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 (vQ) 8 – 6 = 2 vp + 3 vQ 2 vp + 3 vQ = 2 --- (1) For elastic collision, relative speed of approach = relative speed of separation: 4.0-(-2.0) = vQ- vP vQ – vP = 6 3 vQ – 3 vP = 18 --- (2) (1) – (2): 5 vP = -16 vP = - 3.2 m s-1 Speed of vP = 3.2 m s-1

10 B Taking moments about the 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(4.00 l)ρwood g(0.90 l) = A(l)ρrubber g(1.10 l + 0.50 l) 4.00 ρwood (0.90) = ρrubber (1.60)    2.25rubber wood 11 D A solid body experiencing three forces, whose lines of action are not parallel is in equilibrium if 1. The lines of action are coplanar 2. The vector sum of the forces is equal to the zero vector The lines of action intersect at a point 12 C The force to compress the spring is equal to the weight of the passengers, F = W = m g = (450) (9.81) = 4414.5 N. The compression of the spring is equal to the difference in height, x = 0.10 m. Hence, the spring constant is k = F / x = (4414.5 N) / (0.10 m) = 44100 N m-1 (to 3 s.f.). shelf metal cable pivot wall W Lines of action of the forces intersect at this point T 2.90 l 1.10 l l

13 D No kinetic energy at drop-point and at maximum compression, x. Comparing the total energy at the initial and final positions, gain in elastic potential energy = total loss in gravitational potential energy   21 0.152 kx mg x   21(85) (0.20 9.81) 0.152 x x   242.5 1.962 0.2943 0x x   0.109 mx  14 C At maximum speed, engine force = drag force of kv. Power of boat, P = (engine force) v = kv2 Hence, 2 one engine one engine two engines two engines P v P v            2 one engine32 64 14 v  -1 one engine 9 9 m sv . 15 D The useful work done by the engine is the increase in kinetic energy, ∆KE = KE୤୧୬ୟ୪ − KE୧୬୧୲୧ୟ୪ = ଵ ଶ݉ݒଶ − ଵ ଶ݉ݑଶ = ݉ ൫௩మି ௨మ൯ ଶ

16 A There are two forces acting on the person, the force by cage on him, R, and his weight, W. Since the man is in uniform (constant speed) circular motion, the net force on him is directed toward the centre of the circle, i.e., towards the right. Hence, the vector sum of R and W must also point towards the right. 17 C Option A: Satellites can only remain vertically above fixed points on the equator. Option B: Geostationary satellites have to remain vertically above a fixed point on the Earth. Hence, the angular speed ω is equal to the speed of a point on the Earth’s equator . Since v = r ω, this means that the linear speed v cannot be equal to the speed of a point on the Earth’s equator. Option C: This is correct. Option D: The satellite’s motion has to match that of the Earth. Since the Earth is rotating from west to east (which is why the Sun is “rising” in the east: the Earth is rotating in that direction to meet it), geostationary satellites must be travelling from west to east. 18 B The gravitational force between the Sun and the Earth provides the centripetal force. ܨ୥ = ܨୡ ܩ݉ܯ ݎଶ =݉൬2ߨ ܶ൰ ଶ ݎ ܯ= 4ߨଶݎଷ ܩܶଶ = 4ߨଶ(1.50 × 10ଵଵ)ଷ (6.67 × 10ିଵଵ)(365 × 24 × 3600)ଶ = 2.01 × 10ଷ଴ kg where G is the gravitational constant, M is the mass of the Sun, m is the mass of the Earth, r is the distance between the Sun and the Earth, and T is the period of rotation. 19 D Volume V is constant, express cross-sectional area A in terms of L : AL = V  A = V/L Hence, the resistance variation with L is given by : 2 2R L LA V L    20 B Resistance of each of the lamp : 2 1rating rating rating VR P P  since, both of them have the same voltage rating: , , 40 4 10 1 rating BA B rating A PR R P   Since, both lamps are connected in series, the same current passes both lamps and power emitted by each lamp : 2 emittedP I R  .    , , 4 41 emitted A A A B emitted R B P R P PP R

21 D e,m,f, E is defined as the work done per unit charge in driving it through the entire circuit, including the cell and external circuit. Electrical energy is converted to other forms of energy. EQ therefore represents the energy dissipated to other forms of energy in the cell and external circuit. 22 C When no current is flowing, the potential at point Z is equal to the potential at the negative terminal of the source. Hence, the voltmeter will measure zero potential difference between point Z and the negative terminal. Similarly, when no current is flowing, the potentials at X and Y are equal to the potential at the positive terminal of the source. Hence, when connected at X or Y, the voltmeter will measure the potential difference over the sources, which is equal to its e.m.f. of 12 V. 23 B The effective resistance of the parallel part is 10 / 2 = 5 Ω. The effective resistance of the whole circuit is 10 + 5 = 15 Ω. The voltmeter will measure the potent