Cambridge A Level Physics 9702 — 2018 May/June Paper 4 · Variant 2

1 (a) (i) A gravitational field may be represented by lines of gravitational force. State what is meant by a line of gravitational force. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (ii) By reference to lines of gravitational force near to the surface of the Earth, explain why the gravitational field strength g close to the Earth’s surface is approximately constant. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (b) The Moon may be considered to be a uniform sphere of diameter 3.4 × 103 km and mass 7.4 × 1022 kg. The Moon has no atmosphere. During a collision of the Moon with a meteorite, a rock is thrown vertically up from the surface of the Moon with a speed of 2.8 km s–1. Assuming that the Moon is isolated in space, determine whether the rock will travel out into distant space or return to the Moon’s surface. [4] [Total: 8]

Mark scheme: 1(a)(i) direction of force on a (small test) mass or path in which a (small test) mass will move B1 1(a)(ii) (at surface,) lines (of force) are radial B1 Earth has large radius/height above surface is small so lines are (approximately) parallel B1 parallel lines → constant field strength B1 1(b) (change in) KE of rock = (change in) PE or ½mv2 = GMm / R C1 (m)v2 = (m)(2 × 6.67 × 10–11 × 7.4 × 1022) / (1.7 × 103 × 103) C1 v = 2.4 × 103 m s–1 A1 correct conclusion based on comparison of v with 2.8 km s–1 B1 or (change in) KE of rock = (change in) PE (C1) (at infinity) EP = (6.67 × 10–11 × 7.4 × 1022 × m) / (1.7 × 103 × 103) = 2.9 × 106 m (C1) EK of rock = ½ × m × (2.8 × 103)2 = 3.9 × 106 m (A1) correct conclusion based on comparison of EK and EP values (B1) or Question Answer Marks (change in) KE of rock = (change in) PE or ½mv2 = GMm / R (C1) (m) (2800)2 = (m) (2 × 6.67 × 10–11 × 7.4 × 1022) / R (C1) R = 1.3 × 103 km (A1) correct conclusion based on comparison of R with 1.7 × 103 km (B1) or (change in) KE of rock = (change in) PE or ½mv 2 = GMm / R (C1) (m) (2800)2 = (m) (2 × 6.67 × 10–11 × M) / (1.7 × 106) (C1) M = 1.0 × 1023 kg (A1) correct conclusion based on comparison of M with 7.4 × 1022 kg (B1)

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2 (a) Use one of the assumptions of the kinetic theory of gases to explain why the potential energy of the molecules of an ideal gas is zero. ................................................................................................................................................... ...............................................................................................................................................[1] (b) The average translational kinetic energy EK of a molecule of an ideal gas is given by the expression 1 3 EK = 2m 〈c2〉 = 2 kT where m is the mass of a molecule and k is the Boltzmann constant. State the meaning of the symbol (i) 〈c2〉, .......................................................................................................................................[1] (ii) T. .......................................................................................................................................[1] (c) A cylinder of constant volume 4.7 × 104 cm3 contains an ideal gas at pressure 2.6 × 105 Pa and temperature 173 °C. The gas is heated. The thermal energy transferred to the gas is 2900 J. The final temperature and pressure of the gas are T and p, as illustrated in Fig. 2.1. 4.7 × 104 cm3 4.7 × 104 cm3 2900 J 2.6 × 105 Pa p 173 °C T Fig. 2.1 (i) Calculate 1. the number N of molecules in the cylinder, N = ...........................................................[3] 2. the increase in average kinetic energy of a molecule during the heating process. increase = ....................................................... J [1] (ii) Use your answer in (i) part 2 to determine the final temperature T, in kelvin, of the gas in the cylinder. T = ....................................................... K [3] [Total: 10]

Mark scheme: 2(a) no intermolecular forces (so no potential energy) B1 2(b)(i) mean square speed (of molecule(s)) B1 2(b)(ii) kelvin/thermodynamic/absolute temperature B1 2(c)(i)1. pV = NkT C1 4.7 × 10–2 × 2.6 × 105 = N × 1.38 × 10–23 × 446 C1 or pV = nRT and N = nNA 4.7 × 10–2 × 2.6 × 105 = n × 8.31 × 446 n = 3.3 (mol) (C1) N = 3.3 × 6.02 × 1023 (C1) N = 2.0 × 1024 A1 2(c)(i)2. average increase = 2900 / (2.0 × 1024) = 1.5 × 10–21 J A1 2(c)(ii) ∆EK = (3/2)k (∆)T 1.5 × 10–21 = (3/2) × 1.38 × 10–23 × (∆)T C1 (∆)T in range 70–72 K C1 T = 173 + 273 + 70 = 520 K A1

