Cambridge A Level Physics 9702 — 2014 Oct/Nov Paper 4 · Variant 1

1 An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet. (a) Show that the gravitational field strength at the surface of the planet is 3.7 N kg−1. [2] (b) A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m. Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone. Explain your working. change in energy = ..................................................... J [3] (c) A rock, initially at rest at infinity, moves towards the planet. At point P, its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig. 1.1. D 3.5 D path of rock P planet Fig. 1.1 Calculate the speed of the rock at point P, assuming that the change in gravitational potential energy is all transferred to kinetic energy. speed = ............................................... m s−1 [4]

Mark scheme: 1 (a) g = GM / R2 C1 = (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106)2 = 3.7 N kg–1 A1 [2] (b) ∆EP = mg∆h because ∆h ≪ R (or 1800 m ≪ 3.4 × 106 m) g is constant B1 ∆EP = 2.4 × 3.7 × 1800 C1 = 1.6 × 104 J A1 [3] (use of g = 9.8 m s–2 max. 1 for explanation) (c) gravitational potential energy = (–)GMm / x C1 v2 = 2GM / x C1 x = 4D = 4 × 6.8 × 106 C1 v2 = (2 × 6.67 × 10–11 × 6.4 × 1023) / (4 × 6.8 × 106) = 3.14 × 106 v = 1.8 × 103 m s–1 A1 [4] (use of 3.5 D giving 1.9 × 103 m s–1, allow max. 3)

More questions on Gravitational potential energy and kinetic energy

2 A large bowl is made from part of a hollow sphere. A small spherical ball is placed inside the bowl and is given a horizontal speed. The ball follows a horizontal circular path of constant radius, as shown in Fig. 2.1. ball 14 cm Fig. 2.1 The forces acting on the ball are its weight W and the normal reaction force R of the bowl on the ball, as shown in Fig. 2.2. wall of R ball bowl e W Fig. 2.2 The normal reaction force R is at an angle θ to the horizontal. (a) (i) By resolving the reaction force R into two perpendicular components, show that the resultant force F acting on the ball is given by the expression W = F tan θ. [2] (ii) State the significance of the force F for the motion of the ball in the bowl. ........................................................................................................................................... ...................................................................................................................................... [1] (b) The ball moves in a circular path of radius 14 cm. For this radius, the angle θ is 28°. Calculate the speed of the ball. speed = ............................................... m s−1 [3]

Mark scheme: 2 (a) (i) F = R cosθ M1 W = R sinθ M1 dividing, W = F tanθ A0 [2] (max. 1 if derivation to final line not shown) (ii) provides the centripetal force B1 [1] (b) either F = mv2 / r and W = mg or v2 = rg / tan θ C1 v2 = (14 × 10–2 × 9.8) / tan 28° C1 = 2.58 v = 1.6 m s–1 A1 [3]

More questions on Kinematics of uniform circular motion

3 (a) State what is meant by an ideal gas. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A storage cylinder for an ideal gas has a volume of 3.0 × 10−4 m3. The gas is at a temperature of 23 °C and a pressure of 5.0 × 107 Pa. (i) Show that the amount of gas in the cylinder is 6.1 mol. [2] (ii) The gas leaks slowly from the cylinder so that, after a time of 35 days, the pressure has reduced by 0.40%. The temperature remains constant. Calculate the average rate, in atoms per second, at which gas atoms escape from the cylinder. rate = .................................................. s−1 [4]

Mark scheme: 3 (a) obeys the equation pV / T = constant B1 [1] (accept pV = nRT) (b) (i) pV = nRT C1 5.0 × 107 × 3.0 × 10–4 = n × 8.31 × 296 giving n = 6.1 mol A1 [2] (ii) pressure ∝ amount of substance loss = 0.40 / 100 × 6.1 mol = 0.0244 mol C1 = 0.0244 × 6.02 × 1023 (atoms) C1 = 1.47 × 1022 atoms C1 rate = (1.47 × 1022) / (35 × 24 × 60 × 60) = 4.9 × 1015 s–1 A1 [4]

