VJC_H1_PHY_Solutions

1 2016 VJC Prelim H1 Paper 1 Suggested Solutions 1 A 11 C 21 C 2 B 12 C 22 B 3 B 13 D 23 C 4 A 14 C 24 C 5 B 15 C 25 D 6 B 16 D 26 A 7 D 17 C 27 C 8 B 18 C 28 B 9 B 19 B 29 C 10 C 20 C 30 B 1 Ans: A Energy produced = Pt = 3.0 x 109 x 2.0 x 10-12 = 6.0 x 10-3 J = 6.0 x 10-15 TJ 2 Ans: B Minimum hypotenuse length is given by sides of 8.8 cm and 6.8 cm. Minimum hypotenuse length = 22 8.68.8  = 11.12 cm Maximum hypotenuse length is given by sides of 9.2 cm and 7.2 cm. Maximum hypotenuse length = 22 2.72.9  = 11.68 cm Average = (11.12 + 11.68)/2 = 11.40 cm Uncertainty = extreme – average = 11.68 – 11.40 = 0.28  0.3 cm

2 3 Ans: B Uncertainty of period = 20 5.05.0  = 0.05 s 4 Ans: A  The ball starts with zero speed. So when s = 0, v = 0.  The ball can move either up or down. So v should change sign.  But the ball is always below the point P. So s should always have the same sign. 5 Ans: B Remember: v = vf - vi The horizontal component of vi = vf since there’s no horizontal acceleration, so v is vertical. v = 12 sin25o = 5.1 m s-1 6 Ans: B Take upwards as positive. Vertically, s = (u sin)t + ½at2 -15.0 = (8.50 x sin40.0o)t – (½ x 9.81)t2 Solving, t = 2.39 s (reject negative answer) 25o vf -vi = 12 m s-1 v  = 40.0o s= -15.0 m u= 8.50 m s-1

3 7 Ans: D Taking moments about pivot: Wrubber x (1.90 L) = Wsteel x (0.10 L) 1.0Lx A x rubber x 1.90Lg = 3.00L x A x steel x 0.10L  𝜌𝑠𝑡𝑒𝑒𝑙 𝜌𝑟𝑢𝑏𝑏𝑒𝑟 = 1.0𝐿𝐴×1.90𝐿 3.00𝐿𝐴×0.10𝐿 = 6.33 8 Ans: B Torque by a couple = F x perpendicular distance between them = F x L sin 9 Ans: B Weight of helicopter is the gravitational force of Earth on the helicopter. So by N3L, the helicopter will exert an equal and opposite gravitational force on the Earth. 10 Ans: C Statement C is false because the collision forces are internal, not external, forces, and so the total momentum of the system should be conserved throughout the whole duration of the collision. 11 Ans: C Area under F-t graph = change in momentum ½ x (30 + 15)x 1500 – ½ x15 x 1500 = m(v – 10) v = 32.5 m s-1 12 Ans: C Since object is at constant speed up the inclined plane: Total w.d by 50 N force = Gain in GPE + w.d against friction 1500 = 50 x 12 + w.d against friction  w.d against friction = 1500 – 600 = 900 J 13 Ans: D When boat is travelling at constant speed v, the driving force is equal to the drag force. Then the power P is given by: P = Fv = Dv = (kv2)v Ie. P = kv3 When both engines are working: 2 x 30 kW = k(10)3 - - - - (1) When only one engine is working: 30 kW = kv3 - - - - (2) (2)/(1): and solve for v: v = 7.9 m s-1

4 14 Ans: C The electrical power is the input power = IV 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 = 𝑃𝑜𝑢𝑡 𝐼𝑉 0.80 = 4.0×1