2014 H2 P1 solutions vjc

1 2014 GCE A-Level H2 Paper 1 Suggested Solutions 1 C 9 C 17 A 25 A 33 D 2 A 10 D 18 C 26 D 34 B 3 A 11 B 19 A 27 D 35 D 4 D 12 B 20 D 28 B 36 C 5 D 13 B 21 D 29 A 37 A 6 C 14 C 22 C 30 D 38 D 7 C 15 A 23 B 31 D 39 A 8 D 16 B 24 D 32 C 40 C 1. Ans: C Absolute uncertainties are generally expressed to 1sf. Hence Δv = ±3 m s-1. The value of the measurement must match the number of decimal places (d.p.) of the absolute uncertainty. Hence, v = (348 ± 3) m s-1. 2. Ans: A This is mainly book work. Visible light has approximate wavelengths 400 nm - 700 nm, which includes 0.5 µm. Ultraviolet is around 10 nm – 400 nm, hence it includes 0.05 µm. Infrared is approximately from 700 nm – 0.1mm, which includes 5 µm and 50 µm. 3. Ans: A The gravitational field strength is given by the gradient of the curve at t = 0 s where motion of the object is just about to start and air friction has not affected motion yet. Draw a tangent at this instant, calculate the gradient, we find the value to be around 6 N kg-1. 4. Ans: D Let the total time taken be t, and total distance be s. Hence, 2)(2 1 tas= -- Eqn (1) From rest to time = (t - 1), the stone covered 0.25s. 2)1(2 125.0 −= tas --Eqn(2) Take Eqn(1) divided by Eqn(2) 2 2 )1(4 −= t t Solving, we find t = 2.0s Alternative: velocity time t-1 t g(t-1) gt Note: v = u+at

2 Let s be total distance covered and t the total time taken for the fall. Then the distance covered is the area under the velocity-time graph. We have 2 2 1 gts= …(1) Now considering the distance covered in the second last second of the fall. ( )ggtttgttgs tts −=−−+−= −= 22 1)]1()[)1((2 1 4 3 to1 from trapeziumof area4 3 …..(2) Substitute (1) into (2), we have ( )ggtgt −= 22 1)2 1(4 3 2 0483 2 =+− tt giving t = 2.00 s (The other answer t = 2/3 s is not applicable) 5. Ans: D D is the answer because since MV > mv, the net momentum is always non-zero, which means that both velocities cannot be zero at the same time. 6. Ans: C The tension in the coupling between wagons 2 and 3 pulls and accelerates wagons 3 to 6. By N2L, maFnet = mafT =− fmaT += Since total friction action on the 4 wagons = 4000 x 4 = 16000 N 16000)15.0)(100.64( 4 +=T 52000=T N 7. Ans: C P is the work done on the fibre by an external force. Q is work done by the fibre. Hence net work done on the fibre is P – Q. This probably means that there could be deformation to the wire or increase in internal energy in the fibre. 8. Ans: D Since it is falling with constant speed, the net force is zero. Wagons 3 to 6 Tension, T friction, f Weight, W Upthrust, U Viscous force, kv W = U + kv v

3 9. Ans: C Since Fnet = ma Ftrain – friction = ma where Ftrain is the driving or forward force of the train. Ftr