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3 (a) During melting, a solid becomes liquid with little or no change in volume. Use kinetic theory to explain why, during the melting process, thermal energy is required although there is no change in temperature. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) An aluminium can of mass 160 g contains a mass of 330 g of warm water at a temperature of 38 °C, as illustrated in Fig. 3.1. ice warm water aluminium can Fig. 3.1 A mass of 48 g of ice at –18 °C is taken from a freezer and put in to the water. The ice melts and the final temperature of the can and its contents is 23 °C. Data for the specific heat capacity c of aluminium, ice and water are given in Fig. 3.2. c / J g–1 K–1 aluminium 0.910 ice 2.10 water 4.18 Fig. 3.2 Assuming no exchange of thermal energy with the surroundings, (i) show that the loss in thermal energy of the can and the warm water is 2.3 × 104 J, [2] (ii) use the information in (i) to calculate a value L for the specific latent heat of fusion of ice. L = .................................................. J g–1 [2] [Total: 7]

Mark scheme: 3(a) (during melting,) bonds between atoms/molecules are broken B1 potential energy of atoms/molecules is increased B1 no/little work done so required input of energy is thermal B1 3(b)(i) (∆Q =) mc∆θ C1 loss = (160 × 0.910 × 15) + (330 × 4.18 × 15) = 2.3 × 104 J A1 3(b)(ii) 2.3 × 104 = (48 × 2.10 × 18) + 48L + (48 × 4.18 × 23) C1 48L = 1.66 × 104 L = 350 J g–1 A1

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4 (a) State two conditions necessary for a mass to be undergoing simple harmonic motion. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) A trolley of mass 950 g is held on a horizontal surface by means of two springs attached to fixed points P and Q, as shown in Fig. 4.1. trolley mass 950 g spring P Q Fig. 4.1 The springs, each having a spring constant k of 230 N m–1, are always extended. The trolley is displaced along the line of the springs and then released. The variation with time t of the displacement x of the trolley is shown in Fig. 4.2. x 0 0 t1 t Fig. 4.2 (i) 1. State and explain whether the oscillations of the trolley are heavily damped, critically damped or lightly damped. .................................................................................................................................... .................................................................................................................................... 2. Suggest the cause of the damping. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... [3] (ii) The acceleration a of the trolley of mass m may be assumed to be given by the expression 2 k a = – x . d m n 1. Calculate the angular frequency ω of the oscillations of the trolley. ω = ............................................... rad s–1 [3] 2. Determine the time t1 shown on Fig. 4.2. t1 = ....................................................... s [2] [Total: 10]

Mark scheme: 4(a) acceleration proportional to displacement B1 acceleration directed towards fixed point or displacement and acceleration in opposite directions B1 4(b)(i) 1. amplitude decreases gradually so light damping or oscillations continue so light damping B1 2. loss of energy B1 due to friction in wheels or due to friction between wheels and surface (during slipping) or due to air resistance (on trolley) B1 4(b)(ii)1. ω2 = 2k / m C1 = (2 × 230) / 0.950 C1 ω = 22 rad s–1 A1 4(b)(ii)2. T = 2π / ω C1 T = (2π / 22) = 0.286 s time = 1.5T = 0.43 s A1

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5 (a) In radio communication, the bandwidth of an FM transmission is greater than the bandwidth of an AM transmission. State (i) what is meant by bandwidth, ........................................................................................................................................... .......................................................................................................................................[1] (ii) one advantage and one disadvantage of a greater bandwidth. advantage: ........................................................................................................................ ........................................................................................................................................... disadvantage: .................................................................................................................... ........................................................................................................................................... [2] (b) A carrier wave has a frequency of 650 kHz and is measured to have an amplitude of 5.0 V. The carrier wave is frequency modulated by a signal of frequency 10 kHz and amplitude 3.0 V. The frequency deviation of the carrier wave is 8.0 kHz V–1. Determine, for the frequency modulated carrier wave, (i) the measured amplitude, amplitude = ....................................................... V [1] (ii) the maximum and the minimum frequencies, maximum frequency = ........................................................ kHz minimum frequency = ........................................................ kHz [2] (iii) the minimum time between a maximum and a minimum transmitted frequency. time = ........................................................s [1] [Total: 7]

Mark scheme: 5(a)(i) range of frequencies (of signal) B1 5(a)(ii) advantage: e.g. better quality (of reproduction) greater rate of transfer of data less distortion B1 disadvantage: e.g. fewer stations (in any frequency range) B1 5(b)(i) 5.0 V A1 5(b)(ii) maximum: 674 kHz A1 minimum: 626 kHz A1 5(b)(iii) T = 1 / (10 × 103) = 1.0 × 10–4s minimum time = T / 2 = 5.0 × 10–5 s A1

