SJI 2018 Prelim P1 and P2 ANS

2018 Secondary 4 Prelim Exam Physics 6091 Suggested Solutions Paper 1 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 C C C D D D D C B A Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 D B A D D B B B A B Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 A B C A C C B D C B Q31 C32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 C B D D A D C A A A Paper 2 Section A 1 (a) Speed increases at a constant rate of 2.0 m/s2. For every 1s, the speed increases constantly by 2.0 m/s. (b) Car A – Straight line starting from 0 with a gradient of 2.0 (+value) Car B – Horizontal line at 2.0 m/s (-value) (c)(i) d = 0.5(t)(2.0t) = t2 (c)(ii) t2 + 2t = 1000 t = 30.6 s = 31 s (ecf) 2 (a) 5.0 N acting on ball C and 5.0 N acting on ball A in the opposite direction. (b)  0. 66 N ± 0.06 N.  Correct arrow direction  Correct diagram (c) Fr = ma 0.50N = (0.10 kg)(a) a = 5.0 m/s2 (d) As the resultant force is not too large and due to inertia, ball A continue to move in the general forward direction. 3 (a)(i) M = W x d = 10 N x 1.0 m = 10 Nm (a)(ii) Taking moment about P, CW moment = ACW moment 10 N x 1.0 m + 15 N x 2.5 m = Q x 1.8 m + 5 N x 0.5 m Q = 25 N

(b)  As the weight of L increases, the clockwise moment about P increases,  Hence anticlockwise moment by tension in Q must increase also (to keep the rod in equilibrium). (When the tension in Q is > than the maximum tension, Q will break) 4  As air is pumped in, the number of air particles per unit volume in the balloon increases  Frequency of collision (between air particles and the wall of the balloon) will increase (the force of each collision remains constant as temp, hence speed is constant)  As the pressure inside the balloon is greater than the pressure outside, there is a net outward force (which cause the balloon to expand.) 5 (a) John’s loss in GPE is equal to gain in KE and work done against friction (b) GPE = mgh = (70kg)(10 N/kg)(5.0m) = 3500 J (c) work done against = net loss in Energy = loss in GPE – gain in KE = 3500 J – ½(70kg)[(8.0m/s)2-(2.0m/s)2] = 1400 J (d)  Since there is less water, the frictional force is larger, resulting in a larger work done against friction.  (Hence there is lesser gain in KE and hence the velocity will be lower than 8.0 m/s.) 6 (a) Q = mcΔθ = (100g)[ 2.4 J/(gK)](50°C - 17°C) = 7900 J (b) Qfreeze = mlf = (0.100kg)(4.2 x 105 J/kg) = 42000 J time = 100 s / 7900 J x 42000 J = 530 s t = 630 s (ecf) (c) In real life the difference between the temperature of the glycerol and ice (its surrounding) is getting smaller, hence the rate of heat transfer will decrease with time. 7 (a) The angle of incidence of ray 2 at S is 0o. (b) sin c = 1/n = 1/1.47 c=42.8o (c) Ray 1: The angle of incidence of ray 1 in the Pyrex glass < critical angle of Pyrex Ray 3: It is travelling from a