2025 ASRJC Physics Prelim P3 MS and ans

1 9749/03/ASRJC/2025PRELIM [Turn Over Anderson Serangoon Junior College 2025 H2 Physics Prelim P3 Mark Scheme Paper 3 (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 sx = vxt vx = 24 / 1.5 = 16 m s-1 tan 28° = vY / vX or vX = v cos 28° and vY = v sin 28° vY = 16 tan 28° or vY = 16 × (sin 28° / cos 28°) so vY = 8.507 = 8.5 m s–1 Examiner’s comments: Most students were able to solve. Some workings did not show complete substitution of values to meet the demand of a “Show” question. There were still some students who did not show more s.f. before final answer which was required in a “Show” question. A C1 C1 A0 1b Taking upwards as positive vY = uY + at t = (0 – 8.5) / (– 9.81) = 0.87 s Examiner’s comments: Students did not substitute the sign of the direction correctly in their Kinematics questions which resulted in wrong answer. Some students wrongly assumed initial velocity as zero. C1 A1 1c straight line from positive vY at t = 0 to negative vY at t =1.5 s line starts at (0, 8.5) and crosses t-axis at (0.87, 0) and does not go beyond t = 1.5 s. M1 A1

2 9749/03/ASRJC/2025PRELIM Examiner’s comments: Those who drew a straight line graph often did not ensure the x-intercept was within half smallest square and extended line beyond t = 1.5 s. 1d acceleration (of freefall) is unchanged / not dependent on mass, and so no effect (on maximum height and time taken) Examiner’s comments: Many did not pinpoint acceleration as the quantity that was unchanged / independent of mass. Even though net force (= weight) was larger, the acceleration of free fall remained as g, so this showed it was independent of mass. E B1 1e Since air resistance acts downwards, net downward force is larger hence, shorter time taken Examiner’s comments: Students need to be explicit about the direction of the net force / net deceleration as downwards. Stating AR is directed opposite to the motion of the object was too generic and not contextualizing. Students need to illustrate their awareness about the net deceleration or net force is greater than before. Many students used inappropriate term such as “increases” to illustrate greater / larger. Some did not conclude for the time taken. D M1 A1 2ai Use A ρR l= =       = − − 350 4 )10x020.0( )0.2)(10x5.5( 23 8 power dissipated = I2R = (0.42)2(350) = 62 W Examiner’s comments: Most students performed the correct calculation and obtained the correct answer. A