TMJC 2023 H2 Physics P3 MS

1 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2023 JC2 H2 Physics Preliminary Examination Paper 3 Suggested Solution 1 (a) (i) Elastic collision (as the total kinetic energy of the system (car and wall) is conserved before and after collision) [B1] (ii) Considering the toy car and the wall as one system, The total momentum is conserved as there is no external resultant force acting on this system. There is a corresponding change of momentum to the fixed wall. Or Considering the toy car as the system only, The momentum is not conserved as there is an external resultant force acting on the toy car by the wall. [B1] [B1] [B1] [B1] (b) (i) -1 area under - graph 1 (0.18)( 15)2 1.35 kg m s p Ft∆= = − =− -1 () () 1.35 2 1.35 2 0.675 ms kg fi fi i ppp mv v mv v mv p mv ∆= − = − = −− −= − = = = [C1] [A1] (ii) max 11 ( 15) 7.5 N N 22 or 1.35 (0.18) 7.5 average average average average FF pF t F F = = −= − ∆= −= =− -2 7.5 0.65 11.5 m s average average average average F ma a a = −= =− [C1] [A1]

2 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Or -1 -2 0.675 0.65 1.04 m s 1.04 1.04 0.18 11.6 m s ii i fi p mv v vva t = = = −= −−= =− [C1] [A1] (iii) [B1] shape with the time increased and magnitude of max force decreased. force / N time / s 0 0.18 −15

3 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 2 (a) A field of force is a region of space in which a body experiences a force without physical contact with another body. [The field is produced by the presence of another body (or bodies)]. [B1] (b) State one similarity and one difference between gravitational field and electric field. Similarity [B1]: o The field strength due to a point charge/mass follows an inverse square law with distance from the charge/mass. o The potential due to a point charge/mass varies inversely with distance from the charge/mass. o (Both G-fields and E-fields can be represented by field lines) In notes, we accept for students who memorised. Difference [B1]: o Gravitational fields are always attractive, whereas electric fields can be attractive or repulsive. o Gravitational potential is always negative, whereas electric potential can be positive or negative. o Source of gravitational field is a mass while source of electric field is a charge. (c) (i) At point P, it is a gravitational neutral point where net gravitational field strength g is zero. [B1] It is closer to the moon since the mass of the moon is smaller than the mass of the Earth. [B1] (ii) At point P, gravitational neutral point. ( ) ( ) 22 2 8 28 2 7 24 22 [M1- apply ] 3.80 10 3.80 1 5.98 10 7.35 10 0 [B1] 3.79 10 m [A0] ME ME gg MM MG G gGrr r r r r = = = × × ×− ×− = = × OR using potential method to solve, Sum of potentials at point P (give method mark if approach is shown)

4 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics (iii) By conservation of energy 2 26 41 loss in KE gain in GPE 1 ( )( ) 0 ( )2 1 ( ) 0 ( 1.3 ( 62.3)) 10 [C1]2 11018 1.10 10 m s [A1] fimv m v v φφ − = −= − − =− −− × = = × The space probe need s to have sufficient energy from the Earth’s surface to just reach the point P with maximum potential. Thereafter, the net gravitational force on the space probe (pointing towards Moon) will enable it to reach the Moon. [B1] (d) (i) P is negative. Near P, the electron has a positive potential energy. [M1] Being very close to P, potential energy due to P on t he electron is more significant than potential energy due to Q on the electron. (i.e. approximately 04 P P QeE rπε≈ ) For potential energy to be positive, P and electron must be of the same sign. Hence P is negative. [A1] OR P is negative. Potential energy increases towards P.[M1] Since positive work mus t be done on electron to move it towards P. Therefore, P and electron must be repelling each other. Hence P is negative. [A1] OR P is negative. Gradient of the graph is negative. [M1] Since force points in the direction of decreasing potential energy ( dUF dr=− ) electrostatic force on electron must be in the direction of positive r (right). Hence P and electron must be repelling each other Hence P is negative. [A1] (ii) Magnitude of force = gradient of potential energy graph [C1] Gradient of graph (at x = 3.0 cm)

5 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 2.3 ( 1.8) 124(0.0115 0.0445) −−= =− eV m−1 [C1: gradient within range] Accept gradient between 110 to 135 eV m−1 Convert the unit to SI unit, force = 124 x 1.6 x 10 −19 = 1.98 x 10−17 N [A1:conversion to SI] Accept a range of force due to gradient. 1.76 x 10 −17 N 2.16 x 10−17 N (iii) Curve with value of Ep near P to be lower & steeper gradient nearer to Q. -5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 (0.0445, -1.8) (0.0115, 2.3) Ep / eV x / m

6 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 3 (a) (i) ( )( ) -1 max 2 max 2 1.4 m s 1 2 1 0.50 1.4 [C1]2 0.49 J [A1] K v E mv = = = = (ii) max 0 0 1.4 [C1]2 0.50 0.11 m [A1] vx x ω π = =   = (b) Damped oscillation. [B1] Air resistance [B1] on the wooden board. (c) (i) (ii) Correct plot (2.0, 3.2) [B1] Correct curve of best fit [B1] (iii) 1.8 Hz (accept 1.75, 1.8, 1.85) [B1] 0.0 2.0 4.0 6.0 0.0 1.0 2.0 3.0 x

7 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 4 (a) (i) ( ) o 3 7 sin sin 0.16 0.24 10 6.70 10 m [B1] b λθ λ λ − − = = × = × (ii) Distance from centre to first minimum, y: ( ) ( ) o3 tan 1.8 tan 0.16 5.0 10 m 5.0 mm [M1] y D y θ − = = = × = Width of central maximum = 2y = 10.0 mm [A1] (iii) [B1] correct shape [B1] correct labels on horizontal axis + distance between minima = distance from centre to 1 st minimum (iv) The minimum will be further away from the centre / The central maximum will spread out more. [B1] The intensity will be reduced. [B1] (b) (i) Light waves from the sources have a constant phase difference (and same frequency) [B1] (ii) Fringe separation, x: ( ) ( ) 9 3 3 638 10 1.8 [M1]0.82 10 1.40 10 m 1.40 mm [M1] Dx a λ − − − × = = × = ×= distance to 2nd dark fringe = 1.5x = 2.10 mm [A1] distance from centre of screen / mm intensity of light 0 Fig. 4.2 5.0 10.0 –10.0 –5.0

8 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics (iii) Resultant amplitude at bright fringe = A + A = 2A [C1] ( ) 2 22 22 2 [M1 - awarded for concept of proportionality] 4 [A0] fringe fringe fringe IA IA A IA A II ∝ = = = (iv) Although intensity of bright fringe is 4I, intensity at dark fringes are zero. [B1] Hence average intensity of the interference pattern is 2I. [B1] Principle of conservation of energy is not violated. 5 (a) (i) 2 377 [M1] 60 Hz [A1] f f π = = (ii) 0 240 [M1] 22 170 V [A1] rms VV = = = (iii) Alternating magnetic field is created in the coil. Hence there is an alternating magnetic flux linkage through the ring [B1] By Far aday’s law , an e.m.f. will be induced in the ring [B1 – link to emf to flux linkage] Since the ring is a closed loop, c urrent is induced in the ring. [B1] By