NJC_H1_PHY_Solutions

2016 NJC SH2 H1 Physics Prelim Paper Suggested Solutions Paper 1 1 C 11 A 21 B 2 D 12 B 22 B 3 B 13 A 23 B 4 C 14 C 24 B 5 D 15 C 25 C 6 D 16 A 26 B 7 C 17 A 27 C 8 A 18 A 28 B 9 A 19 B 29 D 10 C 20 A 30 B Paper2 Section A 1(a)(i) Now consider a body of mass m at rest brought to velocity v over a distance s by a constant force F. The final velocity v is given by sauv 222  , where a is constant acceleration given by m Fa  . We have sm Fuv 222  . Rearranging, 22 2 1 2 1 mumvsF  The change in Kinetic Energy of an object equals the net work done on the object. Since the body starts from rest ( 0u ), its final kinetic energy of body = work done on body by F = 212kE mv . 1(a)(ii) Since the speed is constant, the change in KE is zero. Hence, by COE, the chemical energy is all used to do work against air resistance and thus no such transformation taking place. 1(b)(i) P = W / t 36.6 x 103 = W / 300 W = 1.1x107 J 1(b)(ii) Not worthwhile, since the KE =3.46 x 105 J even when it is moving at 31 m s-1 and thus will be even lower when it slows down. Hence, since the KE will be so much lower than the WD against the resistive force, a large part of the energy would still be provided by the fuel of the car.

2(a) Consider a collision that occurs when A collides with B in a straight line. By Newton’s second law the change in momentum for A, ΔpA = FBA*Δt, where FBA is the force B exerts on A and Δt is the duration the force is exerted while the change in momentum for B, ΔpB = FAB*Δt, where FAB is the force A exerts on B. By Newton’s third law, FBA = - FAB since they are an action-reaction pair. Hence, ΔpA = - ΔpB. This implies pAF - pAI = - (pBF – pBI), where pAF is the final momentum of A, pAI is the initial momentum of A, pBF is the final momentum of B and pBI is the initial momentum of B. Rearranging, pAI + pBI = pAF + pBF. This implies the total initial momentum is the same as the total final momentum if no external force acts on this system. 2(b)(i) The total initial momentum of the system is zero. Hence the total final momentum is zero, implying that the final momentum of magnet A = - (final momentum of magnet B). If the mass of both magnets are the same, then the final velocity of magnet A = - (final velocity of magnet B). 2(b)(ii) Work was done on both magnets since an external force was applied to oppose the repulsive magnetic force between the two magnets. This was stored as potential energy between the two magnets resulting in an increase in kinetic energy when the external force was removed. 3(a) (i) Electric current is rate of flow of charged particles. The resistance of a body is defined as the ratio of potential difference across it to the current passing through it. (ii) Q = It = 25000 (40 x 10-6) = 1.00 C (iii) V = IR = 5 x10-3(1000) = 5.00 V 3(b) (i) R = ρ 2)05.0( 6.15  A L = 1018 = 1020  (ii) Yes. When