2025 VJC H2 Phy Prelim P1 Solution

VJC 2025 9749/01/J2Prelim [Turn over VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SOLUTION CLASS TUTOR’S NAME PHYSICS 9749/01 Paper 1 Multiple Choice 22 September 2025 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 and tutor name in the spaces on 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 VJC 2025 9749/01/J2Prelim 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

3 VJC 2025 9749/01/J2Prelim [Turn over 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  = IB d 0 2 magnetic flux density due to a flat circular coil = NIB r 0 2 magnetic flux density due to a long solenoid =B nI 0 radioactive decay 0 exp( )x x t =− decay constant 1 2 ln2 t =

4 VJC 2025 9749/01/J2Prelim 1 The e.m.f. induced in a coil by a changing magnetic flux is equal to the rate of change of flux with time. Which is a unit for magnetic flux? A kg m2 s−2 A−1 B kg m2 s−2 A C kg m2 s2 A−1 D m2 s−2 A−1 Ans: A Using Faraday’s Law, 221 2 2 1 Φ Φ Φ Unit for Φ dE dt Pd I dt P tI kg m s s s kg m s AA −− −− =− =−  = − == 2 What is a reasonable estimate for the volume of a wooden metre rule found in a school laboratory? A 1.5 cm3 B 15 cm3 C 150 cm3 D 1500 cm3 Ans: C Volume of wooden ruler = l x b x h = (100cm)(3 cm)(0.5 cm) = 150 cm3 3 A student carried out an experiment to determine the resistivity  of copper using a copper wire. The uncertainties in the measurements are shown. uncertainty in length l of wire = 0.2% uncertainty in diameter d of wire = 1.6% The equation for resistivity  is 2 .4 dR= l  He obtains a resistivity value of ( ) 81.71 0.07 10 −  m with its associated uncertainty. What is the uncertainty in the measurement of resistance R of the wire? A 0.007% B 0.7% C 0.9% D 7% Ans: B 0.07 1.6 0.221.71 100 100 0.007 0.7% R R R R = + +  ==

5 VJC 2025 9749/01/J2Prelim [Turn over Option A: If student forgets to convert to percentage. Option C: If student includes 4 in percentage uncertainty to give 0.0009 = 0.9% Option D: If student makes R the subject first and adds all the percentage uncertainty. 4 X and Y are vectors. The magnitude of X is less than the magnitude of Y. The vectors are initially in opposite directions. As Y is rotated through 180°, how does the magnitude of the vector (X − Y) vary? Ans: A At an angle of 0°, X and Y are opposite in directions. Vector sum (X − Y) is maximum. At an angle of 180°, X and Y are in the same direction. Vector sum (X − Y) is minimum.

6 VJC 2025 9749/01/J2Prelim 5 A car, starting from rest at time t = 0, travels along a road. The distance travelled from the starting point is measured over the next 25 seconds. Which best describes the motion of the car? A The maximum speed during the first 20 seconds is 10 m s−1. B At some instant during the first 20 seconds the speed is exactly 20 m s−1. C The average speed for the first 200 m of the journey is 20 m s−1. D The average speed between 20 and 25 seconds is greater than that between 15 and 20 seconds. 6 A boy with a ball was in a stationary lift. When the lift starts to accelerate upwards at 1.2 m s−2, the boy released the ball from a height of 1.5 m above the floor of the lift. What is the time taken by the ball to hit the floor of the lift? A 0.27 s B 0.52 s C 0.55 s D 0.59 s Ans: B The relative acceleration of the ball to the lift 1 2 9 81 11 01. . .= + = m s−2 Using 2 1 1 2 22 Eqn1: Eqn2 y y s t st = , we get ( ) 211 5 11 012. . t= 0 52 st.= Ans: B Option A is wrong as it covered 400 m in 20 s. The average speed is already at 20 m s −1. Thus the max speed must be higher. Option C is wrong. It took 15 s to cover 200 m. Thus, the average speed is 13.3 m s−1. Option D is wrong. From 20 to 25 s, 100 m was covered. From 15 to 20 s, 200 m was covered, thus the average speed from 15 to 20 s must be higher. Option B is correct. The average speed over 20 s is 20 m s−1. At 0 s, it started at a low speed, at 20 s the instantaneous speed is higher than 20 m s−1. As speed is a continuous variable (no abrupt change, jumping from one value to another without going through values inbetween), thus at some instant during the 20 s, the speed must have been 20 m s−1.

7 VJC 2025 9749/01/J2Prelim [Turn over 7 Two blocks of masses m1 = 4.0 kg and m2 = 1.0 kg are connected by a cord of negligible mass that passes over a frictionless pulley of negligible mass. The blocks slide on frictionless planes inclined at angles θ1 = 30o and θ2 = 60o. What is the tension in the cord? A 2.3 N B 5.8 N C 8.0 N D 10.7 N Ans: D Apply F = ma to both masses: 11 sin30 om g T m a −= ---(1) 22 sin60 oT m g m a−= ---(2) Sub. T from (2) into (1): 1 2 2 1sin30 ( sin60 )oom g m g m a m a− + = oom m a g m m1 2 1 2( ) ( sin30 sin60 )+ = − ooa(4.0 1.0) 9.81 (4.0sin30 1.0sin60 )+ = − a = 2.22 m s−2 Sub. Into (2): oT 9.81sin60 2.22−= T = 10.7 N 8 Two steel balls A and B of masses 2 M and 1M respectively move towards each other with the same speed v and collide elastically. What are the final velocities of the two balls in terms of v? Take the rightward direction as positive. final velocity of ball A final velocity of ball B 30o 60 o m1 m2 30 o m1g T T m2g v v 2M M A B

8 VJC 2025 9749/01/J2Prelim A 4 3 v 7 3 v B v1 3− v5 3 C v1 3 v2 3 D v− v2 Ans: B Let v1 = final velocity of the ball A Let v2 = final velocity of the ball B By Law of Conservation of Momentum, 2𝑀𝑣 + 𝑀(−𝑣) = 2𝑀𝑣1 + 𝑀𝑣2 2𝑣1 + 𝑣2 = 𝑣 ---(1) Since the collision is elastic, initial velocity of B relative to A = −( final velocity of B relative to A ) (−𝑣) − 𝑣 = −(𝑣2 − 𝑣1) 𝑣2 = 𝑣1 + 2𝑣 ---(2) Sub. into (1): 2𝑣1 + (𝑣1 + 2𝑣) = 𝑣 𝑣1 = − 1 3 𝑣 Sub. into (2):