ASRJC Prelim H2P2 MS_Final (sharing)

1 9749/02/ASRJC/2023Prelim [Turn Over Anderson Serangoon Junior College 2023 H2 Physics Prelim P2 Exam Mark Scheme Paper 2 (80 marks) E – Easy, A – Average, D – Difficult ECF Error carried forward SF Significant figures error M0 No A marks awarded AE Arithmetic error BOD Benefit of doubt ^ More is needed in answer POT Power of ten error CON Contradictory response XP Wrong physics TE Transcription error IR Irrelevant (part) response 1a = = = –1 0.180 9.81 0.036 4 m9 N  Fk e A C1 A1 1b = Fk e   = + = + 2 0.0236 k F e k F e  = +  = = –1 2( 0.02) 49 . N 6 37 4  m 3k Accept −= max min 2 kkk A C1 A1 1ci F is weight of the column of liquid above the area A    == = = = () () fluid fluid fluid fluid mgFp AA Vg A hA g A hg A B1 B1 A0 1cii U + ke = mg U = mg − ke = (0.180)(9.81) − 49(0.030) = 0.2958 N      = = = = -3 –50.2958 (2.0 10 )( 0 9.81) 15 0 kg m p A h A g U V g A C1 C1 A1 2a The total momentum before collision is non-zero. By COM, (total) momentum is never zero, so not possible for both blocks to be at rest simultaneously. A M1 A1

2 9749/02/ASRJC/2023Prelim bi By COLM, taking rightwards as positive (3M x 0.40) – (M x 0.25) = (3M x 0.20) + Mv v =0.35 m s-1 E A1 bii To right / away from block A, as direction to the right is taken as positive in (b)(i) E B1 c relative speeds of approach is non-zero, and relative speeds of separation is zero Relative speed of approach is not equal the relative speed of separation, Hence inelastic collision. A M1 A1 3ai work done per unit mass bringing (small test) mass from infinity (to the point) E B1 3aii (near Earth’s surface change in) height ≪ radius or height much less than radius potential inversely proportional to radius and radius approximately constant (so potential approximately constant) A B1 B1 3b curve from r to 4r, with gradient of decreasing magnitude and starting at (r, ±) and line passing through (2r, ±0.5) and (4r, ±0.25) line showing potential is negative throughout A B1 B1 3ci Gain in KE = loss in GPE ½m v2 = 0−(– GMm / R) At distance R = 3r, –11 24 6 6.67 10 6v r .0 10 6 2G 2 .4 13 0 M 3     == = 6.46 × 103 m s-1 At distance R = 4r, 6 –11 242GM 2v 6.67 10 6.0 10 614r .4 04 =  =  = 5.59 × 103 m s-1 Change in speed = 6.46 × 103 −5.59 × 103 = 8.7 × 102 m s-1 D C1 C1 A1 × × ×

3 9749/02/ASRJC/2023Prelim [Turn Over 4a a = –  2x a = acceleration, x = displacement from equilibrium position and  = angular frequency E B1 4bi  = 2 / T = 2 / 4.0 = 1.57 = 1.6 rad s–1 A B1 A0 4bii E = ½m2x02 Or E = max Ek = ½mv2max= ½m2x02 = ½ × 36