2018 9749 H2 Physics MS EJC Suggested_ver Jun2022

©EJC 2021 9749/H2 PHYSICS 2018 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2018 Paper 1 Multiple Choice Question Key Question Key Question Key 1 A 6 A 11 C 2 A 7 C 12 D 3 B 8 B 13 D 4 C 9 D 14 C 5 A 10 A 15 D 16 C 21 A 26 A 17 D 22 A 27 C 18 C 23 D 28 A 19 C 24 A 29 C 20 A 25 A 30 B Notes: Candidates found Questions 1, 2, 5, 9, 11, 18, 24, 26 and 27 relatively challenging. Question 1 Percentage uncertainty in L needed to be subtracted from, rather than added to, the percentage uncertainty in g. Question 2 Distance is the total magnitudes of the areas of the triangles defined by the graph lines and the time axes. Question 9 Those who chose option C confused between angular displacement and linear displacement. Question 11 Those who chose option D assumed wrongly that the spacecraft must start from rest. Question 17 Those who chose B found the centre-to-fringe distance instead of fringe-fringe distance. This requires a doubling of the new centre-fringe distance. Question 23 Those who chose either B or C assumed wrongly that current must split equally at the junction Question 24 Those who chose option B, omitted a factor of 2 either from omitting the sin 30° factor or from missing that the magnetic force is exerted on the coil at double the distance away from the pivot as is the weight. Those who chose option D treated the coil as consisting of a single turn rather than 50. Question 27 Those who chose option D neglected to consider the effect of half-wave rectification.

©EJC 2021 9749/H2 PHYSICS 2018 Qns Marks 1(a) when a body is in rotational equilibrium, the sum of clockwise moments about any point must be equal to the sum of anticlockwise moments about the same point. B1 1(b)(i) Solving (1) and (2): 31 NAT = 13 NBT = Notes: Be careful not to get the = 3 7 B A T T values the wrong way around. A1 1(b)(ii) Taking moments about the centre of mass, T A (d – 0.20) = TB (1.00 – d – 0.20) 31 (d – 0.20) = 13 (1.00 – d – 0.20) d = 0.38 m C1 A1 Paper 2 Structured Questions TA TB mg As metal sign is in translational equilibrium, TA + TB = mg = (4.5)(9.81) TA + TB = 44.145 ---- (1) Since the perpendicular distance of T A is smaller than that of TB from the centre of mass, TA > TB therefore ---- (2) M1

3 ©EJC 2021 9749/H2 PHYSICS 2018 [Turn over Qns Marks 1(c) wind exerts a horizontal force on the sign to the right to maintain translational equilibrium of the metal sign, force that each support provides on the sign must have a horizontal component to balance the rightward force. Notes: Do not simply state that a horizontal force is needed as wind is horizontal. Remember to state explicitly that this force is to maintain t