NYJC 2021 H2 Physics 9749 P3 Answer

NYJC 2021 9749/03/J2Prelim/21 [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SOLUTION CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9749/03 Paper 3 Longer Structured Questions 20 September 2021 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class, Centre number and index number 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 or rough working. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. 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. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 / 8 2 / 10 3 / 8 4 / 10 5 / 8 6 / 7 7 / 9 Section B 8 / 20 9 / 20 Total / 80 This document consists of 22 printed pages.

2 NYJC 2021 9749/03/J2Prelim/21 Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space  = 4 × 10−7 H m−1 permittivity of free space  = 8.85 × 10−12 F m−1 (1 / (36)) × 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

3 NYJC 2021 9749/03/J2Prelim/21 [Turn over Formulae uniformly accelerated motion 21 2s ut at 22 2v u as work done on / by a gas W p V hydrostatic pressure p gh gravitational potential /Gm r  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 sinx x t  velocity of particle in s.h.m. 0 cosv v t  22 0xx  electric current I Anvq resistors in series 12 . . .R R R   resistors in parallel 121/ 1/ 1/ . . .R R R   electric potential 04 QV r alternating current/voltage 0 sinx x t  magnetic flux density due to a long straight wire   0 2 IB d magnetic flux density due to a flat circular coil  0 2 NIB r magnetic flux density due to a long solenoid  0B nI radioactive decay 0 exp( )x x t  decay constant 1 2 ln2 t 

4 NYJC 2021 9749/03/J2Prelim/21 Section A Answer all the questions in the spaces provided. 1 The Poiseuille equation relating the volume flow rate V t of a fluid under laminar conditions through a horizontal tube of length L and internal radius r is 4 8 V pr tL   where p is the pressure difference between the two ends of the tube and  is the viscosity of the fluid. (a) Show that the SI base units for  is kg m-1 s-1.        4 -1 -2 4 3 -1 -1 -1 kg m s m m m s kg m s (shown) pr VL t       [2] (b) In an experiment to determine  for water, a student recorded the following measurements in SI units, as shown in Table 1.1. Table 1.1 quantity magnitude in SI units percentage uncertainty / % V t 1.0 × 10–6 3 p 500 2 L 0.20 0.5 The internal diameter of the tube was measured and recorded as (0.200 ± 0.002) cm. (i) Calculate the percentage uncertainty in the internal radius r of the tube. 2 0.002% uncertainty 100%0.200 1% or 1.0% (1 or 2 s.f.) A [ 1] dr dr dr      percentage uncertainty = % [1]

5 NYJC 2021 9749/03/J2Prelim/21 [Turn over (ii) Using the results in Table 1.1 and (b)(i), determine  with its associated uncertainty. Give your answer to an appropriate number of significant figures.  = ± kg m-1 s-1 [4] (iii) State and explain which measured quantity has the greatest contribution to the uncertainty of  . [1] [Total: 8]     4 42 6 4 -1 -1 4 5 5 -1 -1 8 500 0.10 10 8 0.20 1.0 10 9.817 10 kg m s 4 0.02 4 0.01 0.005 0.03 0.095 0.095 9.817 [ 10 9.326 10 9 10 kg m s (1 s.f.) C1] [M1] [ A pr VL t Vp r L t Vp r L t                                              4 -1 -19.8 0.9 10 kg m s 1] recognise to 1 s.f. [A1] recognise to same d.p. as          Internal diameter or radius, because of the power of 4, so greatest contribution to the uncertainty of viscosity [B1]

6 NYJC 2021 9749/03/J2Prelim/21 2 Two atoms X and Y, have masses 3m and 2m respectively. The 2 atoms move head-on towards each other with the same speed v as shown in Fig. 2.1. Fig. 2.1 Fig 2.2 comprises two velocity-time graphs A and B, which show how the velocity of each atom varies. The interaction between the atoms is elastic. Fig. 2.2 (not to scale) (a) (i) Explain why it is not possible for the atoms to stop at the same instant. [1] (ii) At one instant during the interaction between the atoms, they are both traveling in the same direction with the same speed. Calculate this speed, in terms of v. speed = …………………….. [2] X v v Y time velocity v v 0 From the principle of Conservation of momentum: Total initial momentum = total final momentum 3 2 3 2 0.2 mv mv mu mu u mv   Since the total momentum before the collision is mv, it is not possible for both atoms to stop at the same instant as then the total momentum of the system at that instant will be zero. This will violate the principle of conservation of momentum. A B T

7 NYJC 2021 9749/03/J2Prelim/21 [Turn over (b) (i) State and explain, which of the curves A or B is the velocity-time sketch for atom Y. [3] (ii) On Fig. 2.2, mark the instant in time at which the atoms are at their distance of closest approach. Label this point T. [1] (iii) Determine the final speed of each atom in terms of v. final speed of X = …………………….. [3] final speed of Y = …………………….. [3] [Total: 10] Let Vx be the final speed of X and Vy be the final speed of Y. Since the collision is elastic: relative speed of approach = relative speed of separation,      () 2 .....(1) YX YX v v V V v V V By the principle of Conservation of momentum:     3 2 3 2 32 ......(2) XY XY mv mv mV mV v V V Solving equation (1) and (2):   0.6 1.4 X Y Vv Vv Curve A is the velocity-time sketch for atom Y. During the collision, by Newton’s third law, X and Y will experience a force of equal magnitude and in opposite direction. Since the force on X and Y are the same, and the mass of Y is smaller, Y will experience a larger acceleration and the change in velocity for Y will be larger, which is curve A. For momentum to be conserved, the magnitude of the change in momentum of X is equal to the magnitude of change in momentum of Y. Since the mass of Y is smaller, the change in velocity for Y will be larger, which is the curve A.

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