Cambridge A Level Physics 9702 — 2017 May/June Paper 4 · Variant 2

1 (a) Define gravitational field strength. ................................................................................................................................................... ...............................................................................................................................................[1] (b) The mass of a spherical comet of radius 3.6 km is approximately 1.0 × 1013 kg. (i) Assuming that the comet has constant density, calculate the gravitational field strength on the surface of the comet. field strength = ............................................... N kg–1 [2] (ii) A probe having a weight of 960 N on Earth lands on the comet. Using your answer in (i), determine the weight of the probe on the surface of the comet. weight = ...................................................... N [2] (c) A second comet has a length of approximately 4.5 km and a width of approximately 2.6 km. Its outline is illustrated in Fig. 1.1. Fig. 1.1 Suggest one similarity and one difference between the gravitational fields at the surface of this comet and at the surface of the comet in (b). similarity: ................................................................................................................................... ................................................................................................................................................... difference: ................................................................................................................................. ................................................................................................................................................... [2] [Total: 7]

Mark scheme: 1(a) force per unit mass B1 1(b)(i) g = GM / r 2 = (6.67 ×10–11 × 1.0 × 1013) / (3.6 × 103)2 C1 = 5.1 × 10–5 N kg–1 A1 1(b)(ii) mass = (960 / 9.81) kg weight on comet = (960 / 9.81) × 5.1 ×10–5 C1 = 5.0 × 10–3 N A1 1(c) similarity: e.g. both attractive/pointed towards the comet e.g. same order of magnitude B1 difference: e.g. radial/non-radial e.g. same (over surface)/varies (over surface) B1

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2 (a) The pressure p and volume V of an ideal gas are related to the density ρ of the gas by the expression 1 p = ρ 〈c 2〉. 3 (i) State what is meant by the symbol 〈c 2〉. ........................................................................................................................................... .......................................................................................................................................[1] (ii) Use the expression to show that the mean kinetic energy EK of a gas molecule is given by 3 EK = 2 kT where k is the Boltzmann constant and T is the thermodynamic temperature. [3] (b) (i) An ideal gas containing 1.0 mol of molecules is heated at constant volume. Use the expression in (a)(ii) to show that the thermal energy required to raise the 3 temperature of the gas by 1.0 K has a value of R, where R is the molar gas constant. 2 [3] (ii) Nitrogen may be assumed to be an ideal gas. The molar mass of nitrogen gas is 28 g mol–1. Use the answer in (b)(i) to calculate a value for the specific heat capacity, in J kg–1 K–1, at constant volume for nitrogen. specific heat capacity = .......................................... J kg–1 K–1 [2] [Total: 9]

Mark scheme: 2(a)(i) mean/average square speed/velocity B1 2(a)(ii) pV = NkT or pV = nRT B1 ρ = Nm / V or ρ = nNAm / V and k = nR / N B1 EK = ½ m〈c2〉 with algebra to (3 / 2)kT B1 2(b)(i) no (external) work done or ∆U = q or w = 0 B1 q = NA × (3 / 2)k × 1.0 M1 NAk = R so q = (3 / 2)R A1 2(b)(ii) specific heat capacity = {(3 / 2) × R} / 0.028 C1 = 450 J kg–1 K–1 A1

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3 A bar magnet of mass 250 g is suspended from the free end of a spring, as illustrated in Fig. 3.1. spring magnet coil Fig. 3.1 The magnet hangs so that one pole is near the centre of a coil of wire. The coil is connected in series with a resistor and a switch. The switch is open. The magnet is displaced vertically and then allowed to oscillate with one pole remaining inside the coil. The other pole remains outside the coil. At time t = 0, the magnet is oscillating freely as it passes through its equilibrium position. At time t = 6.0 s, the switch in the circuit is closed. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 3.2. 2.0 1.5 y / cm 1.0 0.5 0 0 2 4 6 8 10 12 14 16 t / s –0.5 –1.0 –1.5 –2.0 Fig. 3.2 (a) For the oscillating magnet, use data from Fig. 3.2 to calculate, to two significant figures, (i) the frequency f, f = .................................................... Hz [2] (ii) the energy of the oscillations during the time t = 0 to time t = 6.0 s. energy = ....................................................... J [3] (b) (i) State Faraday’s law of electromagnetic induction. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Use Faraday’s law and energy conservation to explain why the amplitude of the oscillations of the magnet reduces after time t = 6.0 s. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] [Total: 10]

