2024 JPJC H1 Prelim P1 Solutions

Answers to 2024 JC2 Preliminary Examination H1 Paper 1 1 C 6 C 11 B 16 D 21 B 26 D 2 B 7 C 12 C 17 A 22 D 27 D 3 B 8 A 13 C 18 B 23 A 28 D 4 A 9 C 14 A 19 B 24 D 29 B 5 A 10 A 15 A 20 C 25 A 30 C Suggested Solutions: 1 Assume that the diameter of the cross-section of the wire in a paper clip is 1 mm. Cross-sectional area of the wire in a paper clip           2 7 2 7 20.001 7.85 10 m 8 10 m2 Answer: C 2 nano should be 10−9 Answer: B 3 Answer: B 4    2 2Using 2 0 2(9.81)(2.5)v u as   1 7.0 m sv 21 2 21 2 Using 0.12 7.0 (9.81) 0.0167 s s ut at t t t        Answer: A change in velocity

2 2024/JPJC/PHYSICS/8867 5   21 2Horizontally, using s ut at 900 (450 cos31.6 ) 0 2.35 s t t               21 2 21 2 Vertically, using (450 sin31.6 )(2.35) ( 9.81)(2.35) 527 m s ut at s Answer: A 6 In throwing vertically upward, the ball has the same horizontal velocity as the thrower. So the ball would be caught if the thrower maintain a constant velocity. Answer: C 7 At equilibrium, force by air on four propellers = weight of drone        air drone4 0 dmv m g dt          air4 0 0.40 1.2 9.81v  1 air 7.4 m sv Answer: C 8 Let T be the tension in the spring. At equilibrium,    0.30 9.81T Immediately after the thread is cut,             2 0.30 9.81 0.20 9.81 0.20 4.9 m s T mg ma a a Answer: A 9 Momentum has to be conserved since there is no external net force acting on the system (comprising the two asteroids). As the collision is inelastic, some of the kinetic energy of the system could be converted to other forms of energy such as thermal energy which raises their temperature. Answer: C 10 Torque of couple  5.0 (0.60sin40 ) 1.9 N m   Answer: A

3 2024/JPJC/PHYSICS/8867 11 Area under the F-x graph =   2 1 2 1 1 2 T T x x  Answer: B 12 Resolving the forces,        2 1 2 1 1 sin30 (1) cos30 (2) (1): tan30(2) tan30 15tan30 8.66 8.7 N (2s.f.) N N N W N W N W    1 (Alternative: Using vector triangle) tan30 15tan30 8.66 8.7 N (2s.f.)N W Answer: C 13 By conservation of energy, loss in K.E. = work done against frictional force.      2 22 1 02 800 30 7200 N2 2 50 mu fd muf d      Answer: C 14 v u at gt     222 21 1Kinetic energy 2 2 2 mgmv m gt t         . This is given by Option A or C. Based on conservation of energy, Potential Energy = Total Energy – Kinetic energy Answer: A N1 W N2 N1 W N2 30° 30°

4 2024/JPJC/PHYSICS/8867 15 Using conservation of energy, loss in G.P.E. = gain in K.E. + work done against frictional force            2 2 1 2002 160 50 60 20 2002 87 N mgh mv f g f f      Answer: A 16 The horizontal component of T which is directed towards the centre of the circle provides the centripetal force. The vertical component of T is equal to the weight. Applying Newton’s second law of motion, 2 sin mvT r  - - - - (1) cosT mg   - - - - (2) 2 2(1) : tan(2) v r rg g    Answer: D 17 When the passengers feel weightless, the centripetal acceleration is equal to the gravitational acceleration at that point.    2 11000 9.81 99 m s v gr v rg      Answer: A  W r T

5 2024/JPJC/PHYSICS/8867 18 For circular motion at the highest point, 2 2 2 2 3 3 2 3 2 3 mvmg N r W mvW r mg mvmg r mg mv r grv         Answer: B 19 R A L b ac    Answer: B 20 From Fig. 20.1, when the current is 5 mA, the p.d. across the diode is 0.8 V The p.d. across the 50  resistor is given by V = (5 mA)(50 ) = 0.25 V So the p.d. across the supply = 0.8 + 0.25 = 1.05 V Answer: C 21 Using the equation P = IV, Current in circuit total power dissipated in circuit e.m.f. P p E   Answer: B 22 The variable resistor is set such that current can pass through the whole resistor (i.e. resistance = R), or that current can totally bypass the resistor (resistance = 0). Hence, the maximum p.d. happens when the resistance of variable resistor is zero  maximum p.d. = 6 V Minimum p.d happens when the resistance of the variable resistor is maximum  minimum p.d. (6) 3 VR R R  Answer: D

6 2024/JPJC/PHYSICS/8867 23 Effective resistance across YZ YZ XZ YZ XZ 10 6(10 10) .667 10 (10 10) R R R R        P.d. across YZ = p.d across XZ 6.667 10 10 6. 12 4.80 67 V6 eff eff R R E     Voltmeter reading = pd across WX = half of pd across XZ = 2.4 V Answer: A 24 By Fleming’s left-hand rule, F is perpendicular to B. Answer: D 25 sin30 (2 )(0.5 )BF Bqv B e v Bev   BF Beva m m  Answer: A 26 At equilibrium, mgd qVF E When d is twice, FE is reduced by half, mgF 2 1 E  ga mamgmgmgF 5.0 5.02 1 resultant   Oil drop accelerates downwards as weight is now greater than upward electric force. Answer: D 12 V 10 Ω 10 Ω 10 Ω 10 Ω V W X Y Z

7 2024/JPJC/PHYSICS/8867 27 Using Right Hand Grip rule, the direction of flux density at O due to the currents in each option are: A: leftwards (XO) B: rightwards (OX) C: upwards (OY) D: downwards (YO) Answer: D 28 Majority of the -particles pass through the gold foil without being deflected since the atom consists of mostly empty space . S ome are deflected by small angles . Very few are detected on the same side as the alpha beam. Answer: D 29 An α-particle is a helium nucleus 4 2He . Hence the answer can only be either option A or B. The symbol of a neutron is 1 0n and not 1 1n as depicted in option A. Answer: B 30 Total mass of nuclides after 12 days             6 3 1 1200 100 15.6 g2 2 Answer: C