Cambridge A Level Physics 9702 — 2010 May/June Paper 4 · Variant 1

1 (a) Define the radian. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown in Fig. 1.1. glue ω P 85 cm stone, C weight 3.0 N Fig. 1.1 The rod is rotated about end C so that the stone moves in a vertical circle of radius 85 cm. The angular speed ω of the rod and stone is gradually increased from zero until the glue snaps. The glue fixing the stone snaps when the tension in it is 18 N. For the position of the stone at which the glue snaps, (i) on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone, [1] (ii) calculate the angular speed ω of the stone. angular speed = ................................... rad s–1 [4]

Mark scheme: 1 (a) angle (subtended) at centre of circle B1 (by) arc equal in length to radius B1 [2] (b) (i) point S shown below C B1 [1] (ii) (max) force / tension = weight + centripetal force C1 centripetal force = mrω2 C1 15 = 3.0/9.8 × 0.85 × ω2 C1 ω = 7.6 rad s–1 A1 [4]

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2 (a) Some gas, initially at a temperature of 27.2 °C, is heated so that its temperature rises For to 38.8 °C. Examiner’s Calculate, in kelvin, to an appropriate number of decimal places, Use (i) the initial temperature of the gas, initial temperature = ............................................. K [2] (ii) the rise in temperature. rise in temperature = ............................................ K [1] (b) The pressure p of an ideal gas is given by the expression 1 2 p = 3ρ c where ρ is the density of the gas. (i) State the meaning of the symbol c 2 . .................................................................................................................................. ..............................................................................................................................[1] (ii) Use the expression to show that the mean kinetic energy <EK> of the atoms of an ideal gas is given by the expression 3 <EK> = 2 kT. Explain any symbols that you use. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [4] (c) Helium-4 may be assumed to behave as an ideal gas. For A cylinder has a constant volume of 7.8 × 103 cm3 and contains helium-4 gas at a Examiner’s pressure of 2.1 × 107 Pa and at a temperature of 290 K. Use Calculate, for the helium gas, (i) the amount of gas, amount = ......................................... mol [2] (ii) the mean kinetic energy of the atoms, mean kinetic energy = .............................................. J [2] (iii) the total internal energy. internal energy = .............................................. J [3]

Mark scheme: 2 (a) (i) 27.2 + 273.15 or 27.2 + 273.2 C1 300.4 K A1 [2] (ii) 11.6 K A1 [1] (b) (i) (<c2> is the) mean / average square speed B1 [1] (ii) ρ = Nm/V with N explained B1 so, pV = 1/3 Nm<c2> B1 and pV = NkT with k explained B1 so mean kinetic energy / <EK> = ½m<c2> = 3/2 kT B1 [4] (c) (i) pV = nRT 2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 C1 n = 68 mol A1 [2] (ii) mean kinetic energy = 3/2 kT = 3/2 × 1.38 × 10–23 × 290 C1 = 6.0 × 10–21 J A1 [2] (iii) realisation that total internal energy is the total kinetic energy C1 energy = 6.0 × 10–21 × 68 × 6.02 × 1023 C1 = 2.46 × 105 J A1 [3]

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3 (a) State what is meant by For Examiner’s (i) oscillations, Use .................................................................................................................................. ..............................................................................................................................[1] (ii) free oscillations, .................................................................................................................................. ..............................................................................................................................[1] (iii) simple harmonic motion. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) Two inclined planes RA and LA each have the same constant gradient. They meet at their lower edges, as shown in Fig. 3.1. ball L R A Fig. 3.1 A small ball moves from rest down plane RA and then rises up plane LA. It then moves down plane LA and rises up plane RA to its original height. The motion repeats itself. State and explain whether the motion of the ball is simple harmonic. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

Mark scheme: 3 (a) (i) to-and-fro / backward and forward motion (between two limits) B1 [1] (ii) no energy loss or gain / no external force acting / constant energy / constant amplitude B1 [1] (iii) acceleration directed towards a fixed point B1 acceleration proportional to distance from the fixed point / displacement B1 [2] (b) acceleration is constant (magnitude) M1 so cannot be s.h.m. A1 [2] GCE AS/A LEVEL – May/June 2010 9702 41

