Prelim SAJC (H1) P1 2024 Ans

1 SAJC 2024 Prelim / 8867 [Turn Over] SAJC JC2 H1 Physics Prelims 2024 Paper 1 Solutions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 C A C D C B D B D C Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 D A C A A D B C B B Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 B B C A B C B C D D 1 Ans: C Wire diameter  0.1 cm (can estimate by using your ruler thus estimate in cm) Area  r2 = (0.05/100)2 = 7.8 x 10-7 m2 2 Ans: A Charge. Current not charge is the SI base quality for electricity quantities. 3 Ans: C Since the plane is moving at constant speed, a = 0 Fnet = 0 Along the path, W sin + R = T Perpendicular to path, W cos = L 4 Ans: D Average value = 48.00 mm is far from the true diameter of 42.03 mm thus it is NOT ACCURATE The set of repeated readings are close to each other thus PRECISE. 5 Ans: C s = ut + ½ at2 s = ½ (9.81)TE 2 --- (1) s = ½ (3.71)TM 2 --- (2) (2) / (1): TM / TE = ( 9.81/3.71)1/2 = 1.63

2 SAJC 2024 Prelim / 8867 [Turn Over] 6 Ans: B acc = gradient of v-t graph Initially, gradient is positive and decrease to zero. Then gradient is negative, much steeper than the positive gradient and then become zero. C, D – incorrect as initial acceleration should be positive A – incorrect as the negative acceleration is not bigger than the positive acceleration 7 Ans D The force the Earth acts on sky-diver is equal to the force the sky-diver acts on Earth as they are action-reaction pair but they will not cancel each other as they are not acting on the same body. 8 Ans: B X: Air resistance increases with velocity till a constant value Y: Fnet = weight – air resistance decreases till zero as weight = constant 9 Ans: D For m, KEf = ¼ KEi ½ m vm2 = ¼ ( ½ mu2) vm = u / 2 By COM, mu = m (- u /2) + 4mv4m 3u / 8 = v4m 10 Ans: C Since p i is not zero as the momenta of the spheres are not equal, p2kg = 4x 2  p3kg = 6 x 3 thus by COM, pf  0 as pf = pi A – Impulse Ft = p is equal and opp when COM can be applied as Fnet = 0 (N3L) B, D – momentum not mass or speed is considered in COM 11 Ans: D Torque of a couple = 15sin 65 (0.4) = 5.4 Nm

3 SAJC 2024 Prelim / 8867 [Turn Over] 12 Ans: A For 0.50 kg, 0.50(9.81) – T = 0.50 (2.0) T = 3.9 For 0.20 kg, T – f = 0.20(2.0) f = 3.9 – 0.4 = 3.5 N 13 Ans: C Horizontally, F = p/t =(200/1000) [ 7 – (-14)]/0.6 = 7.0 N 14 Ans: A Initially, F = kx for each wire 1/3 W = kx (Weight of lamp W is shared equally by the 3 wires) W = 3kx Finally, ½ W = kx’ (W is now shared equally by the 2 wires) ½ (3 kx) = kx’ x ‘ = 3/2 x difference between h = x’ – x = 3/2 x – x = 0.5 x = 0.5 (0.4 cm) = 0.20 cm 15 Ans: A Since the car is travelling at constant speed, friction f =driving force F P = Fv = fv Pt = fvt E = fvt