NYJC 2022 Prelim P2 Guide

NYJC 2022 9749/02/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 9749/02 Paper 2 Structured Questions 13 September 2022 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and tutor’s name in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 8 2 / 8 3 / 8 4 / 8 5 / 9 6 / 11 7 / 8 8 / 20 Total / 80 This document consists of 22 printed pages.

2 NYJC 2022 9749/02/J2PRELIM/22 Answer all the questions in the space provided. 1 (a) A ball leaves the edge of a table with a horizontal velocity v, as shown in Fig. 1.1. The height of the table is 1.25 m. The ball travels a distance of 1.50 m horizontally before hitting the floor. Air resistance is negligible. Calculate, for the ball, (i) the horizontal velocity v as it leaves the table, v = m s−1 [2] (ii) the velocity just before it hits the floor. velocity = m s−1 [2] Fig. 1.1 ux = v , uy = 0 =xxs u t = vt [1] =+ 210 2 ys at  = + 211.25 0 (9.81)2 t [1] 1.50 = v (0.505) Hence v = 2.97 m s-1[1] =+ 22( ) ( )yxv v v [1] = + = + 0 (9.81)(0.505)yyv u at = 4.95 m s-1[1] = 5.77 m s-1[1]

3 NYJC 2022 9749/02/J2PRELIM/22 [Turn over (b) A second ball leaves the edge of the table with a horizontal velocity 2v. (i) State and explain whether the time take n to hit the floor is the same or different compared to the first ball. [2] (ii) Describe the variation of the vertical component of velocity if air resistance is not negligible. [2] [Total: 8] Since both ball initial velocity for the vertical component is zero, the height is the same and same acceleration of free fall, the time taken is the same. The vertical component of velocity is increasing at a decreasing rate.

4 NYJC 2022 9749/02/J2PRELIM/22 2 (a) State Newton’s Second Law of motion. [1] (b) A jet of water hits a vertical wall at right angles, as shown in Fig. 2.1. The jet of water has density ρ, cross-sectional area A, and hits the vertical wall with impact velocity u. The water then runs down the wall after impact with the wall. Fig. 2.1 (i) Using Newton’s Law of motion, show that the magnitude of the average force exerted on the water by the wall is F= ρAu2 . [3] Horizontal jet of water, with density ρ impact velocity u cross-sectional area A From Newton’s Second Law, Force on water, F = rate of change of momentum of water Hence, magnitude of F = ρAu2 Equation for N2L [1] Sub in final velocity = 0 [1] Sub in (M/t) = ρAu [1] The rate of change in the momentum of a body is proportional to the resultant external force that acts on it, and the change in momentum is in the direction of the force.

5 NYJC 2022 9749/02/J2PRELIM/22 [Turn over (ii) The density of water ρ is 1 000 kg m –3. Given that the jet of water in (b) has cross-sectional area A of 1.5 cm2 and impact velocity u of 5.0 m s–1, 1. Calculate the magnitude of the average force exerted on the wall by the water. Explain your answer. magnitude of average force = N [2] 2. On Fig 2.2, ske tch a graph to show the variation of pressure p on the wall with impact velocity u. [1] Fig. 2.2 3. Suggest the change, if any, to pressure P if the cross-sectional area A of the water jet is doubled. [1] [Total: 8] 0 u p Force by wall on water = ρAu2 = (1000)(1.5 x 10-4)(5.02) = 3.75 = 3.8 N [1] By Newton’s 3rd Law, magnitude of force by water on wall is equal to that by wall on water [1] Hence magnitude of average force by water on wall = 3.8 N No change. (Since Pressure = Force/Area = F/A = (ρAu2)/A = ρu2, pressure is independent of A)

6 NYJC 2022 9749/02/J2PRELIM/22 3 A ball of mass M of 750 g is held on a smooth horizontal surface between two identical springs at their natural lengths as shown in Fig. 3.1. Fig. 3.1 One spring is attached to a fixed point while the other spring is attached to a mechanical oscillator. At t = 0 the ball is displaced to its amplitude position. The variation with time t of the displacement L of the ball is shown in Fig. 3.2. Fig. 3.2 (a) For the first 12 s of the oscillations, (i) state one time at which the ball is moving with maximum speed, time = s [1] (ii) state one time at which the springs have maximum elastic potential energy, time = s [1] (iii) calculate the angular frequency  of the ball,  = rad s−1 [1] Equilibrium: 1.0 s, 3.0 s, 5.0 s, 7.0 s, 9.0 s, 11.0 s. Max displacement: 0.0 s, 2.0 s, 4.0 s, 6.0 s, 8.0 s, 12.0 s.  − == = 1 22 4.0 1.57 rad s T

7 NYJC 2022 9749/02/J2PRELIM/22 [Turn over (iv) calculate the maximum acceleration of the ball. maximum acceleration = m s−2 [2] (b) Some salt is sprinkled on the horizontal surface at t = 12.0 s. Calculate the loss in total energy of the oscillations during the first 24 s of the oscillations. Show your working clearly. [3] [Total: 8] At t = 0 s, xo = 0.05 m ( )( ) ( ) 2 2 2 max 22 3 11Total 22 1 0.750 1.5708 0.052 2.3132 10 J oE mv m x  − == = = At t = 24 s, xo = 0.0125 m ( )( ) ( ) 2 2 2 max 22 4 11Total 22 1 0.750 1.5708 0.01252 1.4457 10 J oE mv m x  − == = = Loss in Total E = 32.17 10 J− ( ) ( ) 2 max 2 2 1.57 0.05 0.123 m s oax  − = = =

8 NYJC 2022 9749/02/J2PRELIM/22 4 Microwaves of the same wavelength and amplitude are emitted in phase from two point sources X and Y, as shown in Fig. 4.1. Fig. 4.1 (not to scale) (a) State and explain along which of the lines XY and OA do the microwaves superpose to produce a stationary wave. [2] (b) A microwave detector is moved along a line from A to C. The microwave detector gives a maximum intensity reading at A and the first minimum reading at B. The microwaves have a wavelength of 4.0 cm. For the waves arriving at B, determine the path difference. path difference = m [1] XY. [A1] Only along XY the microwaves travel in a direction opposite to one another. [M1] Do not accept: “Different direction” B is at position of 1st minima. Hence path difference is 0.5λ. Path difference = 0.020 m [A1]

9 NYJC 2022 9749/02/J2PRELIM/22 [Turn over (c) Describe the effect, if any, on the intensity of the microwave detected at A and B when the following changes are made, separately to the sources X and Y: (i) when the amplitude of both source X and Y is doubled. [2] (ii) the amplitude of one of the sources is halved. [2] (iii) the sources are now anti-phase. [1] [Total: 8] Resultant amplitude at maxima is 1.5 x0 and minima at 0.5 x0. [B1] Maximum intensit