Cambridge A Level Physics 9702 — 2014 May/June Paper 4 · Variant 2

1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet. (a) A stone, travelling at speed v, is in a circular orbit of radius r about the planet, as illustrated in Fig. 1.1. stone v planet r Fig. 1.1 Show that the speed v is given by the expression GM v = r where G is the gravitational constant. Explain your working. [2] (b) A second stone, initially at rest at infinity, travels towards the planet, as illustrated in Fig. 1.2. stone V0 planet x Fig. 1.2 (not to scale) The stone does not hit the surface of the planet. (i) Determine, in terms of the gravitational constant G and the mass M of the planet, the speed V0 of the stone at a distance x from the centre of the planet. Explain your working. You may assume that the gravitational attraction on the stone is due only to the planet. [3] (ii) Use your answer in (i) and the expression in (a) to explain whether this stone could enter a circular orbit about the planet. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

Mark scheme: 1 (a) gravitational force provides/is the centripetal force B1 GMm / r2 R mv2 / r M1 v R √(GM / r) A0 [2] allow gravitational field strength provides/is the centripetal acceleration (B1) GM / r2 R v2 / r (M1) (b) (i) kinetic energy increase/change R loss / change in (gravitational) potential energy B1 ½mV02 R GMm / x C1 V02 R 2GM / x V0 R √(2GM / x) A1 [3] (max. 2 for use of r not x) (ii) V0 is (always) greater than v (for x = r) M1 so stone could not enter into orbit A1 [2] (expressions in (a) and (b)(i) must be dimensionally correct)

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2 A constant mass of an ideal gas has a volume of 3.49 × 103 cm3 at a temperature of 21.0 °C. When the gas is heated, 565 J of thermal energy causes it to expand to a volume of 3.87 × 103 cm3 at 53.0 °C. This is illustrated in Fig. 2.1. 3.49 × 103 cm3 3.87 × 103 cm3 565 J 21.0 °C 53.0 °C Fig. 2.1 (a) Show that the initial and final pressures of the gas are equal. [2] (b) The pressure of the gas is 4.20 × 105 Pa. For this heating of the gas, (i) calculate the work done by the gas, work done = ..................................................... J [2] (ii) use the first law of thermodynamics and your answer in (i) to determine the change in internal energy of the gas. change in internal energy = ..................................................... J [2] (c) Explain why the change in kinetic energy of the molecules of this ideal gas is equal to the change in internal energy. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3]

Mark scheme: 2 (a) use of kelvin temperatures B1 both values of (V / T) correct (11.87), V / T is constant so pressure is constant M1 [2] (allow use of n R1. Do not allow other values of n.) (b) (i) work done R p∆V R 4.2 × 105 × (3.87 – 3.49) × 103 × 10–6 C1 = = = = = R 160 J A1 [2] (do not allow use of V instead of ∆V) (ii) increase / change in internal energy R heating of system N work done on system C1 R 565 – 160 = = = = R 405 J A1 [2] (c) internal energy R sum of kinetic energy and potential energy / EK N EP B1 no intermolecular forces M1 no potential energy (so ∆U R ∆EK) A1 [3]

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3 A microwave cooker uses electromagnetic waves of frequency 2450 MHz. The microwaves warm the food in the cooker by causing water molecules in the food to oscillate with a large amplitude at the frequency of the microwaves. (a) State the name given to this phenomenon. .............................................................................................................................................. [1] (b) The effective microwave power of the cooker is 750 W. The temperature of a mass of 280 g of water rises from 25 °C to 98 °C in a time of 2.0 minutes. Calculate a value for the specific heat capacity of the water. specific heat capacity = ....................................... J kg−1 K−1 [3] (c) The value of the specific heat capacity determined from the data in (b) is greater than the accepted value. A student gives as the reason for this difference: ‘heat lost to the surroundings’. Suggest, in more detail than that given by the student, a possible reason for the difference. ................................................................................................................................................... .............................................................................................................................................. [1]

Mark scheme: 3 (a) resonance B1 [1] (b) Pt R mc ∆θ C1 750 × 2 × 60 R 0.28 × c × (98 – 25) C1 c R 4400 J kg–1 K–1 A1 [3] (use of ∆θ R 73 N 273 max. 1 / 3) (use of t R 2 s not 120 s max. 2 / 3) GCE A LEVEL – May/June 2014 9702 42 (c) e.g. some microwave leakage from the cooker e.g. container for the water is also heated (any sensible suggestion) B1 [1] 2