Mark scheme: 3(a)(i) e.g. period = 6 / 2.5 C1 frequency = 0.42 Hz A1 3(a)(ii) energy = ½ m × 4π2f 2y0 2 C1 = ½ × 0.25 × 4π2 × 0.422 × (1.5 × 10–2)2 C1 = 2.0 × 10–4 J A1 3(b)(i) (induced) e.m.f. proportional to rate of M1 change of magnetic flux (linkage) or cutting of magnetic flux A1 3(b)(ii) coil cuts flux/field (of moving magnet) inducing e.m.f. in coil B1 (induced) current in resistor causes heating (effect) M1 thermal energy/heat derived from energy of oscillations (of magnet) A1

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4 (a) Explain the main principles behind the use of ultrasound to obtain diagnostic information about internal body structures. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[6] (b) A parallel beam of ultrasound has intensity I0 as it enters a muscle of thickness 4.6 cm, as illustrated in Fig. 4.1. 4.6 cm muscle beam of I0 IT ultrasound Fig. 4.1 The intensity of the beam just before it leaves the muscle is IT. The ratio I0 / IT is found to be 2.9. Calculate the linear attenuation (absorption) coefficient μ of the ultrasound in the layer of muscle. μ = ................................................. cm–1 [3] [Total: 9]

Mark scheme: 4(a) pulse (of ultrasound) B1 * produced by quartz crystal/piezo-electric crystal * gel/coupling medium (on skin) used to reduce reflection at skin reflected from boundaries (between media) B1 reflected pulse/wave detected by (ultrasound) transmitter B1 reflected wave processed and displayed B1 * intensity of reflected pulse/wave gives information about boundary * time delay gives information about depth of boundary max. 2 of additional detail points marked * B2 4(b) IT = I0 exp (–µx) C1 2.9 = exp (4.6µ) C1 µ = 0.23 cm–1 A1

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5 (a) State two advantages of the transmission of data in digital form rather than in analogue form. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) An analogue signal SI is converted into a digital signal D using an analogue-to-digital converter (ADC). After transmission of the digital signal, it is converted back to an analogue signal ST using a digital-to-analogue converter (DAC), as illustrated in Fig. 5.1. digital signal analogue signal analogue signal ADC DAC SI D ST Fig. 5.1 (i) Outline the process by which the ADC converts the analogue signal SI into the digital signal D. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) The ADC and the DAC operate with the same sampling rate and the same number of bits in each digital number. State the effect on the transmitted analogue signal ST when, for the ADC and the DAC, 1. the sampling rate is increased, .................................................................................................................................... .................................................................................................................................... 2. the number of bits in each digital number is increased. .................................................................................................................................... .................................................................................................................................... [2] [Total: 6]

Mark scheme: 5(a) any two reasonable suggestions e.g. • signal can be regenerated/noise removed (not “no noise”) • circuits more reliable • circuits cheaper to produce • multiplexing (is possible) • error correction/checking • easier encryption/better security B2 5(b)(i) samples the analogue signal M1 at regular intervals and converts it (to a digital number) A1 5(b)(ii) 1. smaller step depth B1 2. smaller step height B1

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6 (a) State Coulomb’s law. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Two charged metal spheres A and B are situated in a vacuum, as illustrated in Fig. 6.1. 6.0 cm sphere A sphere B P x Fig. 6.1 The shortest distance between the surfaces of the spheres is 6.0 cm. A movable point P lies along the line joining the centres of the two spheres, a distance x from the surface of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 6.2. 10 E / 103 V m–1 5 0 0 1 2 3 4 5 6 x / cm –5 –10 –15 Fig. 6.2 (i) Use Fig. 6.2 to explain whether the two spheres have charges of the same, or opposite, sign. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) A proton is at point P where x = 5.0 cm. Use data from Fig. 6.2 to determine the acceleration of the proton. acceleration = ................................................. m s–2 [3] (c) Use data from Fig. 6.2 to state the value of x at which the rate of change of electric potential is maximum. Give the reason for the value you have chosen. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 9]

