〉 C1 〈c2〉 = 1.89 × 106 C1 or ½ m〈c2〉 = (3 / 2) kT 0.5 × (4.00 × 10–3 / 6.02 × 1023) × 〈c2〉 = 1.5 × 1.38 × 10–23 × 300 (C1) 〈c2〉 = 1.87 × 106 (C1) or nRT = ⅓ Nm〈c2〉 1.00 × 8.31 × 300 = ⅓ × 4.00 × 10–3 × 〈c2〉 (C1) 〈c2〉 = 1.87 × 106 (C1) cr.m.s. = 1.37 × 103 m s–1 A1 Question Answer Marks 4(c)(ii) 〈c2〉 ∝ T C1 〈c2〉 at 177 °C = 1.89 × 106 × (450 / 300) C1 cr.m.s. at 177 °C = 1.68 × 103 m s–1 A1

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5 An α-particle is travelling in a vacuum towards the centre of a gold nucleus, as illustrated in Fig. 5.1. gold nucleus α-particle × 10–13 J charge 79e energy 7.7 Fig. 5.1 The gold nucleus has charge 79e. The gold nucleus and the α-particle may be assumed to behave as point charges. At a large distance from the gold nucleus, the α-particle has energy 7.7 × 10–13 J. (a) The α-particle does not collide with the gold nucleus. Show that the radius of the gold nucleus must be less than 4.7 × 10–14 m. [3] (b) Determine the acceleration of the α-particle for a separation of 4.7 × 10–14 m between the centres of the gold nucleus and of the α-particle. acceleration = ................................................. m s–2 [3] (c) In an α-particle scattering experiment, the beam of α-particles is incident on a very thin gold foil. Suggest why the gold foil must be very thin. ................................................................................................................................................... ...............................................................................................................................................[1] [Total: 7]

Mark scheme: 5(a) or 7.7 × 10–13 = Qq / 4πε0r C1 7.7 × 10–13 = 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2/ r M1 r = 4.7 × 10–14 m r is closest distance of approach so radius less than this A1 5(b) force = Qq / 4πε0r 2 = 4u × a C1 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2/ (4.7 × 10–14)2 = 4 × 1.66 × 10–27 × a C1 a = 2.5 × 1027 m s–2 A1 5(c) so that single interactions between nucleus and α-particle can be studied or so that multiple deflections with nucleus do not occur B1

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6 A comparator circuit is designed to switch on a mains lamp when the ambient light level reaches a set value. An incomplete diagram of the circuit is shown in Fig. 6.1. +5 V D – 6 V + RV –5 V Fig. 6.1 (a) (i) A relay is required as part of the output device. This is not shown in Fig. 6.1. Explain why a relay is required. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) On Fig. 6.1, draw the symbol for a relay connected in the circuit as part of the output device. [2] (b) Describe the function of (i) the variable resistor RV, ........................................................................................................................................... .......................................................................................................................................[1] (ii) the diode D. ........................................................................................................................................... .......................................................................................................................................[1] (c) State whether the lamp will switch on as the light level increases or as it decreases. Explain your answer. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 9]

Mark scheme: 6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1 op-amp can deliver only a small current/small voltage B1 6(a)(ii) correct symbol for relay coil connected between output and earth B1 switch between mains supply and lamp B1 6(b)(i) vary light intensity at which lamp is switched on/off B1 6(b)(ii) so that relay operates for only one current/voltage direction or so that relay/lamp operates for either dark or light conditions B1 6(c) when light level increases, LDR resistance decreases B1 (RLDR low,) so V – > V+, so VOUT negative/–5 V (must be consistent with B1 mark) M1 or when light level decreases, LDR resistance increases (B1) (RLDR high,) so V – < V+, so VOUT is positive/+5 V (must be consistent with B1 mark) (M1) lamp comes on as light level decreases or lamp goes off as light level increases A1

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7 An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V. The electron then enters a region of uniform magnetic field of magnetic flux density B, as shown in Fig. 7.1. uniform magnetic field into plane of paper path of electron Fig. 7.1 The direction of the uniform magnetic field is into the plane of the paper. The velocity of the electron as it enters the magnetic field is normal to the magnetic field. The radius of the circular path of the electron in the magnetic field is r. (a) Explain why the path of the electron in the magnetic field is the arc of a circle. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) Show that the magnitude p of the momentum of the electron as it enters the magnetic field is given by p = (2mqV ). [2] (c) The potential difference V is 120 V. The radius r of the circular arc is 7.4 cm. Determine the magnitude B of the magnetic flux density. B = ....................................................... T [3] (d) The potential difference V in (c) is increased. The magnetic flux density B remains unchanged. By reference to the momentum of the electron, explain the effect of this increase on the radius r of the path of the electron in the magnetic field. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 10]

Mark scheme: 7(a) (magnetic) force (always) normal to velocity/direction of motion M1 (magnitude of magnetic) force constant or speed is constant/kinetic energy is constant M1 so provides the centripetal force A1 7(b) increase in KE = loss in PE or ½ mv 2 = qV M1 p = mv with algebra leading to p = √(2mqV) A1 7(c) Bqv = mv2 / r mv = Bqr or p = Bqr C1 (2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 = B × 1.60 × 10–19 × 0.074 C1 B = 5.0 × 10–4 T A1 7(d) greater momentum M1 (p = Bqr and) so r increased A1

