NYJC 2022 Prelim P1 Guide

NYJC 2022 9749/01/J2Prelim/22 [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 20 September 2022 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 16 printed pages.

2 NYJC 2022 9749/01/J2Prelim/22 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 2022 9749/01/J2Prelim/22 [Turn over 1 Which of the following is not a reasonable estimate? A Sound of frequency in the order of 109 Hz is audible to human beings. B Light of wavelength in the order of 10−7 m is visible to human beings. C The volume of an average-sized orange is in the order of 10−4 m3. D The speed of beta particles emitted from a nucleus is in the order of 107 m s−1. 2 The kinetic energy EK of a rotating solid sphere is given by the following expression 2 k 2 2 I=E T where I is the moment of inertia and T is the period of rotation of the solid sphere. What is the unit for the moment of inertia I, expressed in SI base units? A kg m2 s−1 B kg m2 C J s2 D N m s2 Ans: A Options B and D are typical values that candidates should be familiar with. Using = 34 3Vr , Option C works out for an orange having radius of about 5 cm. The frequency of sound (or even ultrasound) cannot hit GHz range. Ans: B ( )( )( ) I I I   − = = = = = = 2 2 2 22 2 2 2 2 2 2 unit for (unit for )(unit for ) {unit for ( )}(unit for ) kg m s s kg m 2 2K K K T E ET ET mv T

4 NYJC 2022 9749/01/J2Prelim/22 3 A ruler is supported horizontally by two pivots as shown. The vertical displacement y at the centre of the ruler can be used to measure the mass loaded on it and is given by the equation 3 3 kMLy wt= where k is a constant, L is the distance between the pivots, M is the mass loaded onto the ruler, t is the thickness of the ruler and w is the width of the ruler. When a particular M is loaded onto the ruler, the following results are obtained: ( ) ( ) ( ) ( ) y . . . t . . w . . L.= =  = = 0 25 0 01 mm 0 2 cm80 0 6 0 0 1 mm 23 0 0 5 mm Which measurement contributes the most to the uncertainty of M? A y B L C t D w Ans: C Make M subject of formula: =     = + + +  ==  ==  ==  == 3 3 33 0 01 0 0400 25 05 0 02223 0 013 3 0 05060 023 3 0 007580 0 ywtM kL M y w t L M y w t L y. .y. w. .w. t. .t. L. .L. y

5 NYJC 2022 9749/01/J2Prelim/22 [Turn over 4 The speed of an aeroplane in still air is 200 km h−1. The wind pushes it from the west at a speed of 85.0 km h−1. In which direction must the pilot steer in order to fly due north? A 23.0 east of north B 23.0 west of north C 25.2 east of north D 25.2 west of north 5 A radio-controlled toy car travels along a straight line for a time of 15 s. The variation with time t of the velocity v of the car is shown below. What is the average velocity of the toy car for the 15 s journey? A −1.5 m s−1 B 0.0 m s−1 C 4.0 m s−1 D 4.5 m s−1 Area under the graph = s = zero vave = s/t = 0 Ans: D vPE = vPW + vWE where vPE is velocity of plane relative to Earth vPW is velocity of plane relative to Wind vWE is velocity of wind relative to Earth sin θ =85/200 θ = 25.2 west of north t / s v / m s−1 6.0 −6.0 −3.0 3.0 0 0 5 10 15

6 NYJC 2022 9749/01/J2Prelim/22 Answer: B 6 A tennis ball is thrown horizontally in air from the top of a tall building. The effect of air resistance is not negligible. Which of the following correctly describes the subsequent change in the horizontal and vertical components of the ball’s velocity? Since the building is tall, the time of flight is long enough for the vertical component of velocity to reach a maximum value. Answer: D 7 A rock of mass 2m in deep space, initially travelling at velocity v, explodes into two parts of equal mass, one of which becomes stationary immediately after the explosion. What is the kinetic energy of the moving part after the explosion? A mv 21 2 B mv 2 C mv 23 2 D 2mv2 By COM, momentum after explosion = 2mv, which is only carried by moving part. So KE of moving part = (2mv)2/(2m) =2mv2 Answer: D horizontal component of velocity vertical component of velocity A B C D constant constant decreases to zero decreases to zero constant increases at a constant rate increases at a constant rate increases to a maximum value

7 NYJC 2022 9749/01/J2Prelim/22 [Turn over 8 A single horizontal force F is applied to a block X which is in contact with a separate block Y, as shown. The blocks remain in contact as they accelerate along a horizontal frictionless surface. X has a greater mass than Y. Which statement is correct? A The acceleration of X is equal to force F divided by the mass of X. B The force that X exerts on Y is equal to F. C The force that X exerts on Y is less than F. D The force that X exerts on Y is less than the force that Y exerts on X. Answer: C A: wrong. Acceleration of X = (F – Fby Y on X) divided by mass of X B: wrong. Fby X on Y is equal in magnitude with Fby Y on X. If Fby Y on X = F, net force on X = 0 and X will not accelerate. C: correct. Fby X on Y is equal in magnitude with Fby Y on X. If Fby Y on X is less than F, net force on X acts in