Cambridge A Level Physics 9702 — 2016 Feb/March Paper 4 · Variant 2

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A satellite of mass m has a circular orbit of radius r about a planet of mass M. It may be assumed that the planet and the satellite are uniform spheres that are isolated in space. Show that the linear speed v of the satellite is given by the expression GM v = r where G is the gravitational constant. Explain your working. [2] (c) Two moons A and B have circular orbits about a planet, as illustrated in Fig. 1.1. vA B A vB rA rB planet Fig. 1.1 (not to scale) Moon A has an orbital radius rA of 1.3 × 108 m, linear speed vA and orbital period TA. Moon B has an orbital radius rB of 2.2 × 1010 m, linear speed vB and orbital period TB. (i) Determine the ratio vA 1. , vB ratio = ...........................................................[2] TA 2. . TB ratio = ...........................................................[3] (ii) The planet spins about its own axis with angular speed 1.7 × 10–4 rad s–1. Moon A is always above the same point on the planet’s surface. Determine the orbital period TB of moon B. TB = ........................................................s [2] [Total: 11]

Mark scheme: 1 (a) force proportional to product of the (two) masses and inversely proportional to the square of their separation M1 either reference to point masses or separation << ‘size’ of masses A1 [2] (b) gravitational force provides / is the centripetal force B1 GMm / r 2 = mv 2 / r or GMm / r 2 = mrω 2 and v = rω and algebra leading to v = (GM / r )1 / 2 B1 [2] (c) (i) 1. vA / vB = (rB / rA)1 / 2 = (2.2 × 1010 / 1.3 × 108)1 / 2 C1 = 13 (13.0) A1 [2] 2. v = 2πr / T or v ∝ r / T or vT / r = constant C1 TA / TB = (rA / rB) × (vB / vA) = (1.3 × 108 / 2.2 × 1010) × (1 / 13) C1 = 4.5 (4.54) × 10–4 A1 or T 2 = 4π2r 3 / GM or T 2 ∝ r 3 or T 2 / r 3 = constant (C1) TA / TB = (rA3 / rB3)1 / 2 = [(1.3 × 108)3 / (2.2 × 1010)3]1 / 2 (C1) = 4.5 (4.54) × 10–4 (A1) [3] (ii) T = 2π / 1.7 ×10–4 = 3.70 × 104 s C1 TB = 3.70 × 104 / 4.54 × 10–4 = 8.1 × 107 s A1 [2] If identifies TA as TB then 0 / 2

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2 (a) State (i) what is meant by internal energy, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) the basic assumption of the kinetic theory of gases that leads to the conclusion that there is zero potential energy between the molecules of an ideal gas. ........................................................................................................................................... .......................................................................................................................................[1] (b) The pressure p and volume V of an ideal gas are related by 1 pV = Nm 〈c2〉 3 where N is the number of molecules, m is the mass of a molecule and 〈c2〉 is the mean-square speed of the molecules. Use this equation to show that the mean kinetic energy 〈EK〉 of a molecule is given by 3 〈EK〉 = kT 2 where k is the Boltzmann constant and T is the thermodynamic temperature. [3] (c) A cylinder contains 17 g of oxygen gas at a temperature of 12 °C. The mass of 1.0 mol of oxygen gas is 32 g. It may be assumed that the oxygen behaves as an ideal gas. Calculate, for the oxygen gas in the cylinder, (i) the mean kinetic energy of a molecule, mean kinetic energy = ........................................................J [2] (ii) the number of molecules, number = ...........................................................[2] (iii) the total internal energy. internal energy = ........................................................J [1] [Total: 11]

Mark scheme: 2 (a) (i) sum of kinetic and potential energy of atoms/molecules M1 reference to random (distribution) A1 [2] (ii) no forces (of attraction or repulsion) between molecules B1 [1] (b) pV = NkT or pV = nRT and R = kNA , n = N / NA B1 1/3 Nm<c 2> = NkT or 1/3 m<c 2> = kT B1 <EK> = 1/2 m<c 2> so <EK> = 3/2 kT B1 [3] (c) (i) <EK> = 3/2 × 1.38 × 10–23 × (273 + 12) C1 = 5.9 (5.90) × 10–21 J A1 [2] (use of T = 12 K not T = 285 K scores 0 / 2) (ii) number = (17 / 32) × 6.02 × 1023 C1 = 3.2 (3.20) × 1023 A1 [2] (iii) internal energy = 5.9 × 10–21 × 3.2 × 1023 = 1900 (1890) J A1 [1]