Mark scheme: 6(a) force proportional to product of charges and inversely proportional to the square of the separation M1 reference to point charges A1 6(b)(i) (near to each sphere,) fields are in opposite directions or point (between spheres) where fields are equal and opposite or point (between spheres) where field strength is zero M1 so same (sign of charge) A1 6(b)(ii) (at x = 5.0 cm,) E = 3.0 × 103 V m–1 and a = qE / m C1 E = (1.60 × 10–19 × 3.0 × 103) / (1.67 × 10–27) C1 = 2.9 × 1011 m s–2 A1 6(c) field strength or E is potential gradient or field strength is rate of change of (electric) potential M1 (field strength) maximum at x = 6 cm A1

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7 A capacitor consists of two parallel metal plates, separated by an insulator, as shown in Fig. 7.1. insulator metal plates Fig. 7.1 (a) Suggest why, when the capacitor is connected across the terminals of a battery, the capacitor stores energy, not charge. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Define the capacitance of the capacitor. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) The capacitor is charged so that the potential difference between its plates is V0. The capacitor is then connected across a resistor for a short time. It is then disconnected. 1 The energy stored in the capacitor is reduced to of its initial value. 16 Determine, in terms of V0, the potential difference across the capacitor. potential difference = ...........................................................[2] [Total: 6]

Mark scheme: 7(a) equal and opposite charges on the plates so no resultant charge B1 +ve and –ve charges separated so energy stored B1 7(b) charge / potential difference M1 reference to charge on one plate and p.d. between plates A1 7(c) energy = ½ CV2 or energy = ½ QV and C = Q / V C1 (1 / 16) × ½ CV0 2 = ½ CV2 V = ¼ V0 A1

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8 A student designs a circuit incorporating an operational amplifier (op-amp) as shown in Fig. 8.1. +6 V component C X R +5 V – + Y R –5 V B G RV 0 V Fig. 8.1 (a) (i) On Fig. 8.1, draw a circle around the output device. [1] (ii) State the purpose of this circuit. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) The resistors X and Y each have resistance R. When conducting, the LED labelled B emits blue light and the LED labelled G emits green light. (i) State whether blue light or green light is emitted when the resistance of component C is greater than the resistance RV of the variable resistor. Explain your answer. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (ii) State and explain what is observed as the resistance of component C is reduced. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (c) Suggest the function of the variable resistor. ................................................................................................................................................... ...............................................................................................................................................[1] [Total: 10]

Mark scheme: 8(a)(i) circle around both diodes B1 8(a)(ii) indicates (whether) temperature M1 (is) above or below a set value A1 8(b)(i) (when resistance of C > RV,) V– > V+ or V+ < 3 V or p.d. across RV < p.d. across R/Y/3 V or p.d. across C > p.d. across R/ X/3 V M1 op-amp output is negative M1 (only) green A1 8(b)(ii) resistance of C becomes less than RV or V– < V+ B1 green (LED) goes out A1 blue (LED) comes on A1 8(c) changes/determines temperature at which LEDs switch B1

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9 A Hall probe is placed near to one end of a current-carrying solenoid, as shown in Fig. 9.1. X solenoid Hall probe Y Fig. 9.1 The probe is rotated about the axis XY and is then held in a position so that the Hall voltage is maximum. (a) Explain why (i) a Hall probe is made from a thin slice of material, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) in order for consistent measurements of magnetic flux density to be made, the current in the probe must be constant. ........................................................................................................................................... .......................................................................................................................................[1] (b) The probe is now rotated through an angle of 360° about the axis XY. At angle θ = 0, the Hall voltage VH has maximum value VMAX. On Fig. 9.2, sketch the variation with angle θ of the Hall voltage VH for one complete revolution of the probe about axis XY. + V MAX V H 0 0 90 180 270 360 θ/ ° – Fig. 9.2 [3] [Total: 6]

Mark scheme: 9(a)(i) Hall voltage depends on thickness of slice C1 thinner slice, larger Hall voltage A1 9(a)(ii) Hall voltage depends on current in slice B1 9(b) sinusoidal wave, one cycle B1 at θ = 0 and at θ = 360°, VH = VMAX B1 at θ = 180°, VH = –VMAX B1

