CJC H1 PHY P1 QP and SOL

1 [Turn Over] NAME: ______SOLUTIONS__________ CLASS: ___________ INDEX: __________ CATHOLIC JUNIOR COLLEGE JC2 Preliminary EXAMINATIONS Higher 1 PHYSICS 8866/01 Paper 1 1 September 2015 60 min Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write your name, tutorial group and index number on this cover page. Write and/or shade your name, NRIC / FIN number and HT group on the Answer Sheet (OMR sheet), unless this has been done for you. Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. There are a total of 30 Multiple Choice Questions (MCQs) in this paper. Answer all questions. For each question, there are four possible answers, A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the Answer Sheet (OMR sheet) provided. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. Calculators may be used. This document consists of 12 printed pages. [Turn over]

2 [Turn Over] PHYSICS DATA: speed of light in free space, c = 3.00 x 108 m s-1 permeability of free space, o = 4 x 10-7 H m-1 elementary charge, e = 1.60 x 10-19 C the Planck constant, h = 6.63 x 10-34 J s unified atomic mass constant, u = 1.66 x 10-27 kg rest mass of electron, me = 9.11 x 10-31 kg rest mass of proton, mp = 1.67 x 10-27 kg acceleration of free fall, g = 9.81 m s-2 PHYSICS FORMULAE: uniformly accelerated motion, s = u t + 2 1 a t2 v2 = u2 + 2 a s work done on / by a gas, W = p V Hydrostatic pressure p =  g h resistors in series, R = R1 + R2 + ... resistors in parallel, R 1 = ... 21 11 RR

3 [Turn Over] 1 Estimate the time taken for a laser beam to travel from a laser gun at the grandstand to the goalkeeper. A 3 ns B 300 ns C 3 ps D 300 ps Answer: B Estimate distance = 100 m Speed = 3 x 108 m s-1 time = distance / speed = 100 / 3 x 108 = 3 x 10-7 = 300 x 10-9 = 300 ns 2 A petrol gauge in a car indicates the volume V of fuel in the tank. V is given by the angular deflection θ of the pointer on a dial such hat the scale is linear. The graph below shows the actual variation of the volume V of fuel with the angular deflection θ. Which of the following statements are not correct? P: The petrol gauge is most sensitive when the tank has low fuel level. Q: The petrol gauge is most sensitive when the tank has high fuel level. R: The readings of V from the scale results in a random error. S: The readings of V from the scale results in a systematic error V θ Actual relation between V and θ Empty 0.0 1.0 Full θ 0.2 0.4 0.6 0.8

4 [Turn Over] A P & R B P & S C Q & R D Q & S Answer: C Low fuel level the gauge is sensitive because the change in angle is larger per unit change in volume The volume recorded is always larger than the actual volume, hence there is a systematic error in the reading. 3 Figure below shows the movement of a vehicle. It is initially moving at 10 m s -1 with a bearing of 60° and after 2 seconds it moves in the bearing of 180o at 5 m s-1. What is the direction and magnitude of the change in velocity? Direction Magnitude/ m s-1 A 180° 5.0 B 90° 8.7 C 123° 11 D 221° 13 By scale drawing and measurement ΔV = 13 m s-1 Direction = 220o By calculation: By cosine rule, (ΔV)2 = 102 + 52 – 2 (10)(5) cos (120°) ΔV = √175 = 13.2 m s-1 By sine rule, 13.2 sin 120° = 10 sin 𝜃 θ = 41° Hence the bearing is 180 + 41 = 221° 10 m s-1 5 m s -1 60° 10 m s-1 5 m s -1 120° θ ΔV Before After

5 [Turn Over] 4 A ball, dropped from a building, is timed to take (4.5 ± 0.1) s to fall to the ground. If the acceleration of free fall is taken to be 10 m s-2, the calculated height of the building should be quoted as A (101 + 2 ) m B (101 + 5 ) m C (101.3 + 2.3) m D (101.3 + 4.5) m Ans B H = ut + ½ at2 = 0 + ½ (10)(4.5)2= 101.3 m H/H = a/a + 2t/t H/101.3 = 0 + 2 x (0.1 /4.5)  H = 4.5 = 5 m s-1 (1 sf) H = (101 ± 5) m 5 A body is thrown vertically upwards in a medium in which the viscous drag cannot be neglected. If the times of flight for the upward motion tu and the downward motion td (to return to the same level) are compared, then A td < tu, because the average speed is smaller its downward motion as compared to its upwards motion. B td < tu, because the net accelerating force when the body is moving downwards is greater than the net decelerating force when it is moving upwards C td > tu, because the viscous force is greater in the downward motion as compared to its upwards motion. D td > tu, because the net accelerating force when the body is moving downwards is smaller than the net decelerating force when it is moving upwards. Answer: D For the upward motion, viscous drag and gravitational force are both acting downwards, net force = gravitational force + viscous drag net deceleration = (gravitational force + viscous drag)/mass For the downward motion, viscous drag and gravitational force are in opposite direction and the net acceleration force = gravitational force - viscous drag. net acceleration = (gravitational force - viscous drag)/mass Comparing the acceleration in downwards motion is less than the deceleration during the upwards motion, hence td > tu,

6 [Turn Over] 6 When a man is standing in an ascending lift that has a constant upward acceleration, the magnitude of the force exerted on the man’s feet by the floor is always A equal to the magnitude of his weight. B less than the magnitude of his weight. C greater than the magnitude of his weight. D greater than his weight only when the acceleration is greater than g. Ans C Let N : normal upwards force; W = weight of man ; M: mass of man Net upwards force = N – W Using N2LM “F=ma” N-W = Ma N = W + Ma  N > W 7 Three identical stationary discs, P, Q and R are placed in a line on a horizontal, flat and frictionless surface. Disc P is projected straight towards disc Q. If all consequent collisions are perfectly elastic, what will be the final motion of the three spheres? P Q R A stationary stationary moving right B moving left moving left moving right C moving left stationary moving right D moving right moving right stationary Answer: A Consider the collision between P and Q, After collision, P stops and Q move with speed of P Reason: By conservation of momentum, m1u1 = m1v1 + m2v2  u1 = v1 + v2 Since the collision is elastic,  u1 – u2 = v2 – v1 Solving simultaneously, v1 = 0 and v2 = u1 Similarly, when Q collide with R, Q stops and R move with speed of Q P P Q P R P

7 [Turn Over] 8 Three forces act on different parts of a rigid body as shown in the diagram below. Which of the following vector diagram represents the body in equilibrium? Answer: D In equilibrium, the vectors form a closed vector triangle with their directions in the same sense. A B C D 9 A smooth uniform rod of mass 30 g rests on the rim of a smooth, hemispherical bowl as shown in the diagram below (drawn to scale). What is the magnitude of the normal contact force by the rim of the bowl on the rod? Diagram is drawn to scale

8 [Turn Over] A 0.4 W B 0.6 W C 0.9 W D W Answer B By the principle of moments, Takin