NYJC 2021 H2 Physics 9749 P1 Answer

NYJC 2021 9749/01/J2Prelim/21 [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9749/01 Paper 1 Multiple Choice 23 September 2021 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class, Centre number and index number in the spaces at the top of this page. There are thirty questions on 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 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 booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 15 printed pages.

2 NYJC 2021 9749/01/J2Prelim/21 Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space  = 4 × 10−7 H m−1 permittivity of free space  = 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 21 2s ut at 22 2v u as work done on / by a gas W p V hydrostatic pressure p gh gravitational potential /Gm r  temperature / K / C 273.15TT    pressure of an ideal gas 21 3 Nmpc V   mean translational kinetic energy of an ideal molecule 3 2E kT displacement of particle in s.h.m. 0 sinx x t  velocity of particle in s.h.m. 0 cosv v t  22 0xx  electric current I Anvq resistors in series 12 . . .R R R   resistors in parallel 121/ 1/ 1/ . . .R R R   electric potential 04 QV r alternating current/voltage 0 sinx x t  magnetic flux density due to a long straight wire   0 2 IB d magnetic flux density due to a flat circular coil  0 2 NIB r magnetic flux density due to a long solenoid  0B nI radioactive decay 0 exp( )x x t  decay constant 1 2 ln2 t 

3 NYJC 2021 9749/01/J2Prelim/21 [Turn over 1 Which of the following is considered as a random error? A Error as a result of using g = 10 m s–2, instead of g = 9.81 m s-2. B Error in measuring the time duration of a 100 m sprint using a stopwatch. C Error due to a stopwatch running too fast. D Zero error of a measuring instrument. Ans: B Options A, C and D will cause values to be consistently biased (either always larger or always smaller) and hence causes systematic errors 2 A tennis ball is thrown vertically upward. It hits the ceiling before it falls down. Assuming the effect of air resistance is negligible, which graph best represents the variation of velocity v with time t before and after hitting the ceiling? A B C D Ans: A Before hitting and after hitting ceiling, the ball is under free fall, with a constant downward acceleration of 9.81 m s -2, where straight downward diagonal lines are expected. During the collision with the ceiling, the downward acceleration is greater than 9.81 m s-2 as the ceiling, in addition to the weight of the ball, exerts a downward foce  hence a more negative gradient. 3 An object is projected with velocity 40 m s-1 at an angle of 60° to the horizontal. Air resistance is negligible. What is the speed of the object after 5.0 s? A 14 m s-1 B 25 m s-1 C 35 m s-1 D 45 m s-1 v t 0 v v v t t t

4 NYJC 2021 9749/01/J2Prelim/21 Ans: B The horizontal velocity is constant at vx = ux = 40 cos 60° = 20.0 m s-1. The initial vertical velocity is ux = 40 sin 60° = 34.6 m s-1. After 5.0 s, vy = uy + ay t = 34.6 + (-9.81) (5.0) = -14.4 m s-1. Hence, after 5.0 s, 𝑣 = √𝑣𝑥 2 + 𝑣𝑦 2 = 25 m s−1. 4 A block with a pendulum bob hanging lies on a rough surface. The total mass of the block and the pendulum bob is 100 kg. A force of 250 N is applied to the block and the pendulum bob makes an angle  to the vertical axis. What is the magnitude of  if a constant frictional force of 60 N is acting on the block? A 11o B 14o C 76o D 79o 2 Ans: A Referring to FBD of whole system, 250 60 100 1.9 m s Referring to FBD of bob, sin a----- (1) 0 cos ----- (2) (1)/(2): tan / 1.9 / 9.81 11 x bob bob y bob F ma a a F m a Tm F T m g ag               5 Which one of the following pair of forces is not an example of Newton’s third law action and reaction pair? A The thrust on the rocket and expulsion of hot air due to the burning of fuel. B The gravitational force by Earth on man standing on E arth and the gravitational force on Earth by man. C The upthrust on a piece of wood in water and the weight of the wood. D The magnetic force on a magnet by a long wire carrying current and the magnetic force on the wire by the magnet. Ans: C  250 N block rough surface 60 N

5 NYJC 2021 9749/01/J2Prelim/21 [Turn over 6 A cylindrical block of wood has a cross-sectional area A and weight W. It is completely immersed in water with its axis vertical. The block experiences pressures pt and pb at its top and bottom surfaces respectively. Which of the following expressions is equal to the upthrust on the block? A (pb - pt)A + W B (pb - pt) C (pb - pt)A D (pb - pt)A - W Ans: C 7 The engine of a boat supplies a constant power of 110 kW to propel the boat forward. The boat attains a maximum speed of 21.0 m s1. If the magnitude of the resistive force acting on the boat is proportional to the square of the boat’s speed, what is the resultant force acting on the boat when it is moving at the instant when its speed is 15.0 m s1? A 2.7 kN B 3.6 kN C 4.7 kN D 7.3 kN Ans: C Drag @ 21.0 m s-1: 110 5.24 kN21 Df  Drag @ 15.0 m s-1: 2 ' 15.0 ' 2.67 kN21.0 D D D f ff    Force of engine @ 15.0 m s-1: 110 7.33 kN15 engineF  Hence resultant force = 7.33 – 2.67 = 4.66 kN 8 A crane is used to raise a weight of 200 N at a constant speed through a vertical height of 8.0 m in 4.0 s. The efficiency of the crane is 20%. What is the electrical power needed to be supplied to the crane? A 80 W B 400 W C 1600 W D 2000 W Ans: D Power to raise the weight  200 8.0 400 W4.0 Electrical power needed 400 2000 W0.2

6 NYJC 2021 9749/01/J2Prelim/21 9 Two discs P and Q of mass m and 2m respectively are placed on a rough, horizontal and level turntable as shown in the diagram. P and Q are at a distance of r and 2r from the centre of the turntable respectively. The turntable starts rotating from rest with gradually increasing angular velocity ω. Top View Given that the maximum frictional force acting on P is half of that on Q, which of the following is correct? A P will slip first. B Q will slip first. C P and Q will slip at the same time. D Neither P nor Q will slip. Ans: B The frictional force provides the centripetal force  f = MRω2. When P is about to slip, fmax P = mrωmax P2 ωmax P2 = fmax P / mr ----- eqn (1) When Q is