IJC H1 PHY P2 SOLUTIONS

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2 © IJC 2011 8866/Prelim2 [Turn over For Examiner’s Use Data speed of light in free space, c = 3.00 x 108 m s-1 elementary charge, e = 1.60 x 10-19 C the Planck constant, h = 6.63 x 10-34 J s unified atomic mass constant, u = 1.66 x 10-27 kg rest mass of electron, me = 9.11 x 10-31 kg rest mass of proton, mp = 1.67 x 10-27 kg acceleration of free fall, g = 9.81 m s-2 Formulae uniformly accelerated motion, s= u t + ½ a t2 v2 = u2+ 2as work done on/by a gas, W = pV hydrostatic pressure, p = gh resistors in series, R= R 1 + R2 + … resistors in parallel, 1/R = 1/R 1 + 1/R2 + …

3 © IJC 2011 8866/Prelim2 [Turn over For Examiner’s Use Section A Answer all the questions in this section. 1 (a) Complete Fig. 1.1 to show each quantity and its base units. Quantity Base Unit speed m s -1 density kg m-3 power kg m 2 s-3 voltage kg m2 s-3 A-1 [3] Fig. 1.1 (b) Two parallel strings S 1 and S2 are attached to a disc of diameter 12 cm, as shown in Fig. 1.2. Fig. 1.2 The disc is free to rotate about an axis normal to its plane. The axis passes through the centre C of the disc. A lever of length 30 cm is attached to the disc. When a force F is applied at right angles to the lever at its end, equal forces are produced in S 1 and S 2. The disc remains in equilibrium. (i) On Fig. 1.2, show the direction of the force in each string that acts on the disc. [1] Force is leftward for string S1 and rightward for string S2. (B1) string S2 string S1 force F lever C 30 cm 12 cm disc

4 © IJC 2011 8866/Prelim2 [Turn over For Examiner’s Use (ii) For a force F of magnitude 150 N, determine 1. the moment of force F about the centre of the disc, Moment = F x d = 150 x 0.30 = 45 N m (C1) moment = N m [1] 2. the torque of the couple produced by the forces in the strings, By Principle of Moments, torque of couple produced by strings = Clockwise moment caused by F = 45 N m (C1) torque = N m [1] 3. the force in S1. Torque of couple in strings = S 1 x 0.06 = 45 (C1) Hence S1 = 45 / 0.06 = 750 N (A1) force = N [2]

5 © IJC 2011 8866/Prelim2 [Turn over For Examiner’s Use 2 A steady stream of water strikes a wall horizontally without rebounding, and, as a result, exerts a force on the vertical wall. (a) With reference to Newton’s Laws of motion, (i) state and explain why the momentum of the water changes as it strikes the wall, The momentum of the water changes because the velocity of the water changes when the water was brought to rest by the wall. [B1] This is because when the water strikes the wall, it exerts a force on the wall and by Newton’s third law, the wall exerts an equal but opposite force on the water. [B1] By Newton’s second law, the force experienced by the water means that the water undergoes a rate of change of momentum . Thus the momentum of the water changes. [B1] [3] (ii) explain why the water exerts a constant force on the wall. The water exerts a constant force because the water flows at a constant rate. [B1] [1] (b) Water arrives at the wall at a rate of 18 kg s -1. It strikes the wall horizontally, at a speed of 7.2 m s-1 without rebounding. Calculate (i) the change in momentum of the water in one second, change in momentum in one second = mv [C1] = 18(0 – 7.2) = - 130 kg m s -1 [A1] change in momentum = kg m s -1 [2] (ii) the force exerted by the water on the wall. Force by the water on the wall = - force on water by wall = 130 kg m s -2 [A1] force = kg m s -2 [1] (c) State and explain the effect on the magnitude of the force if the water rebounds after striking the wall. The magnitude is greater [A1] because there is a bigger rate of change of momentum of the water when the water rebounds. [M1] [2]

6 © IJC 2011 8866/Prelim2 [Turn over For Examiner’s Use 3 (a) What do you understand by progressive wave? [1] (b) A radar transmitter produces pulses of microwaves each with a mean power P = 2.0 MW which are emitted uniformly in all directions. A small spherical target of effective area S is placed at a distance of 50 km from the transmitter. The target reflects k of the energy incident on it uniformly in all directions as shown in Fig 3.1. Fig. 3.1 (i) Calculate the mean intensity of the emitted pulse at a range of 50 km. 2 x 106 / (4 x  x 502 x 106) = 6.4 x 10-5 W m-2 intensity = W m -2 [2] (ii) Given that S = 1.0 m2. Calculate the power received by a target. 6.4 x 10-5 x 1 = 6.4 x 10-5 W power = W m -2 [2] (iii) Assuming that the fraction of energy reflected is k = 0.50, calculate the mean intensity of the reflected pulse when received back at the transmitter. I = P / (4r2) = {0.50 x 6.4 x 10-5} / {4 x 502 x 106} = 1.0 x 10-15 W m-2 intensity = W m -2 [3] transmitter target 50 km

7 © IJC 2011 8866/Prelim2 [Turn over For Examiner’s Use 4 A household electric lamp is rated as 240 V, 60 W. The filament of the lamp is made from tungsten and is a wire of constant radius 6.0 x 10 6 m. The resistivity of tungsten at the normal operating temperature of the lamp is 7.9 x 107  m. (a) State Ohm’s law. Ohm’s law states that for an ohmic conductor, the ratio of its potential difference to its current is constant, at constant temperature. [C1] (b) For the lamp at its normal operating temperature, (i) calculate the current in the lamp, Current = P / V = (60) / (240) = 0.25 A [A1] current = A [1] (ii) show that the resistance of the filament is 960 . Resistance, R = V / I = (240) / (0.25) [C1] = 960  current = A [1] (c) Calculate the length of the filament. Using, R =  l / A l = R A /  = (960) () (6.0 x 106 m)2 / (7.9 x 107  m) [M1] = 0.137 m [A1] length = m [2] (d) Comment on your answer to (c). The length of the filament must be coiled to fit into the light bulb. [A1]

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