Cambridge A Level Physics 9702 — 2012 May/June Paper 4 · Variant 2

1 (a) State Newton’s law of gravitation. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) The Earth and the Moon may be considered to be isolated in space with their masses concentrated at their centres. The orbit of the Moon around the Earth is circular with a radius of 3.84 × 105 km. The period of the orbit is 27.3 days. Show that (i) the angular speed of the Moon in its orbit around the Earth is 2.66 × 10–6 rad s–1, [1] (ii) the mass of the Earth is 6.0 × 1024 kg. [2] (c) The mass of the Moon is 7.4 × 1022 kg. For Examiner’s (i) Using data from (b), determine the gravitational force between the Earth and the Use Moon. force = .............................................. N [2] (ii) Tidal action on the Earth’s surface causes the radius of the orbit of the Moon to increase by 4.0 cm each year. Use your answer in (i) to determine the change, in one year, of the gravitational potential energy of the Moon. Explain your working. energy change = ............................................... J [3]

Mark scheme: 1 (a) force proportional to product of masses and inversely proportional to square of separation (do not allow square of distance/radius) M1 either point masses or separation @ size of masses A1 [2] (b) (i) ω = 2π / (27.3 × 24 × 3600) or 2π / (2.36 x 106) M1 = 2.66 × 10–6 rad s–1 A0 [1] (ii) GM = r3ω2 or GM = v2r C1 M = (3.84 × 105 × 103)3 × (2.66 × 10–6)2 / (6.67 × 10–11) M1 = 6.0 × 1024 kg A0 [2] (special case: uses g = GM/r2 with g = 9.81, r = 6.4 × 106 scores max 1 mark) (c) (i) grav. force = (6.0 × 1024) × (7.4 × 1022) × (6.67 × 10–11)/(3.84 × 108)2 C1 = 2.0 × 1020 N (allow 1 SF) A1 [2] (ii) either ∆EP = Fx because F constant as x ! radius of orbit B1 ∆EP = 2.0 × 1020 × 4.0 × 10–2 C1 = 8.0 × 1018 J (allow 1 SF) A1 [3] or ∆EP = GMm/r1 – GMm/r2 C1 Correct substitution B1 8.0 × 1018 J A1 (∆EP = GMm/r1 + GMm/r2 is incorrect physics so 0/3) 2 2

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2 A ball of mass 37 g is held between two fixed points A and B by two stretched helical springs, For as shown in Fig. 2.1. Examiner’s Use ball mass 37 g A B Fig. 2.1 The ball oscillates along the line AB with simple harmonic motion of frequency 3.5 Hz and amplitude 2.8 cm. (a) Show that the total energy of the oscillations is 7.0 mJ. [2] (b) At two points in the oscillation of the ball, its kinetic energy is equal to the potential energy stored in the springs. Calculate the magnitude of the displacement at which this occurs. displacement = ............................................ cm [3] (c) On the axes of Fig. 2.2 and using your answers in (a) and (b), sketch a graph to show For the variation with displacement x of Examiner’s Use (i) the total energy of the system (label this line T), [1] (ii) the kinetic energy of the ball (label this line K), [2] (iii) the potential energy stored in the springs (label this line P). [2] 8 6 energy / mJ 4 2 0 –3 –2 –1 0 1 2 3 x / cm Fig. 2.2 (d) The arrangement in Fig. 2.1 is now rotated through 90° so that the line AB is vertical and the ball oscillates in a vertical plane. Suggest one form of energy, other than those in (c), that must be taken into consideration when plotting new graphs to show energy changes with displacement. ......................................................................................................................................[1]

Mark scheme: 2 (a) energy = ½mω2a2 and ω = 2πf C1 = ½ × 37 × 10–3 × (2π × 3.5)2 × (2.8 × 10–2)2 M1 = 7.0 × 10–3 J A0 [2] (allow 2π × 3.5 shown as 7π) Energy = ½ mv2 and v = rω (C1) Correct substitution (M1) Energy = 7.0 × 10–3 J (A0) (b) EK = EP ½mω2 (a2 – x2) = ½mω2x2 or EK or EP = 3.5 mJ C1 x = a/√2 = 2.8 /√2 or EK = ½mω2 (a2 – x2) or EP = ½mω2x2 C1 = 2.0 cm A1 [3] (EK or EP = 7.0 mJ scores 0/3) Allow: k = 17.9 (C1) E = ½ kx2 (C1) x = 2.0 cm (A1) GCE AS/A LEVEL – May/June 2012 9702 42 (c) (i) graph: horizontal line, y-intercept = 7.0 mJ with end-points of line at +2.8 cm and –2.8 cm B1 [1] (ii) graph: reasonable curve B1 with maximum at (0,7.0) end-points of line at (–2.8, 0) and (+2.8, 0) B1 [2] (iii) graph: inverted version of (ii) M1 with intersections at (–2.0, 3.5) and (+2.0, 3.5) A1 [2] (Allow marks in (iii), but not in (ii), if graphs K & P are not labelled) (d) gravitational potential energy B1 [1]