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4 (a) Explain what is meant by the potential energy of a body. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ......................................................................................................................................[2] 2 (b) Two deuterium ( 1 H) nuclei each have initial kinetic energy EK and are initially separated by a large distance. The nuclei may be considered to be spheres of diameter 3.8 × 10–15 m with their masses and charges concentrated at their centres. The nuclei move from their initial positions to their final position of just touching, as illustrated in Fig. 4.1. 2 2 initially 1 H 1 H kinetic energy EK kinetic energy EK 3.8 × 10–15 m 2 2 finally 1 H 1 H at rest Fig. 4.1 (i) For the two nuclei approaching each other, calculate the total change in 1. gravitational potential energy, energy = ............................................ J [3] 2. electric potential energy. energy = ............................................ J [3] (ii) Use your answers in (i) to show that the initial kinetic energy EK of each nucleus For is 0.19 MeV. Examiner’s Use [2] (iii) The two nuclei may rebound from each other. Suggest one other effect that could happen to the two nuclei if the initial kinetic energy of each nucleus is greater than that calculated in (ii). .................................................................................................................................. ..............................................................................................................................[1]

Mark scheme: 4 (a) ability to do work B1 as a result of the position/shape, etc. of an object B1 [2] (b) (i) 1 ∆Egpe = GMm / r C1 = (6.67 × 10–11 × {2 × 1.66 × 10–27}2) / (3.8 × 10–15) C1 = 1.93 × 10–49 J A1 [3] 2 ∆Eepe = Qq / 4πε0r C1 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) C1 = 6.06 × 10–14 J A1 [3] (ii) idea that 2EK = ∆Eepe – ∆Egpe B1 EK = 3.03 × 10–14 J = (3.03 × 10–14) / 1.6 × 10–13 M1 = 0.19 MeV A0 [2] (iii) fusion may occur / may break into sub-nuclear particles B1 [1]

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5 (a) A constant current is maintained in a long straight vertical wire. A Hall probe is positioned For a distance r from the centre of the wire, as shown in Fig. 5.1. Examiner’s Use current-carrying wire Hall probe X Y terminals to r Hall probe circuitry and voltmeter Fig. 5.1 (i) Explain why, when the Hall probe is rotated about the horizontal axis XY, the Hall voltage varies between a maximum positive value and a maximum negative value. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) The maximum Hall voltage VH is measured at different distances r. Data for VH and the corresponding values of r are shown in Fig. 5.2. VH / V r / cm 0.290 1.0 0.190 1.5 0.140 2.0 0.097 3.0 0.073 4.0 0.060 5.0 Fig. 5.2 It is thought that VH and r are related by an expression of the form k VH = r where k is a constant. 1. Without drawing a graph, use data from Fig. 5.2 to suggest whether the For expression is valid. Examiner’s Use [2] 1 2. A graph showing the variation with of VH is plotted. r State the features of the graph that suggest that the expression is valid. .............................................................................................................................. ..........................................................................................................................[1] (b) The Hall probe in (a) is now replaced with a small coil of wire connected to a sensitive voltmeter. The coil is arranged so that its plane is normal to the magnetic field of the wire. (i) State Faraday’s law of electromagnetic induction and hence explain why the voltmeter indicates a zero reading. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3] (ii) State three different ways in which an e.m.f. may be induced in the coil. 1. .............................................................................................................................. .................................................................................................................................. 2. .............................................................................................................................. .................................................................................................................................. 3. .............................................................................................................................. .................................................................................................................................. [3]

Mark scheme: 5 (a) (i) VH depends on angle between (plane of) probe and B-field B1 either VH max when plane and B-field are normal to each other or VH zero when plane and B-field are parallel or VH depends on sine of angle between plane and B-field B1 [2] (ii) 1 calculates VHr at least three times M1 to 1 s.f. constant so valid or approx constant so valid or to 2 s.f., not constant so invalid A1 [2] 2 straight line passes through origin B1 [1] (b) (i) e.m.f. induced is proportional / equal to M1 rate of change of (magnetic) flux (linkage) A1 constant field in coil / flux (linkage) of coil does not change B1 [3] (ii) e.g. vary current (in wire) / switch current on or off / use a.c. current rotate coil move coil towards / away from wire (1 mark each, max 3) B3 [3]