More questions on Equation of state

4 (a) State what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 4.1. curved track, radius r x P Fig. 4.1 The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by the expression gx a = − r where g is the acceleration of free fall. (i) Show that the ball undergoes simple harmonic motion. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) The radius r of curvature of the track is 28 cm. Determine the time interval τ between the ball passing point P and then returning to point P. τ = ..................................................... s [3] (c) The variation with time t of the displacement x of the ball in (b) is shown in Fig. 4.2. x 0 0 t o 2 ot 3 ot 4 ot Fig. 4.2 Some moisture now forms on the track, causing the ball to come to rest after approximately 15 oscillations. On the axes of Fig. 4.2, sketch the variation with time t of the displacement x of the ball for the first two periods after the moisture has formed. Assume the moisture forms at time t = 0. [3]

Mark scheme: 4 (a) acceleration / force proportional to displacement (from a fixed point) M1 either acceleration and displacement in opposite directions or acceleration always directed towards a fixed point A1 [2] (b) (i) g and r are constant so a is proportional to x B1 negative sign shows a and x are in opposite directions B1 [2] (ii) ω 2 = g / r and ω = 2π / T C1 ω 2 = 9.8 / 0.28 = 35 C1 T = 2π / √35 = 1.06 s time interval τ = 0.53 s A1 [3] (c) sketch: time period constant (or increases very slightly) M1 drawn line always ‘inside’ given loops A1 successive decrease in peak height A1 [3]

More questions on Simple harmonic oscillations

5 (a) Define electric potential at a point. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) An isolated solid metal sphere is positively charged. The variation of the potential V with distance x from the centre of the sphere is shown in Fig. 5.1. 200 160 V / V 120 80 40 0 0 2 4 6 8 10 x / cm Fig. 5.1 Use Fig. 5.1 to suggest (i) why the radius of the sphere cannot be greater than 1.0 cm, ........................................................................................................................................... ...................................................................................................................................... [1] (ii) that the charge on the sphere behaves as if it were a point charge. [3] (c) Assuming that the charge on the sphere does behave as a point charge, use data from Fig. 5.1 to determine the charge on the sphere. charge = ..................................................... C [2]

Mark scheme: 5 (a) work done in moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) inside the sphere, the potential would be constant B1 [1] (ii) for point charge, Vx is constant B1 co-ordinates clear and determines two values of Vx at least 4 cm apart M1 conclusion made clear A1 [3] (c) q = 4πε0Vx q = 4π × 8.85 × 10–12 × 180 × 1.0 × 10–2 M1 = 2.0 × 10–10 C A1 [2]

More questions on Electric field of a point charge

6 A stiff straight copper wire XY is held fixed in a uniform magnetic field of flux density 2.6 × 10−3 T, as shown in Fig. 6.1. stiff wire X uniform magnetic field 34° flux density 2.6 × 10–3 T 4.7 cm 34° Y Fig. 6.1 The wire XY has length 4.7 cm and makes an angle of 34° with the magnetic field. (a) Calculate the force on the wire due to a constant current of 5.4 A in the wire. force = ..................................................... N [2] (b) The current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine the total variation in the force on the wire due to the alternating current. variation in force = ..................................................... N [3]

Mark scheme: 6 (a) F = BIL sinθ C1 = 2.6 × 10–3 × 5.4 × 4.7 × 10–2 × sin 34° = 3.69 × 10–4 N A1 [2] (allow 1 mark for use of cos 34°) (b) peak current = 1.7 × √2 C1 = 2.4 A max. force = 2.6 × 10–3 × 2.4 × 4.7 × 10–2 × sin 34° = 1.64 × 10–4 N C1 variation = 2 × 1.64 × 10–4 = 3.3 × 10–4 N A1 [3] 2

More questions on Force on a current-carrying conductor

7 (a) The mean value of an alternating current is zero. Explain (i) why an alternating current gives rise to a heating effect in a resistor, ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) by reference to heating effect, what is meant by the root-mean-square (r.m.s.) value of an alternating current. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) A simple iron-cored transformer is illustrated in Fig. 7.1. primary secondary coil coil iron core Fig. 7.1 (i) State Faraday’s law of electromagnetic induction. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) Use Faraday’s law to explain why the current in the primary coil is not in phase with the e.m.f. induced in the secondary coil. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3]

