EJC 2019 J1H2 Promo MS (Final)

© EJC 2019 9749/01/J1H2PROMO/2019 EUNOIA JUNIOR COLLEGE JC1 PROMOTIONAL EXAMINATIONS 2019 General Certificate of Education Advanced Level Higher 2 PHYSICS Multiple-Choice & Structured Questions MARK SCHEME w/ EXAMINERS’ COMMENTS Maximum Mark 9749 September / October 2019 Paper 1 Answer Key 1 2 3 4 5 6 7 8 9 10 B C D A D C B C D B 11 12 13 14 15 16 17 18 19 20 C B C C D B D B D A 21 22 23 24 25 26 27 28 29 30 C D A C A B A B A C

2 © EJC 2019 9749/J1H2PROMO/2019 Paper 2 Mark Scheme 1 (a) (i) Gradient of tangent (at t = 10 s) = (100-0)/(10-6) = 25 m s-1 M1 A1 (ii) By Principle of conservation of energy Loss in Ep = Gain in Ek (0.55) mgh = ½ mv2 (0.55)(9.81)h = ½ (25)2 h = 57.9 = 58 m (2 s.f) M1 A1 (b) (i) All 3 forces (Normal Contact, Weight, Resistive Force) drawn and labelled correctly. B1 (ii) The resultant force acting on the skier is downwards and parallel along the slope. Hence the skier experienced an acceleration, resulting in an increased in speed. This explains the increasing rate of change of distance with time shown in the graph. B1 B1 (c) (i) s = ut + (0.5)at2 80 = (0.5)(9.81)(t2) t = 4.0 s A1 (ii) x = vt = 25(4.0) = 100 m A1 Normal contact force Friction and/or air resistance Weight

3 © EJC 2019 9749/J1H2PROMO/2019 Examiners’ Comments (a) (i) Many candidates incorrectly used s = ½ (u + v)t to solve the problem. They cannot assume that the curve graph is parabolic and that the acceleration is constant. Some candidates incorrectly used the relationship speed = total distance / total time which gives the average speed. The tangent at t = 10 s were absent in many scripts. Working for the gradient of the tangent are often missing. (ii) Presentations were often poor and lacking in the concept being used. Candidates should write down the physics concepts used and the required equations. Candidate often substituted numbers without the equation. Many candidates missed out or wrongly used the 55% value given in the question. They used 55% of the kinetic energy instead. Candidates should have checked their working again when they get small values of the ramp height. (b) (i) Some candidates missed out the air resistance, friction, and normal contact force. Some of the weight and normal contact force were drawn in the wrong direction. Some wrongly drew the forward force. Some candidates did not label in full. Relative length of the arrows drawn is expected. (ii) Many candidates failed to explain the presence and direction of the resultant force. Explanations were often vague and not coherent. Some used energy concepts to explain the graph which were not the required answers. They should use the idea of the resultant force to bring about the idea of acceleration which then lead to the increasing speed as shown in the graph. (c) (i) Many candidates wrongly used s = ½ (u + v)t. (ii) This part was done well by most.

4 © EJC 2019 9749/J1H2PROMO/2019 2 (a) (i) Linear momentum is the product of mass and velocity. B1 (ii) The Principle of Conservation of Linear Momentum states that the total linear momentum of a system remains constant provided that no resultant external force acts on the system. B1 (b) (i) Total momentum in the horizontal direction = m v =(0.3) (8.0 cos 60 o) = 1.2 kg m s -1 B1 By conservation of momentum, initial momentum = final momentum 1.2 = mxvx + mYvY 1.2 = (0.20)(-4) + (0.10) vY vY = 20 m s-1 M1 A0 (ii) Impulse of Y = m (v f – vi) = (0.1)(20-4) = 1.6 kg m s -1 B1 A1 (iii) 1.6 0.001 1600 N pF t ∆= = = M1 A1 (c) At P: Before the explosion, E k = ½ m v2 = ½ (0.30)(8.0 cos 60 o )2 = 2.4 J After the explosion, E k = ½ m vx2 + ½ m vy2 = ½ (0.20)4 2 + ½ (0.1)(20)2 = 1.6 + 20 = 21.6 J Energy released in the explosion = 21.6 – 2.4 = 19.2 J M1 M1 A1 (d) Point P’ is lower and closer t o start point as compared to P. B1 Examiners’ Comments (a) In stating the Principle of Conservation of Momentum, candidates should not use the any form of the root word ‘conserve’. That fails to demonstrate any understanding. (b) A small proportion of candidates fail to realise that the horizontal velocity of the firework at the top of the projectile is 8 cos 60°, several took the horizontal speed at that point to be 8 m s-1. (c) Candidate are advised that the difference in kinetic energy should be found using (½ mv2 - ½ mu2) rather than ½ m(v -u)2. This issue cropped up mostly when there was an attempt to find the difference in kinetic energy of X and Y separately. (d) Point P’ was generally clearly indicated.

