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 time taken for ball to reach B from A in part (c)) This part is the most poorly done in question 1. A number of students gave up and left this question blank. Again, some students separated the motion into 2 parts for analysis, as in part (c), which is unnecessary. Some students also failed to realize that analysis is needed only in the vertical direction. Misconception: Some students thought that when the rock is thrown, the ball must have travelled up and down back to the same height as the building (i.e. sy of ball is 0) and proceeded to calculate the time when this happens. This is a misconception because at this time the ball has some vertical velocity downwards while the rock is thrown horizontally without any vertical velocity. Hence if the rock is throw n at this time, it would reach the ground only after the ball has reached. A separate mark is awarded for the step ‘time after ball is thrown = 2.17 – 1.28 s’, even when time calculated in the first part is wrong. However marks are only awarded if the time taken for ball to reach B (most calculated this in part (c)) and the time taken for rock to reach B are both clearly stated. (e)(ii) sx = uxt 17.9 = ux (1.28) (17.9 m is from part c) ux = 14.0 ms-1 Generally well done. However some students wrote a value of t in part (e)(i) and substituted into the t in this part to try and get e.c.f.. However, this is wrong as the answer in (e)(i) is the time that rock is thrown after ball is thrown, not time taken for rock to travel from A to B as in this part.

3 ©EJC 2018 9749/02/J1H2PROMO/2018 [Turn over 2 (a) The Principle of Conservation of Momentum states that the total momentum of a system remains constant provided no net external force acts on the system. Underlined words are important words yet easily missed by students. Many students wrote ‘object’ / ‘objects’/ ‘2 bodies’ instead of ‘system’, this is not general enough and not accepted. Some students wrote ‘conserved’ instead of ‘constant’, this is not accepted as the word ‘conserve’ is already present in the name of this principle and needs paraphrasing. Many students forgot to write ‘net’ or ‘external’ as well. (b) (i) ΔpB = mB(vB – uB) = 1.2(-0.8 – 4) = -5.76 kg m s-1 Magnitude = 5.76 kg m s -1 Direction = to the left for correct magnitude and direction A significant number of students read the final velocity of B as -1.0 instead of - 0.8, some students missed out the negative sign in the -0.8. A few students divided mB(vB – uB) by the time of collision, but this formula gives the force (rate of change of momentum), not the change in momentum alone. For direction, ‘negative’, ‘to the left’, ‘opposite to the direction of initial velocity ’, ‘horizontal and away from ball S’ are accepted. Simply writing ‘opposite’ or ‘away from Ball S’ are not accepted as it is unclear what the exact direction is. Some students gave the magnitude as negative, which is wrong. (b) (ii) By principle of conservation of momentum, mBuB + mSuS = mBvB + mSvS 1.2(4) = 1.2(-0.8) + 3.6vS vS = 1.6 m s-1 This question is generally well done. Some students read off the -0.8 as -1.0 again, but this will be ECF if the same value has already been read off wrongly in the previous part. A few students used the wrong mass (3.6 kg instead of 1.2 kg) to calculate. A few students used uB – uS = vS – vB to calculate but this only applies for elastic collisions and it cannot be assumed that the collision is elastic unless explicitly stated. OR change in momentum of ball S = -change in momentum of ball B = 5.76 kg m s-1 mS (vS – uS) = 3.6(vS – 0) = 5.76 kg m s-1 vS = 1.6 m s-1

4 ©EJC 2018 9749/02/J1H2PROMO/2018 (b) (iii) Before and after collision: Correct initial and final speeds and correct shape (horizontal lines) [B1] During collision: Change in momentum occurs between 0.18s to 0.26s and correct shape (straight upward sloping line) This question is generally well done. A significant number of students did not look at the diagram carefully and start drawing the change in momentum from 0.20 s to 0.26 s instead. (b) (iv) time over which collision occurs = Δt = 0.26 – 0.18 = 0.08 s force = ΔpS / Δt = mS (vS – uS) / 0.08 = 3.6 (1.6 – 0) / 0.08 = 5.76 / 0.08 [M1] (5.76 kg m s-1 might have been calculated earlier also) = 72 N Magnitude = 72 N Direction = to the right A significant number of students read 0.18 s and 0.26 s as 1.8 s and 2.6 s, some students read 0.18 s as 0.20 s. Also some students forgot did not divide by Δt as they probably have forgotten that force is the rate of change of momentum. A few students used the wrong mass (1.2 kg instead of 3.6 kg) to calculate. A significant number of students read o