Cambridge A Level Physics 9702 — 2012 Oct/Nov Paper 4 · Variant 1

1 (a) State Newton’s law of gravitation. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M. For this planet, the product GM is 4.00 × 1014 N m2 kg–1, where G is the gravitational constant. The planet may be assumed to be isolated in space. (i) By considering the gravitational force on the satellite and the centripetal force, show that the kinetic energy EK of the satellite is given by the expression GMm EK = . 2r [2] (ii) The satellite has mass 620 kg and is initially in a circular orbit of radius 7.34 × 106 m, as illustrated in Fig. 1.1. initial orbit 7.34 × 106 m 7.30 × 106 m new orbit Fig. 1.1 (not to scale) Resistive forces cause the satellite to move into a new orbit of radius 7.30 × 106 m. For Examiner’s Determine, for the satellite, the change in Use 1. kinetic energy, change in kinetic energy = ............................................. J [2] 2. gravitational potential energy. change in potential energy = ............................................. J [2] (iii) Use your answers in (ii) to explain whether the linear speed of the satellite increases, decreases or remains unchanged when the radius of the orbit decreases. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

Mark scheme: 1 (a) force is proportional to the product of the masses and inversely proportional to the square of the separation M1 either point masses or separation >> size of masses A1 [2] (b) (i) gravitational force provides the centripetal force B1 mv2/r = GMm/r2 and EK = ½mv2 M1 hence EK = GMm/2r A0 [2] (ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 9.26 × 107 J (ignore any sign in answer) A1 [2] (allow 1.0 × 108 J if evidence that EK evaluated separately for each r) 2. ∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 1.85 × 108 J (ignore any sign in answer) A1 [2] (allow 1.8 or 1.9 × 108 J) (iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased M1 speed has increased A1 [2]

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2 A student suggests that, when an ideal gas is heated from 100 °C to 200 °C, the internal For energy of the gas is doubled. Examiner’s Use (a) (i) State what is meant by internal energy. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) By reference to one of the assumptions of the kinetic theory of gases and your answer in (i), deduce what is meant by the internal energy of an ideal gas. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3] (b) State and explain whether the student’s suggestion is correct. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

Mark scheme: 2 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles M1 reference to random A1 [2] (ii) no intermolecular forces B1 no potential energy B1 internal energy is kinetic energy (of random motion) of molecules B1 [3] (reference to random motion here then allow back credit to (i) if M1 scored) (b) kinetic energy ∝ thermodynamic temperature B1 either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled B1 [2]

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3 (a) Two metal spheres are in thermal equilibrium. For State and explain what is meant by thermal equilibrium. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) An electric water heater contains a tube through which water flows at a constant rate. The water in the tube passes over a heating coil, as shown in Fig. 3.1. water out heating coil tube water in Fig. 3.1 The water flows into the tube at a temperature of 18 °C. When the power of the heater is 3.8 kW, the temperature of the water at the outlet is 42 °C. The specific heat capacity of water is 4.2 J g–1 K–1. (i) Use the data to calculate the flow rate, in g s–1, of water through the tube. flow rate = ........................................ g s–1 [3] (ii) State and explain whether your answer in (i) is likely to be an overestimate or an underestimate of the flow rate. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

Mark scheme: 3 (a) temperature of the spheres is the same B1 no (net) transfer of energy between the spheres B1 [2] (b) (i) power = m × c × ∆θ where m is mass per second C1 3800 = m × 4.2 × (42 – 18) C1 m = 38 g s–1 A1 [3] (ii) some thermal energy is lost to the surroundings M1 so rate is an overestimate A1 [2]

