NYJC 2025 JC1 Promo Past Year Package Solutions

9478 H2 PHYSICS JC 1 PROMO EXAM PAST YEAR PAPERS (SOLUTION) MEASUREMENTS KINEMATICS DYNAMICS FORCES WORK, ENERGY POWER CIRCULAR MOTION GRAVITATIONAL FIELD OSCILLATIONS TEMPERATURE & IDEAL GASES FIRST LAW OF THERMODYNAMICS WAVE MOTION NANYANG JUNIOR COLLEGE JC1 PHYSICS TEAM SCIENCE DEPARTMENT 2025

NYJC 2024 9749/01/J1EOY/24 [Turn over NANYANG JUNIOR COLLEGE JC 1 END OF YEAR EXAMINATION Higher 2 CANDIDATE NAME Solution CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9749/01 Paper 1 Multiple Choice 1 October 2024 40 minutes 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 twenty 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. 1 2 3 4 5 6 7 8 9 10 C D C B D D C D C B 11 12 13 14 15 16 17 18 19 20 A C A B C A B D C B This document consists of 12 printed pages.

2 NYJC 2024 9749/01/J1EOY/24 Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space µ0 = 4 π × 10−7 H m−1 permittivity of free space ε0 = 8.85 × 10−12 F m−1 (1 / (36π)) × 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 pV= ∆ hydrostatic pressure p ghρ= gravitational potential /Gm rφ = − 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 sinxx t ω= velocity of particle in s.h.m. 0 cosvv t ω= 22 0xxω= ±− electric current =I Anvq resistors in series 12 . . .RR R=++ resistors in parallel 121/ 1/ 1/ . . .RR R=++ electric potential 04 QV rπε= alternating current/voltage 0 sinxx t ω= magnetic flux density due to a long straight wire µ π= 0 2 IB d magnetic flux density due to a flat circular coil µ= 0 2 NIB r magnetic flux density due to a long solenoid µ= 0B nI radioactive decay 0 exp( )xx t λ= − decay constant 1 2 ln2 tλ =

3 NYJC 2024 9749/01/J1EOY/24 [Turn over 1 Which of the following is the best estimate of the weight of a basketball? A 5000 mg B 0.500 kg C 5.0 × 10-6 MN D 5.0 × 1011 nN 2 In an experiment, the external diameter d 1 and internal diameter d2 of a glass tube are found to be 82 ± 2 mm and 53 ± 1 mm respectively. What is the percentage uncertainty of the term (d1 – d2)? A 0.7% B 3% C 4% D 10% 3 A cyclist moving horizontally takes off from a point 3.0 m above the river surface, landing 4.5 m away as shown. What was the speed at take-off? A 3.1 m s –1 B 3.2 m s–1 C 5.8 m s–1 D 7.4 m s–1 3.0 m 4.5 m cross-section of river channel Ans: C Option A and B are masses, not weights. Option C: 5.0 × 10−6 × 106 = 5.0 N ( ≈ 0.5 kg equivalent) Option D: 5.0 × 1011 × 10−9 = 500 N ( ≈ 50 kg equivalent) Ans: D 12 1 2 12 12 Δ Δ Δ 2 1 3 mm ΔPercentage uncertainty 100 3 10082 53 10 (2 sig. fig.) () () %() % % − = + = += −= × − = × − = dd d d dd dd Ans: C Use vertical component to determine time of flight: 21 2 yy ys ut at= + 213.0 9.812 t= ×× 0.782t ≈ s Use horizontal component to determine speed at take off xxs ut= 4.5 0.782xu= × 5.75xu ≈ m s−1 Option D looks possible, but question mentioned that cyclist landed 4.5m as shown. The question is NOT asking what are the possible speeds that the cyclist can have to clear the channel.

4 NYJC 2024 9749/01/J1EOY/24 4 A trolley moves down a slope with a constant acceleration a. The mass of the trolley is now doubled and the trolley is allowed to move down the same slope. In both cases, effects of friction and air resistance are negligible. Which statement is correct for the second experiment? A The acceleration is 1 2 a. B The acceleration is a. C The acceleration is 2a. D The resultant force is the same. 5 Two steel balls X and Y are suspended on strings. Ball X is pulled to one side as shown. After ball X is released, the balls collide. Which quantities must be conserved in the collision? A kinetic energy, total energy and momentum B kinetic energy and momentum only C kinetic energy and total energy only D total energy and momentum only Ans: B sin sin Acceleration is independent of mass. Σ= = = F ma mg ma ag θ θ Ans: D Some kinetic energy may be converted to other forms of energy such as sound energy. However, by conservation of energy, total energy MUST be conserved. Momentum is also conserved since there is no net external force acting on the system. Take note that the question is only concern IN THE COLLISION, hence we do not need to look at the processes before and after the collision.

5 NYJC 2024 9749/01/J1EOY/24 [Turn over 6 A mass is tied to the end of a string. The mass is pulled horizontally to one side by a spring of spring constant 25 N m−1. The mass is in equilibrium and the extension of the spring is 0.060 m. The string is at an angle of 36° to the vertical, as shown. What is the weight of the mass? A 1.1 N B 1.5 N C 1.9 N D 2.1 N Ans: D With reference to the FBD of the mass, 0 sin sin --- (1) 0 cos --- (2) (1)/(2): tan (25)(0.06) 2.1 Ntan tan36 x spring y F FT kx T F WT kx W kxW θ θ θ θ θ ∑= = = ∑= = = = = =

6 NYJC 2024 9749/01/J1EOY/24 7 As a gun fires a shot horizontally, a constant force F is exerted on the bullet of mass m over a time interval t. During this period, the bullet accelerates from rest to a speed v in the gun’s barrel with an acceleration a. The length of the gun’s barrel is L. Which of the following gives the expression for the average power of the bullet in the described motion? A Fv B t maL2 C t mv 2 2 D t mv 2 8 A body moves from X to Y along a track. At point Y, its kinetic energy is 25 kJ and its potential energy is 30 kJ more than that at point X. If the work done against friction along XY is 10 kJ, what is the kinetic energy of the body at X? A 35 kJ B 45 kJ C 55 kJ D 65 KJ X Y Ans: C Average power = energy gained time taken = 21 2 mv t Ans: D By Principle of conservation of energy, 10 25 30 25 30 10 65 kJ X X byfriction Y Y XY X KE GPE W KE GPE KE KE GPE KE ++ = + − = +∆ = + =++=

7 NYJC 2024 9749/01/J1EOY/24 [Turn over 9 A pendulum bob of mass 1.27 kg is supported by a string so that the radius of its path is 0.600 m. It is moving with velocity 0.575 m s-1 horizontally at the lowest point of its motion when the string is vertical. What is the tension in the string