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 an optically less dense to an optically denser medium . 8 (a) i. Loudness decreases as amplitude of sound decreases with time.

ii. pitch remains constant as frequency of sound remains constant as period is constant. (b) 9 (a) As the dust particle comes into contact with the rods, electrons from the rods are transferred to the dust particles. (b)(i) As the metal plates are positively charged, the dust particles are attracted to it as unlike charges attract. (b)(ii) 10 (a)(i)  As the current enters the solenoid, the solenoid becomes an electromagnet.  The bolt, which is made of magnetic material iron, becomes an induced magnet.  It is attracted to the solenoid. (a)(ii) To push the iron bolt back to its original position (the striker plate) when the switch is open (a) (iii) The magnetic force generated by the solenoid is weaker and its magnitude is less than the tension of the spring. negatively charged dust particle charged metal plate

10 (b) (i)  Current in CD produces a magnetic field, interacts with the magnetic field of the permanent magnet.  Combined magnetic field at the bottom of CD is stronger than at the top of CD. This produces a resultant upward force on CD (b)(ii)  Between 0 and t1: Direction of F is upward. Hence positive value.  Between t1 and t2: Direction of F is downward. Hence negative value Magnitude of F is constant, because current in CD and external magnetic field strength are constant. (b)(iii) As the coil CDEF spins, it cuts the magnetic field lines. There is a rate of change of magnetic field lines linking the coil. By Faraday’s law, an induced emf in CDEF. By Lenz’s law, the direction of the induced emf in the coil will produce a magnetic field to oppose the magnetic field linkage created by the rotating coil with the external magnetic field. Section B 11 (a)(i) (a)(ii) the cross sectional area of the wire is inversely proportional to its potential difference. As the cross sectional area of the wire is doubled when wire K and wire L is used, the p.d is halved. (a)(iii) V = 0.11/2 = 0.055 A Or V = 0.028 x 2 = 0.056 A Or AV = constant = 0017 V = 0.053 A (a)(iv) Can choose any of the wires R = V/I = 0.85 /2.0 = 0.425 Ω R= ρL/A ρ = 0.425 x(0.002 x10-6)/0.050 = 1.7 x 10-8 Ωm (a)(v) Rheostat is needed to change the effective resistance in the circuit so that current can be kept constant. (b)(i) VXY = 0.50 V Emf = 0.50 +1.00 = 1.50 V Alternatively,

I = 1.00/0.200 = 5.00 A V= RxI = 5.00 x 0.300 Ω = 1.50 V (allow ecf from (i)) (b)(ii) Total effective resistance across XY decreases. Hence, by potential divider principle, more of the e.m.f. will go towards YZ. Or Total effective resistance of whole circuit decrease, hence current flowing through YZ is higher. (b)(iii) Voltmeter has a very high resistance/infinite resistance. 12 (a) Direction is from D to A. (b) - The induced current in the coil produces a magnetic field that interacts with the external magnetic field, producing an unbalanced resultant magnetic field that give rise to an induced force - By Lenz’s law, the direction of the force will oppose the motion that is producing it. (c) (d)(i) Np/Ns = Vp/Vs 1/5 = 6V/Vs Vs=30 V Voltage across one bulb = 15 V (d)(ii) A direct supply current produces a steady magnetic field hence there will be no change in magnetic flux/field lines linking the secondary coil and no e.m.f and current is induced emf time ¼T ½T ¾T T 0

(d)(iii) This is to ensure that the current is kept low so that it reduces the power loss due to heating in the cables. (P loss = I2 R) 13E (a)(i) The earth wire is not connect. If there is leakage of current, the current will not be earth. When the user touches the kettle, he will experience electric shock. (a)(ii) P = VI (100W) = (240V)(I) I = 0.42 A or 0.417 A fuse = 1 A (a)(iii)  The red wire is the live wire which has to be connected to the fuse.  When there is excessive current, it will melt the fuse thus disconnect the kettle from the high electrical potential. (b) P = V2/R 100 W = (240V)2/R R = 576 Ω V = IR 110 V = I(576 Ω) I = 0.191 A (c)(i)  Electical fire may occur.  Thinner wire has higher resistance, as air conditioner unit draws high current too, it will results in a lot of heat produced causing electrical fire to occur. (c)(ii) Resistance per unit metre = 0.050 Ω / (2.0)2 = 0.013 Ω 13 O (a) λ = 60 cm / 3 = 20 cm (b) 3T = 30 s T = 10 s f = 1/T = 1/10s = 0.1 Hz. (c)(i) There is a change in speed (c)(ii) Perpendicular to the wave direction and continue from the incident wavefront. Scale: 0.80 cm representing 20 cm

(c)(iii) λ = 7.5 ± 0.75 cm (remember to multiply the scale) λ/2 = 3.75 = 3.8 cm