2025 EJC H2 Prelim P2 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2025 9749 PHYSICS MARK SCHEME Paper 2 – Structured Qns Answer Marks 1(a) k = [(3)(1.29) (3.3 × 102)2] / (9.9 × 104) = 4.3 A1 1(b) Method 1: fractional uncertainty = 8 0.092 0.07 0.30100 1.29 + + = 0.3 4.2 1.3 1k =  = = (allow 1 s.f.) Method 1: ( ) ( ) ( ) ( ) ( ) 22 max 4 22 max 4 3 3.3 10 1.08 1.29 0.09 5.7129.9 10 0.93 3 3.3 10 0.92 1.29 0.09 3.1329.9 10 1.07 1 5.712 3.132 1.29 12 k k k   + ==    − ==   = − = = 41k = C1 C1 C1 (calc of both kmax and kmin) C1 A1 1(c) unit of fA = unit of speed unit of A = unit of (speed/f) unit of A = 1 1 m s s − − = m M1 A1

2 ©EJC 2025 9749/J2HPRELIM/2025 Qns Answer Marks 2(a)(i) 21 2s ut at=+ (Taking upwards as positive) ( ) ( ) 217.8 5.9sin60 9.81 2 1.9 s (or 1.89 s) o tt t − = + − = C1 A1 2(a)(ii) Vertically: (Taking upwards as positive) v2 = u2 + 2as vv 2 = (5.9sin60)2 +2(–9.81)(1.2–9.0) vv = -13.4 m s-1 or v = u + at = -5.9 sin60 + 9.81 (1.885) = -13.4 m s-1 (Correct substitution and answer) horizontally: vh = 5.9cos60 vh = 2.95 (Correct substitution and answer) 13.4tan 2.95 78o   = = C1 C1 A1 2(b)(i) speed decrease, (So) viscous force / drag (force) decrease (resultant force decreases as) upthrust and weight remain the same B1 B1 2(b)(ii) U Vg= 2(1000)(9.81)(7.5 10 )−=U = 736 N = 740 N M1 2(b)(iii) resultant force = 740 + 950 – (78)(9.81) = 925 acceleration = F / m = 925 / 78 = 12 m s–2 (vertically) upwards C1 A1 A1

3 ©EJC 2025 9749/J2PRELIM/2025 Qns Answer Marks 3(a) For a system in equilibrium, sum of clockwise moments about a (or any) point equals sum of anticlockwise moments about the same point B1 B1 3(b) Let the length of the rod be L. Take moment about the point of contact between the rod and the ground. sum of clockwise moments 1000 cos302   oL OR 1000 sin602   oL sum of anticlockwise moments ( )sin17oTL OR ( )cos73 oTL OR sum of clockwise moments ( )1000 cos30 cos47 sin302  + o o oL TL OR ( )1000 sin60 sin43 cos602  + o o oL TL sum of anticlockwise moments ( )sin47 cos30ooTL OR ( )cos43 sin60ooTL T = 1500 N (or 1480 N) C1 C1 (C1) (C1) A1

4 ©EJC 2025 9749/J2HPRELIM/2025 Qns Answer Marks 4(a) By Principle of Conservation of Linear Momentum (0.50)(2.0) + (0.20)(0) = (0.50)(v1) + (0.20)(v2) [1] By Relative speed of approach = relative speed of separation u1 − u2 = v2 − v1 2.0 − 0 = v2 − v1 [2] Solve equations [1] and [2], v2