Cambridge A Level Physics 9702 — 2016 May/June Paper 2 · Variant 2

1 (a) Define acceleration. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A man travels on a toboggan down a slope covered with snow from point A to point B and then to point C. The path is illustrated in Fig. 1.1. man toboggan, at rest A 40° horizontal B 20° horizontal C Fig. 1.1 (not to scale) The slope AB makes an angle of 40° with the horizontal and the slope BC makes an angle of 20° with the horizontal. Friction is not negligible. The man and toboggan have a combined mass of 95 kg. The man starts from rest at A and has constant acceleration between A and B. The man takes 19 s to reach B. His speed is 36 m s–1 at B. (i) Calculate the acceleration from A to B. acceleration = ................................................. m s–2 [2] (ii) Show that the distance moved from A to B is 340 m. [1] (iii) For the man and toboggan moving from A to B, calculate 1. the change in kinetic energy, change in kinetic energy = ....................................................... J [2] 2. the change in potential energy. change in potential energy = ....................................................... J [2] (iv) Use your answers in (iii) to determine the average frictional force that acts on the toboggan between A and B. frictional force = ...................................................... N [2] (v) A parachute opens on the toboggan as it passes point B. There is a constant deceleration of 3.0 m s–2 from B to C. Calculate the frictional force that produces this deceleration between B and C. frictional force = ...................................................... N [2] [Total: 12]

Mark scheme: 1 (a) acceleration = change in velocity / time (taken) or rate of change of velocity B1 [1] (b) (i) v = 0 + at or v = at C1 (a = 36 / 19 =) 1.9 (1.8947) m s–2 A1 [2] (ii) s = ½(u + v)t or s = v2 / 2a or s = ½at2 = ½ × 36 × 19 = 362 / (2 × 1.89) = ½ × 1.89 × 192 = 340 m (342 m / 343 m / 341 m) M1 [1] (iii) 1. (∆KE =) ½ × 95 × (36)2 C1 = 62 000 (61 560) J A1 [2] 2. (∆PE =) 95 × 9.81 × 340 sin 40° or 95 × 9.81 × 218.5 C1 = 200 000 J A1 [2] (iv) work done (by frictional force) = ∆PE – ∆KE or work done = 200 000 – 62 000 (values from 1b(iii) 1. and 2.) C1 (frictional force = 138 000 / 340 =) 410 (406) N [420 N if full figures used] A1 [2] (v) –ma = mg sin 20° – f or ma = –mg sin 20° + f C1 –95 × 3.0 = 95 × 3.36 – f f = 600 (604) N A1 [2]

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2 (a) Fig. 2.1 shows a liquid in a cylindrical container. F\OLQGULFDO FRQWDLQHU OLTXLG K DUHD $ Fig. 2.1 The cross-sectional area of the container is A. The height of the column of liquid is h and the density of the liquid is ρ. Show that the pressure p due to the liquid on the base of the cylinder is given by p = ρgh. [3] (b) The variation with height h of the total pressure P on the base of the cylinder in (a) is shown in Fig. 2.2. 3.0 3 / 105 Pa 2.0 1.0 0 0 0.5 1.0 1.5 2.0 K / m Fig. 2.2 (i) Explain why the line of the graph in Fig. 2.2 does not pass through the origin (0,0). ........................................................................................................................................... .......................................................................................................................................[1] (ii) Use data from Fig. 2.2 to calculate the density of the liquid in the cylinder. density = .............................................. kg m–3 [2] [Total: 6]

Mark scheme: 2 (a) p = F / A M1 use of m = ρV and use of V = Ah and use of F = mg M1 correct substitution to obtain p = ρgh A1 [3] (b) (i) (when h is zero the pressure is not zero due to) pressure from the air/atmosphere B1 [1] (ii) gradient = ρg or P – 1.0 × 105 = ρgh C1 e.g. ρg = 1.0 × 105 / 0.75 (= 133333) ρ = 133 333 / 9.81 = 14 000 (13 592) kg m–3 A1 [2]

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3 (a) Define the Young modulus. ................................................................................................................................................... ...............................................................................................................................................[1] (b) The Young modulus of steel is 1.9 × 1011 Pa. The Young modulus of copper is 1.2 × 1011 Pa. A steel wire and a copper wire each have the same cross-sectional area and length. The two wires are each extended by equal forces. (i) Use the definition of the Young modulus to determine the ratio extension of the copper wire . extension of the steel wire ratio = ...........................................................[3] (ii) The two wires are each extended by a force. Both wires obey Hooke’s law. On Fig. 3.1, sketch a graph for each wire to show the variation with extension of the force. Label the line for steel with the letter S and the line for copper with the letter C. force 00 extension Fig. 3.1 [1] [Total: 5]

