CJC 2024.H2.Phy.PRELIM.P3 - Solns

CANDIDATE NAME MARK SCHEME CLASS 2T PHYSICS 9749/03 Paper 3 Longer Structured Questions 10 September 2024 2 hours Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your name and class 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 an HB pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. Section A Answer all questions. Section B Answer one question only. You are advised to spend one and a half hours on Section A and half an hour on Section B. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 37 printed pages and one blank page. [Turn over FOR EXAMINER’S USE SECTION A Q1 / 8 Q2 / 9 Q3 / 8 Q4 / 12 Q5 / 7 Q6 / 8 Q7 / 8 SECTION B Q8 / 20 Q9 / 20 PAPER 3 / 80 PAPER 2 / 80 PAPER 1 / 30 PAPER 4 / 55 TOTAL (WEIGHTED) % Catholic Junior College JC2 Preliminary Examinations Higher 2

2 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

3 [Turn over FORMULAE uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as 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 = 1 2 ln2 t

4 Section A Answer all questions in the spaces provided. 1 (a) An object Q of weight 30.0 N is supported by two ropes A and B as shown in Fig. 1.1. Rope A is at an angle  to the vertical and exerts force FA on Q. Rope B is at an angle  to the vertical and exerts a force FB on Q. The angle  of rope B is varied from 0° to 90°. The force FA is varied in magnitude and direction to keep Q in equilibrium. (i) Determine the magnitude of force FA when the angle  is 35° and FB is 20.0 N. magnitude of FA = ……………………………… N [3] L2 Horizontally, no net force: FA sin = 20.0 sin35° -----(1) Vertically, no net force: FA cos + 20.0 cos35° = 30.0 FA cos = 13.617 ------(2) (1) / (2): tan  = 0.84244  = 40.112 = 40.1° ----- sub into (1) or (2) FA = 17.805 = 17.8 N OR The three coplanar forces in equilibrium should form a closed vector triangle: M1 M1 A1 [M1]   FB FA 30.0 N Fig. 1.1 rope B rope A Q

5 [Turn over Use cosine rule (and/or sine rule) to solve, FA = 17.8 N ( = 40.1°) [M1] [A1] (ii) Explain why angles  and θ cannot be 90° at the same time. ………………………………………………….……………………………………………… ………………………………………………….……………………………………………… ………………………………………………….……………………………………………… …………………………………………………….……………………………………. [2] L2 When both angles are 90° at the same time, both FA and FB are horizontal and there is no vertical component of force. There must be a vertically upward force that is equal in magnitude and opposite in direction to the object Q’s weight. to maintain vertical equilibrium. B1 B1 (b) A uniform metal rod AB is freely pivoted at end A as illustrated in Fig. 1.2. The end B is suspended by a light spring. The other end of the spring is supported at Z. The rod is in equilibrium. Fig. 1.2 The spring is now aligned vertically along YB so that the angle between the rod and the spring is no longer 90°. The rod remains in equilibrium in the same position. Explain why the spring force increases. 35°  20.0 N FA 30.0 N  B A Z Y

6 …………………………………………..…………….……………………………………………… …………………………………………..…………….……………………………………………… …………………………………………..…………….……………………………………………… …………………………………………..…………….……………………………………………… …………………………………………..…………….……………………………………………… ………………………….……………………………………………………………………… [3] L2 The total clockwise moment about the pivot A due to the rod’s weight is unchanged. If the spring is aligned vertically along YB, the perpendicular distance of the line of action of the spring force from pivot A will decrease. Therefore, the spring force must increase to maintain the same total anticlockwise moment about the pivot. B1 B1 B1 [Total: 8] 2 Two spheres A and B approach each other as illustrated in Fig. 2.1. Fig. 2.1 Sphere A has a mass of 0.500 kg and moves to the right with a speed of 7.40 m s-1. Sphere B has a mass of 0.350 kg and moves to the left with a speed of 9.60 m s-1. The spheres collide and are in contact for a time of 0.400 s. Sphere B reverses its direction of motion and moves off with a speed of 10.4 m s-1. (a) Using momentum consideration, e xplain quantitatively why spheres A and B cannot be at rest at the same instant. …………………………………………..……………….…………………………………………….. …………………………………………..…………….……………………………………………….. 7.40 m s-1 9.60 m s-1 0.350 kg 0.500 kg A B 10.4 m s-1 A B Before collision After collision

7 [Turn over …………………………………………..…………….……………………………………………….. .………………………………………………………………………………………………….. [2] L2 Take right as positive direction. Total initial momentum of A and B = (0.500)(+7.40) + (0.350)(-9.60) = +0.34 kg m s-1 No net external force acts on system of A and B , therefore total momentum is conserved at all times and always equal to the total initial momentum. Total initial momentum equals 0.34 kg m s-1 to the right, which is non-zero. Therefore both A and B cannot be at rest at the same time. M1 M1 A0 (b) For the time during the collision, calculate the average force between the spheres. average force = ……………………………… N [2] L2 Take right as positive direction. Considering sphere B’s rate of change in momentum, Average force = 𝛥𝑝 𝛥𝑡 = 0.350(10.4 − (−9.60)) 0.400 = 𝟏𝟕. 𝟓 N (i.e. to the right) *Mark for magnitude only. M1 A1 (c) Use your answer in (b) to determine the magnitude of the velocity of sphere A after the collision. Explain your working. magnitude of velocity = ……….………………… m s-1 [3] L2 Take right as positive direct