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4 A helium nucleus contains two protons. In a model of the helium nucleus, each proton is considered to be a charged point mass. The separation of these point masses is assumed to be 2.0 × 10−15 m. (a) For the two protons in this model, calculate (i) the electrostatic force, electrostatic force = ..................................................... N [2] (ii) the gravitational force. gravitational force = ..................................................... N [2] (b) Using your answers in (a), suggest why (i) there must be some other force between the protons in the nucleus, ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) this additional force must have a short range. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

Mark scheme: 4 (a) (i) FE R Q1Q2 / 4πε0r2 C1 = = = = R 8.99 × 109 × (1.6 × 10–19)2 / (2.0 × 10–15)2 R 58 N A1 [2] (ii) FG R Gm1m2 / r2 C1 = = = R 6.67 × 10–11 × (1.67 × 10–27)2 / (2.0 × 10–15)2 = = = R 4.7 × 10–35 N A1 [2] (b) (i) force of repulsion (much) greater than force of attraction B1 must be some other force of attraction M1 to hold nucleus together A1 [3] (Do not allow if FG > FE in (a) or one of the forces not calculated in (a)) (ii) outside nucleus there is repulsion between protons B1 either attractive force must act only in nucleus or if not short range, all nuclei would stick together B1 [2]

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5 A Hall probe is placed a distance d from a long straight current-carrying wire, as illustrated in Fig. 5.1. current-carrying 4.0 A wire Hall probe X Y d Fig. 5.1 The direct current in the wire is 4.0 A. Line XY is normal to the wire. The Hall probe is rotated about the line XY to the position where the reading VH of the Hall probe is maximum. (a) The Hall probe is now moved away from the wire, along the line XY. On the axes of Fig. 5.2, sketch a graph to show the variation of the Hall voltage VH with distance x of the probe from the wire. Numerical values are not required on your sketch. VH 0 0 d x Fig. 5.2 [2] (b) The Hall probe is now returned to its original position, a distance d from the wire. At this point, the magnetic flux density due to the current in the wire is proportional to the current. For a direct current of 4.0 A in the wire, the reading of the Hall probe is 3.5 mV. The direct current is now replaced by an alternating current of root-mean-square (r.m.s.) value 4.0 A. The period of this alternating current is T. On the axes of Fig. 5.3, sketch the variation with time t of the reading of the Hall voltage VH for two cycles of the alternating current. Give numerical values for VH, where appropriate. 6 VH / mV 4 2 0 0 T 2T t –2 –4 –6 Fig. 5.3 [3] (c) A student suggests that the Hall probe in (a) is replaced with a small coil connected in series with a millivoltmeter. The constant current in the wire is 4.0 A. In order to obtain data to plot a graph showing the variation with distance x of the magnetic flux density, the student suggests that readings of the millivoltmeter are taken when the coil is held in position at different values of x. Comment on this suggestion. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

Mark scheme: 5 (a) only curve with decreasing gradient M1 acceptable value near xR 0 and does not reach zero A1 [2] (if graph line less than 4.0 cm do not allow A1 mark) (no credit if graph line has positive and negative values of VH) (b) graph: from 0 to 2T, two cycles of a sinusoidal wave M1 all peaks above 3.5 mV C1 peaks at 4.95 / 5.0 mV (allow 4.8 mV to 5.2 mV) A1 [3] (c) e.m.f. induced in coil when magnetic field / flux is changing / cutting B1 either at each position, magnetic field does not vary so no e.m.f. is induced in the coil / no reading on the millivoltmeter or at each position, switch off current and take millivoltmeter reading or at each position, rapidly remove coil from field and take meter reading B1 [2]

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6 (a) Explain the use of a uniform electric field and a uniform magnetic field for the selection of the velocity of a charged particle. You may draw a diagram if you wish. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Ions, all of the same isotope, are travelling in a vacuum with a speed of 9.6 × 104 m s−1. The ions are incident normally on a uniform magnetic field of flux density 640 mT. The ions follow semicircular paths A and B before reaching a detector, as shown in Fig. 6.1. detector A B vacuum uniform magnetic field, flux density 640 mT Fig. 6.1 Data for the diameters of the paths are shown in Fig. 6.2. path diameter / cm A 6.2 B 12.4 Fig. 6.2 The ions in path B each have charge +1.6 × 10−19 C. (i) Determine the mass, in u, of the ions in path B. mass = ..................................................... u [4] (ii) Suggest and explain quantitatively a reason for the difference in radii of the paths A and B of the ions. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3]