Mark scheme: 3 (a) Young modulus = stress / strain B1 [1] (b) (i) E = (F × l) / (A × e) or e = (F × l) / (A × E) B1 e ∝ 1 / E or ratio eC / eS = ES / EC or (1.9 × 1011) / (1.2 × 1011) or 19 / 12 C1 (ratio =) 1.6 (1.58) A1 [3] (ii) two straight lines from (0,0) with S having the steepest gradient B1 [1]

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4 (a) By reference to the direction of the propagation of energy, state what is meant by a longitudinal wave and by a transverse wave. longitudinal: ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... transverse: ................................................................................................................................ ................................................................................................................................................... ................................................................................................................................................... [2] (b) The intensity of a sound wave passing through air is given by Ι = Kvρf 2A2 where Ι is the intensity (power per unit area), K is a constant without units, v is the speed of sound, ρ is the density of air, f is the frequency of the wave and A is the amplitude of the wave. Show that both sides of the equation have the same SΙ base units. [3] (c) (i) Describe the Doppler effect. ........................................................................................................................................... .......................................................................................................................................[1] (ii) A distant star is moving away from a stationary observer. State the effect of the motion on the light observed from the star. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (d) A car travels at a constant speed towards a stationary observer. The horn of the car sounds at a frequency of 510 Hz and the observer hears a frequency of 550 Hz. The speed of sound in air is 340 m s–1. Calculate the speed of the car. speed = ................................................ m s–1 [3] [Total: 10]

Mark scheme: 4 (a) longitudinal: vibrations/oscillations (of the particles/wave) are parallel to the direction or in the same direction (of the propagation of energy) B1 transverse: vibrations/oscillations (of the particles/wave) are perpendicular to the direction (of the propagation of energy) B1 [2] (b) LHS: intensity = power / area units: kg m s–2 × m × s–1 × m–2 or kg m2 s–3 × m–2 B1 RHS: units: m s–1 × kg m–3 × s–2 × m2 M1 LHS and RHS both kg s–3 A1 [3] (c) (i) change/difference in the observed/apparent frequency when the source is moving (relative to the observer) B1 [1] (ii) wavelength increases/frequency decreases/red shift B1 [1] (d) observed frequency = vfS / (v – vS) C1 550 = (340 × 510) / (340 – vS) C1 vS = 25 (24.7) m s–1 A1 [3]

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5 (a) Light of a single wavelength is incident on a diffraction grating. Explain the part played by diffraction and interference in the production of the first order maximum by the diffraction grating. diffraction: ................................................................................................................................. ................................................................................................................................................... interference: .............................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... [3] (b) The diffraction grating illustrated in Fig. 5.1 is used with light of wavelength 486 nm. second order first order light wavelength 486 nm 59.4° zero order diffraction grating first order second order screen Fig. 5.1 (not to scale) The orders of the maxima produced are shown on the screen in Fig. 5.1. The angle between the two second order maxima is 59.4°. Calculate the number of lines per millimetre of the grating. number of lines per millimetre = ................................................ mm–1 [3] [Total: 6]

Mark scheme: 5 (a) diffraction: spreading/diverging of waves/light (takes place) at (each) slit/ element/gap/aperture B1 interference: overlapping of waves (from coherent sources at each element) B1 path difference λ/phase difference of 360(°)/2π (produces the first order) B1 [3] (b) d sinθ = nλ or sinθ = Nnλ C1 d = (2 × 486 × 10–9) / sin 29.7° (= 1.962 × 10–6) C1 number of lines = 510 (509.7) mm–1 A1 [3]

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6 Two parallel vertical metal plates are connected to a power supply, as shown in Fig. 6.1. metal plate metal plate 16 mm + – Fig. 6.1 The separation of the plates is 16 mm. (a) On Fig. 6.1, draw at least six field lines to represent the electric field between the plates. [1] (b) An α-particle travels in a vacuum between the two plates. The electric field does work on the α-particle. The gain in kinetic energy of the α-particle is 15 keV. Calculate the electric field strength between the plates. electric field strength = ................................................ V m–1 [4] [Total: 5]