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10 (a) Briefly describe two phenomena associated with the photoelectric effect that cannot be explained using a wave theory of light. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) The maximum energy EMAX of electrons emitted from a metal surface when illuminated by light of wavelength λ is given by the expression 1 1 EMAX = λ – hc ( λ0) where h is the Planck constant and c is the speed of light. (i) Identify the symbol λ0. .......................................................................................................................................[1] 1 (ii) The variation with of EMAX for the metal surface is shown in Fig. 10.1. λ 4 EMAX / 10–19 J 3 2 1 0 1.5 2.0 2.5 3.0 3.5 4.0 1 / 106 m–1 λ Fig. 10.1 1. Use Fig. 10.1 to determine the magnitude of λ0. λ0 = ...................................................... m [1] 2. Use the gradient of Fig. 10.1 to determine a value for the Planck constant h. h = ..................................................... J s [3] (c) The metal surface in (b) becomes oxidised. Photoelectric emission is still observed but the work function energy is increased. 1 On Fig. 10.1, draw a line to show the variation with of EMAX for the oxidised surface. [2] λ [Total: 9]

Mark scheme: 10(a) two from: • frequency below which electrons not ejected • maximum energy of electron depends on frequency • maximum energy of electrons does not depend on intensity • instantaneous emission of electrons B2 10(b)(i) (λ0 is the) threshold wavelength or wavelength corresponding to threshold frequency or maximum wavelength for emission of electrons B1 10(b)(ii)1. intercept = 1 / λ0 = 2.2 × 106m–1 λ0 = 4.5 × 10–7 m or 450 nm A1 10(b)(ii)2. gradient = hc C1 gradient = 2.0 × 10–25 or correct substitution into gradient formula C1 h = (2.0 × 10–25) / (3.0 × 108) = 6.7 × 10–34 J s A1 10(c) line: same gradient B1 straight line, positive gradient, intercept at greater than 2.2 × 106 when candidate’s line extrapolated B1

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11 An electron has charge –q and mass m. It is accelerated from rest in a vacuum through a potential difference V. (a) Show that the momentum p of the accelerated electron is given by p = (2 mqV ) . [2] (b) The potential difference V through which the electron is accelerated is 120 V. (i) State what is meant by the de Broglie wavelength. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Calculate the de Broglie wavelength of the electron. wavelength = ...................................................... m [3] (c) The separation of copper atoms in a copper crystal is approximately 2 × 10–10 m. By reference to your answer in (b)(ii), suggest whether electron diffraction could be observed using a beam of electrons that have been accelerated through a potential difference of 120 V and are then incident on a thin copper crystal. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 9]

Mark scheme: 11(a) loss of (electric) potential energy = gain in kinetic energy or qV = ½ mv2 or EK = p2 / 2m = qV B1 p = mv with algebra leading to p = √(2mqV) B1 11(b)(i) particle/electron has a wavelength (associated with it) B1 dependent on its momentum or when/because particle is moving B1 11(b)(ii) p = (2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 C1 λ = (6.63 × 10–34) / (5.91 × 10–24) C1 = 1.12 × 10–10 m A1 11(c) wavelength is similar to separation of atoms M1 so diffraction observed A1

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12 One nuclear reaction that can take place in a nuclear reactor may be represented, in part, by the equation 23592 U + 10 n 9542 Mo + 13957 La + 210 n + …………. + energy Data for a nucleus and some particles are given in Fig. 12.1. nucleus or particle mass / u 13957 La 138.955 10 n 1.00863 11 p 1.00728 –1 e0 5.49 × 10–4 Fig. 12.1 (a) Complete the nuclear reaction shown above. [1] (b) (i) Show that the energy equivalent to 1.00 u is 934 MeV. [3] (ii) Calculate the binding energy per nucleon, in MeV, of lanthanum-139 (13957 La). binding energy per nucleon = ................................................ MeV [3] Question 12 continues on the next page. (c) State and explain whether the binding energy per nucleon of uranium-235 (23592 U) will be greater, equal to or less than your answer in (b)(ii). ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 10]

Mark scheme: 12(a) 7 e 0 1 − A1 12(b)(i) E = mc2 C1 = 1.66 × 10–27 × (3.00 × 108)2 M1 = 1.494 × 10–10 J division by 1.60 × 10–13 clear to give 934 MeV A1 12(b)(ii) ∆m = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u = (–) 1.16762 (u) C1 energy = 1.16762 × 934 C1 energy per nucleon = (1.16762 × 934) / 139 = 7.85 MeV A1 12(c) above A = 56, binding energy per nucleon decreases as A increases B1 U-235 has larger nucleon number M1 so less (binding energy per nucleon) A1 or fission takes place with uranium (B1) fission reaction releases energy (M1) binding energy per nucleon less (for uranium than for products) (A1)

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