SCGS 2025 PHY PRELIM P2 MS

1 2025 PRELIMINARY EXAMINATION SUGGESTED ANSWER SCHEME PAPER 2 SECTION A ( 50 marks) Qn Part Answer Remark 1 (a)(i) 1. Loss of GPE = 92 kg x 10 N/kg x 1500 m = 1380000 J = 1.4 x 106 J ( 2 s.f.) 2. Kinetic energy = ½ (92kg)(52 m/s)2 = 124384 J = 1.2 x 105 J (ii) Some energy in the gravitational potential store is transferred to the internal store of the surroundings due to air resistance. (b)(i) The air resistance is increasing (ii) W = mg = 92 kg x 10N/kg = 920 N 2 (a)(i) ▪ Block moves with uniform acceleration. ▪ Resultant force of 0.6 N acting to the right. (ii) ▪ Block moves at constant velocity/same speed same direction ▪ Force X = total resistive force. Zero resultant force acting on the block. (b) Power = Force x velocity 2.0 W = 0.80 N x v v = 2.5 ms-1 Distance = 2.5 ms-1 x 3.0 s = 7.5 m (c) During the first three seconds, the power delivered is not constant since the velocity is changing while in (b), the velocity is constant and the power delivered is constant. 3 (a) 1 Frictional force acting at the rod 2 Normal (Reaction) Force acted on card by rod (b) Since the direction and magnitudes of the two forces In (b) are different/ are of different nature/acting on the same body they do not satisfy the conditions for action-reaction pair. 4 (a) ▪ The uneven bombardment on all sides by air particles ▪ The random/ erratic and continuous movement is caused by to the resultant force on the smoke particle at any position (b)(i) The number of erratic paths compared to that at the lower temperature will be more (ii) ▪ Air particles impacted the smoke particles at higher velocities. The frequency of collisions also increased. ▪ The greater force of impact and frequent collisions resulted in more frequent change in direction of the particle in shorter intervals compared to that at lower temperature, thus explaining the more erratic movement * Reject in increase in pressure / average force per unit area as these leads to pressure of the gas. 5 (a)(i) The angle of incidence in the glass block has exceeded critical angle of glass. (ii) When the critical angle is reached, the beam is at B. Using trigonometry, Critical angle, c = 90o - 𝑡𝑎𝑛−1( 1.68 1.50) = 90o – 48.2o = 41.8o Refractive index = 1 sin 𝑐 = 1 sin 41.8𝑜 = 1.50 (iii) Higher refractive index→smaller critical angle. Hence AB will be shorter.

2 Qn Part Answer Remark (b)(i) (ii) To enable the formation of an upright image on the screen/to give an image that is right side up. 6 (a) From 𝐼 = 𝑉 𝑅 = 3.6 𝑉 5800 = 6.2 x 10-4 A (b) From R = V/ I 3400 + Rthermistor = 6.0 V/6.2x 10-4 A