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3 (a) State what is meant by the internal energy of a system. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ......................................................................................................................................[2] (b) State and explain qualitatively the change, if any, in the internal energy of the following systems: (i) a lump of ice at 0 °C melts to form liquid water at 0 °C, .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3] (ii) a cylinder containing gas at constant volume is in sunlight so that its temperature rises from 25 °C to 35 °C. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3]

Mark scheme: 3 (a) sum of potential energy and kinetic energy of atoms/molecules/particles M1 reference to random (distribution) A1 [2] (b) (i) as lattice structure is ‘broken’/bonds broken/forces between molecules reduced (not molecules separate) B1 no change in kinetic energy, potential energy increases M1 internal energy increases A1 [3] (ii) either molecules/atoms/particles move faster/ <c2> is increasing or kinetic energy increases with temperature (increases) B1 no change in potential energy, kinetic energy increases M1 internal energy increases A1 [3]

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4 A charged point mass is situated in a vacuum. A proton travels directly towards the mass, as For illustrated in Fig. 4.1. Examiner’s Use charged proton point mass r Fig. 4.1 When the separation of the mass and the proton is r, the electric potential energy of the system is UP . The variation with r of the potential energy UP is shown in Fig. 4.2. r / cm 0 2 4 6 8 10 0 –10 UP / 10–26 J –20 –30 –40 –50 Fig. 4.2 (a) (i) Use Fig. 4.2 to state and explain whether the mass is charged positively or For negatively. Examiner’s Use .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) The gradient at a point on the graph of Fig. 4.2 is G. Show that the electric field strength E at this point due to the charged point mass is given by the expression Eq = G where q is the charge at this point. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) Use the expression in (a)(ii) and Fig. 4.2 to determine the electric field strength at a distance of 4.0 cm from the charged point mass. field strength = ........................................ V m–1 [4]

Mark scheme: 4 (a) (i) as r decreases, energy decreases/work got out (due to) M1 attraction so point mass is negatively charged A1 [2] (ii) electric potential energy = charge × electric potential B1 electric field strength is potential gradient B1 field strength = gradient of potential energy graph/charge A0 [2] (b) tangent drawn at (4.0, 14.5) B1 gradient = 3.6 × 10–24 A2 (for < ±0.3 allow 2 marks, for < ±0.6 allow 1 mark) field strength = (3.6 × 10–24) / (1.6 × 10–19) = 2.3 × 10–5 V m–1 (allow ecf from gradient value) A1 [4] (one point solution for gradient leading to 2.3 × 10–5 Vm–1 scores 1 mark only) GCE AS/A LEVEL – May/June 2012 9702 42

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5 (a) Define the tesla. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3] (b) A horseshoe magnet is placed on a balance. A stiff metal wire is clamped horizontally between the poles, as illustrated in Fig. 5.1. horseshoe magnet stiff metal wire balance pan Fig. 5.1 The magnetic flux density in the space between the poles of the magnet is uniform and is zero outside this region. The length of the metal wire normal to the magnetic field is 6.4 cm. When a current in the wire is switched on, the reading on the balance increases by 2.4 g. The current in the wire is 5.6 A. (i) State and explain the direction of the force on the wire due to the current. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[3] (ii) Calculate the magnitude of the magnetic flux density between the poles of the For magnet. Examiner’s Use flux density = ...............................................T [2] (c) A low frequency alternating current is now passed through the wire in (b). The root-mean-square (r.m.s.) value of the current is 5.6 A. Describe quantitatively the variation of the reading seen on the balance. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2]

Mark scheme: 5 (a) (long) straight conductor carrying current of 1 A M1 current/wire normal to magnetic field M1 (for flux density 1 T,) force per unit length is 1 N m–1 A1 [3] (b) (i) (originally) downward force on magnet (due to current) B1 by Newton’s third law (allow “N3”) M1 upward force on wire A1 [3] (ii) F = BIL 2.4 × 10–3 × 9.8 = B × 5.6 × 6.4 × 10–2 C1 B = 0.066 T (need 2 SF) A1 [2] (g missing scores 0/2, but g = 10 leading to 0.067T scores 1/2) (c) new reading is 2.4√2 g C1 either changes between +3.4 g and –3.4 g or total change is 6.8 g A1 [2]