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6 A student is asked to design a circuit by which a direct voltage of peak value 9.0 V is obtained For from a 240 V alternating supply. Examiner’s The student uses a transformer that may be considered to be ideal and a bridge rectifier Use incorporating four ideal diodes. The partially completed circuit diagram is shown in Fig. 6.1. 240 V + load – Fig. 6.1 (a) On Fig. 6.1, draw symbols for the four diodes so as to produce the polarity across the load as shown on the diagram. [2] (b) Calculate the ratio number of turns on the secondary coil . number of turns on the primary coil ratio = ................................................ [3]

Mark scheme: 6 (a) all four diodes correct to give output, regardless of polarity M1 connected for correct polarity A1 [2] (b) NS / NP = VS / VP C1 V0 = √2 × Vrms C1 ratio = 9.0 / (√2 × 240) = 1/38 or 1/37 or 0.027 A1 [3] GCE AS/A LEVEL – May/June 2010 9702 41

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7 Negatively-charged particles are moving through a vacuum in a parallel beam. The particles For have speed v. Examiner’s The particles enter a region of uniform magnetic field of flux density 930 μT. Initially, the Use particles are travelling at right-angles to the magnetic field. The path of a single particle is shown in Fig. 7.1. negatively-charged arc of radius 7.9 cm particles, speed v uniform magnetic field, flux density 930 μT Fig. 7.1 The negatively-charged particles follow a curved path of radius 7.9 cm in the magnetic field. A uniform electric field is then applied in the same region as the magnetic field. For an electric field strength of 12 kV m–1, the particles are undeviated as they pass through the region of the fields. (a) On Fig. 7.1, mark with an arrow the direction of the electric field. [1] (b) Calculate, for the negatively-charged particles, (i) the speed v, v = ....................................... m s–1 [3] charge (ii) the ratio . mass ratio = .................................... C kg–1 [3]

Mark scheme: 7 (a) arrow pointing up the page B1 [1] (b) (i) Eq = Bqv C1 v = (12 × 103) / (930 × 10–6) C1 = 1.3 × 107 m s–1 A1 [3] (ii) Bqv = mv 2 / r C1 q/m = (1.3 × 107) / (7.9 × 10–2 × 930 × 10–6) C1 = 1.8 × 1011 C kg–1 A1 [3]

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8 A π0 meson is a sub-atomic particle. For A stationary π0 meson, which has mass 2.4 × 10–28 kg, decays to form two γ-ray photons. Examiner’s The nuclear equation for this decay is Use π0 γ + γ. (a) Explain why the two γ-ray photons have the same energy. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) Determine, for each γ-ray photon, (i) the energy, in joule, energy = .............................................. J [2] (ii) the wavelength, wavelength = ............................................ m [2] (iii) the momentum. For Examiner’s Use momentum = ........................................... N s [2]

Mark scheme: 8 (a) momentum conservation hence momenta of photons are equal (but opposite) M1 same momentum so same energy A1 [2] (b) (i) (∆)E = (∆)mc2 C1 = 1.2 × 10–28 × (3.0 × 108)2 = 1.08 × 10–11 J A1 [2] (ii) E = hc / λ λ = (6.63 × 10–34 × 3.0 × 108) / (1.08 × 10–11) C1 = 1.84 × 10–14 m A1 [2] (iii) λ = h / p p = (6.63 × 10–34) / (1.84 × 10–14) C1 = 3.6 × 10–20 N s A1 [2] Section B

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9 The circuit diagram of Fig. 9.1 is an amplifier circuit incorporating an operational amplifier (op-amp). 4.2 kΩ +9 V 1.0 kΩ – + + 1.5 V –9 V V – Fig. 9.1 (a) (i) On Fig. 9.1, mark, with the letter X, the virtual earth. [1] (ii) Explain what is meant by a virtual earth. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3] (b) In bright sunlight, the light-dependent resistor (LDR) has resistance 200 Ω. (i) Calculate, for the LDR in bright sunlight, the voltmeter reading. reading = ............................................ V [3] (ii) The sunlight incident on the LDR becomes less bright. For State and explain the effect on the voltmeter reading of this decrease in Examiner’s brightness. Use .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3]

