HCI 2024 H2 Physics Paper 1 Answers

1 Solutions to Paper 1 1 D Temperature difference = 𝜃௙ − 𝜃௜ = 80 – 20 = 60 oC Absolute uncertainty, = ∆𝜃௙ + ∆𝜃௜ = 0.5 + 0.5 = 1 oC Percentage uncertainty = 100% 1.7%1 C60 C 2 B The object starts from rest and moves along a straight line. Area under the acceleration-time graph represents change in velocity. The triangle below the time axis shows that the change in velocity is in the negative direction. Since the initial velocity is zero, that triangle represents the increasing velocity (in – ve direction). This happens up to the time at A. Beyond A, the acceleration changes direction (from – ve to + ve dir). However, the velocity still points in the – ve direction after A i.e. object continues to travel in the same direction (still getting further away from the starting point), except that since the acceleration and velocity point in opposite directions, the object is slowing down. At point B the object comes to a momentary stop and it is also the furthest from the starting point. Between B and C, the object changes its velocity direction and starts to speed up back towards the starting point. At C, the object possesses maximum speed as it moves towards the starting point. Beyond C, the object slows down but it is still moving towards the starting point. At D, the object is back at the starting point. 3 D 𝑆௕ − 𝑆௘ = 2 2.0 = [𝑢𝑡 + (0.5)(9.81) 𝑡ଶ] − [𝑢𝑡 + (0.5)(5.8) 𝑡ଶ] 𝑡 = 1.00 s Or recognise that the effective acceleration = 9.81 – 5.8 = 4.01 m s-2. Using 212s ut at ,  212.0 4.012t, giving t = 1.00 s. 4 A The question asks for the HORIZONTAL forces only. The horizontal force the propels the man forward is the frictional force of pool’s floor on his feet. The horizontal resistive force on his motion comes from the drag due to water. Using Newton's second law along the horizontal direction: Net force on man = (mass of man)(acceleration of man) i.e. f – fD = ma

2 5 A Magnitude of the change in momentum Δ𝑃 = area under graph = ଵଶ (150)(40 × 10ିଷ) = 3 𝑁𝑠. This change in momentum takes place in the direction of the force, i.e. opposite to the initial momentum. In the vector diagram below, the ball is approaching the racket from right to left. 𝑃ሬ⃗௙ − 𝑃ሬ⃗௜ = Δ𝑃ሬ⃗ 𝑃௙ + 𝑃௜ = Δ𝑃 (0.060)(30) + 𝑃௜ = 3 𝑃௜ = 1.2 𝑁𝑠 6 C The thrust must provide an upward force to propel the rocket upward. This force must be least at equal to its weight. Thrust = 𝑣ቀௗ௠ௗ௧ቁ 𝑣ቀௗ௠ௗ௧ቁ = 𝑀𝑔 ௗ௠ௗ௧ = ெ௚௩ = (ହ଴଴)(ଽ.଼ଵ)ଵ଴଴଴ = 4.9 𝑘𝑔 𝑠ିଵ 7 C Taking moments about the CG (pivot) and letting A be the x-sectional area, Wwood (distance of center-of-mass of wood from pivot) = Wrubber (distance of center-of-mass of rubber from pivot) A(5.00 l)ρwood g(1.10 l) = A(l)ρrubber g(1.90 l) 5.00 ρwood (1.10) = ρrubber (1.90) 5.50 2.891.90rubberwood  Δ𝑃ሬ⃗ 𝑃ሬ⃗௜ 𝑃ሬ⃗௙ 5.00 l l 2.40 l 1.10 l 1.90 l 2.50 l

3 8 D Lines of action o