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6 (a) Explain what is meant by the capacitance of a parallel plate capacitor. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) Three parallel plate capacitors each have a capacitance of 6.0 μF. Draw circuit diagrams, one in each case, to show how the capacitors may be connected together to give a combined capacitance of (i) 9.0 μF, [1] (ii) 4.0 μF. [1] (c) Two capacitors of capacitances 3.0 μF and 2.0 μF are connected in series with a battery of electromotive force (e.m.f.) 8.0 V, as shown in Fig. 6.1. 3.0 μF 2.0 μF 8.0 V Fig. 6.1 (i) Calculate the combined capacitance of the capacitors. capacitance = ..................................................... μF [1] (ii) Use your answer in (i) to determine, for the capacitor of capacitance 3.0 μF, 1. the charge on one plate of the capacitor, charge = ..........................................................μC 2. the energy stored in the capacitor. energy = ............................................................ J [4] [Total: 10]

Mark scheme: 6(a) capacitance = charge / potential M1 charge is (numerically equal to) charge on one plate A1 potential is potential difference between plates A1 6(b)(i) two in series, in parallel with the other (correct symbols) A1 6(b)(ii) two in parallel connected to one in series (correct symbols) A1 6(c)(i) capacitance = 1.2 µF A1 6(c)(ii) 1. Q = CV C1 = 1.2 × 8.0 = 9.6 µC A1 2. E = ½QV and V = Q / C or E = ½CV2 and V = Q / C or E = ½Q2 / C C1 E = ½ (9.6 × 10–6)2 / (3.0 × 10–6) = 1.5 × 10–5 J A1

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7 (a) Negative feedback is often used in amplifiers. State (i) what is meant by negative feedback, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) two effects of negative feedback on the gain of an amplifier. 1. ....................................................................................................................................... ........................................................................................................................................... 2. ....................................................................................................................................... ........................................................................................................................................... [2] (b) An ideal operational amplifier (op-amp) is incorporated into the circuit shown in Fig. 7.1. 6400 Ω +9.0 V – + –9.0 V VOUT VIN 800 Ω Fig. 7.1 (i) Calculate the gain G of the amplifier circuit. G = ...........................................................[1] (ii) Determine the output potential difference VOUT for an input potential difference VIN of 1. +0.60 V, VOUT = ............................................................ V 2. –2.1 V. VOUT = ............................................................ V [2] (iii) The gain of the amplifier shown in Fig. 7.1 is constant. State one change that may be made to the circuit of Fig. 7.1 so that the amplifier circuit monitors temperature with the gain decreasing as the temperature rises. ........................................................................................................................................... .......................................................................................................................................[1] [Total: 8]

Mark scheme: 7(a)(i) (fraction of) output is combined with the input M1 output (fraction) subtracted/deducted from input A1 7(a)(ii) Any two valid points e.g. • greater bandwidth/gain constant over a larger range of frequencies • smaller gain B2 7(b)(i) gain = 1 + (6400 / 800) = 9.0 A1 7(b)(ii) 1. (+)5.4 V A1 2. –9.0 V A1 7(b)(iii) replace the 6400 Ω resistor with a thermistor B1

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8 (a) Explain how a uniform magnetic field and a uniform electric field may be used as a velocity selector for charged particles. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) Particles having mass m and charge +1.6 × 10–19 C pass through a velocity selector. They then enter a region of uniform magnetic field of magnetic flux density 94 mT with speed 3.4 × 104 m s–1, as shown in Fig. 8.1. path of charged particle velocity selector 15.0 cm uniform magnetic field into page Fig. 8.1 The direction of the uniform magnetic field is into the page and normal to the direction in which the particles are moving. The particles are moving in a vacuum in a circular arc of diameter 15.0 cm. Show that the mass of one of the particles is 20 u. [4] (c) On Fig. 8.1, sketch the path in the uniform magnetic field of a particle of mass 22 u having the same charge and speed as the particle in (b). [2] [Total: 9]

Mark scheme: 8(a) electric and magnetic fields at right-angles to one another (may be shown on a clearly labelled diagram) B1 particle enters fields (with velocity) normal to the (two) fields (may be shown on a clearly labelled diagram) B1 no deviation for particles with selected velocity B1 8(b) magnetic force equals/is the centripetal force C1 Bqv = mv2 / r C1 M = Bqr / v = (94 × 10–3 × 1.6 × 10–19 × 0.075) / (3.4 × 104) M1 division by 1.66 × 10–27 shown, to give m = 20 u A1 8(c) sketch: semicircle clear (in same direction) B1 with larger radius B1