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3 (a) Define specific heat capacity. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1. liquid out, tube temperature 25.5 °C heating liquid in, coil temperature 19.5 °C Fig. 3.1 Liquid enters the tube at a constant temperature of 19.5 °C and leaves the tube at a temperature of 25.5 °C. The mass of liquid flowing through the tube per unit time is m. Electrical power P is dissipated in the heating coil. The student changes m and adjusts P until the final temperature of the liquid leaving the tube is 25.5 °C. The data shown in Fig. 3.2 are obtained. m / g s–1 P/ W 1.11 33.3 1.58 44.9 Fig. 3.2 (i) Suggest why the student obtains data for two values of m, rather than for one value. ........................................................................................................................................... .......................................................................................................................................[1] (ii) Calculate the specific heat capacity of the liquid. Show your working. specific heat capacity = .......................................... J kg–1 K–1 [3] (c) When the heating coil in (b) dissipates 33.3 W of power, the potential difference V across the coil is given by the expression V = 27.0 sin (395t). The potential difference is measured in volts and the time t is measured in seconds. Determine the resistance of the coil. resistance = .......................................................Ω [3] [Total: 9]

Mark scheme: 3 (a) the (thermal) energy per unit mass to raise the temperature M1 of a substance by one degree A1 [2] (If ratio not clear for M1 mark, allow 1 / 2 marks for an otherwise correct answer) (b) (i) to allow for / determine / cancel heat transfer to / from tube / surroundings B1 [1] (do not allow ‘to stop / prevent’ heat loss) (ii) either P = mc∆θ ± h or 44.9 = 1.58 × 10–3 × c × (25.5 – 19.5) ± h or 33.3 = 1.11 × 10–3 × c × (25.5 – 19.5) ± h B1 (44.9 – 33.3) = (1.58 – 1.11) × 10–3 × c × (25.5 – 19.5) C1 c = 4100 (4110) J kg–1 K–1 A1 [3] (allow 1 / 3 for use of only 33.3 W, 1.11 g s–1 leading to 5000 J kg –1 K –1) (allow 1 / 3 for use of only 44.9 W, 1.58 g s–1 leading to 4740 J kg –1 K –1) (c) V0 = 27 or Vrms = 19.1 C1 33.3 = 272 / 2R or 33.3 = 19.12 / R C1 R = 11 Ω A1 [3]

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4 An object of mass 80 g oscillates with simple harmonic motion. The variation with time t of the displacement x of the object is shown in Fig. 4.1. 2.0 x / cm 1.0 0 00 0.1 0.2 0.3 0.4 0.5 t / s –1.0 –2.0 Fig. 4.1 (a) Use Fig. 4.1 to determine the amplitude and the period of the oscillations. amplitude = ..........................................................cm period = .............................................................s [1] (b) Use Fig. 4.1 and your answers in (a) to calculate the kinetic energy of the object at time t = 0.19 s. kinetic energy = ........................................................J [3] [Total: 4]

Mark scheme: 4 (a) amplitude = 1.8 cm and period = 0.30 s A1 [1] (b) EK = ½m ω 2 (x02 – x2) or EK = ½mv2 and v = ± ω √(x02 – x2) C1 = ½ × 0.080 × (2π / 0.30)2 × [(1.8 × 10–2)2 – (1.2 × 10–2)2] C1 = 3.2 × 10–3 J A1 [3]

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5 (a) A digital signal is produced by sampling an analogue signal and passing the samples through an analogue-to-digital converter (ADC). (i) State what is meant by a digital signal. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) State one change to the sampling or to the ADC that will improve the accuracy of reproduction of the original analogue signal. ........................................................................................................................................... .......................................................................................................................................[1] (b) The least significant bit of the four-bit digital number 1100 represents a signal voltage of 2.5 mV. Determine the signal voltage, in mV, represented by this digital number. voltage = .................................................... mV [1] [Total: 4]

Mark scheme: 5 (a) (i) (series of) ‘highs’ and ‘lows’ / ‘on’ and ‘off’ / 1’s and 0’s / two values M1 with no intermediate values / the values are discrete A1 [2] (ii) either use higher sampling frequency / rate or use more bits in each sample / each digital number or use more levels in each sample B1 [1] (b) voltage = 30 mV A1 [1]