5 © EJC 2019 9749/J1H2PROMO/2019 3 (a) (i) ( ) ρ − − = = ×× × = Ω 7 6 169.8 10 0.20 10 78.4 LR A M1 A1 (ii) ( ) I= = = 2 2 2 240 78.4 VPR R = 734.7 W = 735 W (3 s.f) M1 A1 (b) switch A switch B switch C Lowest OFF ON OFF Highest ON OFF ON A1 A1 Examiners’ Comments Several candidates had problems with part (a) due to the conversion of mm2 to m2. For part (b), many candidates failed to realise that the maximum operating power occurs when there is minimum resistance (when the resistance are in parallel) and minimum operating power occurs when there is maximum resistance (when the resistance are in series).

6 © EJC 2019 9749/J1H2PROMO/2019 4 (a) It is the change of phase per unit time of an oscillation. OR rate of change of phase of the oscillating mass B1 (b) (i) 1. 2.0 cm A1 2. 1.5 cm A1 (ii) reference to displacement from s = 2.0 cm or equilibrium position is at s = 2.0 cm B1 straight line indicates that acceleration is directly proportional to displacement. B1 negative gradient indicates that acceleration and displacement are in opposite directions. B1 (iii) 1. 2 1 gradientω = − 2a ωx= − C1 ( ) 2 1 0.035 0.005 0.9 0.9 60 ω ω = − − − +− = ( ) 20.9 0.015 60 ω ω −= − = M1 2 60 1.2 Hz πf f = = A1 2. ( ) 160 0.015 0.116 m soov ωx −= = = (allow e.c.f.) A1 (c) () () 22 () 22 () () 3 4 1 32 1 4 2 3 0.015 4 Kf Ki of oi of E E mωx mωx x = = = C1 () 0.013 m 1.3 cmofx = = (allow e.c.f.) A1 Examiners’ Comments (a) A large number of candidates provided the definition of frequency instead of angular frequency as required. (b) A large number of candidates were not able to answer (i)1 correctly. Those who were able did not seem to realize that that was the equilibrium position of the oscillations, which had an implication for their answers in part (ii); many candidates incorrectly stated that the graph passed through the origin, which was obviously incorrect. For part (iii), a large number of candidates made powers of ten (P.O.T.) errors in their substitutions. (c) Candidates are reminded that for questions involving comparison of 2 similar situations, calculations using the numerical ratio will normally suffice. Power of Ten (P.O.T.) errors are ignored

7 © EJC 2019 9749/J1H2PROMO/2019 5. (a) (i) (i) Us = UA + UB = − 𝐺𝐺𝐺𝐺𝐴𝐴𝐺𝐺𝑠𝑠 𝑟𝑟 + (− 𝐺𝐺𝐺𝐺𝐵𝐵𝐺𝐺𝑠𝑠 𝑥𝑥−𝑟𝑟 ) = −𝐺𝐺𝑚𝑚𝑠𝑠( 𝐺𝐺𝐴𝐴 𝑟𝑟 + 𝐺𝐺𝐵𝐵 𝑥𝑥−𝑟𝑟) B1 (ii) B1