2021 RI Promo Sect B QP

© Raffles Institution 9749   Name: ( ) CT Group: 22S0 RAFFLES INSTITUTION 2021 YEAR 5 PROMOTIONAL EXAMINATION 30 September 2021 H2 PHYSICS RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFL ES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFL ES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFL ES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFL ES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFL ES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES Section B INSTRUCTIONS TO CANDIDATES Write your name, index number and CT Group. Write your answers to Section B in the spaces provided on the question paper. For Examiner’s Use Section A MCQ / 15 Section B 1 / 10 2 / 10 3 / 10 4 / 8 5 / 12 Section C 6 / 15 7 / 15 Deductions Total / 95 % This document consists of 12 printed pages.

2 © Raffles Institution 9749 [Turn over     Section B (50 marks) 1 A sphere of mass 0.040 kg and density 38910 kg m is suspended by a thin string and submerged in a tall container of liquid of density 3 k11 0 g m4  as shown in Fig. 1.1. Fig. 1.1 The string is cut and the sphere begins falling from rest through the liquid, where it experiences a drag force F D such that DFv  where v is the speed of the sphere and  is a constant with a value of 10.26 kg s . After falling for 2.0 s, the sphere reaches terminal velocity. The sphere then continues to fall at terminal velocity for another 2.5 s. (a) (i) Show that the upthrust on the sphere is 0.062 N. [1] (ii) Determine the terminal velocity of the sphere. terminal velocit y = m s –1 [2] sphere liquid string container

3 © Raffles Institution 9749 [Turn over     (iii) Determine the magnitude of the average fo rce acting on the sphere during the first 4.5 s of its motion through the liquid. average force = N [2] (b) On Fig. 1.2, sketch the variation with time t of the displacement s of the sphere for the first 4.5 s of the sphere’s motion through the liquid. At t = 0, the string is cut and s = 0. Take the downwards direction to be positive. Fig. 1.2 [2] (c) The setup shown in Fig. 1.1 is placed on top of a mass balance. Immediately after the string is cut, the reading on the mass balance is X. When the sphere is moving at terminal velocity, the reading on the mass balance is Y. State and explain if Y is smaller, equal to, or larger than X. [3] s / m t / s 0 1.0 2.0 3.0 4.0 5.0

4 © Raffles Institution 9749 [Turn over     2 A small smooth ring of mass m is threaded on a light inextensible string of length 8L . The two ends of the string are fixed at a distance of 4L apart on a vertical rod as shown in Fig. 2.1. The ring is then made to move in a horizontal circle around the rod at constant speed with the string taut and the lower portion of the string horizontal. Fig. 2.1 (a) Determine, in terms of m, L and the acceleration of free-fall g, where appropriate, (i) the radius of the circular motion of the ring, radius = [2] (ii) the tension in the string, tension = [3] rod ring 4L

5 © Raffles Institution 9749 [Turn over     (iii) the angular speed of the ring, angular speed = [2] (iv) the linear speed of the ring. linear speed = [1] (b) State and explain in terms of the forces acting on the ring, whether the ring will rise or fall while undergoing circular motion if its angular speed is increased slightly. [2]

6 © Raffles Institution 9749 [Turn over     3 (a) A ball thrown vertically upwards from the edge of a tabletop reaches its highest point, reverses direction and bounces off the ground below the table, before returning to its initial position. Fig. 3.1 shows this path of the ball. Fig. 3.1 Fig. 3.2 shows the variation with vertical displacement s from the tabletop of the acceleration a of the ball, with the position of the tabletop taken as s = 0. Ignore all energy losses. Fig. 3.2 With reference to Fig. 3.2 and the definition for simple harmonic motion, discuss whether the motion of this bouncing ball is considered simple harmonic. [3] a s highest point ground 0 ground tabletop s = 0 highest point

7 © Raffles Institution 9749 [Turn over     (b) A pendulum of length L is suspended from a fixed point as shown in Fig. 3.3. Fig. 3.3 (not drawn to scale) Keeping the string taut, the pendulum of mass m is displaced by a small angle such that its centre of mass rises vertically by 0.40 cm. When the pendulum is released from this position, it performs simple harmonic motion. The total potential energy of the pendulum is zero at the equilibrium position. The variation with horizontal displacement x from the equilibrium position of the kinetic energy EK of this pendulum is as shown in Fig. 3.4. Fig. 3.4 0.40 cm L x / cm 5.0 0 10.0 15.0 10.0 5.0 15.0 EK / J

8 © Raffles Institution 9749 [Turn over     (i) Determine an expression, in terms of m and the acceleration of free-fall g, for the total energy of the pendulum. Explain your working clearly. total energy = J [2] (ii) If the angular frequency  of the pendulum is g L , use your expression in (b)(i) to deduce the length L of the pendulum. L = m [3] (iii) The pendulum is now displaced such that it s centre of mass rises vertically by 0.30 cm instead and released. Without any further calculations, use Fig. 3.4 to determine the new amplitude of the oscillations. Show your construction clearly on Fig. 3.4. amplitude = cm [2]

9 © Raffles Institution 9749 [Turn over     4 (a) State what is meant by polarisation with respect to an electromagnetic wave. [2] (b) Two sheets of polarisers P and Q are placed close to each other with their polarising axes vertical. A parallel beam of unpolarised light passes through polariser P. After passing through polariser P, the light beam has amplitude 0A and intensity 0I . The light beam then passes through polariser Q. Polariser Q is now ro