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4 A ball is held between two fixed points A and B by means of two stretched springs, as shown For in Fig. 4.1. Examiner’s Use ball A B Fig. 4.1 The ball is free to oscillate horizontally along the line AB. During the oscillations, the springs remain stretched and do not exceed their limits of proportionality. The variation of the acceleration a of the ball with its displacement x from its equilibrium position is shown in Fig. 4.2. 15 a / m s–2 10 5 0 –3 –2 –1 0 1 2 3 x / cm –5 –10 –15 Fig. 4.2 (a) State and explain the features of Fig. 4.2 that indicate that the motion of the ball is For simple harmonic. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[4] (b) Use Fig. 4.2 to determine, for the oscillations of the ball, (i) the amplitude, amplitude = .......................................... cm [1] (ii) the frequency. frequency = ........................................... Hz [3] (c) The arrangement in Fig. 4.1 is now rotated through 90° so that the line AB is vertical. The ball now oscillates in a vertical plane. Suggest one reason why the oscillations may no longer be simple harmonic. .......................................................................................................................................... ......................................................................................................................................[1]

Mark scheme: 4 (a) straight line through origin M1 shows acceleration proportional to displacement A1 negative gradient M1 shows acceleration and displacement in opposite directions A1 [4] GCE AS/A LEVEL – October/November 2012 9702 41 (b) (i) 2.8 cm A1 [1] (ii) either gradient = ω2 and ω = 2πf or a = –ω2x and ω = 2πf C1 gradient = 13.5 / (2.8 × 10–2) = 482 ω = 22 rad s–1 C1 frequency = (22/2π =) 3.5 Hz A1 [3] (c) e.g. lower spring may not be extended e.g. upper spring may exceed limit of proportionality / elastic limit (any sensible suggestion) B1 [1]

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5 (a) (i) Define capacitance. For Examiner’s .................................................................................................................................. Use ..............................................................................................................................[1] (ii) A capacitor is made of two metal plates, insulated from one another, as shown in Fig. 5.1. insulation metal plate Fig. 5.1 Explain why the capacitor is said to store energy but not charge. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[4] (b) Three uncharged capacitors X, Y and Z, each of capacitance 12 μF, are connected as shown in Fig. 5.2. Y X 12 ȝF A B Z 12 ȝF 12 ȝF Fig. 5.2 A potential difference of 9.0 V is applied between points A and B. (i) Calculate the combined capacitance of the capacitors X, Y and Z. For Examiner’s Use capacitance = ........................................... μF [2] (ii) Explain why, when the potential difference of 9.0 V is applied, the charge on one plate of capacitor X is 72 μC. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (iii) Determine 1. the potential difference across capacitor X, potential difference = ............................................. V [1] 2. the charge on one plate of capacitor Y. charge = ........................................... μC [2]

Mark scheme: 5 (a) (i) ratio of charge and potential (difference) / voltage (ratio must be clear) B1 [1] (ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1 total charge on capacitor is zero (so does not store charge) B1 (+)ve and (-)ve charges to be separated M1 work done to achieve this so stores energy A1 [4] (b) (i) capacitance of Y and Z together is 24 µF C1 1 / C = 1 / 24 + 1 / 12 C = 8.0 µF (allow 1 s.f.) A1 [2] (ii) some discussion as to why all charge of one sign on one plate of X B1 Q = (CV =) 8.0 × 10–6 × 9.0 M1 = 72 µC A0 [2] (iii) 1. V = (72 × 10–6) / (12 × 10–6) = 6.0 V (allow 1 s.f.) (allow 72/12) A1 [1] 2. either Q = 12 × 10–6 × 3.0 or charge is shared between Y and Z C1 charge = 36 µC A1 [2] Must have correct voltage in (iii)1 if just quote of 36 µC in (iii)2.

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6 (a) (i) State the condition for a charged particle to experience a force in a magnetic field. For Examiner’s .................................................................................................................................. Use .................................................................................................................................. ..............................................................................................................................[2] (ii) State an expression for the magnetic force F acting on a charged particle in a magnetic field of flux density B. Explain any other symbols you use. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) A sample of a conductor with rectangular faces is situated in a magnetic field, as shown in Fig. 6.1. direction of magnetic field B C direction of F G movement A D of electrons E H Fig. 6.1 The magnetic field is normal to face ABCD in the downward direction. Electrons enter face CDHG at right-angles to the face. As the electrons pass through the conductor, they experience a force due to the magnetic field. (i) On Fig. 6.1, shade the face to which the electrons tend to move as a result of this force. [1] (ii) The movement of the electrons in the magnetic field causes a potential difference between two faces of the conductor. Using the lettering from Fig. 6.1, state the faces between which this potential difference will occur. face ................................. and face .................................[1] (c) Explain why the potential difference in (b) causes an additional force on the moving electrons in the conductor. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

