EJC 2020 J1H2 Promo Mark Scheme

© EJC 2020 9749/01/J1H2PROMO/2020 EUNOIA JUNIOR COLLEGE JC1 PROMOTIONAL EXAMINATIONS 2020 General Certificate of Education Advanced Level Higher 2 PHYSICS Multiple-Choice, Structured Questions and Practical MARK SCHEME w/ EXAMINERS’ COMMENTS Maximum Mark 9749 October 2020 Paper 1 Solution 1 A 𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸 = 𝑘𝑔 𝑚2 𝑠−2 𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐 = 𝑚𝑠−1 For equation to homogeneous, units of (𝛼𝑐2)2 and (𝛽𝑐)2 must be equal to units of 𝐸2 (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸)2 = (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝛼)2(𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐)4 (𝑘𝑔 𝑚2 𝑠−2)2 = [𝛼]2(𝑚2 𝑠−1)4 [𝛼] = 𝑘𝑔 Therefore, 𝛼 represents mass. (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸)2 = (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝛽)2(𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐)2 (𝑘𝑔 2 𝑠−2)2 = [𝛽]2(𝑚 𝑠−1)2 [𝛽] = 𝑘𝑔 𝑚 𝑠−1 Therefore, 𝛽 represents momentum. 2 B 𝜌 = 𝑀 𝑉 = 𝑀 𝑙𝑏𝑡 Δ𝜌 𝜌 = Δ𝑀 𝑀 + Δ𝑙 𝑙 + Δ𝑏 𝑏 + Δ𝑡 𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 𝑜𝑓 𝜌 = 1 + 1 + 2 + ( 0.4 11.2 × 100%) = 7.6% Refer to EJC 2020 Measurement tutorial Qn 7 for deeper understanding on handling percentage uncertainty when “stacking” is involved. 3 B B will result in the voltmeter giving readings that are larger than the true value. 4 D Option C is actually the velocity-time graph of the car. Note that the gradient at t = 0 for the displacement-time graph should be zero as the car starts from rest. Given that the area under the acceleration-time graph cancels out one another, the final velocity of the car is zero as change in velocity is zero. Hence gradient of displacement-time graph is zero at the end. Lastly, when acceleration is 0 between t and 2t, the car is travelling at constant speed. Hence gradient of displacement-time graph is constant.

2 © EJC 2020 9749/J1H2PROMO/2020 5 B As both the hammer and spanner are released from the same vertical height and have the same initial vertical velocity relative to the first floor they will land in the first floor at the same time as they experience the same acceleration. 6 C Solution: the 12 N force will be exerted equally o n all springs as they are being connected in series. Using Hooke’s Law, F = kx, xA = 12/2 = 6 m xB = 12/3 = 4 m xC = 12/6 = 2 m Therefore answer is 3 : 2 : 1. 7 D Pressure at point A = 𝜌𝑤𝑎𝑡𝑒𝑟𝑔ℎ + 𝑃𝑎𝑡𝑚 = (1000)(9.81)(ℎ) + 𝑃𝑎𝑡𝑚 = 9810ℎ + 𝑃𝑎𝑡𝑚 Pressure at point B = 𝜌𝐻𝑔𝑔ℎ𝐻𝑔 + 𝜌𝑜𝑖𝑙𝑔ℎ𝑜𝑖𝑙 + 𝑃𝑎𝑡𝑚 = (13600)(9.81)(0.05) + (800)(9.81)(ℎ − 0.05) + 𝑃𝑎𝑡𝑚 = 6278.4 + 7848ℎ + 𝑃𝑎𝑡𝑚 Pressure at the same depth are equal at equilibrium. 6278.4 + 7848ℎ + 𝑃𝑎𝑡𝑚 = 9810ℎ + 𝑃𝑎𝑡𝑚 ℎ = 3.20 𝑚 8 A Team C and F alone gives a resultant force of 300 N along OF Team A and D gives a resultant force of 300 along OD Team B and E gives a resultant force of 300 N along OE Drawing the vector diagram gives resultant force of all teams to be 600 N along OE. Question can also be d one by resolving resultant forces due to CF and AD along the direction OE.

3 © EJC 2020 9749/J1H2PROMO/2020 9 B Taking 3M and 2M as one system, Resultant force upslope, 5ma = F – 5mgsin30 a = F/(5m) –gsin30 Co