EJC 2017 J1H2 Promo MS (Final)

Paper 1: Multiple Choice 1 2 3 4 5 6 7 8 9 10 A A C D Voided D B A D C 11 12 13 14 15 16 17 18 19 20 B B C A B C A C C D

Paper 2: Structured Questions 1 (a) Taking upwards as positive, 𝑢𝑦 = 20.0 sin 30° (substitution of values must be shown) = 10.0 𝑚 𝑠−1 A number of candidates did not provide an explanation on how they arrived at their final answer. A1 (b) (i) Using 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 𝑠 = (10.0)(4.0) + 1 2 (−9.81)(4.0)2 = −38.5 𝑚 The bomb will be 38.5 m below the point of release. A handful of candidates were not careful when using the kinematics equation and neglected the signs of the values involved. The M1 (b) (ii) When the bomb is falling, the helicopter is rising at the same time, In 4.0 s, the helicopter rise a vertical distance = (10.0)(4.0) = 40.0 𝑚 above the point where the bomb is released. Total distance from the helicopter = 40.0 + 38.5 = 78.5 m The majority of the candidates got this question correct. M1 (c) Correct Gradient of graph Correct starting points of the graphs Correct axis labels A number of candidates did not read the question properly and did not start drawing the graph for B at t = 0s. Many also go t the gradient of the graph wrong or did not label the axis of the graph, which result ed in a loss of marks. B1 B1 B1

(d) For first bomb, after 𝑡 = 5.0 𝑠 Using 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 𝑠1 = (10.0)(5.0) + 1 2 (−9.81)(5.0)2 = −72.6 𝑚 For second bomb, Displacement, 𝑠2 = area under v-t graph. = (10.0)(2.0) + 1 2 (10.0)(1.0) − 1 2 (19.4)(2.0) = 5.6 𝑚 Hence total distance between them = 72.6 + 5.6 =78.2 m. The m ajority of the candidates did badly for this question by being unable to apply the concept that the area under a v-t graph gives the change in displacement. A large number of candidates did not realise that the area below the t-axis is actually considered negative displacement. M1 A1 (e) As the bombs are being released at different timings, the second bomb will continue to travel forward after the first bomb hits the ground. Hence, it will land at a spot further away from where the first bomb lands. The bombs will not hit the same spot when they hit the ground. A large number of candidates incorrectly mentioned that the helicopter and the first bomb will have different horizontal displacement s; the 2 bombs will have the same horizontal displacement in flight. M1 A1 Total: 10 marks

2 (a) The Principle of Conservation of Linear Momentum states that the total linear momentum of a system of interacting particles remains unchanged provided no net external force acts on the system. The majority of the candidates lost marks for this question because of missing key words such as “total momentum” and “ex ternal resultant/net force”. B1 (b) (i) By conservation of momentum, (2.0)(3.5) = (2.0 + 5.0)𝑣0 𝑣0 = 1.0 𝑚 𝑠−1 The majority of th