Cambridge A Level Physics 9702 — 2016 May/June Paper 4 · Variant 1

1 (a) By reference to the definition of gravitational potential, explain why gravitational potential is a negative quantity. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Two stars A and B have their surfaces separated by a distance of 1.4 × 1012 m, as illustrated in Fig. 1.1. 1.4 = 1012 m star A star B P x Fig. 1.1 Point P lies on the line joining the centres of the two stars. The distance x of point P from the surface of star A may be varied. The variation with distance x of the gravitational potential φ at point P is shown in Fig. 1.2. x / 1012 m 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 –2 –4 –6 –8 q/ 108 J kg–1 –10 –12 –14 –16 Fig. 1.2 A rock of mass 180 kg moves along the line joining the centres of the two stars, from star A towards star B. (i) Use data from Fig. 1.2 to calculate the change in kinetic energy of the rock when it moves from the point where x = 0.1 × 1012 m to the point where x = 1.2 × 1012 m. State whether this change is an increase or a decrease. change = ............................................................. J .................................................................................. [3] (ii) At a point where x = 0.1 × 1012 m, the speed of the rock is v. Determine the minimum speed v such that the rock reaches the point where x = 1.2 × 1012 m. minimum speed = ............................................... m s−1 [3] [Total: 8]

Mark scheme: 1 (a) (gravitational) potential at infinity defined as/is zero B1 (gravitational) force attractive so work got out/done as object moves from infinity (so potential is negative) B1 [2] (b) (i) ∆E = m∆φ = 180 × (14 – 10) × 108 C1 = 7.2 × 1010 J A1 increase B1 [3] (ii) energy required = 180 × (10 – 4.4) × 108 or energy per unit mass = (10 – 4.4) × 108 C1 ½ × 180 × v2 = 180 × (10 – 4.4) × 108 or ½ × v2 = (10 – 4.4) × 108 C1 v = 3.3 × 104 m s–1 A1 [3]

More questions on Energy conservation

2 (a) An ideal gas is assumed to consist of atoms or molecules that behave as hard, identical spheres that are in continuous motion and undergo elastic collisions. State two further assumptions of the kinetic theory of gases. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... [2] (b) Helium-4 (42He) may be assumed to be an ideal gas. (i) Show that the mass of one atom of helium-4 is 6.6 × 10−24 g. [1] (ii) The mean kinetic energy EK of an atom of an ideal gas is given by the expression EK = 32 kT. Calculate the root-mean-square (r.m.s.) speed of a helium-4 atom at a temperature of 27 °C. r.m.s. speed = ............................................... m s−1 [3] [Total: 6]

Mark scheme: 2 (a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules (total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules any two B2 [2] 2 (b) (i) mass = 4.0 / (6.02 × 10 23) = 6.6 × 10–24 g or mass = 4.0 × 1.66 × 10–27 × 103 = 6.6 × 10–24 g B1 [1] 3 1 (ii) kT = m <c 2> C1 2 2 3 1 × 1.38 × 10–23 × 300 = × 6.6 × 10–27 × <c 2> 2 2 <c 2> = 1.88 × 106 (m2 s–2) C1 r.m.s. speed = 1.4 × 103 m s–1 A1 [3]

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3 (a) State, by reference to displacement, what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A mass is undergoing oscillations in a vertical plane. The variation with displacement x of the acceleration a of the mass is shown in Fig. 3.1. a 0 0 x Fig. 3.1 State two reasons why the motion of the mass is not simple harmonic. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... [2] (c) A block of wood is floating in a liquid, as shown in Fig. 3.2. oscillation block of block liquid Fig. 3.2 The block is displaced vertically and then released. The variation with time t of the displacement y of the block from its equilibrium position is shown in Fig. 3.3. 2.0 y / cm 1.5 1.0 0.5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 t / s –0.5 –1.0 –1.5 –2.0 Fig. 3.3 Use data from Fig. 3.3 to determine (i) the angular frequency ω of the oscillations, ω = ............................................. rad s−1 [2] (ii) the maximum vertical acceleration of the block. maximum acceleration = ............................................... m s−2 [2] (iii) The block has mass 120 g. The oscillations of the block are damped. Calculate the loss in energy of the oscillations of the block during the first three complete periods of its oscillations. energy loss = ...................................................... J [3] [Total: 11]

