(VJC) 2024 H2 Prelim P2_Soln

2024 Physics Prelim Exam H2 Paper 2 suggested solutions 1(a) Moment of a force: The moment of a force about an axis is the product of the force and the perpendicular distance [1] from the line of action of the force to the axis [1]. Minus 1 mark if no diagram given or diagram does not show corresponding d and F correctly. Torque of a couple: Torque of couple is defined as the product of one force F [1] and the perpendicular distance 𝑑⊥ between the two forces [1]. Minus 1 mark if no diagram given or diagram does not show corresponding d and F correctly. (b)(i) F d F d F T: Tension T: Tension W: Weight of truck N: Normal reaction force

(b)(ii) Considering the truck in equilibrium: sin40 2Tm g T = [1] Taking moments about the centre of the pulley, Sum of clockwise moments = sum of anti-clockwise moments 3mg r T r =  sin403 2 Tmgmg r r  =  [1m for correct clockwise expression, 1m for correct anti - clockwise expression] 1500sin40 2(3)m = m = 161 kg [1] 2(a) (i) 1m for correct graph P 1m for correct graph K (ii) 1m for correct graph F Explanation: Some potential energy lost will be converted to do work against friction [1], hence the gain in kinetic energy will be less as compared to the situation without friction. [1] Energy/ J h / m P K F

3(a) momentum after elastic collision = - mu [1] time between collisions = 2L/u [1] number of collisions per unit time = u/2L [1] rate of change of momentum = - mu2/L [1] average force = mu2/L [1] (b)(i) Gas molecules collide with one another elastically. [1] (ii) Since collisions are elastic, the average speed of gas molecules incident on the wall remains the same. [1] So frequency of collision between each wall and the molecule also remains unchanged. [1] Therefore the pressure exerted on the wall is not affected. (c) Given in (b) pV = 1/3 Nm<c2> - - - - (1) By Ideal Gas Law: pV = NkT - - - - (2) [1] Equating (1) = (2) 1/3 Nm<c2> = NkT => ½ m<c2> = 3/2 kT [1] Hence average ke, ½ m<c2> = 3/2 kT  Ave KE  T (because k is a constant) (d)(i)1. Shade the area under p-V graph [1]

2. Since the process A -> B is such that U = 0 (isothermal), and U = 3 2 pV then pAVA = pBVB. [1] Hence there is no difference in the product pAVA and pBVB. [1] (ii) If the change takes place very quickly, there is no time for heat transfer with the surrounding. So it is an adiabatic change. i.e. Q = 0. [1] Fig. 3.2 shows a compression. i.e. W is positive. So U = Q + W is also positive. i.e. U increases. [1] Since U  crms2 of the molecules, it means the mean square speed of the molecules increases. [1] , 4(a) Since the tube is in equilibrium, Weight of (tube and ball bearings = Upthrust [1] (m + M)g = AHg ( )MmH A += [1]