Cambridge A Level Physics 9702 — 2012 May/June Paper 2 · Variant 1

1 (a) (i) State the SI base units of volume. base units of volume ................................................. [1] (ii) Show that the SI base units of pressure are kg m–1 s–2. [1] (b) The volume V of liquid that flows through a pipe in time t is given by the equation V π Pr 4 = t 8Cl where P is the pressure difference between the ends of the pipe of radius r and length l. The constant C depends on the frictional effects of the liquid. Determine the base units of C. base units of C ................................................. [3]

Mark scheme: 1 (a) (i) V units: m3 (allow metres cubed or cubic metres) A1 [1] (ii) Pressure units: kg m s–2 / m2 (allow use of P = ρgh) M1 Units: kg m–1 s–2 A0 [1] (b) V / t units: m3 s–1 B1 Clear substitution of units for P, r4 and l M1 π P r 4 kg m −1 s −2 m 4 C = = 8 V t −1 l m 3 s −1 m Units: kg m–1 s–1 A1 [3] (8 or π in final answer –1. Use of dimensions max 2/3)

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2 A ball is thrown vertically down towards the ground with an initial velocity of 4.23 m s–1. The For ball falls for a time of 1.51 s before hitting the ground. Air resistance is negligible. Examiner’s Use (a) (i) Show that the downwards velocity of the ball when it hits the ground is 19.0 m s–1. [2] (ii) Calculate, to three significant figures, the distance the ball falls to the ground. distance = ............................................. m [2] (b) The ball makes contact with the ground for 12.5 ms and rebounds with an upwards velocity of 18.6 m s–1. The mass of the ball is 46.5 g. (i) Calculate the average force acting on the ball on impact with the ground. magnitude of force = .................................................. N direction of force ...................................................... [4] (ii) Use conservation of energy to determine the maximum height the ball reaches after it hits the ground. height = ............................................. m [2] (c) State and explain whether the collision the ball makes with the ground is elastic or inelastic. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[1]

Mark scheme: 2 (a) (i) v = u + at C1 = 4.23 + 9.81 × 1.51 M1 = 19.0(4) m s–1 (Allow 2 s.f.) A0 [2] (Use of –g max 1/2. Use of g = 10 max 1/2. Allow use of 9.8. Allow 19 m s–1) (ii) either s = ut + ½ at2 (or v2 = u2 + 2as etc.) = 4.23 × 1.51 + 0.5 × 9.81 × (1.51)2 C1 = 17.6 m (or 17.5 m) A1 [2] (Use of –g here wrong physics (0/2)) (b) (i) F = ∆P / ∆t need idea of change in momentum C1 = [0.0465 × (18.6 + 19)] / 12.5 × 10–3 C1 = 140 N A1 (Use of – sign max 2/4. Ignore –ve sign in answer) Direction: upwards B1 [4] (ii) h = ½ × (18.6)2 / 9.81 C1 = 17.6 m (2 s.f. –1) A1 [2] (Use of 19 m s–1, 0/2 wrong physics) (c) either kinetic energy of the ball is not conserved on impact or speed before impact is not equal to speed after hence inelastic B1 [1]