Mark scheme: 7 (a) (i) either heating effect in a resistor ∝ (current)2 B1 square of value of an alternating current is always positive B1 so heating effect A0 or current moves in opposite directions in resistor during half-cycles (B1) heating effect is independent of direction (B1) [2] (ii) that value of the direct current M1 producing the same heating effect (as the alternating current) in a resistor A1 [2] (b) (i) induced e.m.f. proportional to the rate M1 of change of (magnetic) flux (linkage) A1 [2] (ii) flux in core is in phase with current in the primary coil B1 (induced) e.m.f. in secondary because coil cuts the flux B1 flux and rate of change of flux are not in phase B1 [3]

More questions on Characteristics of alternating currents

8 White light is incident on a cloud of cool hydrogen gas, as illustrated in Fig. 8.1. cool hydrogen gas incident emergent white light light Fig. 8.1 The spectrum of the light emerging from the gas cloud is found to contain a number of dark lines. (a) Explain why these dark lines occur. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Some electron energy levels in a hydrogen atom are illustrated in Fig. 8.2. –0.38 –0.55 –0.85 –1.51 energy / eV –3.41 Fig. 8.2 One dark line is observed at a wavelength of 435 nm. (i) Calculate the energy, in eV, of a photon of light of wavelength 435 nm. energy = ................................................... eV [4] (ii) On Fig. 8.2, draw an arrow to indicate the energy change that gives rise to this dark line. [1]

Mark scheme: 8 (a) photon ‘absorbed’ by electron B1 photon has energy equal to difference in energy of two energy levels B1 electron de-excites emitting photon (of same energy) in any direction B1 [3] (b) (i) E = hc / λ C1 = (6.63 × 10–34 × 3 × 108) / (435 × 10–9) C1 = 4.57 × 10–19 J (allow 2 s.f.) C1 = (4.57 × 10–19) / (1.6 × 10–19) (eV) = 2.86 eV (allow 2 s.f.) A1 [4] (ii) arrow pointing in either direction between –3.41 eV and –0.55 eV B1 [1]

More questions on Energy and momentum of a photon

9 One likely means by which nuclear fusion may be achieved on a practical scale is the D-T reaction. (a) State what is meant by nuclear fusion. ................................................................................................................................................... .............................................................................................................................................. [1] (b) In the D-T reaction, a deuterium (21H) nucleus fuses with a tritium (31H) nucleus to form a helium-4 (42He) nucleus. The nuclear equation for the reaction is 21H + 31H 42He + 10n + energy Some data for this reaction are given in Fig. 9.1. mass / u deuterium (21H) 2.01356 tritium (31H) 3.01551 helium-4 (42He) 4.00151 neutron (10n) 1.00867 Fig. 9.1 (i) Calculate the energy, in MeV, equivalent to 1.00 u. Explain your working. energy = ................................................ MeV [3] (ii) Use data from Fig. 9.1 and your answer in (i) to determine the energy released in this D-T reaction. energy = ................................................ MeV [2] (iii) Suggest why, for the D-T reaction to take place, the temperature of the deuterium and the tritium must be high. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

Mark scheme: 9 (a) ‘light’ nuclei combine to form ‘heavier’ nuclei B1 [1] (b) (i) either energy = c2∆m or energy = (3.00 × 108)2 × 1.66 × 10–27 C1 energy = 1.494 × 10–10 J C1 = (1.494 × 10–10) / (1.60 × 10–13) = 934 MeV (3 s.f.) A1 [3] (ii) ∆m = (2.01356 + 3.01551) – (4.00151 + 1.00867) = 5.02907 – 5.01018 = 0.01889 u C1 energy = 0.01889 × 934 = 17.6 MeV (allow 2 s.f.) A1 [2] (iii) high temperature means high speeds / kinetic energy of nuclei B1 D and T nuclei collide despite repelling one another B1 [2] Section B

