2024 HCI H1 Physics Paper 1 and Paper 2 Solutions

Hwa Chong Institution H1 Physics Prelim Exam Suggested Solutions Paper 1 1 2 3 4 5 6 7 8 9 10 D D B D D D C D C B 11 12 13 14 15 16 17 18 19 20 D C D C D A C B D B 21 22 23 24 25 26 27 28 29 30 D C B B D A D C A B 1 D Temperature difference = ߠ௙ − ߠ௜ = 80 – 20 = 60 °C Absolute uncertainty, = ∆ߠ௙ + ∆ߠ௜ = 1 °C Percentage uncertainty =    100% 1.7%1 C 60 C 2 D Let [x] denote the units of x. [intensity] = [power] / [area] = [energy] / [time x area] = J s-1 m-2 = N m s-1 m-2 = kg m s-2 m s-1 m-2 Hence, the unit of intensity expressed in SI base units is kg s-3. 3 B A is incorrect, as energy is a scalar. B is correct. C is incorrect, as power is a scalar. D is incorrect, as energy is a scalar. 4 D The object starts from rest and moves along a straight line. Area under the a-t graph represent 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 the negative direction). This happens up to the time at B. Beyond B, the acceleration changes direction (from negative to positive direction) However, the velocity still points in the negative direction after B i.e. object continues to travel in the same direction (still getting further away from the starting point). Since acceleration and velocity point in opposite direction, this object is slowing down. The upper triangle BCD represents the change in velocity from B onward, i.e., the velocity is decreasing in magnitude. Since the area of BCD is smaller than the area of the other triangle, the velocity of the object sti ll points in the same direction up to the time at D.

5 D ܵ௕ − ܵ௘ = 2 2.0 = [(0) ݐ + (0.5)(9.81) ݐଶ] − [(0) ݐ + (0.5)(5.8) ݐଶ] ݐ =1.00 s 6 D The horizontal velocity is constant at 10.0 cos 30° = 8.6603 m s-1. Hence, the ball takes t = (10.0 m) / (8.6603 m s-1) = 1.1547 s to cover the distance to the wall. For the vertical velocity, v = u + a t = (10.0 sin 30°) + (-9.81) (1.1547) = -6.3276 m s-1. This means that, when the ball hits the wall, it is travelling at a downward angle. The speed when the ball hits the wall is [(8.6603)2 + (-6.3276)2]1/2 = 10.7 m s-1 7 C The thrust must provide an upward force to propel the rocket upward. This force must be least at equal to its weight. Thrust = ݒቀௗ௠ ௗ௧ ቁ ݒቀௗ௠ ௗ௧ ቁ = ݃ܯ ௗ௠ ௗ௧ = ெ௚ ௩ = (ହ଴଴)(ଽ.଼ଵ) ଵ଴଴଴ = 4.9 kg sିଵ 8 D Momentum and total energy are always conserved. However, some of the initial kinetic energy of ball X might be converted to other forms of energy, e.g., in the production of sound. 9 C Assume velocity of P and Q immediately after the collision is vP and vQ respectively in the direction to the right. Conservation of linear momentum: 2.0 (4.0) + 3.0 (-2.0) = 2.0 (vP) + 3.0 (v