EJC Physics 2024 J2 H1 MYE MS

©EJC 2024 9749/J1H2MYE/2024 EUNOIA JUNIOR COLLEGE JC2 MIDYEAR EXAMINATIONS 2024 8867 H1 PHYSICS MARK SCHEME Qns Answer Marks 1(a) ( )( ) ( ) ( ) ( ) -3 222 -3 -3 0.170density 2164 kg m11 3.14 2.50 10 0.1233 0.001 0.01 0.12 2 0.02220.170 2.50 12.0 0.0222 2160 48 50 kg m (1 s.f.) 2160 50 kg m ρ π ρ ρ ρ − = = = = × ∆∆ ∆ ∆   = + += + + =     ∆= = ≈ ∴± MM V rh M rh M rh Examiner’s Comments Candidates are advised to be familiar with the Mathematical requirement of the H1 Physics syllabus. For example, if the shape in question was a rectangular, cylinder or sphere, the formula for volume would not be given. If remembering “give value to the same d.p. as uncertainty” gives problem, candidates might benefit from remembering “give value to the same place value as the significant number”. In this case, the uncertainty was 5 tens. Hence density was given to the nearest tens. C1 density C1 Fractional uncertainty formula C1 uncertainty A1 1(b) No. The two masses are individually pulled down the slope with have the same acceleration down the slope. Whether the masses are allowed to slide alone or with the other mass, they would still experience the same acceleration. Examiner’s Comments The key word in the question was “are allowed to slide down”. This suggests that there is no applied force on the masses, and the only force causing the motion down the slope is the component of their weight down the slope, and their acceleration down the slope is g sinθ. B1 1(c)(i) Consider both masses (as a system), -1 350 (10 11) 16.7 m s = = + = F ma a a Examiner’s Comments Taking both masses as 1 system, the acceleration of the system can be found. And each of the mass would have the same acceleration as the system. C1 A1 1(c)(ii) Consider mass A (as a system), ( )(10) 16.7 167 N = = = F ma N N Examiner’s Comments In this question, it was deliberate that the force asked for is the contact force by mass B on mass A. If the force that was asked for the contact force my mass A on mass B, Newton’s 3rd Law would need to be invoked. C1 A1

2 ©EJC 2024 8867/J2H1MYE/2024 Qns Answer Marks 2(a) Rate of change of velocity. Examiner’s Comments Well done! B1 2(b)(i) ( )( )1Area under graph 0.6 6 122 5.4 m = + = Examiner’s Comments The key to this question is to identify the point at which the ball hit the floor, which is at t = 0.6 s, when there is a sudden change in velocity due to the bounce on the floor. C1 A1 2(b)(ii) The ball experienced an inelastic collision with the floor and lost energy. With less kinetic energy at the point it leaves the floor , it has a smaller rebound speed. Examiner’s Comments Quite a few key phrases were expected for this question for the answer to be clear. Candidates are advised to review the answers to this question carefully. B1 B1 2(b)(iii) The gradients of the