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 the circuit is closed, there will be a potential difference drop due the internal resistance (key point), thus the pd across his hand will be lower than 14 kV. (iv) Current through the mechanic, I = mAk k R V 00.7)10182000( 14  [1] Since current through mechanic, I = 7.00 mA is higher than maximum safe current which is 5.0 mA, it will kill the mechanic. [1] 4(a) Principle of Superposition states that when two or more waves meet at a point, the resultant displacement at that point is equal to the vector sum of the displacements due to the individual waves at that point. 4(b)(i) Wave travel down the tube and gets reflected by the water. The incident and the reflected waves, both having the same amplitude, frequency(or wavelength) and speed travelling in opposite directions superpose to form standing wave. OR The incident sound wave travels along the tube and is reflected by the water. The superposition of the incident and reflected wave of same amplitude, speed and wavelength(or frequency) but travelling in opp osite directions creates a standing wave in the pipe.

4(b)(ii) The length of the a ir column in tube will limit the type of stationary wave which can be formed within the air column as there is the boundary condition that it has to be a node formed at the closed end of the air column and a antinode at the open end of the air column. In general, the length of the air column L needs to be equal to (2n+1)/4, where  is the wavelength of the incident wave and n is an integer. If this condition is met, the loudness of the sound will be maximum, otherwise the sound will be of lower amplitude. 4(b)(iii) cL  1 4  -------------(1) c is the end correction cL  2 4 3 --------------(2) (2) – (1)  4.322 1 12  LL cm   = 64.8 cm 332648.0512  fv m s-1 4(b)(iv) ccL  7.154 1 1  c= 7.158.644 1  = 0.50 cm Therefore, antinode is 0.50 cm above the top of the tube OR antinode is 16.2 cm above water surface. There is a presence of end correction. The region of lowest pressure is not at the mouth of the tube but is actually a distance away from end of tube. 5 (a) The waves from the loudspeakers must have about the same amplitude. [1] 5(b)(i) wavelength of sound  = v / f = 340/ 680 = 0.5 m Fringe separation x = D/ a = (0.5)(20) / 3 = 3.33 m Hence for the 3rd minimum, distance = 2.5 x 3.33 = 8.33 m 5(b)(ii) When the two loud speakers are connected, there is destructive interference of the sound waves at point Z. When L2 is disconnected, there is no longer any interference, hence there is no destructive interference at point Z and the intensity of sound will be higher than before. Though there is an increase in intensity of sound at point Z, the total energy of the system is still equal to the output energy of speaker L2, hence it does not conflict with the law of conservation of energy. [When L2 is connected, the total energy of the system was equal to the sum of the output energy of speaker L1 and L2. The energy has been redistributed due to interference as there will be points of constructive and destructive interference.] 5(c) The signal received by the microphone will not be a minimum at places which was previously of maximum intensity and vise-versa.

Paper2 Section B 6 (a)(i)1. [2] 6(a)(i)2. Distance travelled by car A = ½ x 4.0 x 32.0 = 64.0 m Distance travelled by car B = ½ x 4.0 X 27.8 = 55.6 m Distance between the cars = 64.0 - 55.6 = 8.4 m 6(a)(ii)1. Acceleration of car A, aA = 27.8/3.5 = 7.94 ms-2 Acceleration of car B, aB = 27.8/4.0 = 6.95 ms-2 Effective acceleration of car A = a’A = (aA x 0.8) – gsin 30o = 7.94 x 0.8 -9.81 (0.5) =1.447 =1.45 m s-2 Effective acceleration of car B = a’B = aB – gsin 30o = 6.95 - 9.81 (0.5) = 2.045 = 2.05 m s-2 Fig.1. 1 10.0 20.0 30.0 0 1.0 2.0 3.0 4.0 v/ms- 1 t/s Car A Car B

6(a)(ii)2. VA = 40 + 1.45 (10) = 54.5 ms-1 VB = 37 + 2.05 (10) = 57.5 ms-1 6(a)(ii)3. The area under the velocity-time graph will give the displacement of the car. From the graph, at t = 8.0 s, the area under the graph of Car A is bigger than the area under the graph by Car B. From t = 0 to t = 5.4 s, Car A covered a larger distance. From t = 5.8 s to t = 8.0 s, Car B covered a larger distance. Comparing the 2 regions, area under the graph by car A is still more than shaded area B. Hence Car A has covered more distance than Car B at t = 8.0 s. Or The 2 cars’ velocities are equal at t = 5.8 s. Car A is ahead of Car B. Hence for the displacemen