CJC_H1_PHY_P2_QP_and_SOL

1 CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 1 CANDIDATE NAME CLASS 2T INDEX NUMBER PHYSICS 8866/02 Paper 2 28 August 2015 2 hours Additional Materials: Answer Paper READ THESE INSTRUCTIONS FIRST Write your index number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer any two questions. Circle the 2 questions that you answered in the table below. At the end of the examination, fasten all work securely together. The number of marks is given in brackets [ ] at the end of each question or part of the question. This document consists of 24 printed pages [Turn over] FOR EXAMINER’S USE SECTION A (40 MARKS) 1 /7 2 / 8 3 /7 4 /8 5 /3 6 /7 SECTION B (40 MARKS) 7 / 20 8 / 20 9 / 20 TOTAL /80

2 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 = gh resistors in series, R = R1 + R2 + ... resistors in parallel, R 1 = ... 21 11 RR

3 SECTION A (40 marks) Answer all questions in Section A. 1 A car of mass 1380 kg, travelling at 31.1 m s -1, is brought to rest by applying the brakes. The average braking force is estimated to be 1.38 x 104 N. Calculate (a) the initial kinetic energy of the car, kinetic energy = ………………………..J [1] K.E. = ½ m v2 = ½ (1380)(31.1)2 = 667400 J A1 (b) the average deceleration of the car, deceleration = ……………………… m s-2 [1] F = ma  a = F/m = 1.38 x 104 /1380 = 10.0 m s-2 A1 (c) the distance travelled before it comes to rest. braking force = …………………………N [2] Work done = KE loss Fd = KE d = KE loss / F = 667400/1.38 x 104 = 48.2 m M1 A1 (d) Suggest whether the answer in (c) is an over-estimation or under-estimation. [2] In practice, air resistance and rolling friction of the road are presence. The total decelerating force is larger and hence the distance travel will be shorter. The value is an overestimation. B1