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6 A parallel beam of ultrasound is incident normally on the surface of a layer of fat of thickness 1.1 cm, as shown in Fig. 6.1. fat muscle ultrasound I1 I2 beam I4 I3 1.1 cm Fig. 6.1 For the ultrasound, I1 is the intensity just after entering the surface of the fat layer, I2 is the intensity incident on the fat-muscle boundary, I3 is the intensity reflected from the fat-muscle boundary, I4 is the intensity received back at the surface of the fat layer. Some data for the fat are given in Fig. 6.2. specific acoustic impedance Z 1.4 × 106 kg m–2 s–1 density ρ 940 kg m–3 absorption (attenuation) coefficient μ 48 m–1 Fig. 6.2 (a) Calculate the time interval between a short pulse of ultrasound initially entering the layer of fat and then returning back to the surface of the fat layer. time = ........................................................s [3] I2(b) Calculate the ratio . I1 ratio = ...........................................................[2] (c) Intensity I4 is 0.33% of intensity I1. I3 Determine the ratio . I2 ratio = ...........................................................[2] (d) The specific acoustic impedance of the muscle is greater than that of the fat. I3 State the effect, if any, on the value of the ratio of an increase in the difference between the I2 specific acoustic impedance of the muscle and that of the fat. ................................................................................................................................................... ...............................................................................................................................................[1] [Total: 8]

Mark scheme: 6 (a) speed = Z / ρ = 1.4 × 106 / 940 (=1490) C1 time = (1.1 × 10–2 × 2) / 1490 C1 = 1.5 × 10–5 s A1 [3] (time of 7.4 × 10–6 s is one way only and scores 2 / 3 marks) (use of speed of light is wrong physics and scores 0 / 3 marks) (b) I = I0 exp (–µx) or I2 = I1 exp(–µx) C1 ratio = exp (– 48 × 1.1 × 10–2) = 0.59 A1 [2] (c) 0.33 / 100 = 0.59 × (I3 / I2) × 0.59 C1 ratio = 9.5 × 10–3 A1 or 0.33 / 100 = exp (– 48 × 2.2 × 10–2) × (I3 / I2) (C1) ratio = 9.5 × 10–3 (A1) [2] (d) ratio I3 / I2 increases B1 [1] (accept: “there is an increase in the proportion of the intensity that is reflected”)

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7 (a) Define capacitance. ................................................................................................................................................... ...............................................................................................................................................[1] (b) Three capacitors of capacitances C1, C2 and C3 are initially uncharged. They are then connected in series to a battery, as shown in Fig. 7.1. C1 C2 C3 V Fig. 7.1 The battery applies a potential difference V across the three capacitors. Show that the combined capacitance C of the capacitors is given by 1 1 1 1 = + + . C C1 C2 C3 [2] (c) A battery of e.m.f. 12 V and negligible internal resistance is connected to a network of two capacitors and a resistor, as shown in Fig. 7.2. 200 +F A B 12 V 600 +F Fig. 7.2 The capacitors have capacitances of 200 μF and 600 μF. The switch has two positions, A and B. (i) The switch is moved to position A. Calculate 1. the combined capacitance of the two capacitors, combined capacitance = ..................................................... μF [1] 2. the charge on the 600 μF capacitor, charge = .......................................................C [1] 3. the potential difference across the 600 μF capacitor. potential difference = ....................................................... V [1] (ii) The switch is now moved from position A to position B. Calculate the potential difference across the 600 μF capacitor when it has discharged 50% of its initial energy. potential difference = ....................................................... V [3] [Total: 9]

Mark scheme: 7 (a) (capacitance =) charge / potential (difference) B1 [1] (b) V = V1 + V2 + V3 B1 either Q / C = Q / C1 + Q / C2 + Q / C3 or V / Q = V1 / Q + V2 / Q + V3 / Q and so 1 / C = 1 / C1 + 1 / C2+ 1 / C3 B1 [2] (c) (i) 1. 1 / CT = (1 / 200) + (1 / 600) CT = 150 µF A1 [1] 2. Q = CV = 150 × 10–6 × 12 or 600 × 10–6 × 3.0 or 200 × 10–6 × 9.0 = 1.8 × 10–3 C A1 [1] 3. V = Q / C = 1.8 × 10–3 / 600 × 10–6 or V = [200 / (200 + 600)] × 12 = 3(.0) V A1 [1] (ii) energy = ½CV 2 or energy = ½QV and C = Q / V C1 1/2 × C × 32 = 2 × 1/2 × C × V 2 C1 V = 2.1 V A1 [3]

