SRJC H2 PHY P1 SOLUTION

SRJC 2011 9646/Prelim/2011 [Turn Over SERANGOON JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION General Certificate of Education Advanced Level Higher 2 PHYSICS 9646/01 Paper 1 Multiple Choice 24 August 2011 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, Civics Group and index number on the Answer Sheet in the spaces provided. There are forty questions on this paper. Answer all questions. For each question, there are four possible answers labeled A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. 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 question paper. This document consist of 24 printed pages and no blank page. For Examiner’s Use Section A Total / 40 CIVICS GROUP CANDIDATE NAME INDEX NUMBER SERANGOON JUNIOR COLLEGE Science Department Physics Unit

1 SRJC 2011 9646/Prelim/2011 [Turn Over DATA AND FORMULAE Data speed of light in free space, c = 3.00  108 m s1 permeability of free space, 0 = 4   107 H m1 permittivity of free space, 0 = 8.85  1012 F m1 (1 / (36 ))  109 F m1 elementary charge, e = 1.60  1019 C the Planck constant, h = 6.63  1034 J s unified atomic mass constant, u = 1.66  1027 kg rest mass of electron, me = 9.11  1031 kg rest mass of proton, mp = 1.67  1027 kg molar gas constant, R = 8.31 J K 1 mol1 the Avogadro constant, NA = 6.02  1023 mol1 the Boltzmann constant, k = 1.38  1023 J K1 gravitational constant, G = 6.67  1011 N m2 kg2 acceleration of free fall, g = 9.81 m s 2 Formulae uniformly accelerated motion, s = ut + ½ at 2 v2 = u 2 + 2as work done on/by a gas, W = p V hydrostatic pressure, p = gh gravitational potential,  = – r Gm displacement of particle in s.h.m., x = x 0 sin t velocity of particle in s.h.m., v = v o cost v =  22 0 xxω  resistors in series, R = R 1 + R2 + … resistors in parallel, 1/ R = 1/R1 + 1/R2 + … electric potential, V = Q / 4or alternating current/voltage, x = x 0sin t transmission coefficient, T  exp(2kd) where k = 2 2 h E)m(U8 π radioactive decay, x = x0 exp(t) decay constant,  = 2 1 693.0 t

3 SRJC 2011 9646/Prelim/2011 [Turn Over 1 In an attempt to find the spring constant of a spring whose theoretical value is 3.0 N m -1, a student attached different weights to a sp ring, measured the corresponding extensions, and plotted his results on a force-extension gr aph. It was noted that wh ile the gradient of the graph was 2.98 N m -1, the line obtained was vertically displaced from its theoretical position. 5 out of the 6 points were also fo und to be located exactly on the best fit line. What kind of error could be present? A Systematic error, due to consistent under-estimation of the extensions. B Systematic error, due to consistent over-estimation of the extensions. C Random error, due to limited sensitivity of ruler in measuring extensions. D Both random and systematic errors due to poor experimental technique and limited sensitivity of ruler in measuring extensions. . Answer: A Since most of the points lie on the best fi t line, there is minimum random error. The vertical displacement of the experimental gr aph from its theoretical value indicates that for a particular force applied to the spring, the extension measured was an under-estimate of its theoretical value. 2 Which of the following is not a reasonable estimate? A The volume of a laptop is 1800 cm3. B The number of seconds taken to drive from East to West by the Pan-Island Expressway is 1800 s. C The average pressure on the seat of a chair when an SRJCian sits on it is 5 kPa. D The rate at which a student loses gravit ational potential energy in walking down the stairs from the fourth floor to the first floor of B block is 1500 W. Answer: D Option A: Volume = 30 cm x 20 cm x 3 cm = 1800 cm3. Option B: Time = Distance / Speed = 42 km / 80 km h-1 = 0.525 h = 1890 s. Option C: Pressure = Force / Area = (60 x 9.81) / (0.35 x 0.35) = 4800 kPa. Option D: Power = GPE / t = (60 x 9.81 x 7.5) / 30 = 147 W. Force Extension

4 SRJC 2011 9646/MYE/2011 3 A stone released from rest from a height of 20.0 m on the surf ace of planet Earth reaches the ground level after a time To. The same stone is thrown from the same height at another planet and it also reaches the ground in a time To. By taking the acceleration due to gravity on the surface of Earth and planet to be 10.0 m s1 and 15.0 m s2 respectively, what is a possible value for its initial velocity on the planet? A 2.2 m s 1 upwards B 2.2 m s 1 downwards C 5.0 m s 1 upwards D 5.0 m s 1 downwards Ans: C Consider Earth, taking +, s = ut + ½ at 2 20 = 0 + ½ (10)t 2 T o = 2 s Consider Planet X, taking +, s = ut + ½ at 2 20 = ( u)(2) + ½ (15)t2 u = 5 m s 1

5 SRJC 2011 9646/Prelim/2011 [Turn Over 4 A ball is released from rest above a horizont al surface and bounces several times. The graph shows how, for this ball, a quantity y varies with time. What are the possible quantities of y1 and y2? y1 y2 A Displacement Momentum B Velocity Acceleration C Work done against gravity Displacement D Kinetic energy Velocity Ans: A 5 A light spring of natural length 25.0 cm is suspended from the ceili ng of a lift. A mass is hung from the end of the spring, as shown in the figure below. When the lift is moving downwards at a constant speed, the length of the spring is 50.0 cm. The lift then slows down with a constant acceleration of 2.0 m s−2. Which of the following is correct? (Take g = 10.0 m s−2) A The spring shortens by a length of 5.0 cm. B The spring lengthens by a length of 5.0 cm. C The spring shortens by a length of 10.0 cm. D The spring lengthens by a length of 10.0 cm. Ans: B lift y1 time time y2

6 SRJC 2011 9646/MYE/2011 6 Two carts of different masses m1 and m2 move towards each other at different speeds u1 and u2 respectively and collide, producing a loud sound. Subsequently, the two carts move in opposite directions. Which of the following must be correct? A The collision was elastic since t he carts move in opposite directions. B Each cart experienced a change in moment um, and thus the tota l momentum of the system must have changed. C The sum of kinetic energies of the carts a fter the collision is less than that before the collision. D The relative speed of approach and the relati ve speed of separation of the carts must be the same. Ans: C 7 A 5.0 kg mass is placed at the end of a spring. The restoring force in the spring F varies with its extension x as shown below. At the origin, the mass is gi ven a sudden initial push such that it has a speed of 4.0 m s −1. The mass comes to a stop at a distance of 2.0 m away from the origin. What is the speed of the mass when the mass is at x = 1.0 m? A 2.0 m s−1 B 2.8 m s−1 C 3.0 m s−1 D 3.5 m s−1 Ans: D Total initial KE = Total final EPE ½ (5.0) (4.0) 2 = 40 J = Total final EPE Considering area under graph, EPE at x = 1.0 will be ¼ of the final EPE. Therefore KE at x = 1.0 will be ¾ of the final EPE = 30 J v = 3.46 m s-1 8 Which of the fol