NJC 2024 H2 Physics Prelim P3 Ans

NJC Preliminary Examination 2024 H2 Physics Paper 3 Solutions and Mark Scheme Section A 1 (a) change in velocity of the body is always perpendicular to velocity when speed is constant. B1 acceleration and so resultant force [Newton’s second] is always perpendicular to the velocity B1 velocity is tangent to circular path , so resultant force (perpendicular to the velocity) directed towards centre of circle B1 (b) (i) centripetal acceleration = (25×103 60×60 ) 2 7.0 M1 = 6.9 m s–2 or 6.89 m s–2 A1 (ii) force on mass = 0.50 × 6.889 = 3.4445 N C1 displacement = 3.4445 × 5.0 = 17 mm or 17.2 mm A1 (iii) extension of spring at B > spring at A / spring at B is extended while spring at A is compressed to provide this resultant force (towards B or centre) M1 pointer moves towards A. A1

2 2 (a) The gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point. B1 (b) (i) Increase in potential energy = final potential energy – initial potential energy B1 1 mark for correct potential energy formula ( WITH negative sign ) and substituted correctly. (ii) Work is done by thrusters M1 Hence total energy increases / not constant M1 Increase in potential energy not equal to decrease in kinetic energy A0 (c) Decrease in potential energy = increase of KE 0 − (− 𝐺𝑀(𝑚𝑟) 𝑟2 ) = 1 2 (𝑚𝑟)𝑣2 − 0 or equate total energy M1 for correct decrease in potential energy, M1 for correct increase in KE 𝑣 = √2𝐺𝑀 𝑟2 A1 for the correct final expression for v.

3 3 (a) (i) Kinetic energy of one gas particle (atom / molecule) = 1 2 𝑚𝑐𝑟𝑚𝑠2 = 3 2 𝑘𝑇 where 𝑚 is the mass of one gas particle/atom/molecule. B1 For one mole of gas containing 𝑁𝐴 particles, the total kinetic energy is given by: 1 2 𝑁𝐴𝑚𝑐𝑟𝑚𝑠2 = 3 2 𝑁𝐴𝑘𝑇 ------------------ (1) From the equation of state of an ideal gas for 1 mole of ideal gas: B1 𝑝𝑉 = (1)𝑅𝑇 = 𝑁𝐴𝑘𝑇 Substituting 𝑁𝐴𝑘𝑇 = 𝑅𝑇 into (1) gives: 1 2 𝑁𝐴𝑚𝑐𝑟𝑚𝑠2 = 3 2 𝑅𝑇 Simplifying to get: 𝑐𝑟𝑚𝑠 = √ 3𝑅𝑇 𝑁𝐴𝑚 = √3𝑅𝑇 𝑀 Where 𝑁𝐴𝑚 is the mass of 6.02 x 1023 particles = molar mass 𝑀. (ii) 𝑐𝑟.𝑚.𝑠. ∝ 1 √𝑀 𝑐𝑟.𝑚.𝑠. 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑐𝑟.𝑚.𝑠. 𝑜𝑓 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = √28 32 = 0.935 or 0.94 A1 (b) (i) 𝑝 = 1 3 𝑁𝑚 𝑉 ⟨𝑐2⟩ = 1 3 𝜌⟨𝑐2⟩ = 1 3 × 1.50 × 105 × (4.85 × 105)2 M1 = 1.18 x 1016 Pa A1 (ii) Forces between nuclei/particles are not negligible. (ignore “attractive”) M1 Forces are repulsive (at that density) contributing to an increase in pressure A1 OR Volume of particles is not negligible (at that density) M1 Resulting in a higher rate (or frequency) of collision compared to the expected rate, causing the actual pressure to be higher than the expected value in (b)(i). A1 (c) Process w / kJ q / kJ U / kJ A to B 19.2 67.2 – 48.0 [B1] B to C 0 48.0 48.0 [B1] C to A 31.6 31.6 0

4 4 (a) (i) (ii) Summing forces vertically: mg = Upthrust = ALg M1 mass of loaded test tube, m = AL A1 (b) (i) Resultant force (in vector notation): F = Upthrust + Weight M1 Since mg + ( ALg) = 0 M1 A0 (ii) M1 M1 Since loaded test-tube is in SHM: a = - ω2x M1 , so (iii) Total energy = ½ m ω2 x02 = ½ (0.050)(9.81/0.125)(0.015)2 M1 = 0.00044 J A1 W: Weight of loaded test-tube or WL: weight of lead shots and WT: weight of test-tube U: Upthrust / Force by fluid on loaded test-tube Legend for both W and U B1 W and U of same length and act along the same vertical line B1 (Note that arrow of W should originate from C.G. of loaded test-tube while arrow of U should originate from centre of mass of the displaced fluid.)

