2025 TJC H2 Phy Prelim P3 Solution

2025 Physics Prelim P3 Solutions 1 (a) (i) yas2+= 2 y 2 y uv = 0 + 2(9.81)(32) vy = 25.1 m s-1 M1 A1 (ii) M1 A1 (b) With splashing, there is a transfer of KE of stone to KE of water. Less KE of stone available to do work against resistive forces in the water. B1 (c) (i) Parabola with initial gradient = 0 [1], GPE = 0 at tS [1] B1 B1 (ii) B1 2 (a) Net force on body proportional to rate of change of momentum of the body (allow eqn with symbols if defined?) (b) (i) ∆p = 140  10–3  [(5.5 - (-4.0)] = 1.33 kg m s–1 (ii) Fnet = 1.33 / 0.04 = 33.3 N Average force on ball due to bar = Fnet + Wball = (1.33 / 0.04) + (9.81 x 140 x 10−3) = 34.6 N By N3 Fbar =Fball. =34.6 N = 35 N 25.1sin 34 47.6   = = Ep tS t Ek tS t

2 (iii) Taking moments about B (35  0.75) + (0.450  9.81  0.25) = FA  0.20 FA = 137 N (iv) loss = ½ (0.140) [ 5.52 - 4.02 ] = 0.998 J (c) Since initial and final velocities remain the same, the change in momentum of the ball on hitting the bar is unchanged. However the duration of contact increases, average contact force on bar at point D decreases, leading to lower average contact force at D And hence a lower exerted force at A 3 (a)(i) An ideal gas is one that obeys the ideal gas equation, pV = nRT at all pressures, volumes and temperatures. B1 (ii) pV = nRT (1.00 × 105) (750 × 10-6) = n (8.31) (300) M1 n = 0.030 A1 (b) work done on gas / J heat supplied to gas / J increase in internal energy of gas / J Ao B +360 0 +360 B to C 0 +670 +670 C to D -810 0 -810 D to A 0 -220 -220 Minus 1 per mistake. (c) the gas molecules bounce off the receding piston at lower speeds And hence lower kinetic energy. B1 For an ideal gas, the temperature is proportional to the average kinetic energy of the molecules. B1 (d) The net work by the engine is positive and can be used to move the car (turn the wheels) B1

3 4 (a) 4.0 1.30.75 5.5 3.0 R R R =++ = 5.5p.d. across AB 4.00.75 3.0 5.5 2.4 V = ++ = C1 C1 A0 (b)(i) 1.50 0.56 2.41.50 1.5 V E −= = C1 A1 (b)(ii) As C is shifted closer to A, the potential at C increases, thus increasing the potential difference between BC. Since the potential between BC will become larger than the terminal p.d. of cell Q, current will now flow from C to B through cell Q. M1 A1 (c)(i) 15.5 / 150 0.0367 cm 1 150 2 152 15 0.0367 0.0367 55 1.28 1.3 R L R −= =  −