Mark scheme: 3 (a) acceleration/force proportional to displacement (from fixed point) M1 acceleration/force and displacement in opposite directions A1 [2] (b) maximum displacements/accelerations are different B1 graph is curved/not a straight line B1 [2] (c) (i) ω = 2π / T and T = 0.8 s C1 ω = 7.9 rad s–1 A1 [2] (ii) a = (–)ω2 x = 7.852 × 1.5 × 10–2 C1 = 0.93 m s–2 or 0.94 m s–2 A1 [2] (iii) ∆E = ½ mω2 (x02 – x2) C1 = ½ × 120 × 10–3 × 7.852 × {(1.5 × 10–2)2 – (0.9 × 10–2)2} C1 = 5.3 × 10–4 J A1 [3]

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4 (a) (i) State what is meant by the specific acoustic impedance of a medium. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) The intensity reflection coefficient α is given by the expression (Z2 − Z1)2 α = . (Z2 + Z1)2 Explain the meanings of the symbols in this expression. α: ....................................................................................................................................... ........................................................................................................................................... Z2 and Z1: .......................................................................................................................... ........................................................................................................................................... [2] (b) A parallel beam of ultrasound has intensity I0 as it enters a muscle. The beam passes through a thickness of 3.4 cm of muscle before being incident on the boundary with a bone, as shown in Fig. 4.1. muscle bone intensity I0 intensity IT entering muscle entering bone 3.4 cm Fig. 4.1 The intensity of the ultrasound beam as it passes into the bone is IT. Some data for muscle and bone are given in Fig. 4.2. linear absorption specific acoustic impedance coefficient / m−1 / kg m−2 s−1 muscle 23 1.7 × 106 bone 130 6.3 × 106 Fig. 4.2 IT Calculate the ratio . I0 ratio = ......................................................... [5] [Total: 9]

Mark scheme: 4 (a) (i) product of speed and density M1 reference to speed in medium (and density of medium) A1 [2] (ii) α: ratio of reflected intensity and/to incident intensity B1 Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1 [2] (b) in muscle: IM = I0 e–µx = I0 exp(–23 × 3.4 × 10–2) C1 IM / I0 = 0.457 C1 at boundary: α = (6.3 – 1.7)2 / (6.3 + 1.7)2 = 0.33 C1 IT /IM = [(1 – α) =] 0.67 C1 IT / I0 = 0.457 × 0.67 = 0.31 A1 [5]

More questions on Production and use of ultrasound

5 The variation with time t of the voltage level of part of an analogue signal is shown in Fig. 5.1. 16 14 voltage level 12 10 8 6 4 2 0 0 0.25 0.50 0.75 1.00 1.25 1.50 time t / ms Fig. 5.1 The signal is sampled at 0.25 ms intervals. Each sample is converted into a four-bit digital number. Fig. 5.2 lists various times t at which the voltage level is sampled. The digital number for time t = 0 is shown. time t / ms 0 0.25 0.50 0.75 1.00 1.25 1.50 digital number 1011 Fig. 5.2 (a) (i) On Fig. 5.2, underline the most significant bit (MSB) for the digital number at time t = 0. [1] (ii) Complete Fig. 5.2 for the times shown. [2] (b) After transmission of the digital numbers, the signal is passed through a digital-to-analogue converter (DAC). On Fig. 5.3, plot the transmitted analogue signal from the DAC. 16 14 voltage level 12 10 8 6 4 2 0 0 0.25 0.50 0.75 1.00 1.25 1.50 time t / ms Fig. 5.3 [3] (c) The transmitted signal in (b) has less detail than the original signal in Fig. 5.1. Suggest and explain two means by which the level of detail in the transmitted signal could be increased. 1. .............................................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... [4] [Total: 10]

Mark scheme: 5 (a) (i) 1011 A1 [1] (ii) 0 0.25 0.50 0.75 1.00 1.25 1.50 1011 0110 1000 1110 0101 0011 0001 All 6 correct, 2 marks. 5 correct, 1 mark. A2 [2] (b) sketch: 6 horizontal steps of width 0.25 ms shown M1 steps at correct heights and all steps shown A1 steps shown in correct time intervals A1 [3] (c) increase sampling frequency/rate M1 so that step width/depth is reduced A1 increase number of bits (in each number) M1 so that step height is reduced A1 [4]

More questions on Electric current

6 A solid metal sphere of radius R is isolated in space. The sphere is positively charged so that the electric potential at its surface is VS. The electric field strength at the surface is ES. (a) On the axes of Fig. 6.1, show the variation of the electric potential with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Vs 0.8 Vs potential 0.6 Vs 0.4 Vs 0.2 Vs 0 0 R 2R 3R distance x Fig. 6.1 [3] (b) On the axes of Fig. 6.2, show the variation of the electric field strength with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Es 0.8 Es field strength 0.6 Es 0.4 Es 0.2 Es 0 0 R 2R 3R distance x Fig. 6.2 [3] [Total: 6]

