2013_PJC_H2_Physics_P1 (Solutions)

Answers to 2013 JC2 Preliminary Examination Paper 1 (H2 Physics) 1 D 6 C 11 C 16 D 21 B 26 A 31 C 36 D 2 B 7 B 12 A 17 A 22 D 27 D 32 C 37 B 3 B 8 A 13 D 18 D 23 B 28 B 33 C 38 D 4 B 9 B 14 B 19 D 24 A 29 B 34 C 39 D 5 D 10 B 15 C 20 B 25 D 30 D 35 A 40 C 1 Since FB L I , units of B = units of F LI = 2kg m s Am  = kg s −2 A−1 Answer: D 2 fivv v    if vvv      222 if vvv    22 57 v 9v ms −1 f i v vtan 7 5tan   36 Answer: B 3 The acceleration-time graph is obtained from the velocity-time graph. Point Q is the turning point of the velocity-time graph and so is the maximum value. Answer: B

4 Let the initial speed of the javelin be u. At the initial point, K.E. 21 2 mu E At the maximum height, K.E.’  21 cos602 mu 211 42 mu  1 4 E Answer: B 5 Only D is the correct answer because it is moving at constant velocity. Resultant force zero. Answer: D 6 Rate of change of momentum of the water = m vt  =1.0  8.0 = 8.0 kg m s1. From Newton’s second law of motion, force on the water = 8.0 N From Newton’s third law, force on the notice-board = 8.0 N. Hence magnitude of acceleration of the notice-board = F m = 0.2 0.8 = 4.0 m s2. Answer: C 7 Upthrust on cargo = weight of sea- water displaced due to cargo = weight of cargo Depth sank due to cargo(∆h)  cross-section of barge(A)  density of sea-water(ρ) g = mg ∆h = m A = 103097 100.2 4   = 0.20 m = 20 cm Depth of bottom below surface is 70  20 = 50 cm Answer: B 8 Extension, e = 90  60 = 30 mm Tension = k e = 500  0.030 = 15 N Answer: A

9 lost in kinetic energy = work done against resistive force in the plank 221 [(150) (125) ] work done against friction2 m  For the second bullet, the work done against friction stays the same when it passes through the plank. 22 2 211 [(150) (125) ] [(90) ( ) ]22mm v  135 m sv  Answer: B 10 Work done (WD) by applied force = (80)(3) = 240 J WD against frictional force = friction  distance moved = (30)(3) = 90 J = heat generated. Gained in gravitational potential energy = mgh = (40)(3.0 sin 30)= 60 J Using conservation of energy, Gained in K.E. = 240  90  60 = 90 J Answer: B 11 Since mass, radius and speed is the same, then the magnitude of the centripetal force on P and Q is the same, regardless of orientat ion. Centripetal force is provided by the net force towards the centre of the circular path and thus the net forces of P and Q are equal in magnitude. Answer: C 12 For mass m 1: 2 1 2 1 () mv keLe ke L ev m   For mass m2: 2 2 2 2 2 2 (2 2 )2( ) (2 )2( ) 2( 2) ( ) mv kL eLLe mv kL eLe kL e L em v     1 1 2( 2) ( ) k( ) 2( 2 ) kL e L em eL e mL e e    Answer: A

13 2() GMmF Rh   where R  radius of Earth h  height of moun