Cambridge A Level Physics 9702 — 2019 May/June Paper 4 · Variant 3

1 (a) Two point masses are isolated in space and are separated by a distance x. State an expression relating the gravitational force F between the two masses to the magnitudes M and m of the masses. State the name of any other symbol used. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [1] (b) A spacecraft is to be put into a circular orbit about a spherical planet. The planet may be considered to be isolated in space. The mass of the planet, assumed to be concentrated at its centre, is 7.5 × 1023 kg. The radius of the planet is 3.4 × 106 m. (i) The spacecraft is to orbit the planet at a height of 2.4 × 105 m above the surface of the planet. At this altitude, there is no atmosphere. Show that the speed of the spacecraft in its orbit is 3.7 × 103 m s –1. [2] (ii) One possible path of the spacecraft as it approaches the planet is shown in Fig. 1.1. A 3.64 × 106 m B 5.00 × 107 m planet mass 7.5 × 1023 kg Fig. 1.1 (not to scale) The spacecraft enters the orbit at point A with speed 3.7 × 103 m s–1. At point B, a distance of 5.00 × 107 m from the centre of the planet, the spacecraft has a speed of 4.1 × 103 m s–1. The mass of the spacecraft is 650 kg. For the spacecraft moving from point B to point A, show that the change in gravitational potential energy of the spacecraft is 8.3 × 109 J. [3] (c) By considering changes in gravitational potential energy and in kinetic energy of the spacecraft, determine whether the total energy of the spacecraft increases or decreases in moving from point B to point A. A numerical answer is not required. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 8]

Mark scheme: 1(a) (F =) GMm / x2, where G is the (universal) gravitational constant B1 1(b)(i) GMm / x2 = mv2 / x or v2 = GM / x C1 v 2 = (6.67 × 10–11 × 7.5 × 1023) / (3.4 × 106 + 240 × 103) so v = 3.7 × 103 m s–1 A1 1(b)(ii) potential energy = (–)GMm / x C1 EA = (–)(6.67 × 10–11 × 7.5 × 1023 × 650) / (3.64 × 106) or EB = (–)(6.67 × 10–11 × 7.5 × 1023 × 650) / (5.00 × 107) M1 correct substitution and subtraction EB – EA shown, leading to ∆Ep = 8.3 × 109 J A1 or φ = (–)GM / x and potential energy = mφ (C1) ∆φ = (6.67 × 10–11 × 7.5 × 1023) × [(1 / (3.64 × 106)) – (1 / (5.00 × 107))] ( = 1.27 × 107 J kg–1) (M1) ∆Ep = 1.27 × 107 × 650 = 8.3 × 109 J (A1) 1(c) kinetic energy or potential energy decreases B1 kinetic energy and potential energy decrease so total energy decreases B1

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2 A fixed mass of an ideal gas has volume 210 cm3 at pressure 3.0 × 105 Pa and temperature 270 K. The volume of the gas is reduced at constant pressure to 140 cm3, as shown in Fig. 2.1. 210 cm3 140 cm3 3.0 × 105 Pa 3.0 × 105 Pa 270 K T Fig. 2.1 The final temperature of the gas is T. (a) Determine: (i) the amount of gas amount = ................................................... mol [3] (ii) the final temperature T of the gas T = .......................................................K [2] (iii) the external work done on the gas. work done = ....................................................... J [2] (b) For this change in volume and temperature of the gas, the thermal energy transferred is 53 J. Determine ΔU, the change in internal energy of the gas. ΔU = ....................................................... J [3] [Total: 10]

Mark scheme: 2(a)(i) pV = nRT C1 n = (3.0 × 105 × 210 × 10–6) / (8.31 × 270) C1 = 0.028 mol A1 2(a)(ii) V ∝ T or T = pV / nR with value of n from (i) C1 T = (140 / 210) × 270 or T = (3.0 × 105 × 140 ×10–6) / (8.31 × 0.028) = 180 K A1 2(a)(iii) W = p∆V = 3.0 × 105 × (210 – 140) × 10–6 C1 = 21 J A1 Question Answer Marks 2(b) ∆U = w + q C1 = 21 – 53 C1 or ∆U = (nNA) × (3 / 2)k∆T (C1) = (0.0281 × 6.02 × 1023) × (3 / 2) × 1.38 × 10–23 × (180 – 270) (C1) or ∆U = (3 / 2)nR∆T (C1) = (3 / 2) × 0.0281 × 8.31 × (180 – 270) (C1) ∆U = (–)32 J A1