Mark scheme: 6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1 smooth curve through (R, VS) and (2R, 0.5VS) B1 smooth curve continues to (3R, 0.33VS) B1 [3] (b) sketch: from x = 0 to x = R, field strength is zero B1 smooth curve through (R, E) and (2R, 0.25E) B1 smooth curve continues to (3R, 0.11E) B1 [3]

More questions on Electric potential

7 A student sets up the circuit shown in Fig. 7.1 to measure the charge on a capacitor C for different values of potential difference across the capacitor. C meter to V measure charge Fig. 7.1 The variation with potential difference V of the charge Q stored on the capacitor is shown in Fig. 7.2. 15 Q / mC 10 5 0 0 1 2 3 4 5 6 V / V Fig. 7.2 (a) State and explain how Fig. 7.2 indicates that there is a systematic error in the readings of one of the meters. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Use Fig. 7.2 to determine the capacitance, in μF, of capacitor C. capacitance = .................................................... μF [3] (c) Use your answer in (b) to determine the additional energy stored in the capacitor C when the potential difference across it is increased from 6.0 V to 9.0 V. energy = ...................................................... J [3] [Total: 8]

Mark scheme: 7 (a) line has non-zero intercept/line does not pass through origin B1 charge is/should be proportional to potential (difference) or charge is/should be zero when p.d. is zero (therefore there is a systematic error) B1 [2] (b) reasonable attempt at line of best fit B1 use of gradient of line of best fit clear M1 C = 2800 µF (allow ± 200 µF) A1 [3] (c) energy = ½ CV 2 or energy = ½ QV and C = Q / V C1 ∆ energy = ½ × 2800 × 10–6 × (9.02 – 6.02) C1 = 6.3 × 10–2 J A1 [3]

More questions on Capacitors and capacitance

8 The circuit of an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 8.1. R1 +V R2 P – + V IN –V V OUT Fig. 8.1 (a) Explain why point P is known as a virtual earth. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Derive an expression, in terms of the resistances R1 and R2, for the gain of the amplifier circuit. Explain your working. [3] (c) A relay and the output terminals of the amplifier circuit are shown in Fig. 8.2. V OUT Fig. 8.2 On Fig. 8.2, show how the relay may be connected to the amplifier output so that the relay operates only when VOUT is positive. [3] [Total: 9]

Mark scheme: 8 (a) op-amp has infinite/(very) large gain B1 op-amp saturates if V + ≠ V – M1 V+ is at earth potential so P (or V–) must be at earth A1 [3] (b) input resistance to op-amp is very large or current in R2 = current in R1 B1 VIN (– 0) = IR2 and (0) – VOUT = IR1 M1 VOUT / VIN = –R1 / R2 A1 [3] (c) relay coil connected between VOUT and earth M1 correct diode symbol connected between VOUT and coil or between coil and earth M1 correct polarity for diode (‘clockwise’) A1 [3]

More questions on Practical circuits

9 A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1. direction of magnetic field Z Y 0.10 mm 50 mm X current 3.8 A Q R P S 65 mm Fig. 9.1 (not to scale) Some of the corners of the slice are labelled. A current I of 3.8 A is normal to face RSXY of the slice. In aluminium, the number of free electrons per unit volume is 6.0 × 1028 m−3. A uniform magnetic field of magnetic flux density B equal to 0.13 T is normal to face QRYZ of the aluminium slice in the direction from Q to P. A Hall voltage VH is developed across the slice and is given by the expression BI VH = . ntq (a) Use Fig. 9.1 to state the magnitude of the distance t. t = ................................................. mm [1] (b) Calculate the magnitude of the Hall voltage VH. VH = ..................................................... V [2] [Total: 3]

Mark scheme: 9 (a) 0.10 mm B1 [1] (b) VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19) C1 = 5.1 × 10–7 V A1 [2]

More questions on Force on a moving charge

10 (a) A coil of insulated wire is wound on a copper core, as illustrated in Fig. 10.1. insulated copper wire core Fig. 10.1 An alternating current is passed through the coil. The heating effect of the current in the coil is negligible. Explain why the temperature of the core rises. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] (b) Two hollow tubes of equal length hang vertically as shown in Fig. 10.2. magnet A magnet B plastic aluminium tube tube Fig. 10.2 One tube is made of plastic and the other of aluminium. Two small similar bar magnets A and B are held above the tubes and then released simultaneously. The magnets do not touch the sides of the tubes. Explain why magnet B takes much longer than magnet A to fall through the tube. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] [Total: 9]

