Cambridge A Level Physics 9702 — 2017 Oct/Nov Paper 2 · Variant 3

1 (a) (i) Define power. ........................................................................................................................................... .......................................................................................................................................[1] (ii) Show that the SI base units of power are kg m2 s–3. [1] (b) All bodies radiate energy. The power P radiated by a body is given by P = kAT 4 where T is the thermodynamic temperature of the body, A is the surface area of the body and k is a constant. (i) Determine the SI base units of k. base units ...........................................................[2] (ii) On Fig. 1.1, sketch the variation with T 2 of P. The quantity A remains constant. P 0 0 T 2 Fig. 1.1 [1] [Total: 5]

Mark scheme: 1(a)(i) work (done) / time (taken) or energy (transferred) / time (taken) B1 1(a)(ii) Correct substitution of base units of all quantities into any correct equation for power. Examples: (P = E / t or W / t gives) kg m2 s–2 / s = kg m2 s–3 (P = Fs / t or mgh / t gives) kg m s–2 m / s = kg m2 s–3 (P = ½mv2/ t gives) kg (m s–1)2 / s = kg m2 s–3 (P = Fv gives) kg m s–2 m s–1 = kg m2 s–3 (P = VI gives) kg m2 s–2 A–1 s–1 A = kg m2 s–3 A1 1(b)(i) units of A: m2 and units of T: K C1 units of k: kg m2 s–3 / m2 K4 = kg s–3 K–4 A1 1(b)(ii) curve from the origin with increasing gradient B1

More questions on SI units

2 A liquid of density ρ fills a container to a depth h, as shown in Fig. 2.1. container liquid h base area A Fig. 2.1 The base of the container has area A. (a) Derive, from the definitions of pressure and density, the equation p = ρgh where p is the pressure exerted by the liquid on the base of the container and g is the acceleration of free fall. [3] (b) A small solid sphere falls with constant velocity through the liquid. (i) State 1. the names of the three forces acting on the sphere, .................................................................................................................................... .................................................................................................................................... 2. a word equation that relates the magnitudes of these forces. .................................................................................................................................... [2] (ii) State and explain the changes in energy that occur as the sphere falls. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (c) The liquid in the container is liquid L. Liquid M is now added to the container. The two liquids do not mix. The total depth of the liquids is 0.17 m. Fig. 2.2 shows how the pressure p inside the liquids varies with height x above the base of the container. 9.25 p / 104 Pa liquid L 9.20 9.15 liquid M 9.10 0 0.05 0.10 0.15 0.20 x / m Fig. 2.2 Use Fig. 2.2 to (i) state the value of atmospheric pressure, atmospheric pressure = .................................................... Pa [1] (ii) determine the density of liquid M. density = ............................................... kg m–3 [2] [Total: 10]

Mark scheme: 2(a) B1 p = F / A or p = W / A B1 p = [ρAhg] / A or p = [ρVg] / [V / h] (so) p = ρgh A1 Question Answer Marks 2(b)(i) 1. weight/gravitational (force) upthrust (force)/buoyancy (force) drag/viscous/frictional (force)/fluid resistance/resistance B1 2. weight = upthrust + viscous (force) B1 2(b)(ii) • decrease in (gravitational) potential energy (of sphere) due to decrease in height (since Ep = mgh) • increase in thermal energy due to work done against viscous force/drag • loss/change of (total) Ep equal to gain/change in thermal energy Any 2 points. B2 2(c)(i) atmospheric pressure = 9.1(0) × 104 Pa A1 2(c)(ii) (∆)p = ρg(∆)h (9.15 – 9.10) × 104 = ρ × 9.81 × (0.17 – 0.10) C1 ρ = 730 (728) kg m–3 A1