Mark scheme: 6 (a) (i) particle must be moving M1 with component of velocity normal to magnetic field A1 [2] (ii) F = Bqv sin θ M1 q, v and θ explained A1 [2] (b) (i) face BCGF shaded A1 [1] (ii) between face BCGF and face ADHE A1 [1] (c) potential difference gives rise to an electric field M1 either FE = qE (no need to explain symbols) or electric field gives rise to force (on an electron) A1 [2] GCE AS/A LEVEL – October/November 2012 9702 41

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7 (a) State Lenz’s law. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ......................................................................................................................................[2] (b) A simple transformer with a soft-iron core is illustrated in Fig. 7.1. laminated soft-iron core input output primary coil secondary coil Fig. 7.1 (i) Explain why the core is 1. made of iron, .................................................................................................................................. ..............................................................................................................................[1] 2. laminated. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) An e.m.f. is induced in the secondary coil of the transformer. Explain how a current in the primary coil gives rise to this induced e.m.f. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[4]

Mark scheme: 7 (a) induced e.m.f./current produces effects / acts in such a direction / tends M1 to oppose the change causing it A1 [2] (b) (i) 1. to reduce flux losses / increase flux linkage / easily magnetised and demagnetised B1 [1] 2. to reduce energy / heat losses (do not allow ‘to prevent energy losses’) M1 caused by eddy currents A1 [2] (allow 1 mark for ‘reduce eddy currents’) (ii) alternating current / voltage B1 gives rise to (changing) flux in core B1 flux links the secondary coil M1 (by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4]

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8 (a) State what is meant by a photon. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ......................................................................................................................................[2] (b) It has been observed that, where photoelectric emission of electrons takes place, there is negligible time delay between illumination of the surface and emission of an electron. State three other pieces of evidence provided by the photoelectric effect for the particulate nature of electromagnetic radiation. 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... 3. ...................................................................................................................................... .......................................................................................................................................... [3] (c) The work function of a metal surface is 3.5 eV. Light of wavelength 450 nm is incident on the surface. Determine whether electrons will be emitted, by the photoelectric effect, from the surface. [3]

Mark scheme: 8 (a) discrete quantity / packet / quantum of energy of electromagnetic radiation B1 energy of photon = Planck constant × frequency B1 [2] (b) threshold frequency (1) rate of emission is proportional to intensity (1) max. kinetic energy of electron dependent on frequency (1) max. kinetic energy independent of intensity (1) (any three, 1 each, max 3) B3 [3] (c) either E = hc/λ or hc/λ = eV C1 λ = 450 nm to give work function of 3.5 eV energy = 4.4 × 10–19 or 2.8 eV to give λ = 355 nm M1 2.8 eV < 3.5 eV so no emission 355 nm < 450 nm so no A1 [3] or work function = 3.5 eV threshold frequency = 8.45×1014 Hz C1 450 nm = 6.67×1014 Hz M1 6.67 × 1014 Hz < 8.45 × 1014 Hz A1 GCE AS/A LEVEL – October/November 2012 9702 41 Section B

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9 An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor. (a) State three properties of an ideal op-amp. 1. ...................................................................................................................................... 2. ...................................................................................................................................... 3. ...................................................................................................................................... [3] (b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1. +5 V – + V1 VOUT –5 V V2 Fig. 9.1 (i) In one application of the comparator, V2 is kept constant at +1.5 V. The variation with time t of the potential V1 is shown in Fig. 9.2. The potential V2 is also shown. 10 For Examiner’s Use 8 potential / V 6 4 2 V2 0 t –2 V1 –4 –6 –8 –10 Fig. 9.2 On Fig. 9.2, show the variation with time t of the output potential VOUT . [4] (ii) Two light-emitting diodes (LEDs) R and G are connected to the output of the op-amp in Fig. 9.1 such that R emits light for a longer time than G. On Fig. 9.1, draw the symbols for the two diodes connected to the output of the op-amp and label the diodes R and G. [3]

