2021 J2 H2 Physics Prelim P2 MS Sharing

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 1 of 9 Annotations used in marking BOD - Benefit of doubt ECF - Error carried forward POT - Powers of ten error TE - Transfer error CE - Calculation error XP - Wrong physics ENG - Generally bad english, phrasing and expression PP - Poor presentation of answers Note: For POT and TE, we can award the M mark, not the A mark. Qn Suggested MS 1 (a)(i) 1 1 1 2 2 2 1 0.1 2 0.2 2 1.8 126 5.1 v T m v T m                         l l 0.055 5.5%   OR Find max and min value Take average and correct answer (a)(ii)      3 1 1.8 1.26 5.1 10 21.08805 m s v        1 0.055 0.055 21.00805 1 m s v v v         121 1 m sv  (b)(i) Average value of 20.7 m s–1 falls within the range due to experimental error. (b)(ii) Random error Repeat the experiment more times so that the average value will tend towards the true value.

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 2 of 9 2 (a)   24 1.5 x x sv t  116 m s (b)   tan28 16 tan28 8.51 y x y v v v     18.5 m s (c)(i) Straight line with negative gradient Gradient = 9.81 m s–2 , t = 0.87 s  cut at 0.84 s < t < 0.90 s Stop at t = 1.5 s, magnitude of vy must be less than initial (c)(ii) maximum height = area under the v-t graph   1 0.866 8.52 3.7 m   (c)(iii)         2 2 1k.e. 2 g.p.e 1 162 9.81 3.7 xmv mgh  k.e. 3.5g.p.e  (iv) Steeper initial gradient, which decreases over time. Area above and below must be the same. Tangent to curve where it cuts the time-axis is parallel to original graph and must be the correct shape. 8.5 0.87

Anglo-Chinese Junior College 2021 H2 Preliminary Exam Paper 2 MS H2 (9749) Physics JC2 2021 Page 3 of 9 3 (a) Upthrust is a force on a partially or fully submerged object , acting in the opposite direction to weight Equal in magnitude to the weight of the fluid displaced (b)      Upthrust 1000 0.40 0.50 9.81 1962 N waterVg      weight upthrust 950 9.81 1962 7357.5 N BET    7360 N (shown) (c)(i) Beam is in equilibrium. Taking moments about A, Total clockwise moment = total anticlock