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3 A hollow tube, sealed at one end, has a cross-sectional area A of 24 cm2. The tube contains sand so that the total mass M of the tube and sand is 0.23 kg. The tube floats upright in a liquid of density t, as illustrated in Fig. 3.1. tube, area of cross-section A liquid, density t h sand Fig. 3.1 The depth of the bottom of the tube below the liquid surface is h. The tube is displaced vertically and then released. The variation with time t of the depth h is shown in Fig. 3.2. 8 h / cm 6 4 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 t / s Fig. 3.2 (a) Determine: (i) the amplitude, in metres, of the oscillations amplitude = ......................................................m [1] (ii) the frequency of oscillation of the tube in the liquid frequency = .....................................................Hz [2] (iii) the acceleration of the tube when h is a maximum. acceleration = .................................................m s–2 [2] (b) The frequency f of oscillation of the tube is given by the expression tg 1 A f = 2π c M m where g is the acceleration of free fall. Calculate the density t of the liquid in which the tube is floating. t = ...............................................kg m–3 [2] (c) The oscillations illustrated in Fig. 3.2 are undamped. In practice, the liquid does cause light damping. On Fig. 3.2, draw a line to show light damping of the oscillations for time t = 0 to time t = 1.4 s. [3] [Total: 10]

Mark scheme: 3(a)(i) amplitude = 0.020 m A1 3(a)(ii) T = 0.60 s C1 f = 1 / T = 1.7 Hz A1 3(a)(iii) a = (–)ω2x and [ω = 2πf or ω = 2π / T] C1 a = (4π2 / 0.602) × 2.0 × 10–2 = 2.2 m s–2 A1 3(b) 1.67 = (1 / 2π) × [(24 × 10–4 × ρ × 9.81) / 0.23]1/2 C1 ρ = 1.1 × 103 kg m–3 A1 3(c) wave starting with a peak at (0,6) B1 wave with same period (or slightly greater) B1 peak height decreasing successively B1

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4 (a) During the transmission of a signal, attenuation occurs and noise is picked up. State what is meant by: (i) attenuation ........................................................................................................................................... ..................................................................................................................................... [1] (ii) noise. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) By reference to (a)(ii), explain the advantage of the transmission of the signal in digital form rather than in analogue form. ................................................................................................................................................... ............................................................................................................................................. [1] (c) Part of an analogue signal is shown in Fig. 4.1. 10 signal voltage / mV 8 6 4 2 0 0 1 2 3 4 5 6 7 time t / ms Fig. 4.1 The signal is to be transmitted in digital form. The analogue signal is sampled at a frequency of 1.0 × 103 Hz using an analogue-to-digital converter (ADC). The ADC produces 4-bit numbers. The times t at which the analogue signal is sampled are shown in Fig. 4.2. time t / ms 0 1.0 2.0 3.0 4.0 5.0 6.0 digital number 0010 0110 0100 0101 ……… ……… ……… Fig. 4.2 On Fig. 4.2: (i) for the digital number at time t = 3.0 ms, underline the least significant bit (LSB) [1] (ii) state the digital numbers corresponding to the sampling times between time t = 4.0 ms and time t = 6.0 ms. [2] (d) The transmitted digital signal is converted back to an analogue signal using a digital-to- analogue converter (DAC). On Fig. 4.3, show the variation with time t of the output levels of the DAC for time t = 0 to time t = 4.0 ms. Assume that there is negligible time delay in the transmission line. 8 output level 6 4 2 0 0 1 2 3 4 time t / ms Fig. 4.3 [3] [Total: 10]