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8 (a) State two effects of negative feedback on the gain of an amplifier incorporating an operational amplifier (op-amp). 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) An incomplete diagram of an amplifier circuit incorporating an ideal operational amplifier is shown in Fig. 8.1. +8.0 V – + 7.2 k1 –8.0 V VIN VOUT Fig. 8.1 The amplifier has a voltage gain of +5.0. (i) Complete the circuit diagram of Fig. 8.1. [2] (ii) Calculate the resistance of any additional resistor you have drawn on Fig. 8.1. resistance = ..................................................... kΩ [2] (iii) Fig. 8.2 shows the variation with time of the input potential VIN. 15 potential / V 10 5 VIN 0 0 1 2 3 4 5 time / ms Fig. 8.2 On Fig. 8.2, draw the variation with time of the output potential VOUT. [2] [Total: 8]

Mark scheme: 8 (a) decreases gain B1 increases bandwidth / decreases distortion / increases (operating) stability B1 [2] (b) (i) additional resistor connected between 7.2 kΩ resistor and earth B1 V – joined to lower end of 7.2 kΩ resistor and V + joined to VIN B1 [2] (ii) either 5 = 1 + (7.2 / R) or 5 = 1 + (7200 / R) C1 R = 1.8 kΩ A1 [2] (iii) horizontal line from (0, 8.0) to (1.8, 8.0) B1 straight line from (1.8, 8.0) to (5.0, 0) B1 [2] (allow a tolerance of ± ½ small square when marking the graph)

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9 A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s–1 in a vacuum. The particle enters a uniform magnetic field of flux density 9.7 × 10–2 T, as shown in Fig. 9.1. uniform magnetic field out of the plane of particle the paper charge +q flux density 9.7 × 10–2 T mass m speed 1.6 × 105 m s–1 path of particle Fig. 9.1 The magnetic field direction is perpendicular to the initial velocity of the particle and perpendicular to, and out of, the plane of the paper. A uniform electric field is applied in the same region as the magnetic field so that the particle passes undeviated through the fields. (a) State and explain the direction of the electric field. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Calculate the magnitude of the electric field strength. Explain your working. electric field strength = .................................................V m–1 [3] (c) The electric field is now removed so that the positively-charged particle follows a curved path in the magnetic field. This path is an arc of a circle of radius 4.0 cm. q Calculate, for the particle, the ratio m. ratio = ................................................C kg–1 [3] (d) The particle has a charge of 3e where e is the elementary charge. (i) Use your answer in (c) to determine the mass, in u, of the particle. mass = ........................................................u [2] (ii) The particle is the nucleus of an atom. State the number of protons and the number of neutrons in this nucleus. number of protons = ............................................................... number of neutrons = ............................................................... [1] [Total: 11]

Mark scheme: 9 (a) direction of force due to electric field opposite to force due to magnetic field B1 electric field is up the page B1 [2] (b) force due to electric field = force due to magnetic field or Eq = Bqv B1 E = Bv C1 = 9.7 × 10–2 × 1.6 × 105 = 1.6 (1.55) × 104 V m–1 A1 [3] (c) q / m = v / Br C1 = 1.6 × 105 / (9.7 ×10–2 × 4.0 × 10–2) C1 = 4.1 (4.12) × 107 C kg–1 A1 [3] (d) (i) m = (3 × 1.60 × 10–19) / (4.12 × 107) C1 m = 1.16 × 10–26 / 1.66 × 10–27 = 7(.0) u (allow 7.1 u) A1 [2] (ii) 3 protons, 4 neutrons A1 [1]

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10 A small coil of wire is situated in a non-uniform magnetic field, as shown in Fig. 10.1. coil, constant 40 turns non-uniform velocity magnetic field P x Fig. 10.1 The coil consists of 40 turns of wire and moves with a constant speed in a straight line. The coil has displacement x from a fixed point P. The variation with x of the magnetic flux Φ in the coil is shown in Fig. 10.2. 8 7 \ / 10–6 Wb 6 5 4 3 2 0 1 2 3 4 5 6 x / cm Fig. 10.2 (a) The coil is moved at constant speed between point P and the point where x = 3.0 cm. (i) Calculate the change in magnetic flux linkage of the coil. change in flux linkage = .................................................... Wb [1] (ii) The e.m.f. induced in the coil is 5.0 × 10–4 V. Determine the speed of the coil. speed = ................................................. m s–1 [2] (b) On Fig. 10.3, sketch the variation with x of the e.m.f. E induced in the coil for values of x from x = 0 to x = 6.0 cm. E 0 0 1 2 3 4 5 6 x / cm Fig. 10.3 [2] [Total: 5]

