CJC 2024 A level H2 Physics Answers

2024_A_Level_Solutions prepared by CJC 1 H2 Physics 9749 – 2024 A Level Exam Paper 1 1 Essential Question(s):  How to calculate the area of the circular orbit of the Moon around the Earth? Solution: The radius of the circular orbit is 4 × 108 m. The area of the circular orbit is π × (4 × 108)2 = 50 × 1016 m2 = 0.5 × 1018 m2 = 0.5 × (109 m)2 = 5 × 10−1 (Gm)2 Answer: A 2 Essential Question(s):  How to relate velocity (goal) to the a-t graph (given info)? Solution: Change in velocity = Area under a-t graph Note that the positive area (positive change in velocity) is lesser than the negative area (negative change in velocity). Answer: B 3 Essential Question(s):  What kinematics equation can be used to determine the time of flight of a projectile?  What do I know about v y or sy for a 2-D projectile motion where the object returns to the initial height? Solution: Resistance from the gases on the Moon is negligible. Take upwards as positive direction. Consider the vertical component of the initial speed u y = 15 sin 40° and the final speed of the vertical component vy = 0 when the golf reaches its maximum height. Using vy = uy + ayt, t = (vy – uy) ÷ ay = (0 – 15 sin 40°) ÷ (-1.62) = 5.95 s Time of flight of the golf ball = 2 × t = 11.9 s OR At maximum height, sy = 0 sy = uyt + ½ayt2 0 = (15 sin 40°)t + ½ (-1.62)t2 t = 0 (at the start) or t = 11.9 s (returns to the ground) Answer: C

2024_A_Level_Solutions prepared by CJC 2 4 Essential Question(s):  What is the force vector diagram that represents the equilibrium of forces acting on the cylinder? Solution: Smooth  No friction. Only contact force by the planes on the cylinder is the normal contact force. The angle between the planes X and Y is 90°. The force acting on the cylinder by plane X, FX, points to the centre of the cylinder at 30° from the vertical. The force acting on the cylinder by plane Y, FY, points to the centre of the cylinder at 60° from the vertical. The force acting on the cylinder by gravity, W, points vertically downward. For 3 non-parallel coplanar forces in equilibrium, their lines of action should intersect at a common point, and, these 3 forces when joined head-to-tail should form a closed polygon vector diagram that represents a zero net force. Answer: B 5 Essential Question(s):  What is the change in the setup?  What determines the extension of each spring?  What is the original force in each spring? What is the new force in each spring? Solution: The 3 springs are identical. From the diagram, the force acting on each spring is W/3 and the extension of each spring is x. When the middle spring is removed and a weight of 2W is hanging from the middle of the beam, the force acting on each of the spring is (2W) / 2 = W which is 3 times of W/3. Hence the extension of each spring should also be 3 times of x. OR Before: 3kx = W ---(1) After: 2ky = 2W ---(2) where y is the new extension (2)/(1): (2y)/(3x) = 2 y = 3x Most candidates did not realise there were only two springs used in the final setup. Answer: D 6 Essential Question(s):  What is the area from the tensile load-extension graph that represents the net work done on the specimen? Solution: Work done by force on the elastic body, f i x x W F dx  When the tensile load is increased from zero to the maximum value at point Q on the tensile load-extension graph, the total work done on the specimen is represented by the area OPQT. When the tensile load is reduced to zero at point S on the tensile load-extension graph, negative work is done on the specimen because extension is reduced and is represented by the area SRQT. Hence the net work done on the specimen during this process = area OPQT – area SRQT = area OPRS. Answer: B

2024_A_Level_Solutions prepared by CJC 3 7 Essential Question(s):  What determines the power output of the car engine?  What are the forces on the car? / What is the energy transformation as the car moves up the slope? Solution: The power provided by the engine of the car to do work against resistive force = F × v = 600 × 16 = 9600 W Per second the car moves 16 m along the slope, i.e. a height of 16 sin 10°. The gain in gravitational potential energy per second = 900 × 9.81 × 16 sin 10° = 24530 J The power provided by the car engine in increasing the gravitational potential energy of the car = 24530 W The total power provided by the car engine = 24530 + 9600 = 34130 W = 34.13 kW Most candidates did not realise the power provided by the engine is also used to increase the gravitational potential energy of the car. Answer: C 8 Essential Question(s):  Given a p-V graph, how to tell when there is work done BY gas, and, when there is work done ON gas?  Expansion (V increases) implies work done BY gas, and, Compression (V decreases) implies work done ON gas. Solution: The change from P to Q involves a drop in the gas pressure while its volume increases . During this change work is done by the gas. The change from Q to R involves a drop in the gas pressure while its volume increases. During this change work is done by the gas. The change from R to S involves an increase in the gas pressure while its volume is reduced. During this change work is done on the gas. The change from S to P involves an increase in the gas pressure while its volume is reduced. During this change work is done on the gas. Answer: D 9 Essential Question(s):  How to calculate angular velocity of a particle moving through a circular arc?  d dt   , rate of change of angular displacement. Solution: The angle the point mass moves through an arc of length l = l ÷ r, where r is the radius of the circular arc. Angular velocity = change in angle ÷ change in time = (l ÷ r) ÷ t = l ÷ (r × t), where t is change in time. Answer: A 10 Essential Question(s):  How to calculate centripetal acceleration?

2024_A_Level_Solutions prepared by CJC 4 Solution: The angular velocity of the object ω = 2π ÷ (24 x 60 × 60) s−1 The centripetal acceleration of the object = rω2 = (1.3 × 107 ÷ 2) × (2π ÷ (24 x 60 × 60))2 = 0.034 m s−2 Answer: A 11 Essential Question(s):  What determines the gravitational potential at a point from a point mass? How would the gravitational potential of two points P and Q change when they move away from a point mass?  ϕ = -GM/r  For the same change in distance away from the point mass, is the change in gravitational potential lesser, greater or same when r is greater compared to when r is smaller? What is the shape of the ϕ-r graph?  hyperbolic  ∆ϕ is lesser for the same ∆r when r is greater. Solution: As the magnitude of gravitational potential of a point is inversely proportional to its distance from a point mass, the magnitude of gravitational potential of both P and Q decreases when they move away from the point mass. Next, consider the shape of the ϕ-r graph  hyperbolic  ∆ϕ is lesser for the same ∆r when r is greater. Answer: A 12 Essential Question(s):  What determines the average force exerted?  How can the kinetic theory of gases & Newton’s laws of motion be applied to determine the force? Solution: Ideal gas assumption  collision is perfectly elastic, i.e. total KE conserved, so rebound speed is the same as v. Change in the particle’s momentum ∆p = 2mv. The time taken between two successive collisions with the same shaded wall, ∆t = (2x) ÷ v where x is the distance between two sides of the box in the x direction. By Newton’s 2nd law, average force exerted on the particle by the side of the box = rate of change of the particle’s momentum = ∆p ÷ ∆t = mv2 ÷ x. By Newton’s 3rd law, the force exerted on the side of the box by the particle = the force exerted on the par