EJC 2018 J1H2 Promo Solutions and Examiners Report

©EJC 2018 9749/02/J1H2PROMO/2018 [Turn over PAPER 1 PAPER 2 1 (a) ux = 10.8 cos(40.0°) = 8.27 m s-1 Almost all candidates got this question correct. (b) Horizontal component of the velocity remains constant, hence ux = vx vy = vx tan(60.0°) = 8.27 tan(60.0°) = 14.3 m s-1 Many candidates failed to notice that the question is asking for vy at point B, not point A and gave the answer 10.8 sin(40.0°) instead. By considering that the initial horizontal component of velocity ux = final horizontal component of velocity vx. This question could have been solved efficiently. However some candidates used the equations of motion to calculate vy from uy. (c) Initial vertical velocity at point A, uy = -10.8 sin(40.0°) = - 6.94 m s-1 Consider the vertical motion, vy = uy + at 14.3 = - 6.94 + 9.81t t = 2.17 s Consider the horizontal motion, x = sx sx = uxt = 8.27(2.17) = 17.9 m A lot of candidates got the signs wrong in the first step. The initial velocity is upwards while the final velocity and the acceleration is downwards so they should not all be of the same sign. Getting the signs wrong w ould result in loss of all the marks as the candidate is unable to use the equations of motion properly right at the start. Many candidates also separated the motion into 2 parts for analysis, the part where the rock is going up to the highest position an d the subsequent part from the highest position to the ground. Note that this is not required and results in unnecessary steps . The equations of motion can be used throughout the whole motion even when there is a change in direction of motion as long as acceleration is constant and all the signs are correct! The second step sx = uxt is quite straightforward except for some students who used sx = ½ (ux + vx) t without realizing ux and vx are the same. 1 2 3 4 5 6 7 8 9 10 B C C D C C C C A C 11 12 13 14 15 16 17 18 19 20 C A C C C D C A D A vy vx 60.0°

2 ©EJC 2018 9749/02/J1H2PROMO/2018 (d) h = sy sy = uyt + ½ ayt2 = -6.94(2.17) + ½ (9.81)(2.17)2 = 8.00 m OR vy2 = uy2 + 2aysy (14.3)2 = (-6.94)2 + 2(9.81)sy sy = 8.00 m OR sy = ½ (uy + vy)t = ½ (-6.94+14.3)(2.17) = 8.00 m This question is generally well done except for candidates who got the signs wrong. As there are many methods to solve this question, even when part (c) is not done properly, most were still able to identify and use one of the appropriate equations of motion to solve this. (e)(i) Consider rock’s vertical motion, sy = uyt + ½ ayt2 8.00 = 0 + ½ (9.81)t2 (uy = 0 as rock is thrown to the right) t = 1.28 s or t = -1.28 s (reject) time after ball is thrown = 2.17 – 1.28 = 0.89 s after first ball is thrown (1.28 s is time taken for rock to reach B from A) (2.17 s is tim