Mark scheme: 10 (a) (i) change in flux linkage = 40 × (5.0 – 3.0) × 10–6 = 8(.0) × 10–5 Wb A1 [1] (ii) time taken = 8.0 × 10–5 / 5.0 × 10–4 = 0.16 (s) C1 speed = 3.0 × 10–2 / 0.16 = 0.19 (0.188) m s–1 A1 or E = (∆Φ / ∆x) × speed speed = 5.0 × 10–4 / (8.0 ×10–5 / 3.0 ×10–2) (C1) = 0.19 (0.188) m s–1 (A1) [2] (b) a constant non-zero value of E from 0 to 3 cm and a different constant non-zero value of E from 3 to 6 cm M1 E from 3–6 cm has the opposite sign to and larger value than E from 0–3 cm A1 [2]

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11 (a) With reference to the photoelectric effect, state what is meant by the threshold frequency. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Electromagnetic radiation of wavelength λ is incident on a metal surface. Electrons of maximum kinetic energy EMAX are emitted. (i) On Fig. 11.1, sketch the variation with 1/λ of EMAX. EMAX 0 0 1 /h Fig. 11.1 [2] (ii) State an equation relating the gradient of the graph drawn on Fig. 11.1 to the Planck constant h. Explain any symbols you use. ........................................................................................................................................... .......................................................................................................................................[1] (iii) Explain why, for any particular wavelength of electromagnetic radiation, most of the electrons are emitted with kinetic energies less than the maximum value EMAX. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iv) Light of a particular wavelength is incident on a metal surface and gives rise to a photoelectric current. The wavelength is reduced. The intensity of the light is kept constant. State and explain the effect, if any, on the photoelectric current. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] [Total: 10]

Mark scheme: 11 (a) minimum frequency for electron(s) to be emitted (from surface) M1 reference to frequency of electromagnetic radiation / photon A1 or frequency causing emission of electron(s) from surface with zero kinetic energy (M1) reference to frequency of electromagnetic radiation / photon (A1) [2] (b) (i) positive intercept on (1 / λ)-axis (when extrapolated) B1 straight line with positive gradient B1 [2] (ii) gradient = hc where c is the speed of light B1 [1] (iii) maximum kinetic energy when electron emitted from surface B1 energy is required to bring an electron to the surface B1 [2] (iv) each photon has more energy M1 fewer photons per unit time M1 fewer electrons per unit time / less current A1 [3]

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12 (a) In an X-ray tube, the hardness of an X-ray beam may be controlled. (i) State what is meant by the hardness of the beam. ........................................................................................................................................... .......................................................................................................................................[1] (ii) State how the hardness of the beam may be decreased. ........................................................................................................................................... .......................................................................................................................................[1] (b) State one advantage and one disadvantage of producing a CT scan image of a person rather than a standard X-ray image. advantage: ................................................................................................................................ ................................................................................................................................................... disadvantage: ............................................................................................................................ ................................................................................................................................................... [2] [Total: 4]

Mark scheme: 12 (a) (i) the penetration of the beam B1 [1] (ii) either decrease the accelerating voltage or decrease voltage between cathode and anode B1 [1] (b) advantage: image gives depth / image is 3D / final image can be viewed from any angle B1 disadvantage: greater exposure / more risk to health / more expensive / person must remain stationary B1 [2]

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13 Beryllium-7 (7 Be) is produced in the upper atmosphere and then sinks down onto the Earth’s 4 surface. Nuclei of beryllium-7 decay with a half-life of 53.3 days to form stable nuclei. The activity of a sample of beryllium-7 on a tree leaf is 39 mBq. (a) Show that the decay constant of beryllium-7 is 1.5 × 10–7 s–1. [1] (b) Determine the mass of the beryllium-7 on the leaf. mass = ......................................................kg [3] (c) The leaf is covered so that no further beryllium-7 is added to the existing sample from the atmosphere. Calculate the time that must elapse before the activity of the sample is reduced to 2.0 mBq. time = ........................................................s [2] [Total: 6]

Mark scheme: 13 (a) λ = ln2 / T½ = ln2 / (53.3 × 24 × 60 × 60) =1.5 × 10–7 s–1 A1 [1] (b) A = λN C1 N = 39 × 10–3 / 1.5 × 10–7 = 2.6 ×105 m = (2.6 × 105 / 6.0 × 1023) × 7 × 10–3 or 2.6 × 105 × 1.66 × 10–27 × 7 C1 = 3.0 × 10–21 kg A1 [3] (c) 2 / 39 = exp (–1.5 × 10–7 × t ) or 2 / 39 = (1 / 2)[t / (53.3 × 24 × 3600)] C1 t = 2.0 × 107 s A1 [2]

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