Mark scheme: 6 (a) at least six horizontal lines equally spaced and arrow to the right B1 [1] (b) charge used 2e C1 gain in KE = 15 × 1.6 × 10–19 × 103 = 2 × 1.6 × 10–19 × V (p.d.across plates) or F (= W / d) = 15 × 1.6 × 10–19 × 103 / 16 × 10–3 C1 (hence V = 7500 V or F = 1.5 × 10–13 N) E = V / d or E = F / Q C1 E = (7500 / 16 × 10–3) or E = (1.5 × 10–13 / 3.2 × 10–19) E = 4.7 × 105 (468 750) V m–1 A1 [4] or KE (= ½mv2) = 15 × 103 × 1.6 × 10–19 v = [(2 × 15 × 103 × 1.6 × 10–19) / (6.68 × 10–27)]1/2 = 8.5 × 105 m s–1 (C1) a (= v2 / 2s) = (8.5 × 105)2 / 2 × 16 × 10–3 = 2.25 × 1013 m s–2 F (= 6.68 × 10–27 × 2.25 × 10–13) = 1.5 × 10–13 N E = F / Q (C1) Q = 2e (C1) E = 4.7 × 105 V m–1 (A1)

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7 (a) Electric current is a flow of charge carriers. The charge on the carriers is quantised. Explain what is meant by quantised. ...............................................................................................................................................[1] (b) A battery of electromotive force (e.m.f.) 9.0 V and internal resistance 0.25 Ω is connected in series with two identical resistors X and a resistor Y, as shown in Fig. 7.1. battery 9.0 V 0.25 1 X Y X 0.15 1 2.7 1 0.15 1 Fig. 7.1 The resistance of each resistor X is 0.15 Ω and the resistance of resistor Y is 2.7 Ω. (i) Show that the current in the circuit is 2.8 A. [3] (ii) Calculate the potential difference across the battery. potential difference = ...................................................... V [2] (c) Each resistor X connected in the circuit in (b) is made from a wire with a cross-sectional area of 2.5 mm2. The number of free electrons per unit volume in the wire is 8.5 × 1029 m–3. (i) Calculate the average drift speed of the electrons in X. drift speed = ................................................ m s–1 [2] (ii) The two resistors X are replaced by two resistors Z made of the same material and length but with half the diameter. Describe and explain the difference between the average drift speed in Z and that in X. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 10]

Mark scheme: 7 (a) charge exists only in discrete amounts B1 [1] (b) (i) E = I(R + r) or V = IR C1 (total resistance =) 2.7 + 0.30 + 0.25 (= 3.25 Ω) M1 I = 9.0 / (2.7 + 0.30 + 0.25) or 9.0 / 3.25 = 2.8 A A1 [3] (ii) V = IRext C1 = 2.77 × 3.0 or 2.8 × 3.0 or V = E – Ir (C1) = 9.0 – 2.77 × 0.25 or 9.0 – 2.8 × 0.25 V = 8.3 (8.31) V or 8.4 V A1 [2] (c) (i) I = nevA v = 2.77 / (8.5 × 1029 × 1.6 × 10–19 × 2.5 × 10–6) M1 = 8.1 (8.147) × 10–6 m s–1 or 8.2 × 10–6 m s–1 A1 [2] (ii) A reduces by a factor 4 (1/4 less) or resistance of Z goes up by 4× M1 current goes down but by less than a factor of 4 (as total resistance does not go up by a factor of 4) so drift speed goes up A1 [2]

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8 (a) State the name of the class (group) to which each of the following belongs: electron ............................................................... neutron ................................................................ neutrino ............................................................... proton .................................................................. [2] (b) A proton may decay into a neutron together with two other particles. (i) Complete the following to give an equation that represents this proton decay. 11p ................ n + ......................... + ......................... [2] (ii) Write an equation for this decay in terms of quark composition. [1] (iii) State the name of the force responsible for this decay. .......................................................................................................................................[1] [Total: 6]

Mark scheme: 8 (a) both electron and neutrino: lepton(s) B1 both neutron and proton: hadron(s)/baryon(s) B1 [2] (b) (i) 11p → 10 n + 01β + 00ν correct symbols for particles M1 correct numerical values (allow no values on neutrino) A1 [2] (ii) up up down or uud → up down down or udd B1 [1] (iii) weak (nuclear) B1 [1]

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