YIJC 2024 JC2 PRELIM H1 Phy P2 Solutions

1 ©YIJC 8867/02/YIJC/24 2024 YIJC JC2 Prelim Exam H1 Physics Paper 2 Solution 1 (a) 1. Initial speed or velocity is zero 2. (non-zero magnitude of) acceleration is constant (uniform) and in a straight line B1 B1 (b) (i) magnitude of acceleration at t = 8.0 s is less than that at t = 14.0 s (due to gradient) direction of acceleration at t = 8.0 s is opposite that at t = 14.0 s (due to sign of gradient) B1 B1 (ii) a = gradient or a = (v-u)/t or a = v/t <a> = (0– (-10)) / 16 a = 0.625 m s-2 C1 A1 (iii) Net displacement (final position) from X = area under v-t graph s = ½ (-10)(4) + ½ (20)(8) = - 20 + 80 = 60 m C1 A1 2 (a) The principle of conservation of momentum states that the total momentum of a system is conserved throughout a collision if there is no net external force acting on the system. B1 (b) (i)             1 , assume is to the right. 0.100 2.0 0.200 0 0.100 0.200 1.3 0.60 m s 2sf ve v v v       Magnitude = 0.60 m s-1 Since v is negative, the direction is to the left. C1 A1 B1 (ii)     1 1 relative speed of approach 2.0 0 2.0 m s relative speed of separation 1.3 0.6 1.9 m s          Since the relative speed of approach is not equal to the relative speed of separation, the collision is inelastic. OR               2 2 2 2 1 1total initial KE 0.100 2.0 0.200 02 2 0.20 J 1 1total final KE 0.100 0.6 0.200 1.32 2 0.187 J 3sf       Since the total initial kinetic energy is not equal to the total final kinetic energy, the collision is inelastic. M1 M1 A1 (M1) (M1) (A1)

2 ©YIJC 8867/02/YIJC/24 (c) By Newton’s third law, the force that block A exerts on block B is equal in magnitude and opposite in direction to the force that block B exerts on block A. By Newton’s second law, the (net) force (on block A) is the rate of change of momentum (of block A.) Hence, the rate of change of momentum of block A is equal in magnitude and opposite in direction to the rate of change of momentum of block B. M1 M1 A1 3 (a) (i) The sum of forces in all directions is zero, and The sum of moments about any axis is zero. B1 B1 (ii) Taking moments about B, the weight of the table top exerts an anticlockwise moment. In order for the table to be in equilibrium, there must be an upward vertical (frictional) component of force acting at the hinge to produce a clockwise moment about B which is equal to the anticlockwise moment of the weight of the table top. OR If there is only a horizontal normal contact force at hinge A, there will be no moment provided by this force and the table will not be in equilibrium. B1 B1 (B1) (b) Taking moments about A,         130 0.30 sin 20 0.60 190 N 3sf T T    C1 A1 (c)  190sin20 130 65.02 N vertical vertical F F     190cos 20 178.5 N horzontalF      2 265.02 178.5 190 N 3sf F      65.02tan 178.5 20.0 3sf vertical horizontal F F      C1 C1 A1 A1 4 (a) The net loss of the gravitational potential energy of the system is converted to the gain in kinetic energy of the system. or Gain in KE of blocks = Loss in GPE of 4.0 kg block – Gain in GPE of 2.0 kg block Not accepted: The GPE of the block/s is converted to KE of the blocks. (mention of net loss is not stated) due to the fact that the blocks are travelling in opposite directions. B1 (b) (i) To be at the same height, 4.0 kg block will descend by 10.0 cm while the 2.0 kg block will rise by 0.10 m. B1