Mark scheme: 4(a)(i) loss of (signal) power/amplitude/intensity B1 4(a)(ii) unwanted/random signal B1 superposed on (transmitted) signal B1 4(b) noise can be eliminated (from digital signals) or signal can be regenerated (from digital signals) B1 4(c)(i) 0101 A1 4(c)(ii) 1000 at t = 4.0 ms B1 0110 at t = 5.0 ms and 0100 at t = 6.0 ms B1 4(d) series of equally-spaced steps of width 1 ms B1 each step in correct time interval (0–1 ms, 1–2 ms, 2–3 ms, 3–4 ms) B1 correct step heights (2, 6, 4 and 5) B1

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5 (a) State what is meant by electric field strength. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Two point charges A and B are situated a distance 15 cm apart in a vacuum, as illustrated in Fig. 5.1. A P B x 15 cm Fig. 5.1 Point P lies on the line joining the charges and is a distance x from charge A. The variation with distance x of the electric field strength E at point P is shown in Fig. 5.2. 10 8 E / 103 N C–1 6 4 2 0 0 2 4 6 8 10 12 14 x / cm –2 –4 –6 Fig. 5.2 (i) By reference to the direction of the electric field, state and explain whether the charges A and B have the same, or opposite, signs. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) State why, although charge A is a point charge, the electric field strength between x = 3 cm and x = 7 cm does not obey an inverse-square law. ........................................................................................................................................... ..................................................................................................................................... [1] (iii) Use Fig. 5.2 to determine the ratio magnitude of charge A . magnitude of charge B ratio = ......................................................... [3] [Total: 8]

Mark scheme: 5(a) force per unit charge B1 (force on) positive charge B1 5(b)(i) field changes direction (between A and B)/field is zero at a point (between A and B) M1 so charges have same sign A1 5(b)(ii) Any one from: • field is (also) influenced by charge B • charge A is not isolated/is not the only charge present • field is due to two/both charges • field is the resultant of two fields B1 5(b)(iii) E = Q / (4πε0x2) C1 at x = 10 cm, EA = EB C1 QA / 102 = QB / 52 QA / QB = 4.0 A1

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6 (a) State two different functions of capacitors in electrical circuits. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) Three uncharged capacitors of capacitances C1, C2 and C3 are connected in series with a battery of electromotive force (e.m.f.) E and a switch, as shown in Fig. 6.1. E C1 C2 C3 plate P charge +q Fig. 6.1 When the switch is closed, there is a charge + q on plate P of the capacitor of capacitance C1. Show that the combined capacitance C of the three capacitors is given by the expression 1 1 1 1 = + + . C C1 C2 C3 [3] (c) A student has available four capacitors, each of capacitance 20 μF. Draw circuit diagrams, one in each case, to show how the student may connect some or all of the capacitors to produce a combined capacitance of: (i) 60 μF [1] (ii) 15 μF. [1] [Total: 7]

Mark scheme: 6(a) Any valid two points e.g.: • to store (electrical) energy • smoothing/reduce ripple (on direct voltages/currents) • to block d.c. • timing/time delay (circuits) • in oscillator (circuits) • in tuning (circuits) • to prevent arcing/sparks B2 6(b) clear indication of equal charge on each capacitor B1 E = V1 + V2 + V3 and V = Q / C M1 completion of algebra leading to 1 / C = 1 / C1 + 1 / C2 + 1 / C3 A1 6(c)(i) three capacitors connected in parallel B1 6(c)(ii) parallel combination of three capacitors connected in series with one capacitor B1

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7 The circuit for an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. 5.2 k 0.80 k +5 V – P + –5 V R VIN VOUT D Fig. 7.1 (a) For the circuit of Fig. 7.1: (i) explain why point P is known as a virtual earth ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) calculate the gain of the amplifier. gain = ......................................................... [2] (b) When the op-amp is saturated, the potential difference across the LED is 2.3 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 30 mA. resistance = ..................................................... Ω [3] [Total: 8]