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3 One end of a spring is fixed to a support. A mass is attached to the other end of the spring. For The arrangement is shown in Fig. 3.1. Examiner’s Use mass Fig. 3.1 (a) The mass is in equilibrium. Explain, by reference to the forces acting on the mass, what is meant by equilibrium. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) The mass is pulled down and then released at time t = 0. The mass oscillates up and down. The variation with t of the displacement of the mass d is shown in Fig. 3.2. 6.0 d / 10–2 m 4.0 2.0 0 0 0.2 0.4 0.6 0.8 1.0 t / s –2.0 –4.0 –6.0 Fig. 3.2 Use Fig. 3.2 to state a time, one in each case, when (i) the mass is at maximum speed, time = .............................................. s [1] (ii) the elastic potential energy stored in the spring is a maximum, time = .............................................. s [1] (iii) the mass is in equilibrium. time = .............................................. s [1] (c) The arrangement shown in Fig. 3.3 is used to determine the length l of a spring when For different masses M are attached to the spring. Examiner’s Use l mass Fig. 3.3 The variation with mass M of l is shown in Fig. 3.4. 35 30 25 l / 10–2 m 20 15 10 5 0 0 0.10 0.20 0.30 0.40 0.50 M / kg Fig. 3.4 (i) State and explain whether the spring obeys Hooke’s law. For Examiner’s .................................................................................................................................. Use .................................................................................................................................. ..............................................................................................................................[2] (ii) Show that the force constant of the spring is 26 N m–1. [2] (iii) A mass of 0.40 kg is attached to the spring. Calculate the energy stored in the spring. energy = .............................................. J [3]

Mark scheme: 3 (a) Resultant force (and resultant torque) is zero B1 Weight (down) = force from/due to spring (up) B1 [2] (b) (i) 0.2, 0.6, 1.0 s (one of these) A1 [1] (ii) 0, 0.8 s (one of these) A1 [1] (iii) 0.2, 0.6, 1.0 s (one of these) A1 [1] GCE AS/A LEVEL – May/June 2012 9702 21 (c) (i) Hooke’s law: extension is proportional to the force (not mass) B1 Linear/straight line graph hence obeys Hooke’s law B1 [2] (ii) Use of the gradient (not just F = kx) C1 K = (0.4 × 9.8) / 15 × 10–2 M1 = 26(.1) N m–1 A0 [2] (iii) either energy = area to left of line or energy = ½ ke2 C1 = ½ × [(0.4 × 9.8) / 15 × 10–2] × (15 × 10–2)2 C1 = 0.294 J (allow 2 s.f.) A1 [3] 2

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4 (a) The output of a heater is 2.5 kW when connected to a 220 V supply. For Examiner’s (i) Calculate the resistance of the heater. Use resistance = ............................................. Ω [2] (ii) The heater is made from a wire of cross-sectional area 2.0 × 10–7 m2 and resistivity 1.1 × 10–6 Ω m. Use your answer in (i) to calculate the length of the wire. length = ............................................. m [3] (b) The supply voltage is changed to 110 V. (i) Calculate the power output of the heater at this voltage, assuming there is no change in the resistance of the wire. power = ............................................. W [1] (ii) State and explain quantitatively one way that the wire of the heater could be changed to give the same power as in (a). .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

Mark scheme: 4 (a) (i) R = V2 / P or P = IV and V = IR C1 = (220)2 / 2500 = 19.4 Ω (allow 2 s.f.) A1 [2] (ii) R = ρl / A C1 l = [19.4 × 2.0 × 10–7] / 1.1 × 10–6 C1 = 3.53 m (allow 2 s.f.) A1 [3] (b) (i) P = 625, 620 or 630 W A1 [1] (ii) R needs to be reduced C1 Either length ¼ of original length or area 4× greater or diameter 2× greater A1 [2]

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5 (a) (i) State Kirchhoff’s second law. For Examiner’s .................................................................................................................................. Use ..............................................................................................................................[1] (ii) Kirchhoff’s second law is linked to the conservation of a certain quantity. State this quantity. ..............................................................................................................................[1] (b) The circuit shown in Fig. 5.1 is used to compare potential differences. cell A 2.0 V 0.50 Ω C D I R 0.90 m X J Y E r uniform resistance wire length 1.00 m cell B Fig. 5.1 The uniform resistance wire XY has length 1.00 m and resistance 4.0 Ω. Cell A has e.m.f. 2.0 V and internal resistance 0.50 Ω. The current through cell A is I. Cell B has e.m.f. E and internal resistance r. The current through cell B is made zero when the movable connection J is adjusted so that the length of XJ is 0.90 m. The variable resistor R has resistance 2.5 Ω. (i) Apply Kirchhoff’s second law to the circuit CXYDC to determine the current I. I = .............................................. A [2] (ii) Calculate the potential difference across the length of wire XJ. For Examiner’s Use potential difference = .............................................. V [2] (iii) Use your answer in (ii) to state the value of E. E = .............................................. V [1] (iv) State why the value of the internal resistance of cell B is not required for the determination of E. .................................................................................................................................. ..............................................................................................................................[1]

