ACJC 2022 Prelim P2 Guide

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2022 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 Qn Suggested Answer Marker’s Report Q1 (a)(i) The absolute uncertainty of both measurements is the same. OR Fractional/percentage uncertainty is reduced by measuring N coins. Some students were unable to show understanding that the absolute uncertainty for both ways of measures is the same. By measuring N coins, the absolute uncertainty of the thickness of 1 coin is divided by N. OR By measuring N coins, fractional uncertainty is ∆x / T. By measuring 1 coin, fractional uncertainty is ∆x / T/N. (a)(ii) Use a micrometer screw gauge/vernier caliper as the instrument has smaller absolute uncertainty compared to the half metre rule. This part was well done. (b) 3 0.00200 8.31 (273.15 36.7) 4 (0.0250)3 78682 Pa pV nRT p  =   += = Most students were able to do this step correctly with a handful of students forgetting to convert temperature to K. (3 ) 0.1 0.1[3( ) ]50.0 (273.15 36.7) p d T p d T    =  + =  + + A number of students did not convert temperature to K when calculating uncertainty. ±∆p = ±497 Pa p ± ∆p = (78700 ± 500) Pa Some students did not express the absolute uncertainty to 1 s.f. Q2 (a)(i) Constant velocity in the horizontal direction and a constant acceleration in the vertical direction Some students did not mention about the motion in the horizontal direction. (a)(ii) 22 2=+v u as 0 = (20.0 sinθ)2 + 2(-9.81)(15.8) 61.7 = Intermediate value should be given to at least 3 s.f. 62 = (0 d.p.) (shown)

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2022 Page 2 of 9 (a)(iii) 0 20.0sin61.7 9.81 y y yv u a t t =+ =  − Students did not take into account direction of acceleration and hence did not use negative sign for ay. t = 1.80 s (3 s.f.) (b)(i) Since force exerted on the spaceship F is constant but mass of the spaceship decreases with time, given that F = ma, a increases with time. Some students did not indicate that force exerted on spaceship is constant. (b)(ii) on fuel 46(3.0 10 )(1.7 10 ) − = =   rel dmFv dt = 0.051 N on spaceship on fuel=FF Students should be aware to show application of N3L. (b)(iii) Change in velocity = area under graph = ½(9.450 + 8.200)10–5  4.80107 A number of students failed to realise the x-axis did not start from 0. = 4240 m s–1 Final velocity = 4240 – 0 = 4240 m s–1 (b)(iv) Increasing gradient from t = 0 to t = 4.8  107 s, starting from v = 0 to final v at 4240 m s–1. Some students did not indicate value of max v in graph nor time where constant v is attained. Between t = 4.8  107 to t = 6.0  107 s, there is no more force. Therefore, constant v at 4240 m s–1.

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2022 Page 3 of 9 Q3 (a) Correct labelling of forces The free body diagram was not well drawn. Many students missed out drawing viscous force or upthrust in the diagram. As speed of metal ball increases, viscous force increases. As viscous force increases, net acceleration decreases. One common misconception was that upthrust increases as the metal ball falls. When the total upward force (viscous force and upthrust) becomes equal in magnitude to the weight of the ball, there is no net force, and the metal ball reaches terminal velocity. Some did not indicate that the upward force comprise viscous and upthrust. (b)(i) Taking moments about B, 6.0cos30 150 3.0sin30 43.301 N P P   =   = Intermediate value should be given to at least 3 s.f. = 43 N (2 s.f.) shown (b)(ii) The horizontal component of Q which is pointing to the left will balance force P which is pointing to the right. Generally well done with some answers missing out directions such as left, right, up and down. The vertical component of Q which is acting upwards will balance W which is acting downwards. This will allow the ladder to be in equilibrium. (b)(iii) Horizontal component = 43 N Vertical component = 150 N 22Magnitude of 43 150 156 N (3 s.f.)Q = + = This part was well done. (b)(iv) Three forces intersect at same point and force at X is facing upwards diagonally. Some students did not check that the 3 forces drawn intersect at the same point. upthrust viscous force weight X W Y

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2022 Page 4 of 9 4 (a) 2 2 G G F Mmg where Fm r M r == = Some students still did not follow the requirement of the question. 34 3 Mas R =  𝑔 = 4𝜋𝜌 𝐺 𝑅3 3 𝑟2    (b) –R < r < R: straight line passing through the origin r < –R and r > R: inverse square graph +ve and -ve direction of g (according to stated sign convention) Many students did not account for the sign convention of the gravitational field strength. 4 3 ρG Rr R : g == −4R 0 4R +ve g g

Anglo-Chinese Junior College 2022 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2022 Page 5 of 9 (c)(i) Additional Notes: (NOT part of answer) • Straight line passing through the origin indicates that the gravitational force on parcel is directly proportional to the distance from the centre of Earth. • Negative gradient indicates that the gravitational force is always directed towards the centre of the Earth. • Hence, net force for SHM is provided by gravitational force. 2 By N2L, 4 3 (1) For SHM, (2) G m a m g ar ax = = =   2 4 3 Comparing G=  Wrong physics include equating g to centripetal acceleration (circular motion) instead a to and fro acceleration (SHM). 2 4 3 3 2 G G T T     = = A number of correct expressions even though it was not simplified. (c)(ii) Maximum time taken (as assuming released from rest in Singapore) = half a period Many correctly deduced the time to be half a period. ( ) 3 -11 3 3 6.67 10 1 22 1 2 5.51 10 2530 s (3 s.f.) G T     = =  = 5 (a)(i) As the light intensity increases, the resistance of the LDR decreases. This was generally well done. Some students did ignored the fact that the emf is constant and thought t