Mark scheme: 7(a)(i) (amplifier) gain is very large/infinite B1 for amplifier not to saturate, V+ = V– or feedback (loop) ensures V+ = V– B1 V+ is at earth/0 V so V– is (almost) at earth/0 V B1 7(a)(ii) gain = (–)5200 / 800 or (–)5.2 / 0.80 C1 = –6.5 A1 Question Answer Marks 7(b) (at saturation,) VOUT = 5 V C1 p.d. across R = 5 – 2.3 = 2.7 (V) C1 resistance = 2.7 / (30 × 10–3) = 90 Ω A1 or Rdiode = 2.3 / 0.030 = 77 Ω Rtotal = 5.0 / 0.030 = 167 Ω (C1) Rresistor (= 167 – 77) = 90 Ω (A1) or Rdiode = 2.3 / 0.030 = 77 Ω 77 / (Rresistor + 77) × 5 = 2.3 (C1) Rresistor = 90 Ω (A1) or Rdiode = 2.3 / 0.030 = 77 Ω Rresistor = 77 × (2.7 / 2.3) (C1) Rresistor = 90 Ω (A1)

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8 A solenoid is connected in series with a battery and a switch, as illustrated in Fig. 8.1. small coil A solenoid axis of solenoid Fig. 8.1 A small coil, connected to a sensitive ammeter, is situated near one end of the solenoid. As the current in the solenoid is switched on, there is a changing magnetic field inside the solenoid. (a) (i) State what is meant by a magnetic field. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) On Fig. 8.1, draw an arrow on the axis of the solenoid to show the direction of the magnetic field inside the solenoid. Label this arrow P. [1] (b) As the current in the solenoid is switched on, there is a current induced in the small coil. This induced current gives rise to a magnetic field in the small coil. (i) State Lenz’s law. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Use Lenz’s law to state and explain the direction of the magnetic field due to the induced current in the small coil. On Fig. 8.1, mark this direction with an arrow inside the small coil. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (c) The small coil has an area of cross-section 7.0 × 10–4 m2 and contains 75 turns of wire. A constant current in the solenoid produces a uniform magnetic flux of flux density 1.4 mT throughout the small coil. The direction of the current in the solenoid is reversed in a time of 0.12 s. Calculate the average e.m.f. induced in the small coil. e.m.f. = .......................................................V [3] [Total: 10]

Mark scheme: 8(a)(i) region where a force is exerted on: a magnetic pole or a moving charge or a current-carrying wire B1 8(a)(ii) arrow on axis of solenoid pointing downwards labelled P B1 8(b)(i) direction of induced e.m.f./current M1 (tends to) oppose the change causing it A1 8(b)(ii) magnetic field in solenoid is increasing B1 field in coil in opposite direction to oppose increase B1 arrow inside or just above small coil pointing in opposite direction to P B1 8(c) e.m.f. = N∆φ / ∆t C1 = (75 × 1.4 × 10–3 × 2 × 7.0 × 10–4) / 0.12 C1 = 1.2 × 10–3 V A1

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9 Nuclear magnetic resonance imaging (NMRI) is used to obtain diagnostic information about internal body structures. State, during the use of NMRI, the function of: (a) the large constant magnetic field ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) the non-uniform magnetic field. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 5]

Mark scheme: 9(a) nuclei precess B1 precession is about direction of magnetic field B1 frequency of precession depends on field strength or frequency of precession is in radio-frequency range B1 9(b) Any two points from: • frequency (of precession) depends on position • to locate position of (spinning) nuclei • to change region where nuclei are detected B2

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10 A bridge rectifier contains four diodes. The output of the rectifier is connected to a resistor R, as shown in Fig. 10.1. bridge output input rectifier resistor R Fig. 10.1 The variation with time t of the input e.m.f. E to the rectifier is given by the expression E = 15 cos(210t ) where t is measured in seconds and E in volts. The variation with time t of the potential difference V across resistor R is shown in Fig. 10.2. V 0 t1 t2 time t Fig. 10.2 Determine: (a) the maximum potential difference VMAX across resistor R VMAX = .......................................................V [1] (b) the time interval, to two significant figures, between time t1 and time t2. time = ....................................................... s [3] [Total: 4]