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6 (a) Describe the main principles of the determination of the charge on an oil drop by For Millikan’s experiment. You may draw a diagram if you wish. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[7] (b) In an experiment to determine the fundamental charge, values of charge on oil drops were found by a student to be as shown below. 3.2 × 10–19 C; 6.4 × 10–19 C; 16 × 10–19 C; 9.7 × 10–19 C; 12.8 × 10–19 C; 3.1 × 10–19 C; 6.3 × 10–19 C. State the value, to two significant figures, of the fundamental charge that is suggested by these values of charge on oil drops. fundamental charge = .............................................. C [1]

Mark scheme: 6 (a) oil drop charged by friction/beta source B1 between parallel metal plates B1 plates are horizontal (1) adjustable potential difference/field between plates B1 until oil drop is stationary B1 mg = q × V/d B1 symbols explained (1) oil drop viewed through microscope (1) m determined from terminal speed of drop (when p.d. is zero) (1) (any two extras, 1 each) B2 [7] (b) 3.2 × 10–19 C A1 [1]

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7 The photoelectric effect may be represented by the equation For Examiner’s photon energy = work function energy + maximum kinetic energy of electron. Use (a) State what is meant by work function energy. .......................................................................................................................................... ......................................................................................................................................[1] (b) The variation with frequency f of the maximum kinetic energy EK of photoelectrons emitted from the surface of sodium metal is shown in Fig. 7.1. 0.8 0.6 EK / eV 0.4 0.2 0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 f / 1014 Hz Fig. 7.1 Use the gradient of the graph of Fig. 7.1 to determine a value for the Planck constant h. Show your working. h = .............................................J s [2] (c) The sodium metal in (b) has a work function energy of 2.4 eV. The sodium is replaced by For calcium which has a work function energy of 2.9 eV. Examiner’s Use On Fig. 7.1, draw a line to show the variation with frequency f of the maximum kinetic energy EK of photoelectrons emitted from the surface of calcium. [3]

Mark scheme: 7 (a) minimum energy to remove an electron from the metal/surface B1 [1] (b) gradient = 4.17 × 10–15 (allow 4.1 → 4.3) C1 h = 4.15 × 10–15 × 1.6 × 10–19 or h = 4.1 to 4.3 × 10–15 eV s A1 = 6.6 × 10–34 J s A0 [2] (c) graph: straight line parallel to given line with intercept at any higher frequency B1 intercept at between 6.9 × 1014 Hz and 7.1 × 1014 Hz B1 [3] GCE AS/A LEVEL – May/June 2012 9702 42

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8 The element strontium has at least 16 isotopes. One of these isotopes is strontium-89. This For isotope has a half-life of 52 days. Examiner’s Use (a) State what is meant by isotopes. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) Calculate the probability per second of decay of a nucleus of strontium-89. probability = ............................................ s–1 [3] (c) A laboratory prepares a strontium-89 source. The activity of this source is measured 21 days after preparation of the source and is found to be 7.4 × 106 Bq. Determine, for the strontium-89 source at the time that it was prepared, (i) the activity, activity = ............................................ Bq [2] (ii) the mass of strontium-89. mass = ...............................................g [2]

Mark scheme: 8 (a) nuclei having same number of protons/proton (atomic) number B1 different numbers of neutrons/neutron number B1 [2] (allow second mark for nucleons/nucleon number/mass number/atomic mass if made clear that same number of protons/proton number) (b) probability of decay per unit time is the decay constant C1 λ = ln 2 / t½ = 0.693 / (52 × 24 × 3600) C1 = 1.54 × 10–7 s–1 A1 [3] (c) (i) A = A0 exp(–λt) 7.4 × 106 = A0 exp(–1.54 × 10–7 × 21 × 24 × 3600) C1 A0 = 9.8 × 106 Bq A1 [2] (alternative method uses 21 days as 0.404 half-lives) (ii) A = λN and mass = N × 89 / NA C1 mass = (9.8 × 106 × 89) / (1.54 × 10–7 × 6.02 × 1023) = 9.4 × 10–9 g A1 [2] GCE AS/A LEVEL – May/June 2012 9702 42 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 four properties of an ideal operational amplifier. 1. ...................................................................................................................................... 2. ...................................................................................................................................... 3. ...................................................................................................................................... 4. ...................................................................................................................................... [4] (b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1. +5 V – + VIN –5 V VOUT Fig. 9.1 The variation with time t of the input potential VIN is shown in Fig. 9.2. 6 potential / V 4 2 VINVIN 0 t –2 –4 –6 Fig. 9.2 On the axes of Fig. 9.2, draw a graph to show the variation with time t of the output potential VOUT . [3] (c) The output potential VOUT is to be displayed using two light-emitting diodes (LEDs). A For Examiner’s diode emitting red light is to indicate when VOUT is positive and a diode emitting green light is to be used to indicate when VOUT is negative. Use Complete Fig. 9.3 to show the connections of the two LEDs to the output of the op-amp. Label each LED with the colour of light that it emits. VOUT Fig. 9.3 [3]