3 ©YIJC 8867/02/YIJC/24 (ii) Loss in GPE for 4.0 kg block = (4.0)(9.81)(0.100) = 3.92 J Gain in GPE for 2.0 kg block = (2.0)(9.81)(0.100) = 1.96 J Gain in KE = Net loss of GPE = 3.92 – 1.96 = 1.96 J ½ (2.0 + 4.0) v2 = 1.96  v = 0.81 m s1 C1 M1 A1 5 (a) (i)         19 19 15 1.6 10 9.38 10 electrons per second 3sf ne t n t I C1 A1 (ii)      8 23 1.72 10 3.0 1.63 10 2 0.0247 3sf R A                l C1 A1 (iii)      15 0.0247 0.371 V 3sf V IR   B1 (iv)   useful supplied 12.0 0.371efficiency 12.0 96.9% 3sf appliance wire V IP P V I     C1 A1 (b) Temperature increase, the lattice ions vibrate more, causing more frequency of collision. This impedes the flow of electrons. Resistivity of the wire increases. M1 M1 A1

4 ©YIJC 8867/02/YIJC/24 6 (a) 2 2(78000) (2.1)(1030)(14.7) D Fv c A  v = 2.215 m s−1 1 2 1 1 0.1 20 0.3 2 100 2.1 1030 14.7 D D cv F A v F c A                      0.1v  m s−1 v = (2.2 ± 0.1) m s−1 [minus 1m for wrong sf] C1 A1 C1 A1 (b) (i) Since the boat is moving with constant velocity, it experiences no acceleration and the net force on it is zero. Thus, the motor provides a forward force and the boat also experience a drag force. The magnitude of these two forces are equal but opposite in direction, resulting in net force of zero. B1 B1 (ii) Using at least 2 sets of value, When m = 200 kg, P = 0.6 kW, P/m = 3 W kg-1 When m = 300 kg, P = 1.9 kW, P/m = 6.3 W kg-1 If power is proportional to the total mass, P/m = k where k is a constant Since P/m is not a constant, P is not proportional to m M1 A1 (iii)1. Point plotted correctly (0.69,0.69) C1 (iii)2. Best fit straight line drawn Linearising the equation (lnP) = n(lnv) + lnk Since the plotted points follow a straight line trend, lnP and lnv have a linear relationship and therefore the proposed relationship is supported. B1 B1

5 ©YIJC 8867/02/YIJC/24 (iii)3. n = gradient of graph = 1.65 ( 0.5) 2.01.2 0.12    Accepted range: 1.9 – 2.1 B1 (iv) P = Fv 1900 = F(2.5) F = 760 N C1 A1 (e) Total energy generated from 1.1 litres of fuel = 32 x 1.1 = 35.2 MJ Distance travelled for 1h = 2.5 x 3600 = 9000 m Total work done by the forward force = 760 x 9000 = 6.84 MJ Efficiency = 6.84/35.2 x 100% = 19 % C1 C1 A1 7 (a) (i) 2mvF = r = 3 2(800)(144×10 / 3600) 80 = 1.6 × 104 N C1 A1 (ii) To the right. B1 (iii) The direction of velocity is changing, so a net force is required to change the velocity. According to Newton’s first law, in the absence of a net force, a moving object will continue to move with a constant speed in the same direction. Since the car is required to turn round the corner, a force must be acting perpendicular to the velocity to enable that manoeuvrer to change direction without changing the speed. B1 (iv) The centripetal force is provided by the frictional force between the wheel and the ground. B1 (b) (i) For the satellite to always remain at the same place directly above the same spot on the Earth spinning below, the satellite must orbit in sync with the Earth’s rotation on the same plane of rotation. Thus, the gravitational force on the satellite which is directed to the centre of the Earth must also be on the same plane as the equator. At places other than at or near the equator, the satellite will be positioned sometimes at the northern hemisphere and sometimes in the southern hemisphere as the orbital plane and equatorial plane do not match. B1

6 ©YIJC 8867/02/YIJC/24 (ii) The centripetal force is provided by the gravitation force. 2 2 GMmF = = mr ωr  3 2GM = r ω         2 -11 24 3 2π6.67×10 5.98×10 = r 24×3600 r = 4.23×107 m altitude = r-R = 3.59×107 m B1 M1 A1 (c) (i) The centripe