2022 H2 CJC Physics Promo Paper 1 Suggested Solutions

1 PHYSICS 9749/1 Paper 1: Multiple Choice Questions 30 September 2022 30 minutes Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write your name and tutorial group on this cover page. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write and shade your name, NRIC / FIN number and HT group on the Answer Sheet (OMR sheet) , unless this has been done for you. There are fifteen 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 (OMR sheet). Read the instructions on the Answer Sheet 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. MARK SCHEME This document consists of 11 printed pages and 0 blank page. [Turn over NAME CLASS 1T Catholic Junior College JC1 Promotional Examinations Higher 2

2 PHYSICS DATA: speed of light in free space c = 3.00 x 108 m s-1 permeability of free space 0 = 4 x 10-7 H m-1 permittivity of free space 0 = 8.85 x 10-12 F m-1  (1/(36)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2 PHYSICS FORMULAE: uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2 a s work done on / by a gas W = p V hydrostatic pressure P = gh gravitational potential  = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin t velocity of particle in s.h.m. v = v0 cos t = 22 0 xx − electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin t magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-t) decay constant λ = 2 1 2ln t

3 1 The resistance of a resistor is measured using an ohmmeter and its reading is 47.12 Ω. The manufacturer of the ohmmeter specifies that the ohmmeter reading has an uncertainty of 10%. How should the resistance of the resistor be recorded, along with its uncertainty? A (47 ± 5) Ω B (47.1 ± 0.1) Ω C (47.1 ± 4.7) Ω D (47.12 ± 5) Ω L2 Answer: A The uncertainty of the resistance = 47.12 × 10% = 4.71 5 (1.s.f) Therefore, resistance = (47 ± 5) Ω 2 Two trains, Train A and Train B, are initially at rest next to each other at a train station. Train A leaves the station with a uniform acceleration of 0.30 m s -2. 2.0 minutes later, Train B leaves the station in the same direction as Train A with a uniform acceleration of 0.50 m s-2. How long does Train B take to catch up with Train A? A 6.9 s B 180 s C 210 s D 410 s L2 Answer: D Let time t be the time at which Train B catches up with Train A. Distance traveled by Train A = 22 A 11s = ut + at = (0.30)(t+120)22 [t+120 s since Train A travels an additional time of 120 s] Distance traveled by Train B = 22 B 11s = ut + at = (0.50)(t)22 Since ABs = s when B catches up, 2211(0.30)(t+120) = (0.50)(t )22 Solving for t, t= 410 s

4 3 A stone is thrown from X and moves under gravity to Y, at the top of its path. Which row correctly describes its motion horizontally and vertically between X and Y? Neglect air resistance. horizontally vertically A travels at decreasing velocity velocity decreases uniformly B travels at increasing velocity velocity increases uniformly C travels at constant velocity velocity increases uniformly D travels at constant velocity velocity decreases uniformly L1 Answer D Since air resistance is neglected, Horizontal motion – there is no horizontal acceleration, so the stone will travel at constant velocity horizontally. Vertical motion – the acceleration is constant at 9.81 m s -2 (downwards direction), since the stone is moving upwards, its velocity decreases uniformly. 4 Two blocks with masses of 500 g and 400 g are moving towards each other at speeds of 2.0 m s-1 and 3.0 m s-1 as shown. The two blocks stick together after the collision and move off with a common speed. What is the speed and the direction of motion of the blocks after the collision? X Y 500 g 400 g 2.0 m s -1 3.0 m s -1

5 A 0.22 m s-1 to the left B 0.22 m s-1 to the right C 2.4 m s-1 to the left D 2.4 m s-1 to the right L1 Answer: A By the Conservation of Momentum, and taking right as positive (0.5)(2.0) + (0.4)(-3.0) = (0.5+0.4)v v = - 0.22 m s-1 (to the left) 5 A tennis ball of mass 60 g is dropped vertically from a height. It hits the ground with a speed of 21 m s-1 downwards and then bounces upwards with an initial speed of 14 m s- 1. The time of contact between the ball and the ground is 0.30 s. What is the magnitude of the impulse provided by the ground? A 0.42 N s B 2.1 N s C 7.0 N s D 2100 N s L1 Answer: B Impulse = change of momentum of the ball fi=m(v -v ) = 0.060 (14-(-21))=2.1 N s 6 The unstretched length of a spring is 30 cm. It is stretched vertically, with one end fixed on a ceiling, to a length of 40 cm by a weight of 2 N attached to the other end as shown. The weight is at rest. What is the potential energy stored in the spring? 30 cm 40 cm weight of 2 N

6 A 0.1 J B 0.2 J C 0.6 J D 0.8 J L1 Answer: A Hooke’s Law: F = k x e Elastic Potential Energy = 1 2 x k x e2 = 0.5 x F x e = 0.5 x 2 x 0.1 = 0.1 J 7 A wooden block of density 700 kg m-3 is completely submerged below the surface of water by a weight placed on top of it. The block floats to the surface when the weight is removed. The density of water is 1000 kg m-3. What is the ratio upthrust on the wooden block when it is completely submerged upthrust on the wooden block when it is floating ? A 0.143 B 0.700 C 1.43 D 7.00 L2 Answer: C Let ρb = density of wooden block = 700 kg m-3 ρw = density of water = 1000 kg m-3 V = volume of wooden block g = gravitational acceleration Upthrust on the wooden block when it is floating = weight of the wooden block = ρb V g [Principle of floatation] Upthrust on the wooden block when it is completely submerged = weight of water displaced by the wooden block = ρw V g Ratio of upthrust on the wooden block when it is completely submerged upthrust on the wooden block when it is floating = ρw V g ρb V g = ρw ρb = 1000 700 = 1.43 8 An object of mass m is hung at one end of a 3.0 m long uniform rod of mass M and placed on a pivot. For the system to be in equilibrium, the pivot has to be placed 1.2 m from the end where the object is hung. When the object’s mass is doubled to 2m, the pivot has to be moved a distance x to the right of its original position for the system to remain in equilibrium as shown.

7 What is the distance x? A 0.20 m B 0.40 m C 0.50 m D 0.80 m L3 Answer: A Since the rod is uniform, its weight acts at the centre of the rod, which is 0.3 m to the left of the pivot initially