Mark scheme: 9 (a) (i) point X shown correctly B1 [1] (ii) op-amp has very large / infinite gain M1 non-inverting input is at earth (potential) / earthed / at 0 V M1 if amplifier is not to saturate, inverting input must be (almost) at earth potential / 0 (V) same potential as inverting input A1 [3] (b) (i) total input resistance = 1.2 kΩ C1 (amplifier) gain (= –4.2 / 1.2) = –3.5 C1 (voltmeter) reading = –3.5 × –1.5 = 5.25 V A1 [3] (total disregard of signs or incorrect sign in answer, max 2 marks) (ii) (less bright so) resistance of LDR increases M1 (amplifier) gain decreases M1 (voltmeter) reading decreases A1 [3] GCE AS/A LEVEL – May/June 2010 9702 41

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10 (a) Briefly explain the principles of CT scanning. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[6] (b) A simple section through a body consists of four voxels, as illustrated in Fig. 10.1. For Examiner’s section Use directions of viewing Fig. 10.1 An X-ray image of the section is obtained by viewing along each of the directions shown in Fig. 10.1. The detector readings for each direction of viewing are summed to give the pattern of readings shown in Fig. 10.2. 25 22 34 31 Fig. 10.2 For any one direction, the total of the detector readings is 16. (i) For the pattern of readings of Fig. 10.2, state the magnitude of the background reading. background reading = ................................................ [1] (ii) On Fig. 10.1, mark the pattern of pixels for the four-voxel section. [2]

Mark scheme: 10 (a) X-ray taken of slice / plane / section B1 repeated at different angles B1 images / data is processed B1 combined / added to give (2-D) image of slice B1 repeated for successive slices B1 to build up a 3-D image B1 image can be viewed from different angles / rotated B1 max 6 [6] (b) (i) 16 A1 [1] (ii) evidence of deducting 16 then dividing by 3 C1 to give A1 [2] 3 2 6 5

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11 Many radio stations now broadcast on FM rather than on AM. In general, FM is broadcast at For much higher frequencies than AM. Examiner’s Use (a) Explain what is meant by FM (frequency modulation). .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) State two advantages and two disadvantages of FM transmissions when compared with AM transmissions. advantages of FM transmissions 1. ..................................................................................................................................... .......................................................................................................................................... 2. ..................................................................................................................................... .......................................................................................................................................... disadvantages of FM transmissions 1. ..................................................................................................................................... .......................................................................................................................................... 2. ..................................................................................................................................... .......................................................................................................................................... [4]

Mark scheme: 11 (a) frequency of carrier wave varies (in synchrony) with signal M1 (in synchrony) with displacement of signal A1 [2] (b) advantages e.g. less noise / less interference greater bandwidth / better quality (1 each, max 2) disadvantages e.g. short range / more transmitters / line of sight more complex circuitry greater expense (1 each, max 2) B4 [4]

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12 A ground station on Earth transmits a signal of frequency 14 GHz and power 18 kW towards For a communications satellite orbiting the Earth, as illustrated in Fig. 12.1. Examiner’s Use ground station, signal power 18 kW frequency signal 14 GHz satellite Earth Fig. 12.1 The loss in signal power between the ground station and the satellite is 190 dB. (a) Calculate the power of the signal received by the satellite. power = .......................................... W [3] (b) The signal received by the satellite is amplified and transmitted back to Earth. (i) Suggest a frequency for the signal that is sent back to Earth. frequency = ...................................... GHz [1] (ii) Give a reason for your answer in (i). .................................................................................................................................. ..............................................................................................................................[1]

Mark scheme: 12 (a) gain / loss/dB = 10 lg(P1/P2) C1 190 = 10 lg(18 × 103 / P2) or –190 = 10 lg P2 / 18 × 103) C1 power = 1.8 × 10–15 W A1 [3] (b) (i) 11 GHz / 12 GHz B1 [1] (ii) e.g. so that input signal to satellite will not be ‘swamped’ to avoid interference of uplink with / by downlink B1 [1]

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