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 p