More questions on Momentum and Newton’s laws of motion

3 (a) State the principle of conservation of momentum. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Ball A moves with speed v along a horizontal frictionless surface towards a stationary ball B, as shown in Fig. 3.1. 6.0 m s–1 4.0 kg A v initial path θ A B of ball A 30° 4.0 kg 12 kg 12 kg B 3.5 m s–1 after collision before collision Fig. 3.1 Fig. 3.2 (not to scale) Ball A has mass 4.0 kg and ball B has mass 12 kg. The balls collide and then move apart as shown in Fig. 3.2. Ball A has velocity 6.0 m s–1 at an angle of θ to the direction of its initial path. Ball B has velocity 3.5 m s–1 at an angle of 30° to the direction of the initial path of ball A. (i) By considering the components of momentum at right-angles to the direction of the initial path of ball A, calculate θ. θ = ........................................................ ° [3] (ii) Use your answer in (i) to show that the initial speed v of ball A is 12 m s–1. Explain your working. [2] (iii) By calculation of kinetic energies, state and explain whether the collision is elastic or inelastic. ........................................................................................................................................... .......................................................................................................................................[3] [Total: 10]

Mark scheme: 3(a) sum/total momentum (of system of bodies) is constant or sum/total momentum before = sum/total momentum after M1 for an isolated system/no (resultant) external force A1 3(b)(i) p = mv C1 (4.0 × 6.0 × sin θ) – (12 × 3.5 × sin 30°) = 0 or (mAvA × sinθ) – (mBvB × sin 30°) = 0 M1 θ = 61° A1 Question Answer Marks 3(b)(ii) shows the horizontal momentum component of ball A or of ball B as (4.0 × 6.0 × cos θ) or (12 × 3.5 × cos 30°) C1 (4.0 × 6.0 × cos 61°) + (12 × 3.5 × cos 30°) = 4.0v so v = 12 (m s–1) A1 3(b)(iii) initial EK (= ½ × 4.0 × 122) = 290 (288) (J) M1 final EK (= ½ × 4.0 × 6.02 + ½ × 12 × 3.52) = 150 (145.5) (J) M1 (initial EK > final EK) so inelastic [both M1 marks required to award this mark] A1

More questions on Momentum and Newton’s laws of motion

4 (a) By reference to the direction of propagation of energy, explain what is meant by a longitudinal wave. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A car horn emits a sound wave of frequency 800 Hz. A microphone and a cathode-ray oscilloscope (c.r.o.) are used to analyse the sound wave. The waveform displayed on the c.r.o. screen is shown in Fig. 4.1. 1 cm 1 cm Fig. 4.1 Determine the time-base setting, in s cm–1, of the c.r.o. time-base setting = ............................................... s cm–1 [3] (c) The intensity I of the sound at a distance r from the car horn in (b) is given by the expression k I = 2 r where k is a constant. Fig. 4.2 shows the car in (b) on a road. O Y X road 30 m 120 m Fig. 4.2 An observer stands at point O. Initially the car is parked at point X which is 120 m away from point O. The car then moves directly towards the observer and stops at point Y, a distance of 30 m away from O. The car horn continuously emits sound when the car is moving between points X and Y. (i) The sound wave at point O has amplitude AX when the car is at X and has amplitude AY when the car is at Y. AY Calculate the ratio . AX ratio = ...........................................................[3] (ii) When the car is parked at X, the frequency of the sound from the horn that is detected by the observer is 800 Hz. As the car moves from X to Y, the maximum change in the detected frequency is 16 Hz. The speed of the sound in air is 330 m s–1. Determine, to two significant figures, 1. the minimum wavelength of the sound detected by the observer, wavelength = ...................................................... m [2] 2. the maximum speed of the car. speed = ................................................. m s–1 [2] [Total: 11]

Mark scheme: 4(a) displacement of particles/vibration(s)/oscillation(s) is parallel to/along the direction of energy/propagation B1 4(b) period = 1 / 800 (= 1.25 × 10–3 s) C1 time-base setting = 1.25 × 10–3 / 2.5 C1 = 5.0 × 10–4 s cm–1 A1 4(c)(i) I ∝ A2 C1 (IX / IY =) [rY / rX] 2 = [AX / AY]2 C1 ratio AY / AX = 120 / 30 = 4.0 A1 Question Answer Marks 4(c)(ii) 1. v = f λ C1 minimum λ = 330 / (800 + 16) = 0.40 m A1 2. fo / fs = v / (v – vs) 816 / 800 = 330 / (330 – vs) C1 vs = 6.5 m s–1 A1