More questions on Mass defect and nuclear binding energy

10 (a) An ideal operational amplifier (op-amp) has infinite open-loop gain and infinite input resistance (impedance). State three further properties of an ideal op-amp. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... 3. .............................................................................................................................................. ................................................................................................................................................... [3] (b) The circuit of Fig. 10.1 is used to detect changes in temperature. 740 1 +9 V A – + –9 V V Fig. 10.1 The voltmeter has infinite resistance. The variation with temperature θ of the resistance R of the thermistor is shown in Fig. 10.2. 4.0 R / k1 3.0 2.0 1.0 0 5 10 15 20 25 30 e / °C Fig. 10.2 (i) When the thermistor is at a temperature of 1.0 °C, the voltmeter reads +1.0 V. Show that, for the thermistor at 1.0 °C, the potential at A is −0.20 V. [4] (ii) The potential at A remains at −0.20 V. Determine the voltmeter reading for a thermistor temperature of 15 °C. voltmeter reading = ..................................................... V [2] (c) The voltmeter reading for a thermistor temperature of 29 °C is 0.35 V. (i) Assuming a linear change of voltmeter reading with change of temperature over the range 1 °C to 29 °C, calculate the voltmeter reading at 15 °C. voltmeter reading = ..................................................... V [1] (ii) Suggest why your answers in (b)(ii) and (c)(i) are not the same. ........................................................................................................................................... ...................................................................................................................................... [1]

Mark scheme: 10 (a) e.g. zero output resistance / impedance infinite bandwidth infinite slew rate 1 mark each, max. 3 B3 [3] (b) (i) at 1.0 °C, thermistor resistance is 3.7 kΩ B1 amplifier gain = –R / 740 = –3700 / 740 (negative sign essential) C1 = –5.0 C1 potential = 1.0 / –5.0 = –0.20 V A1 [4] (ii) at 15 °C, R = 2.15 kΩ (allow ±0.05 kΩ) C1 reading = (2150 / 740) × 0.2 = 0.58 V (0.59 V → 0.57 V) A1 [2] (c) (i) 0.68 V A1 [1] (ii) resistance (of thermistor) does not change linearly with temperature B1 [1]

More questions on Potential dividers

11 The use of X-rays in medical diagnosis gives rise to an increased exposure of the patient to radiation. Explain why (a) an aluminium filter may be placed in the X-ray beam when producing an X-ray image of a patient, ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) the radiation dose received by a patient is different for a CT scan from that for a simple X-ray image. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4]

Mark scheme: 11 (a) X-ray beam contains many wavelengths B1 aluminium filter absorbs long wavelength X-ray radiation M1 that would be absorbed by the body (and not contribute to the image) A1 [3] (b) CT scan consists of (many) X-ray images of a slice M1 and there are many slices A1 X-ray image is a single exposure B1 (so much) greater exposure with CT scan B1 [4]

More questions on Production and use of X-rays

12 (a) Information may be carried by different channels of communication. State one application, in each case, where information is carried using (i) microwaves, ........................................................................................................................................... ...................................................................................................................................... [1] (ii) coaxial cables, ........................................................................................................................................... ...................................................................................................................................... [1] (iii) wire pairs. ........................................................................................................................................... ...................................................................................................................................... [1] (b) A station on Earth transmits a signal of initial power 3.1 kW to a geostationary satellite. The attenuation of the signal received by the satellite is 190 dB. (i) Calculate the power of the signal received by the satellite. power = .................................................. kW [2] (ii) By reference to your answer in (i), state and explain the changes made to the signal before transmission back to Earth. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3]

Mark scheme: 12 (a) (i) e.g. satellite communication, mobile phones, line of sight communication, wifi B1 [1] (ii) e.g. connection of TV to aerial, loudspeaker, microphone (if clearly identified) B1 [1] (iii) e.g. a.f. amplifier to loudspeaker, landline for phone B1 [1] (b) (i) attenuation / dB = 10 lg (P2 / P1) C1 –190 = 10 lg (P2 / 3.1) P2 = 3.1 × 10–19 kW A1 [2] (ii) signal is amplified M1 frequency is changed M1 to prevent swamping of up-link signal by down-link (signal) A1 [3]

More questions on Potential difference and power

13 A simplified block diagram of a mobile phone handset is shown in Fig. 13.1. aerial switch tuning circuit r.f. amplifier r.f. amplifier modulator oscillator demodulator parallel-to- serial-to- serial parallel converter converter ADC DAC a.f. amplifier a.f. amplifier loudspeaker microphone Fig. 13.1 State the purpose of (a) the switch, ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) the tuning circuit. ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 13 (a) either for transmission and reception of signal or switching between transmitted and received signals M1 either so that one aerial may be used or so that transmission and reception can occur in quick succession A1 [2] (b) gives large signal for one (input) frequency M1 (and) rejects / very small signal for all other frequencies A1 [2]

More questions on Practical circuits