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 lines represent acceleration of the ball, which is a constant at 9.81 m s−2 in the absence of air resistance. Examiner’s Comments Since question explicitly mentioned gradient, candidates are expected to state the concept of the gradient representing acceleration of the ball. B1 B1

3 ©EJC 2024 8867/J2H1MYE/2024 Qns Answer Marks 3(a) 1. The net / resultant / sum of force(s) acting on the window panel is zero. 2. The net / resultant / sum of moment about any point acting on the window panel is zero. Examiner’s Comments Candidates should ensure that they quote according to lecture notes to avoid missing out keywords. Candidates should NOT quote the principle of moments. B1 3(b) The centre of gravity of the window panel is the point where its weight appears to act. Examiner’s Comments A few candidates wrongly used the word “mass” instead of weight. Centre of mass and centre of gravity carries different meaning. B1 3(c) Let the distance be l. sum of clockwise moments = sum of anticlockwise moments 90 120 250 sin30 86.4 cm l l × = ×× ° = Examiner’s Comments Candidates need to learn to resolve forces correctly. C1 A1 window panel wall 90 N hinge 30° Fig. 3.1 (not to scale) 250 N rod

4 ©EJC 2024 8867/J2H1MYE/2024 3(d) Line of action goes through the intersection of weight and force due to rod. Arrow points up and right. Examiner’s Comments Candidates do not appear to be aware that 3 non-collinear forces acting on a body must pass through a single point in space. B1 3(e) Method 1: Resolving of forces Fhori = 90 × cos30° = 77.9 N, Fvert = 250 − 90 × sin30° = 205 N magnitude = 2277.9 205 219.3 219 N+= = Method 2: Vector diagram. window panel wall 90 N hinge 30° Fig. 3.1 (not to scale) 250 N rod window panel wall 90 N hinge 30° Fig. 3.1 (not to scale) 250 N rod 30° H

5 ©EJC 2024 8867/J2H1MYE/2024 𝐻𝐻2 = 902 + 2502 − 2 × 90 × 250 × cos 60𝑜𝑜 H = 219 N 90 N 60° 250 N H

6 ©EJC 2024 8867/J2H1MYE/2024 Qns Answer Marks 4(a) Newton’s Second Law of Motion states that the rate of change of momentum of a body is [magnitude] directly proportional to the resultant force acting on it and [direction] takes place in the direction of the resultant force. Examiner’s Comments Candidates are advised to put effort into memorising definitions word for word to avoid missing out key ideas. B1 4(b)(i) 33 change in momentum of A impulse on A area under - graph 1 (1.0 10 )(4.0 10 )2 2.0 N s − = = ∆= × × = Ft p Examiner’s Comments For those who were able to relate to area under the graph, a notable number missed out one of the powers of tens. C1 Area under graph A1 4(b)(ii) Since the force is acting on cart A (by B) acts leftwards, change in momentum is leftwards. Examiner’s Comments Candidates should be aware that the direction of change of momentum provides the direction of the force. For Learning: This is basically the application of Newton’s 2 nd Law (in terms of direction). Refer to answer for 4(a). B1 4(b)(iii) Take the right direction as positive, [ ] 1 () 2.0 (1.5) ( 3.0) 1.7 m s− ∆= − − = −+ =+ A AA A A p mv u v v Velocity of A after collision is to the right. Examiner’s Comments The most common error is the failure to correctly consider the sign of ∆p. For Learning: Just like Fnet = ma where the signs of both Fnet and a are the same, in this case, the signs of both ∆p and ∆v (hence v and u) need to be considered. Both equations are related to Newton’s Second Law! C1 A1 (both dirn & mag required)

7 ©EJC 2024 8867/J2H1MYE/2024 4(c) Relative speed of approach before collision = rate of decrease of distance between carts = 3.0 m s−1 Relative speed of separation after collision = rate of increase of distance between carts = (2.4 – 1.7) = 0.70 m s−1 Since the relative speed of approach is not equal to the relative speed of separation, the collision is not elastic. Examiner’s Comments Poor presentation was a common problem that presented in this question. When trying to prove that 2 quantities are NOT equal, students should tackle the LHS and RHS of the equations separately and conclude accordingly after that. A handful of students also memorized the relationship incorrectly. It should be “u1 − u2 = v2 − v1” (1-2-2-1). B1 B1

8 ©EJC 2024 8867/J2H1MYE/2024 Qns Answer Marks 5(a) Power is the rate of work done. Examiner’s Comments Generally well done. Candidates should note that they should not say “rate of work done per unit time” as this would suggest dividing by time twice, since rate already presents the idea of “per unit time”. A1 5b(i) ( )( )