Mark scheme: 5 (a) (i) sum of e.m.f.’s = sum of p.d.’s around a loop/circuit B1 [1] (ii) energy B1 [1] (b) (i) 2.0 = I × (4.0 + 2.5 + 0.5) C1 I = 0.286 A (allow 2 s.f.) A1 [2] (If total resistance is not 7 Ω, 0/2 marks) (ii) R = [0.90 / 1.0] × 4 (= 3.6) C1 V = I R = 0.286 × 3.6 = 1.03 V A1 [2] (If factor of 0.9 not used, then 0/2 marks) (iii) E = 1.03 V A1 [1] (iv) either no current through cell B or p.d. across r is zero B1 [1]

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6 (a) A laser is used to produce an interference pattern on a screen, as shown in Fig. 6.1. For Examiner’s Use P2 P1 laser light 0.450 mm wavelength 630 nm screen double slit 1.50 m Fig. 6.1 (not to scale) The laser emits light of wavelength 630 nm. The slit separation is 0.450 mm. The distance between the slits and the screen is 1.50 m. A maximum is formed at P1 and a minimum is formed at P2. Interference fringes are observed only when the light from the slits is coherent. (i) Explain what is meant by coherence. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) Explain how an interference maximum is formed at P1. .................................................................................................................................. ..............................................................................................................................[1] (iii) Explain how an interference minimum is formed at P2. .................................................................................................................................. ..............................................................................................................................[1] (iv) Calculate the fringe separation. fringe separation = ............................................. m [3] (b) State the effects, if any, on the fringes when the amplitude of the waves incident on the For double slits is increased. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]

Mark scheme: 6 (a) (i) coherence: constant phase difference M1 between (two) waves A1 [2] (ii) path difference is either λ or nλ or phase difference is 360° or n × 360° or n2π rad B1 [1] GCE AS/A LEVEL – May/June 2012 9702 21 (iii) path difference is either λ/2 or (n + ½) λ or phase difference is odd multiple of either 180° or π rad B1 [1] (iv) w = λD / a C1 = [630 × 10–9 × 1.5] / 0.45 × 10–3 C1 = 2.1 × 10–3 m A1 [3] (b) no change to dark fringes B1 no change to separation/fringe width B1 bright fringes are brighter/lighter/more intense B1 [3]

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7 (a) The spontaneous decay of polonium is shown by the nuclear equation For Examiner’s 210 84 Po ➞ 20682 Pb + X . Use (i) State the composition of the nucleus of X. .................................................................................................................................. ..............................................................................................................................[1] (ii) The nuclei X are emitted as radiation. State two properties of this radiation. 1. ............................................................................................................................... .................................................................................................................................. 2. ............................................................................................................................... .................................................................................................................................. [2] (b) The mass of the polonium (Po) nucleus is greater than the combined mass of the nuclei of lead (Pb) and X. Use a conservation law to explain qualitatively how this decay is possible. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]

Mark scheme: 7 (a) (i) 2 protons and 2 neutrons B1 [1] (ii) e.g. positively charged 2e mass 4u constant energy absorbed by thin paper or few cm of air (3 cm → 8 cm) (not low penetration) highly ionizing deflected in electric/magnetic fields (One mark for each property, max 2) B2 [2] (b) mass-energy is conserved B1 difference in mass ‘changed’ into a form of energy B1 energy in the form of kinetic energy of the products / γ-radiation photons / e.m. radiation B1 [3]

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