Mark scheme: 6 (a) electric and magnetic fields normal to each other B1 either charged particle enters region normal to both fields or correct B direction w.r.t. E for zero deflection B1 for no deflection, v R E / B B1 [3] (no credit if magnetic field region clearly not overlapping with electric field region) GCE A LEVEL – May/June 2014 9702 42 (b) (i) m R Bqr / v C1 = = = R=(640 × 10–3 × 1.6 × 10–19 × 6.2 × 10–2) / (9.6 × 104) C1 = = = R=6.61 × 10–26 kg C1 = = = R=(6.61 × 10–26) / (1.66 × 10–27) u = = = R=40 u A1 [4] (ii) q / m ∝ 1 / r or m constant and q ∝ 1 / r B1 q / m for A is twice that for B B1 ions in path A have (same mass but) twice the charge (of ions in path B) B1 [3]

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7 (a) Define the radian. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A telescope gives a clear view of a distant object when the angular displacement between the edges of the object is at least 9.7 × 10−6 rad. (i) The Moon is approximately 3.8 × 105 km from Earth. Estimate the minimum diameter of a circular crater on the Moon’s surface that can be seen using the telescope. diameter = .................................................. km [2] (ii) Suggest why craters of the same diameter as that calculated in (i) but on the surface of Mars are not visible using this telescope. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

Mark scheme: 7 (a) angle subtended at the centre of a circle B1 by an arc equal in length to the radius B1 [2] (b) (i) arc R distance × angle C1 diameter R 3.8 × 105 × 9.7 × 10–6 = = = R 3.7 km A1 [2] (ii) Mars is (much) further from Earth / away (answer must be comparative) B1 angle (at telescope is much) smaller B1 [2]

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8 Light of wavelength 590 nm is incident normally on a surface, as illustrated in Fig. 8.1. light surface wavelength 590 nm Fig. 8.1 The power of the light is 3.2 mW. The light is completely absorbed by the surface. (a) Calculate the number of photons incident on the surface in 1.0 s. number = ......................................................... [3] (b) Use your answer in (a) to determine (i) the total momentum of the photons arriving at the surface in 1.0 s, momentum = ........................................... kg m s−1 [3] (ii) the force exerted on the surface by the light. force = ..................................................... N [1]

Mark scheme: 8 (a) photon energy R hc / λ R (6.63 × 10–34 × 3.0 × 108) / (590 × 10–9) C1 R 3.37 × 10–19 J C1 number R (3.2 × 10–3) / (3.37 × 10–19) = = = R 9.5 × 1015 (allow 9.4 × 1015) A1 [3] (b) (i) p R h / λ C1 = = = R (6.63 × 10–34) / (590 × 10–9) = = = R 1.12 × 10–27 kg m s–1 C1 total momentum R 9.5 × 1015 × 1.12 × 10–27 = = = R 1.06 × 10–11 kg m s–1 A1 [3] (ii) force R 1.06 × 10–11 N A1 [1]

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9 Some water becomes contaminated with radioactive iodine-131 (13153I). The activity of the iodine-131 in 1.0 kg of this water is 460 Bq. The half-life of iodine-131 is 8.1 days. (a) Define radioactive half-life. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) (i) Calculate the number of iodine-131 atoms in 1.0 kg of this water. number = ......................................................... [3] (ii) An amount of 1.0 mol of water has a mass of 18 g. Calculate the ratio number of molecules of water in 1.0 kg of water . number of atoms of iodine-131 in 1.0 kg of contaminated water ratio = ......................................................... [2] (c) An acceptable limit for the activity of iodine-131 in water has been set as 170 Bq kg−1. Calculate the time, in days, for the activity of the contaminated water to be reduced to this acceptable level. time = ................................................ days [3]

Mark scheme: 9 (a) time for number of atoms / nuclei / activity (of the isotope) M1 to be reduced to one half (of its initial value) A1 [2] (b) (i) A R λN C1 460 R N × ln 2 / (8.1 × 24 × 60 × 60) C1 N R 4.6 × 108 A1 [3] (ii) number of water molecules in 1.0 kg R (6.02 × 1023) / (18 × 10–3) C1 R 3.3 × 1025 ratio R (3.3 × 1025) / (4.6 × 108) R 7.2 (7.3) × 1016 A1 [2] GCE A LEVEL – May/June 2014 9702 42 (c) A R A0 e–λt and λt½ R ln 2 C1 170 R 460 exp (–{ln 2 t } / 8.1) C1 t R 11.6 days (allow 2 s.f.) A1 [3] Section B

