TMJC 2025 H2 Physics P3 Soln

1 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2025 JC2 H2 Physics Preliminary Examination Paper 3 Suggested Solution 1 (a) Elastic collision: The total kinetic energy (KE of the system) of the colliding bodies is conserved before and after the collision. Or The relative speed of approach of the two bodies is equal to their relative speed of separation. (b) By conservation of linear momentum, ( ) + = + + = + = 2.4 0 4 C1: sub and 4 A A B B A A B B A B p B A m u m u m v m v v v u m m By conservation of kinetic energy ( )   + = + + = + = 2 2 2 2 2 22 1 1 1 1 2 2 2 2 2.4 0 4 C1: sub , 4 A A B B A A B B A B A B A m u m u m v m v v v u m m By relative speeds ( ) − = − − = − 2.4 0 C1: sub A B B A B A p u u v v v v u Note: only two C1 marks available Solving simultaneously with any of the two above equations,   11.44 m s A1Av −=− (c) By conservation of linear momentum, The total momentum of a system of objects remains constant provided no resultant external force acts on the system. Since no external resultant force is acting on the system, ( ) ( ) 1 2.4 0 5 B1: sub and 4 0.48 m s shown A A B B A o B o o p He p o m u m u m v m v v u m m v − + = + + = =  =

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics (d) [B1] Before collision: uA at 2.4 m s−1 and uB at 0 m s−1 [B1] shape: Smooth and continuous curvature, 0.48 m s−1 label [B1] After collision: vA at −1.44 m s−1 and uB at 0 m s−1 2 (a) 2 2 kinetic energy of the child at Q with resistive force kinetic energy of the child at Q if there is no resistive force 1 2 1 (4.8 )2 [M1]9.81 (7.0) 0.168 Qmv mgh= = = [A1] (b) 2 2 2 2 percentage efficiency 100% 1 2 100%1 2 54(2.1 ) 100% [M1]20(4.8 ) 51.7 % [A1] Q EPE KE kx mv = = = = (c) Any 2: - Energy loss due to work done by resistive force / friction - Sound produced due to collision with the soft board - Deformation of the soft board 3 (a) Gravitational field strength at a point is the gravitational force per unit mass [B1] on a small test mass placed at that point. before collision during collision after collision t v 2.4 0.48 0 Object B Object A −1.44

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics (b) Gravitational force provides for centripetal force of satellite. [B1] By Newton’s 2nd Law [B1], gravitational field strength must thus be equal to the centripetal acceleration (OR mg = ma → g = a) (c)(i) If obey inverse square law, ( ) ( )2 lg 2lg lgkg g r kr= → =− + [B1] Consider points (7.95, 1.20) and (8.55 , 0) − =−− 1.20 0Gradient = 2 [B1] working (showing coo rdinates) and answer7.95 8.55 Therefore must obey the inverse square law. (c)(ii) For r = 4.18 × 108 m ➔ lg (r) = 8.62 (note: should not round off to 2 s.f. and use 8.6 for finding corresponding lg g value as it will greatly affect the accuracy of the final answer). From graph, lg (g) = − 0.14 [B1] g = 0.7244 − = =  = 2 2 8 1 0.7244 [C1]4.18 10 17400 m s [A1] vg r v v (c)(iii) Same gradient, lower than original graph (of planet X) [B1] -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 7.9 8.1 8.3 8.5 8.7

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 4 (a) (i) Plane polarisation occurs when particles in a wave oscillate only in a single direction perpendicular to the direction of energy transfer of the wave (or direction of wave propagation). [B1] (ii) Since sound waves are longitudinal waves, particles in the wave oscillate parallel / along to the direction of energy transfer of the wave (or direction of wave propagation). [B1] (b) (i) 1. A [B1] 2. 2I [B1] (ii) By Malus’ law, 2 2 2 cos cos cos 20 [C1] 0.883 [A1] o Y   = = = = II II I I (iii) Minimum angle = 140° [B1] 20°

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 5 (a) The 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 of the individual waves at that point [B1]. (b) (i)  − − ==  = 9 3 (590 10 )(3.2) [C1]1.50 10 0.00126 m [A1] Dx a x (ii) As the slits are widened, the extent of diffraction of the waves when they passes through the slits will become lesser. [M1] This will result in the waves not being able to overlap one another resulting in no interference[M1] and thus no fringes formed [A0] (iii)      − = = = ==  ==  + = 9 1.5tan 3.2 25.1 [C1] sin 0.001sin(25.1)sin 700 [C1]590 10 1.03 1.0 [C1] Max. bright fringes = n 2 1 3 [A1] dn dn n [1] for finding d [1] diffraction formula

6 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 6 (a) = = = + = 5(4.5 10 )(0.075) [C1](8.31)(273 60) 12.2 mol [A1] PV nRT PVn RT (b) (i) = = =  + = 12 12 1 22 1 3 since pressure, and are constant, 0.075 (150 273) [C1]333 0.0953 m [A 1] nR VV TT VVT T (ii)  =  =− = 4 3 2 3 (12.2)(8.31)(150 60) [C1]2 1.37 10 J [A1] U nR T (iii) =  =  − = 5 3 (4.5 10 )(0.0953 0.075) [C1] 9.14 10 J [A1] byWD P V (iv)  = + = − =  − −  = 43 4 1.37 10 ( 9.14 10 ) [C1] 2.28 10 J [A1] U q w q U w

7 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 7 (a) Horizontal lines [B1] For between x = 0 cm to x = 5.0 cm, V value is 380 V For between x = 40.0 cm to x = 50.0 cm, V value is 750 V (b) Electric force will be directed towards sphere A and decreasing in magnitude, from surface of A to position of lowest potential At position of lowest potential, electric force is zero Electric force will be directed towards sphere B and increasing in magnitude, from position of lowest potential to surface of sphere B. ([B1] describing how the magnitude and direction of force is changing from A to lowest potential. [B1] identify point of lowest potential to be zero force/ neutral point [B1] describing how the magnitude and direction of force is changing from lowest point to B.) (c) Position of zero field strength/ neutral point/ minimum potential is closer to sphere A [M1] (or potential at surface of/ close to A is smaller than potential at surface of/ close to B) Therefore sphere A has a smaller magnitude of charge [A1] (d) Method 1: Consider a point of resultant potential Consider x = 30 cm, V = 400 V (or any other appropriate point) 9 00 9 9 9 400 1 1.2 10 1 400 [C1]4 0.30 4 0.50 0.30 8.1 10 C [A1] If consider x = 5.0 cm, V = 380 V, then 8.22 10 C If consider x = 40.0 cm, V = 750 V, the n 8.04 10 C   − − − − +=   +=  − = = = AB B B B B VV Q Q Q Q Method 2: Consider neutral point x = 14.0 cm

8 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics ( )  − − =   = − =   9 22 00 9 1 1.2 10 1 [C1]44 0.14 0.50 0.14 7.9 10 C [A1] allow range of 7.19 nC 8.77 nC based on readoff of position of neutral point to be 13.5 cm 14.5 cm AB B B B B EE Q Q Q Q (e) Method 1 Kinetic energy of electron = ½ m v2 = ½ (9.11 × 10-31)( 9.7 × 106)2 = 4.3 × 10-17 J [M1] Gain in electric potential energy from surface of A to neutral point = e (Vneutral poin