NYJC 2022 Prelim P3 Guide
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Text from the first pagesNYJC 2022 9749/03/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 9749/03 Paper 3 Longer Structured Questions 19 September 2022 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 / 12 3 / 8 4 / 9 5 / 6 6 / 9 7 / 8 Section B 8 / 20 9 / 20 Total / 80 This document consists of 24 printed pages.
2 NYJC 2022 9749/03/J2Prelim/22 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 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 =
3 NYJC 2022 9749/03/J2Prelim/22 [Turn over Section A Answer all the questions in the spaces provided. 1 Fig. 1.1 shows a man standing on a stationary sailboard floating in the sea. The sailboard consists of a surfing board, mast and sail. Fig. 1.1 (a) The total mass of the sailboard and the man is 90 kg. Taking the density of seawater to be 1020 kg m‒3, calculate the volume of seawater displaced by the sailboard. F U mg Vpg mg V = −= = == 3 0 0 90 0.088 m(1020) volume of seawater displaced = m3 [2] mast sail surfing board Comments: A significant number of students did not show evidence of resultant force is zero. Hence presentation of answers is important in this question in general.
4 NYJC 2022 9749/03/J2Prelim/22 (b) The sailboard then cruises at constant speed. Fig. 1.2 shows some of the forces acting on the mast and sail of the sailboard. The uniform mast has length 4.3 m and the base of the mast is connected to the surfing board by a smooth hinge. The wind exerts a force of 200 N on the sail, perpendicular to the mast at a distance 3.5 m away from the hinge. The man pulls the sail with a force T at distance 1.5 m away from the hinge and the weight of the mast and sail is 50 N. (i) Show that pulling force T by the man is 460 N. hinge oo T T = − − = = 0 200(3.5) (50)(2.15cos70 ) sin75 (1.5) 0 460 N [2] pulling force by man, T Fig. 1.2 force by wind = 200 N hinge 1.5 m weight of mast and sail = 50 N 3.5 m 75° 70° 4.3 m
5 NYJC 2022 9749/03/J2Prelim/22 [Turn over (ii) Determine the magnitude of the force, R, exerted by the hinge. oo Fy Ry Ry = − − + = = 0 200cos70 50 460sin5 0 21.7 N oo Fx Rx Rx = − + = =− 0 200sin70 460cos5 0 270 N R = + = 22(270) (21.7) 271 N magnitude of R = N [3] (c) The surfing board is designed with the foot straps at the rear part of the board rather than at the centre part of the board, as shown in Fig. 1.3. Fig. 1.3 Such a design allows the surfing board to move across the water surface while inclined at an angle to the surface. Suggest why when the man is moving horizontally with the board at an incline, the volume of seawater displaced by the sailboard is lower than your answer in (a). [1] [Total: 8] As the surfboard move across the water surface inclined at an angle, there will be water continuously changing direction as it come into contact with the surf board. By Newton 2nd law, there is a rate of change in momentum resulting in a force acting iin the downward direction. Hence by Newton’s 3 rd law, there is upwards force acting acting on the surf board. Hence the upthrust can be smaller causing water displaced to decrease. foot straps water level foot straps board is horizontal board is at an incline Comments: This part prove to be challenging for the large majority of students. Comments: A significant number of students are able to get full credit for this question even though it is quite mathematically challenging.
6 NYJC 2022 9749/03/J2Prelim/22 2 (a) Planets have been observed orbiting a star in another solar system. Measurements are made for the orbital radius r and the time period T of each of these planets. The variation with r3 of T2 is shown in Fig. 2.1. The relationship between T and r is given by rT GM 23 2 4= where G is the gravitational constant and M is the mass of the star. Fig. 2.1 r3 / 1034 m3 T2 / year2
7 NYJC 2022 9749/03/J2Prelim/22 [Turn over (i) Determine the mass M of the star. M = kg [2] (ii) The radius of the star is 700 000 km. Determine the minimum speed with which gas particles from its surface have to be ejected to just escape from the star’s pull of gravity. minimum speed = m s−1 [2] (iii) Hydrogen gas, consisting of hydrogen-2 particles, may be assumed to be an ideal gas. If the surface temperature of the star is 6000 K, determine whether hydrogen gas particles are able to escape the surface. [2] (iv) Some gas particles have very large kinetic energy to be able to escape from the star. Given that t he star is rotating about an axis through its poles , suggest why the gas particles at the equator of the star are more likely to escape the surface than those at the poles. [2] Gradient = ( ) 34 34 2.4 0 2.0 101.2 0 10 −− =− ( ) 2 2344 2.0 10 365 24 60 60GM −= [1] M = 2.98 x 1030 kg or in 2 sf [1] By Conservation of Energy, Initial KE + initial GPE at surface = Final KE + final GPE at infinity ½ mv
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