AHS 2025 PHY PRELIM P2 MS

1 2025 AHS Physics Prelim Paper 2 Answers 1 (a) (Electric) current, (thermodynamic) temperature, amount of substance (any two) [2] (b) (i) Circle the value for the prefix deci. 10-9 10-6 10-3 10-2 10-1 103 106 109 1012 [1] (ii) G , k , m , µ , n [1] 2 (a) Calculate the displacement of the car from point A at t = 90 s. Displacement = area (+) – area (-) = ½ (50 + 70) (30) - ½ (10) (20) [1] = 1700 m [1] (b) Describe the motion of the car from t = 80 s to t = 90 s. Travel backwards ( to the West ) [1] also accept opposite , (negative not accepted) Accelerating at a constant rate / constant acceleration [1] also accept descriptions using speed or velocity 3 Correct shape with correct angles [1] Correct arrows + labels [1] Magnitude = 69 N +/- 1 N [1]

2 4 (a) State the principle of moments. For an object in equilibrium, [1] The sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot. [1] (b) Determine the force exerted by the worker at point X. Sum of anti-clockwise moments = sum of clockwise moments 20 x 1.3 = F x 0.60 + 16 x 0.85 [1] F = 21 N [1] 5 (a) Liquid B is mercury, which has a density of 13 600 kg / m3. The gravitational field strength g is taken to be 10 N / kg. Determine the density of liquid A. Same atmospheric pressure, hence hA ρA g = hB ρB g [1 algebraic or numerical form] (120 -15 – 20) x ρA = (120 – 25 – 20) x 13 600 ρA = (75 / 85) x 13 600 = 12 000 kg / m3 [1] (b) Without using any additional apparatus, suggest how you can check that there is no air that is trapped at the top of any one of the inverted glass tubes, while it is at the same place. Tilt/ lean the inverted tube sideways/push inverted tube A downwards until the closed end is below 85 cm (liquid A) or 75 cm (liquid B) in height [1] Glass tube fully filled with liquid / no air bubble present [1] Alternative answers. Push the tube down/ add liquids into the troughs/ tilt the tube sideways [1] ensure that the height of the liquid column remains constant[1] 6 (a) (i) Calculate the change in energy in the kinetic store of the skier as he travelled from point T to point B. Change in Ek = Ek at B - Ek at T [1 algebraic or numerical must show difference, not just KE formula] = ½ (75) (302 – 2.02) = 33 600 J or 34 000 J [1]

3 (ii) Hence, determine the height of point T from point B. Change in EP = Change in Ek = 34 000 (75) (10) (h) = 34 000 h = 44.8 m or 45 m [1] (b) Upon reaching point B, the skier digs his skis into the snow and comes to a stop at point P after moving horizontally for 33 m. Calculate the average braking force that acts on the skier between points B and P.