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8 Explain the main principles behind the use of nuclear magnetic resonance imaging (NMRI) to obtain information about internal body structures. .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... ......................................................................................................................................................[8] [Total: 8]

Mark scheme: 8 strong (uniform) magnetic field B1 * nuclei precess/rotate about field (direction) radio frequency pulse/RF pulse (applied) B1 * RF or pulse is at Larmor frequency / frequency of precession causes resonance / excitation (of nuclei)/nuclei to absorb energy B1 on relaxation/de-excitation, nuclei emit RF/pulse B1 * (emitted) RF/pulse detected and processed non-uniform field (superposed on uniform field) B1 allows positions of (resonating) nuclei to be determined B1 * allows for position of detection to be changed/different slices to be studied max. 2 of additional detail points marked * B2

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9 A simple transformer is illustrated in Fig. 9.1. laminated iron core input output Fig. 9.1 (a) (i) State why the transformer has an iron core, rather than having no core. ........................................................................................................................................... .......................................................................................................................................[1] (ii) Explain why the core is laminated. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) By reference to the action of a transformer, explain why the input to the transformer is an alternating voltage, rather than a constant voltage. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 6]

Mark scheme: 9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1 9(a)(ii) reduces (size of eddy) currents in core B1 (so that) heating of core is reduced B1 9(b) alternating voltage gives rise to changing magnetic flux in core M1 (changing) flux links the secondary coil A1 induced e.m.f. (in secondary) only when flux is changing/cut B1

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10 (a) State (i) what is meant by the hardness of an X-ray beam, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) how the hardness of an X-ray beam from an X-ray tube is increased. ........................................................................................................................................... .......................................................................................................................................[1] (b) The same parallel beam of X-ray radiation is incident, separately, on samples of bone and of muscle. Data for the thickness x of the samples of bone and of muscle, together with the linear attenuation (absorption) coefficients μ of the radiation in bone and in muscle, are given in Fig. 10.1. x / cm μ/ cm–1 bone 1.5 2.9 muscle 4.0 0.95 Fig. 10.1 Determine the ratio intensity transmitted through bone . intensity transmitted through muscle ratio = .......................................................... [2] [Total: 5]

Mark scheme: 10(a)(i) penetration of beam M1 greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1 10(a)(ii) greater accelerating potential difference or greater p.d. between anode and cathode B1 10(b) I = I0 exp(–µx) ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55}) C1 = 0.58 A1

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11 A beam of light consists of a continuous range of wavelengths from 420 nm to 740 nm. The light passes through a cloud of cool gas, as shown in Fig. 11.1. incident light emergent light cool gas wavelengths 420 nm – 740 nm Fig. 11.1 (a) The spectrum of the light emerging from the cloud of cool gas is viewed using a diffraction grating. Explain why this spectrum contains a number of dark lines. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (b) Some of the electron energy levels of the atoms in the cloud of gas are represented in Fig. 11.2. – 0.38 eV – 0.54 eV – 0.85 eV – 1.5 eV energy – 3.4 eV – 13.6 eV Fig. 11.2 (not to scale) (i) Light of wavelength 420 nm has a photon energy of 2.96 eV. Calculate the photon energy, in eV, of light of wavelength 740 nm. photon energy = .................................................... eV [2] (ii) Use data from (i) and your answer in (i) to show, on Fig. 11.2, the changes in energy levels giving rise to the dark lines in (a). [2] [Total: 8]

Mark scheme: 11(a) electrons (in gas atoms/molecules) interact with photons B1 photon energy causes electron to move to higher energy level/to be excited B1 photon energy = difference in energy of (electron) energy levels B1 when electrons de-excite, photons emitted in all directions (so dark line) B1 11(b)(i) photon energy ∝ 1 / λ C1 energy = 1.68 eV A1 or E = hc / λ E = 6.63 × 10–34 × 3.0 × 108 / (740 × 10–9) = 2.688 × 10–19 J (C1) energy = 1.68 eV (A1) 11(b)(ii) 3.4 eV → 1.5 eV 3.4 eV → 0.85 eV 3.4 eV → 0.54 eV all correct and none incorrect 2/2 2 correct and 1 incorrect or only 2 correctly drawn 1/2 B2

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12 One possible nuclear reaction that takes place in a nuclear reactor is given by the equation 23592U + 10n 9542Mo + 13957La + 210n + x –1e0 Data for the nuclei and particles are given in Fig. 12.1. nucleus or particle mass / u 23592U 235.123 9542Mo 94.945 13957La 138.955 10n 1.00863 –1e0 5.49 × 10–4 Fig. 12.1 (a) Determine, for this nuclear reaction, the value of x. x = ...........................................................[1] (b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3] (ii) Calculate the energy, in MeV, released in this reaction. Give your answer to three significant figures. energy = ................................................. MeV [3] (c) Suggest the forms of energy into which the energy calculated in (b)(ii) is transformed. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 9]

Mark scheme: 12(a) x = 7 A1 12(b)(i) E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 = 1.494 × 10–10 J C1 division by 1.6 × 10–13 clear to give 934 MeV A1 12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10–4) or ∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10–4) C1 = 0.21053 u C1 energy = 0.21053 × 934 = 197 MeV A1 12(c) kinetic energy of nuclei/particles/products/fragments B1 γ–ray photon energy B1

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