Mark scheme: 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate 1 each, max. 3 B3 [3] (b) (i) graph: square wave M1 correct cross-over points where V2 = V1 A1 amplitude 5 V A1 correct polarity (positive at t = 0) A1 [4] (ii) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 correct polarity consistent with graph in (i) A1 [3] (R points ‘down’ if (i) correct)

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10 Outline the principles of CT scanning. For Examiner’s ................................................................................................................................................. Use ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. .............................................................................................................................................[6]

Mark scheme: 10 X-ray images taken from different angles / X-rays directed from different angles B1 of one section / slice (1) all images in the same plane (1) images combined to give image of section / slice B1 images of successive sections / slices combined B1 image formed using a computer B1 image formed is 3D image (1) that can be rotated / viewed from different angles (1) (four B-marks plus any two additional marks) B2 [6]

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11 (a) In modern communications systems, the majority of data is transmitted in digital form For rather than analogue form. Examiner’s Suggest three advantages of the transmission of data in digital form. Use 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... 3. ...................................................................................................................................... .......................................................................................................................................... [3] (b) A recording is made of some music. For this recording, the music is sampled at a rate of 44.1 kHz and each sample consists of a 16-bit word. (i) Suggest the effect on the quality of the recording of 1. sampling at a high frequency rather than a lower frequency, .................................................................................................................................. ..............................................................................................................................[1] 2. using a long word length rather than a shorter word length. .................................................................................................................................. ..............................................................................................................................[1] (ii) The recording lasts for a total time of 5 minutes 40 seconds. Calculate the number of bits generated during the recording. number = ..................................................[2]

Mark scheme: 11 (a) e.g. noise can be eliminated / filtered / signal can be regenerated extra bits can be added to check for errors multiplexing possible digital circuits are more reliable / cheaper data can be encrypted for security any sensible advantages, 1 each, max. 3 B3 [3] (b) (i) 1. higher frequencies can be reproduced B1 [1] 2. smaller changes in loudness / amplitude can be detected B1 [1] (ii) bit rate = 44.1 × 103 × 16 C1 = 7.06 × 105 s–1 number = 7.06 × 106 × 340 = 2.4 × 108 A1 [2]

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12 (a) Wire pairs used for the transmission of telephone signals are subject to cross-linking. For Examiner’s (i) Explain what is meant by cross-linking. Use .................................................................................................................................. ..............................................................................................................................[1] (ii) Suggest why cross-linking in coaxial cables is much less than in wire pairs. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) A wire pair has a length of 1.4 km and is connected to a receiver, as illustrated in Fig. 12.1. wire pair constant noise power 3.8 × 10–8 W input signal receiver power 3.0 × 10–3 W 1.4 km Fig. 12.1 The constant noise power in the wire pair is 3.8 × 10–8 W. For an input signal to the wire pair of 3.0 × 10–3 W, the signal-to-noise ratio at the receiver is 25 dB. Calculate the attenuation per unit length for the wire pair. attenuation per unit length = ................................... dB km–1 [4]

Mark scheme: 12 (a) (i) signal in one wire (pair) is picked up by a neighbouring wire (pair) B1 [1] (ii) outer of coaxial cable is earthed B1 outer shields the core from noise / external signals B1 [2] GCE AS/A LEVEL – October/November 2012 9702 41 (b) attenuation per unit length = 1/L × 10 lg(P2/P1) C1 signal power at receiver = 102.5 × 3.8 × 10–8 = 1.2 × 10–5 W C1 attenuation in wire pair = 10 lg({3.0 × 10–3} / {1.2 × 10–5}) = 24 dB C1 attenuation per unit length = 24 / 1.4 = 17 dB km–1 A1 [4] (other correct methods of calculation are possible)

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