2021 H2 Phy Prelim Paper 2 Solution (forJCs)

2021 JC2 H2 Preliminary Examination Paper 2 Suggested Solutions 1 (a) The principle of conservation of linear momentum states that the total momentum of a system of interacting bodies remain constant provided no external force acts on the system. 1 1 (b) (i) By the principle of conservation of momentum, mXuX + mYuY = mXvX + mYvY 4.0(500) + 28.0(340) = 4.0(220) + 28.0(v) v = 380 m s–1 1 1 (ii) Relative speed of approach = uX – uY = 500 – 340 = 160 m s–1 Relative speed of separation = vY – vX = 380 – 220 = 160 m s–1 Since the relative speeds of approach and separation are the same, the collision is elastic. Or Check whether total final kinetic energy equals the total initial kinetic energy. 1 1 (iii) The forces on the two bodies (or on X and on Y) are equal in magnitude and opposite in direction The duration of impact for both forces is the same and force is change in momentum / time 1 1

2 2 (a) On a banked track, the normal contact force by the road on the car will be tilted to the vertical at an angle equal to the angle of inclination of the track. The horizontal components of both the normal force and the frictional force by the road on the car contributes to the required centripetal force to keep the car moving in a circular path . This increases the centripetal force, which would otherwise be provided by just the frictional force alone. 1 1 (b) (i) [1] – One mark for each correct force. * Allow two normal contact forces and two friction forces between the two wheels and the road. * Minus [1] for not naming any force in full * Minus [1] if centripetal force is drawn into diagram 3 (ii) Horizontally, the resultant force towards the centre of the circle is the centripetal force, Vertically, there is no resultant force, Substituting into (1), 14791.7 = 40v2 – 10392.3 v = 25 or 25.1 m s1 1 1 1 30 Normal force, N Frictional force, F Weight, W

3 3 (a) Internal energy is the sum of microscopic kinetic energy and potential energy of all the molecules in the gas Microscopic potential energy for a real gas is not equal to zero as there is intermolecular bonding between the molecules. So internal energy is not only dependent on temperature. 1 1 (b) p= 1 3 ρ ⟨ c2⟩ = 1 3 M V ⟨ c2⟩ , where M = total mass of the ideal gas V = total volume of the ideal gas 3 2 pV = 1 2 M ⟨ c2⟩ 3 2 nRT = 1 2 M ⟨ c2⟩ = Total KE of gas Since internal energy, U = total KE of ideal gas, as PE of ideal gas = 0  U  T 1 1 1 4 (a) (i) pV = NkT (2.5 × 105) × (3.8 × 10–2) = N × (1.38 × 10–23) × (181 + 273) N = 1.5 × 1024 1 1 (ii) From first law of thermodynamics, increase in internal energy = heat energy supplied + work done on the system Since the process took place at constant volume, work done = 0 Thus, increase in internal energy = 2700 + 0 = 2700 J (Since it is an increase in the internal energy, the change in internal energy is positive 2700 J) 1 1 (b) ∆ EK = 3/2 N k (∆T) 2700 = 3/2 (1.5 × 1024) (1.38 ×