Mark scheme: 10(a) VMAX = 15 V A1 10(b) 210 = 2π / T C1 T = 0.0299 s C1 (t2 – t1) = 0.060 s A1

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11 (a) State three pieces of evidence provided by the photoelectric effect for a particulate nature of electromagnetic radiation. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... 3. ............................................................................................................................................... ................................................................................................................................................... [3] (b) The work function energies of some metals are shown in Fig. 11.1. work function energy / eV sodium 2.4 calcium 2.9 zinc 3.6 silver 4.3 Fig. 11.1 Each metal is irradiated with electromagnetic radiation of wavelength 380 nm. (i) Calculate the energy, in eV, of a photon of electromagnetic radiation of wavelength 380 nm. energy = ......................................................eV [3] (ii) Determine which metals will give rise to the emission of photoelectrons. Explain your answer. ........................................................................................................................................... ..................................................................................................................................... [2] (c) Photons of wavelength 380 nm are incident normally on a metal surface at a rate of 7.6 × 1014 s–1. All the photons are absorbed in the surface and no photoelectrons are emitted. Calculate the force exerted on the metal surface by the incident photons. force = ...................................................... N [3] [Total: 11]

Mark scheme: 11(a) Any three points from: • (max) energy of emitted electrons depends on frequency • (max) energy of emitted electrons does not depend on intensity • rate of emission of electrons depends on intensity (at constant frequency) • existence of frequency below which no emission of electrons • instantaneous emission of electrons • increasing the frequency at constant intensity decreases the rate of emission of electrons B3 11(b)(i) photon energy = hc / λ C1 = (6.63 × 10–34 × 3.0 × 108) / (380 × 10–9) ( = 5.23 × 10–19 J) C1 = 3.3 eV A1 11(b)(ii) photon energy must be greater than work function (energy) B1 so sodium and calcium B1 11(c) λ = h / p C1 p = (6.63 × 10–34) / (380 × 10–9) = 1.74 × 10–27 N s C1 force = 1.74 × 10–27 × 7.6 × 1014 = 1.3 × 10–12 N A1

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12 (a) A sample of a radioactive isotope contains N nuclei of the isotope at time T. At time (T + ΔT ), the sample contains (N – ΔN ) nuclei of the isotope. The time interval ΔT is short. Use the symbols N, ΔN, T and ΔT to give expressions for: (i) the average activity of the sample during the time ΔT ..................................................................................................................................... [1] (ii) the probability of decay of a nucleus in the time ΔT ..................................................................................................................................... [1] (iii) the decay constant λ of the isotope. ..................................................................................................................................... [1] (b) The isotope polonium-208 (20884 Po) is radioactive and decays to form lead-204 ( 20482 Pb). The nuclear equation for this decay is 208 84 Po 20482 Pb + 42 He. Data for nuclear masses are given in Fig. 12.1. mass / u 4 2 He 4.002 603 204 82 Pb 203.973 043 208 84 Po 207.981 245 Fig. 12.1 (i) Determine, for the decay of one nucleus of polonium-208: 1. the change, in u, of the mass mass change = .......................................................u [1] 2. the total energy, in pJ, released. energy = ..................................................... pJ [3] (ii) The polonium-208 nucleus is initially stationary. The initial kinetic energy of the 4 2 He nucleus (α-particle) is found to be less than the energy calculated in (i) part 2. Suggest two possible reasons for this difference. 1. ....................................................................................................................................... ........................................................................................................................................... 2. ....................................................................................................................................... ........................................................................................................................................... [2] [Total: 9]

Mark scheme: 12(a)(i) B1 12(a)(ii) ∆N / N B1 12(a)(iii) ∆N / (N ∆T) B1 12(b)(i) 1. mass change = 5.60 × 10–3 u A1 2. energy = (∆)mc2 C1 = 5.6 × 10–3 × 1.66 × 10–27 × (3.0 × 108)2 ( = 8.36 × 10–13 J) C1 = 0.84 pJ A1 12(b)(ii) kinetic energy (of recoil) of lead (nucleus) B1 energy of γ-ray photon B1

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