5 (iv) correct shape (sin2 not modulus) of 4 “humps” with k.e. starting from zero B1 correct label of max K.E. B1 T 2T 0 4.4 x 10-4

6 5 (a) (i) Waves from the two slits overlap and superpose (debrief point) at points on the screen B1 When path difference from slits to point is multiples of wavelength / phase difference between the two waves is multiples of 2𝜋, constructive interference gives bright fringe B1 (ii) separation = (590×10−9)(2.3) 1.2×10−3 M1 = 1.1 mm or 1.13 mm A1 (b) 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜃 = 𝜆 𝑏 = 590×10−9 0.31×10−3 ≈ 𝑥 𝐷 must see sinθ M1 (𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 = 𝑥 𝐷 = 𝑥 2.3 led to 𝑥 = (590×10−9)(2.3) 0.31×10−3 width = 2x M1 = 8.8 mm A0 (c) Diffraction minimum is (8.8 / 2 =) 4.4 mm from P and the fourth order bright fringe is (1.1 × 4 =) 4.4 mm from P B1 Position of 4 th order interference maximum coincides/overlap with (first order) diffraction minimum. B1

7 6 (a) E = Q / 4πε0r2 or E = kQ / r2 with k defined / substituted in 4.1 × 10–5 = [Q / (4π × 8.85 ×10–12 × 0.0252)] – [Q / (4π × 8.85 × 10–12 × 0.0752)] M1 Q = 3.2 × 10–18 C A1 (b) correct shape (start positive E, gradient trend) B1 through points (0, 4.1 × 10–5) (2.5, 0) & (5.0, - 4.1 × 10–5) B1 (c) Using the graph, E -field strength on the left of d = 2.5 cm is positive, while on the right is negative. Thus, the graph shows E-field is always directed towards the d = 2.5 cm. B1 Electric Force, F = qE, (or acceleration) on positive charge is always opposite to displacement from d = 2.5 cm B1

8 7 (a) V = 2.4 V = Vp Ns / Np = 50 = Vs / Vp M1 Vs = 2.4 x 50 = 120V Max VS = 120 x √2 = 170V A1 (b) Pave = ½ (½) P0 = ¼ V02/R = ¼ 1702 / 47 C1 = 154 W A1 (c) Pnew = 2 Pave = 307 W (allow ecf part b) M1 (d) Direct voltage since the voltage shown is always positive (w.r.t. time) B1 (e) Transformers requires an input voltage that varies with time. B1

9 Section B 8 (a) (i) Area cut in time t is the curved surface area of a cylinder traced out by the falling ring. B1 Flux cut, Comments: Some explanation is expected in the working since this is a “show” question. No credit for candidates who just write (ii) From Faraday’s Law, induced e.m.f. Induced current M1 (iii) Magnetic force exerted by the radial magnetic field on the induced current in the ring, 𝐹𝐵 = 𝐵𝐼𝐿 = 𝐵(⬚)(2𝜋𝑟) M1 From Newton’s 2nd Law: Resultant force on the ring M1 A0 (iii) Maximum speed (when a = 0) is 𝑣𝑚𝑎𝑥 = 𝑚𝑔𝑅 (2𝜋𝑟𝐵)2 C1 = (0.0235)(9.81)(2.30×10−4) (2𝜋×0.03×0.800)2 M1 = 2.33 x 10-3 m s-1 A1 M1

10 (iv) Correct shape B1 label terminal velocity B1 Correct shape for a-t graph B1 (b) (i) Induced e.m.f. = Brv = 0.500 x 0.03 x 0.03 M1 = 4.50 x 10-4 V A1 (ii) Q B1 Explanation for (i) above