Mark scheme: 10 (a) (non-uniform) magnetic flux in core is changing M1 induces (different) e.m.f. in (different parts of) the core A1 (eddy) currents form in the core M1 which give rise to heating A1 [4] (b) as magnet falls, tube cuts magnetic flux M1 e.m.f./(eddy) currents induced in metal/aluminium (tube) A1 (eddy) current heating of tube M1 with energy taken from falling magnet A1 or (eddy) currents produce magnetic field (M1) that opposes motion of magnet (A1) so magnet B has acceleration < g or magnet B has smaller acceleration/reaches terminal speed A1 [5]

More questions on Electromagnetic induction

11 The variation with time t of the sinusoidal current I in a resistor of resistance 450 Ω is shown in Fig. 11.1. 1.0 I / A 0.5 0 0 5 10 15 20 25 30 t / ms –0.5 –1.0 Fig. 11.1 Use data from Fig. 11.1 to determine, for the time t = 0 to t = 30 ms, (a) the frequency of the current, frequency = ................................................... Hz [2] (b) the mean current, mean current = ..................................................... A [1] (c) the root-mean-square (r.m.s.) current, r.m.s. current = ..................................................... A [2] (d) the energy dissipated by the resistor. energy = ...................................................... J [2] [Total: 7]

Mark scheme: 11 (a) period = 15 ms C1 frequency (= 1 / T) = 67 Hz A1 [2] (b) zero A1 [1] (c) Ir.m.s. = I0 / √2 C1 = 0.53 A A1 [2] (d) energy = Ir.m.s. 2 × R × t or ½ I02 × R × t or power = Ir.m.s. 2 × R and energy = power × t C1 energy = 0.532 × 450 × 30 × 10–3 = 3.8 J A1 [2]

More questions on Characteristics of alternating currents

12 Some of the electron energy bands in a solid are illustrated in Fig. 12.1. conduction band (partially filled) forbidden band valence band Fig. 12.1 (a) In isolated atoms, electron energy levels have discrete values. Suggest why, in a solid, there are energy bands, rather than discrete energy levels. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) A light-dependent resistor (LDR) consists of an intrinsic semiconductor. Use band theory to explain the dependence on light intensity of the resistance of the LDR when it is at constant temperature. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] [Total: 8]

Mark scheme: 12 (a) (in a solid electrons in) neighbouring atoms are close together (and influence/interact with each other) M1 this changes their electron energy levels M1 (many atoms in lattice) cause a spread of energy levels into a band A1 [3] (b) photons of light give energy to electrons in valence band B1 electrons move into the conduction band M1 leaving holes in the valence band A1 these electrons and holes are charge carriers B1 increased number/increased current, hence reduced resistance B1 [5]

More questions on Atoms, nuclei and radiation

13 Copper-66 is a radioactive isotope. When a nucleus of copper-66 decays, the emissions include a β− particle and a γ-ray photon. The count rate produced from a sample of the isotope copper-66 is measured using a detector and counter, as illustrated in Fig. 13.1. to counter detector radioactive sample shielding Fig. 13.1 (a) State three reasons why the activity of the sample of copper-66 is not equal to the measured count rate. 1. .............................................................................................................................................. ................................................................................................................................................... 2. .............................................................................................................................................. ................................................................................................................................................... 3. .............................................................................................................................................. ................................................................................................................................................... [3] (b) In a time of 42.0 minutes, the count rate from the sample of copper-66 is found to decrease from 3.62 × 104 Bq to 1.21 × 102 Bq. Calculate the half-life of copper-66. half-life = .......................................... minutes [2] (c) The γ-ray photons emitted from radioactive nuclei have specific energies, dependent on the nucleus emitting the photons. By comparison with emission line spectra, suggest what can be deduced about energy levels in nuclei. ................................................................................................................................................... .............................................................................................................................................. [1] [Total: 6]

Mark scheme: 13 (a) e.g. background count (rate)/radiation multiple possible counts from each decay radiation emitted in all directions dead-time of counter (daughter) product unstable/also emits radiation self-absorption of radiation in sample or absorption in air/detector window three sensible suggestions, 1 each B3 [3] (b) A = A0 exp(– ln 2 × t / T½) 1.21 × 102 = 3.62 × 104 exp(– ln2 × 42.0 / T½) or 1.21 × 102 = 3.62 × 104 exp(–λ × 42.0) C1 T½ = 5.1 minutes (306 s) A1 [2] (c) discrete energy levels (in nuclei) B1 [1]

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