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9 (a) State what is meant by the magnetic flux linkage of a coil. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) A coil of wire has 160 turns and diameter 2.4 cm. The coil is situated in a uniform magnetic field of flux density 7.5 mT, as shown in Fig. 9.1. magnetic field flux density 2.4 cm 7.5 mT coil 160 turns Fig. 9.1 The direction of the magnetic field is along the axis of the coil. The magnetic flux density is reduced to zero in a time of 0.15 s. Show that the average e.m.f. induced in the coil is 3.6 mV. [2] (c) The magnetic flux density B in the coil in (b) is now varied with time t as shown in Fig. 9.2. 10 B / mT 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s –5 –10 Fig. 9.2 Use data in (b) to show, on Fig. 9.3, the variation with time t of the e.m.f. E induced in the coil. 8 E / mV 6 4 2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s –2 –4 –6 –8 Fig. 9.3 [4] [Total: 9]

Mark scheme: 9(a) B1 magnetic flux density normal to area or reference to cross-sectional area or × sin (angle between B and A) B1 × number of turns on coil B1 9(b) e.m.f. = BAN / t or e.m.f = rate of change of flux linkage C1 = (7.5 × 10–3 × π × {1.2 × 10–2}2 × 160) / 0.15 = 3.6 × 10–3 V A1 9(c) sketch: zero for 0–0.10 s, 0.25–0.35 s, and 0.425–0.55 s, and non-zero outside these ranges B1 two horizontal steps, with zero voltage either side B1 with same polarity B1 correct values (1st step 3.6 mV and 2nd step 7.2 mV) B1

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10 (a) Describe the photoelectric effect. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Data for the work function energy Φ of two metals are shown in Fig. 10.1. Φ/ J sodium 3.8 × 10–19 zinc 5.8 × 10–19 Fig. 10.1 Light of wavelength 420 nm is incident on the surface of each of the metals. (i) State what is meant by a photon. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Calculate the energy of a photon of the incident light. energy = ....................................................... J [2] (iii) State whether photoelectric emission will occur from each of the metals. sodium: .............................................................................................................................. zinc: ................................................................................................................................... [1] [Total: 7]

Mark scheme: 10(a) emission of electron B1 when electromagnetic radiation incident (on surface) B1 10(b)(i) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 10(b)(ii) E = hc / λ C1 = (6.63 × 10–34 × 3.00 × 108) / (420 × 10–9) = 4.7 × 10–19 J A1 10(b)(iii) sodium: yes zinc: no B1

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11 (a) Describe the basic principles of CT scanning (computed tomography). ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[5] (b) By reference to your answer in (a), suggest why (i) CT scanning was not possible before fast computers with large memories were available, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (ii) the radiation dose for a CT scan is much larger than for an X-ray image of a leg bone. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] [Total: 7]

Mark scheme: 11(a) X-ray image(s) taken of one slice M1 (many images) taken from different angles A1 (computer) produces 2D image of slice B1 (this is) repeated for (many) slices M1 to build up a 3D image (of structure) A1 11(b)(i) combining of images involves (very) large number of calculations B1 11(b)(ii) CT scan consists of (very) many (single X-ray) images B1

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12 (a) State what is meant by radioactive decay. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) An unstable nuclide P has decay constant λP and decays to form a nuclide D. This nuclide D is unstable and decays with decay constant λD to form a stable nuclide S. The decay chain is illustrated in Fig. 12.1. decay constant decay constant λP λD nuclide P nuclide D nuclide S Fig. 12.1 The symbols P, D and S are not the nuclide symbols. Initially, a radioactive sample contains only nuclide P. The variation with time t of the number of nuclei of each of the three nuclides in the sample is shown in Fig. 12.2. number 00 t Fig. 12.2 (i) On Fig. 12.2, use the symbols P, D and S to identify the curve for each of the three nuclides. [2] (ii) The half-life of nuclide P is 60.0 minutes. Calculate the decay constant λP, in s–1, of this nuclide. λP = .....................................................s–1 [2] (c) In the decay chain shown in Fig. 12.1, λP is approximately equal to 5λD. The decay chain of a different nuclide E is illustrated in Fig. 12.3. decay constant decay constant λE λF nuclide E nuclide F nuclide G Fig. 12.3 The decay constant λF of nuclide F is very much larger than the decay constant λE of nuclide E. By reference to the half-life of nuclide F, explain why the number of nuclei of nuclide F in the sample is always small. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 8]

Mark scheme: 12(a) emission of particles/radiation by unstable nucleus B1 spontaneous emission B1 12(b)(i) P – the curve that starts with a high number D – the curve with the peak S – the curve that increases from zero throughout (one correct 1 mark, all three correct 2 marks) B2 12(b)(ii) λt½ = 0.693 λ = 0.693 / (60.0 × 60) C1 = 1.93 × 10–4 s–1 A1 12(c) half-life of F is much shorter than half-life of E B1 nuclei of F decay (almost) as soon as they are produced B1

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