Mark scheme: 9 (a) e.g. infinite input impedance/resistance zero output impedance/resistance infinite (open loop) gain infinite bandwidth infinite slew rate (any four, one mark each) B4 [4] (b) graph: square wave M1 180° phase change A1 amplitude 5.0 V A1 [3] (c) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 diodes identified correctly A1 [3] (special case: if diode symbol, not LED symbol, allow 2nd and 3rd marks to be scored)

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10 (a) An aluminium block is placed near to a small source of X-ray radiation, as shown in For Fig. 10.1. Examiner’s Use aluminium block X-ray A B source Fig. 10.1 X-rays from the source are detected at point A and at point B. State two reasons why the intensity of the X-ray beam at point B is not as great as the intensity at point A. 1. ...................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... [2] (b) A cross-section through a model of a finger is shown in Fig. 10.2. 2.4 cm 1.1 cm bone C D A B soft tissue Fig. 10.2 The thickness of the model is 2.4 cm and that of the bone in the model is 1.1 cm. Parallel beams of X-rays are incident on the model in the directions AB and CD, as shown in Fig. 10.2. Data for the linear attenuation (absorption) coefficient μ for the bone and the soft tissue For in the model are given in Fig. 10.3. Examiner’s Use μ / cm–1 bone 3.00 soft tissue 0.27 Fig. 10.3 Calculate the ratio intensity of X-ray beam incident on the model intensity of X-ray beam emergent from the model for (i) the beam AB, ratio = ..................................................[2] (ii) the beam CD. ratio = ..................................................[2] (c) Use your answers in (b) to suggest why, for this model, an X-ray image with good contrast may be obtained. .......................................................................................................................................... ......................................................................................................................................[1]

Mark scheme: 10 (a) e.g. beam is divergent/obeys inverse square law absorption (in block) scattering (of beam in block) reflection (at boundaries) (any two sensible suggestions, 1 each) B2 [2] (b) (i) I = I0 exp(–µx) C1 I0/I = exp(0.27 × 2.4) = 1.9 A1 [2] (ii) I0/I = exp(0.27 × 1.3) × exp(3.0 × 1.1) C1 = 1.42 × 27.1 = 38.5 A1 [2] (c) either much greater absorption in bone than in soft tissue or Io / I much greater for bone than soft tissue B1 [1]

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11 A signal that is transmitted over a long distance will be attenuated and it will pick up noise. For Examiner’s (a) State what is meant by Use (i) attenuation, .................................................................................................................................. ..............................................................................................................................[1] (ii) noise. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) Explain why regenerator amplifiers do not amplify the noise that has been picked up on digital signals. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (c) A transmitter on Earth produces a signal of power 2.4 kW. This signal, when received by a satellite, is attenuated by 195 dB. Calculate the signal power received by the satellite. power = ............................................. W [3]

Mark scheme: 11 (a) (i) loss of (signal) power B1 [1] (ii) unwanted power (on signal) M1 that is random A1 [2] (b) for digital, only the ‘high’ and the ‘low’ / 1 and 0 are necessary M1 variation between ‘highs’ and ‘lows’ caused by noise not required A1 [2] (c) attenuation = 10 lg(P2 / P1) C1 either 195 = 10 lg({2.4 × 103} / P) or –195 = 10 lg(P / 2.4 × 103) C1 P = 7.6 × 10–17 W A1 [3] GCE AS/A LEVEL – May/June 2012 9702 42

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12 An incomplete simplified block diagram of the circuitry for a mobile-phone handset is shown For in Fig. 12.1. Examiner’s Use aerial switch tuning circuit amplifier amplifier X oscillator demodulator parallel- Y to-serial converter ADC DAC amplifier amplifier microphone loudspeaker Fig. 12.1 (a) State the name of the block labelled (i) X, ..............................................................................................................................[1] (ii) Y. ..............................................................................................................................[1] Question 12 continues on page 24. (b) Explain the purpose of For Examiner’s (i) the switch, Use .................................................................................................................................. ..............................................................................................................................[1] (ii) the parallel-to-serial converter. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

Mark scheme: 12 (a) (i) modulator B1 [1] (ii) serial-to-parallel converter (accept series-to-parallel converter) B1 [1] (b) (i) enables one aerial to be used for transmission and receipt of signals A1 [1] (ii) all bits for one number arrive at one time B1 bits are sent out one after another B1 [2]

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