ASRJC 2025 Prelim H2 P1 MS and Answers

1 9749/01/ASRJC/2025 Prelim [Turn Over Anderson Serangoon Junior College 2025 JC2 H2 Physics P1 Prelim Mark Scheme Paper 1 (30 marks) E – Easy, A – Average, D – Difficult 1 2 3 4 5 6 7 8 9 10 C B B C C A D B B D 11 12 13 14 15 16 17 18 19 20 B B C A B A A D D A 21 22 23 24 25 26 27 28 29 30 D A B A D A D C C C 1 C Typical mass of a small car = 1500 kg Average KE of car = ½ (1500) (302) = 0.68 × 106 = 1 × 106 J (1 s.f.) E 2 B For a fixed resistor, R = constant or ratio of V/I is a constant. The presence of a non-zero y-intercept indicates a systematic error. There is no random error as all points lie on the best fit line E 3 B Using s = ut + 0.5at2, as u = 0 and a = constant, Hence, s  t2 2 2 (0.5 ) 0.25 xT LT xL = = A 4 C Terminal velocity was reached in water as shown by the horizontal line. Area A (distance travelled in air) > Area B (distance travelled in water). Rate of change of velocity in air (gradient of curve) is decreasing. A area A (in air) area B (in water)

2 9749/01/ASRJC/2025 Prelim 5 C Forces must be of the same type, opposite in direction and acts on different body. A 6 A 90 kg min-1 = 1.5 kg s-1 N 305.120)( ==== t mvt mvF A 7 D Since the rod is in equilibrium, Σ = 0 Taking moment about O, T(2Lsin 30o) = 20g(L 2 cos 60o) + 30g(3L 2 cos 60o) T = 269.78 = 270 N D 8 B WD to stretch = ½ × 0.5 × 35 = 9 J Energy recovered = ½ × 0.6 × 20 = 6 J Total WD in 1 cycle = Energy remaining in 1 cycle = 9 – 6 = 3 J Or by counting the no of squares in the area between the two curves Energy remaining = (5 × 0.1) × 6 = 3 J D 9 B Constant force: Acceleration, a = constant Initial speed, u = 0 Therefore v = u + at = at P = Fv = Fat tP  A 20 g 30 g 120o

3 9749/01/ASRJC/2025 Prelim [Turn Over 10 D The vertical component of the acceleration is the centripetal acceleration which is present since the particle is performing circular motion. The horizontal component of the acceleration causes the speed of the object to increase. E 11 B On P, Since weight of mass is 1.0 N, g = weight/mass = 1.0 / 1.0 = 1.0 m s–2 On Q, Mass of 1.0 kg remains unchanged on Q. g = 1.0 x 10 = 10 m s–2 Weight of mass = 1.0 x 10 = 10 N E 12 B 5 1 2 2 2 2 1 3 1 3 1 31.0 10   1.2 500 ms NmPc V Pc c c  − =   = =       = E 13 C For full cycle processes, the net change in internal energy is zero. 560 [( 4.2) (1.0 10 (20.0 5.0) 10 )] 2.7 J U Q W Q Q −  = + = + − +   −  = E 14 A at equilibrium position, x = 0 mm amplitude = 50 mm period = 2.0 s Time shutter remained open t, from x = 0 s to x = 25 mm, = 225 50sin 2.0 t t = 0.17 s D

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