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10 (a) State the function of a comparator circuit incorporating an operational amplifier (op-amp). ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) An ideal op-amp is incorporated into the circuit of Fig. 10.1. +1.5 V 1.2 k1 +5 V – + G –5 V V IN 2.4 k1 R Fig. 10.1 (i) On Fig. 10.1, draw a circle around the part of the circuit that is being used as an output device. [1] (ii) Show that the potential at the non-inverting input of the op-amp is 1.0 V. [1] (iii) The variation with time t of the potential VIN at the inverting input of the op-amp is shown in Fig. 10.2. 6 potential 4 / V VIN 2 +1.0 0 t 1 t 2 time t –2 –4 –6 Fig. 10.2 1. On the axes of Fig. 10.2, draw the variation with time t of the output potential of the op-amp. [3] 2. State whether each diode is emitting light or is not emitting light at time t1 and at time t2. At time t1, diode R will ................................. and diode G will ................................. . At time t2, diode R will ................................. and diode G will ................................. . [2]

Mark scheme: 10 (a) compares the potentials/voltages at the (inverting and non-inverting) inputs B1 either output (potential) dependent on which input is the larger or V+ > V–, then VOUT is positive B1 states the other condition B1 [3] (b) (i) ring drawn around both the LEDs (and series resistors) B1 [1] (ii) V– R (1.5 × 2.4) / (1.2 N 2.4) R 1.0 V B1 [1] (allow 1.5 × 2.4 / 3.6 R 1.0 V) (iii) 1. VOUT switches at N1.0 V B1 maximum VOUT is 5.0 V B1 when curve is above N1.0 V, VOUT is negative (or v.v.) B1 [3] 2. at time t1, diode R is emitting light, diode G is not emitting B1 at time t2, diode R is not emitting, diode G is emitting B1 [2] (must be consistent with graph line. If no graph line then 0 / 2)

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11 (a) Distinguish between an X-ray image of a body structure and a CT scan. X-ray image: .............................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... CT scan: ................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... [5] (b) Data for the linear absorption (attenuation) coefficient μ of X-ray radiation of energy 80 keV are given in Fig. 11.1. metal μ / mm−1 aluminium 0.46 copper 0.69 Fig. 11.1 A parallel X-ray beam is incident on a copper filter, as shown in Fig. 11.2. copper filter incident beam emergent intensity I0 beam Fig. 11.2 The intensity of the incident beam is I0. (i) Calculate the thickness of copper required to reduce the intensity of the emergent beam to 0.25 I0. thickness = ................................................. mm [2] (ii) An aluminium filter of thickness 2.4 mm is now placed in the X-ray beam, together with the copper filter in (i). Calculate the fraction of the incident intensity that emerges after passing through the two filters. fraction = ......................................................... [2] (iii) Express your answer in (ii) as a gain in decibels (dB). gain = ................................................... dB [3]

Mark scheme: 11 (a) X-ray: flat / shadow / 2D image B1 regardless of depth of object / depth not indicated B1 CT scan: built up from (many) images at different angles B1 image is three-dimensional B1 image can be rotated / viewed at different angles B1 [5] (b) (i) I R I0 e–µx C1 0.25 R e–0.69x x R2.0 mm (allow 1 s.f.) A1 [2] (ii) for aluminium, I / I0 R e–0.46 × 2.4 R 0.33 C1 fraction R 0.33 × 0.25 = = = R 0.083 A1 [2] (iii) gain / dB R 10 lg(I / I0) C1 R 10 lg(0.083) = = = R (–) 10.8 dB (allow 2 s.f.) A1 with negative sign B1 [3]

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12 Two people, living in different regions of the Earth, communicate either using a link provided by a geostationary satellite or using optic fibres. (a) (i) Explain what is meant by a geostationary satellite. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) The uplink frequency for communication with the satellite is 6 GHz and the downlink has a frequency of 4 GHz. Explain why the frequencies are different. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) Comment on the time delays experienced by the two people when communicating either using geostationary satellites or using optic fibres. Explain your answer. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3]

Mark scheme: 12 (a) (i) satellite is in equatorial orbit B1 travelling from west to east B1 period of 24 hours / 1 day B1 [3] GCE A LEVEL – May/June 2014 9702 42 (ii) either uplink signal is highly attenuated or signal is highly amplified (before transmission) as downlink signal B1 prevents downlink signal swamping the uplink signal B1 [2] (b) speed of signal is same order of magnitude in both systems B1 optic fibre link (much) shorter than via satellite M1 time delay using optic fibre is less A1 [3]

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