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5 (a) Define electric field strength. ................................................................................................................................................... ...............................................................................................................................................[1] (b) Two parallel metal plates in a vacuum are separated by 0.045 m. A potential difference V is applied between the plates, as shown in Fig. 5.1. metal plate – 0.045 m V + metal plate proton Fig. 5.1 A proton is initially at rest on the surface of the positive plate. The proton in the uniform electric field takes a time of 1.5 × 10–7 s to reach the negative plate. (i) Show that the acceleration of the proton is 4.0 × 1012 m s–2. [2] (ii) Calculate the electric force on the proton. force = ...................................................... N [1] (iii) Use your answer in (ii) to determine 1. the electric field strength, field strength = ................................................ N C–1 [2] 2. the potential difference V between the plates. V = ...................................................... V [2] (c) An α particle is now accelerated between the two metal plates in (b) by the electric field. Calculate the ratio acceleration of α particle . acceleration of proton ratio = ...........................................................[2] [Total: 10]

Mark scheme: 5(a) force per unit positive charge B1 5(b)(i) s = ½at 2 C1 a = (2 × 0.045) / (1.5 × 10–7)2 = 4(.0) × 1012 m s–2 A1 5(b)(ii) F = 1.67 × 10–27 × 4.0 × 1012 = 6.7 (6.68) × 10–15N A1 5(b)(iii) 1. E = F / Q C1 = 6.68 × 10–15 / 1.6 × 10–19 = 4.2 (4.18) × 104 N C–1 A1 2. E = V / d C1 V = 4.18 × 104 × 0.045 = 1.9 × 103 V A1 Question Answer Marks 5(c) a = Eq / m or F = ma and F = Eq C1 ratio = − − − − × × × × × × × × 19 27 19 27 (2 1.6 10 ) (1.67 10 ) (1.6 10 ) (4 1.66 10 ) or × × 2 1 1 4 = 0.50 A1

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6 A filament lamp is rated as 30 W, 120 V. A potential difference of 120 V is applied across the lamp. (a) For the filament wire of the lamp, calculate (i) the current, current = ....................................................... A [2] (ii) the number of electrons passing a point in 3.0 hours. number = ...........................................................[2] (b) Show that the resistance of the filament wire is 480 Ω. [2] (c) The filament wire has an uncoiled length of 580 mm and is made of metal. The metal has resistivity 6.1 × 10–7 Ω m at the operating temperature of the lamp. Calculate the diameter of the wire. diameter = ...................................................... m [3] (d) The potential difference across the lamp is now reduced. State and explain the effect, if any, on the resistance of the filament wire. ................................................................................................................................................... ...............................................................................................................................................[1] [Total: 10]

Mark scheme: 6(a)(i) C1 I = 30 / 120 = 0.25 A A1 6(a)(ii) Q = 0.25 × 3.0 × 3600 (= 2700) C1 number = (0.25 × 3.0 × 3600) / 1.60 × 10–19 = 1.7 × 1022 A1 6(b) R = V / I or R = P / I 2 or R = V 2/ P C1 = 120 / 0.25 or = 30 / 0.252 or = 1202/ 30 = 480 Ω A1 Question Answer Marks 6(c) R = ρl / A C1 A = (6.1 × 10–7 × 580 × 10–3) / 480 (= 7.37 × 10–10) C1 d = [(4 × 7.37 × 10–10) / π]1/2 = 3.1 × 10–5 m A1 6(d) temperature decreases and so resistance decreases B1

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7 (a) A nucleus X decays by emitting a β+ particle to form a new nucleus, 2311Na. State the number of nucleons and the number of neutrons in nucleus X. number of nucleons = ............................................................... number of neutrons = ............................................................... [2] (b) State one similarity and one difference between a β+ particle and a β– particle. similarity: ................................................................................................................................... difference: ................................................................................................................................. [2] [Total: 4]

Mark scheme: 7(a) nucleons = 23 B1 neutrons = 11 B1 7(b) similarity: same (rest) mass or equal (magnitude of) charge B1 difference: opposite (sign of) charge or one is matter and one is antimatter or one is an electron and one is an antielectron B1

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