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 the candidates got this question correct. B1 (b) (ii) Total kinetic energy = 1 2 𝑚𝑣0 2 = 1 2 (2.0 + 5.0)𝑣0 2 = 1 2 (7.0)(1.0)2 = 3.5 𝐽 The majority of the candidates got this question correct. B1 (b) (iii) Some of the kinetic energy has been converted to elastic potential energy stored in the compressed spring. Candidates lost marks as the idea of kinetic energy being converted to elastic potential energy was not clearly articulated in their answers. Quite a handful gave default answers such as “energy is lost as heat and sound” which is not relevant to the context of the question. B1 (c) As the collision is elastic, relative speed of approach = relative speed of separation Taking → as positive, 𝑢𝐴 − 𝑢𝐵 = 𝑣𝐵 − (−𝑣𝐴) 3.5 = 𝑣𝐵 + 𝑣𝐴 ------------(1) By conservation of momentum, 𝑚𝐴𝑢𝐴 = 𝑚𝐵𝑣𝐵 + 𝑚𝐴(−𝑣𝐴) (2.0)(3.5) = 5.0𝑣𝐵 − 2.0𝑣𝐴 ------------(2) Solving Equation (1) and (2), 𝑣𝐴 = 1.5 𝑚 𝑠−1 𝑣𝐵 = 2.0 𝑚 𝑠−1 Candidates were careless in their algebraic manipulation which resulted in the failure to obtain the correct answers. Many answers also lacked explanations on how the numerical equations used were formed. M1 M1 A1 A1 Total: 8 marks

3 (a) 98.1 N B1 (b) As the cupboard is in equilibrium, the horizontal forces acting on it must be zero. Hence X and P must have equal magnitudes. A handful of candidates mentioned incorrectly that the reason for X and P to be equal in magnitude is to ensure that sum of torque / moment has to be zero. B1 (c) Since the cupboard is in equilibrium,     0 98.1 0.15 0.60X     M1 24.5 NX  A1 (d) Correct shape of 1y x   graph B1 Total 5 marks X h 0 Candidates who lost the mark for this question failed to note the salient point when sketching a 𝑦 = 1 𝑥 graph; the rectangular area of each point on the graph are equal.

4 (a) It means that the oscillation has an acceleration that is always directly proportional to its displacement from the equilibrium position and B1 the acceleration is always directed towards the equilibrium point (or is always directed opposite to its displacement) B1 This part question was well done by the majority of the candidates. A handful of candidates did not include the reference position for which displacement was taken against. (b) (i) 8.0 cm B1 (ii) From the equation, 1220 rad s  C1 Hence, 2 220 35.0 Hz f f    A1 (iii) line drawn midway between AB and CD (allow ± 1mm) B1 (iv)   220 0.040 oovx   M1 18.80 m s A1 (v)      2 2 1.5 220 0.040 ooF m x   M1 2900 N A1 A handful of candidates were not mindful that Xo was in centimetres which resulted in a loss of marks for parts (iv) and (v). (c) (i) line drawn at quarter mark 9 mm below CD (allow ± 1 mm) B1 (ii) arrow pointing upwards B1 The majority of candidates got either or both parts to this question wrong. For part (i), candidates need to be aware that the position needs to be scaled relative to the given diagram. Total 12 marks

5 (a) The frictional force provides the centripetal force for circular motion of the mass. B1    2 2 0.78 0.78 0.35 2 0.74 Hz W md mg m f f      M1 144.6 min A1 (b) As the radius of the circular motion increases, the required centripetal force to keep the mud in circular motion increases. B1 Hence mud at the edge will leave the plate first. A large number of candidates incorrectly mentioned that the centripetal force overcomes the frictional force when it is the latter that provides for the former. B1 Total 5 marks

6 (a) (i) vibrations in one direction (normal to direction of propagation) Many candidates mentioned that the wave prop agates in one plane/direction. What many failed to realize that it is the oscillations that is in one plane. B1 (ii) I α cos2θ We require that 2cos 1( ) 20%cos0 5    0 I I 1cos 5   .63 4  Some candidates incorrectly used I α cosθ instead of I α cos2θ. M1 A1 (b) e.g. both transverse/longitudinal/same type meet at a point, same direction of polarisation or unpolarised, etc.......1 each, max 3 B2 (c) (i) 1. allow 0.3 mm → 3 mm Some stated expressions like a << D or a ≈ λ instead of a value as required by the question. Most got the value wrong. B1 2. λ = ax/D (allow any subject) B1 (ii) 1. separation increased less bright B1 B1 2. separation increased less bright B1 B1 3. separation unchanged fringes brighter B1 B1 further detail, i.e quantitive aspect in (ii)1 or (ii)2 (in (b), do not allow e.c.f. from (b)(i)2) B1 Some students did not consider both factors required by the question, e.g. they missed out on the change in maximum brightness. Some could have assumed that nothing should be w