HKDSE Physics Practice Paper: Mechanics, Momentum And Energy (12 Questions, 100 Marks, 75 Minutes)

HKDSE Physics Practice Paper: Mechanics, Momentum and Energy (12 Questions, 100 Marks, 75 Minutes)

2026-11-04

HKDSE Physics: Mechanics, Momentum and Energy

HKDSE Physics paper on mechanics, momentum and energy: 12 questions, 100 marks, 75 minutes, Sections A to C, with full worked answers and marking points.

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Paper details　Level: S4-S6 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Kinematics and Newton’s Laws of Motion (Questions 1-4, 32 marks)

Each question carries 8 marks; state the formula used, substitute the values with units, and give answers to three significant figures.

#	Question	Answer and marking notes
1	A car starts from rest and accelerates uniformly, reaching 20 m/s in 8 s. It then travels at that speed for 12 s and finally decelerates uniformly to rest in 5 s. (a) Calculate the acceleration in the first stage. (b) Calculate the total distance travelled and give the distance of each of the three stages. (c) Calculate the magnitude of the deceleration and explain how braking distance depends on speed.	(a) a = (v – u) / t = (20 – 0) / 8 = 2.5 m/s². (b) Accelerating stage = 0.5 x 20 x 8 = 80 m; constant-speed stage = 20 x 12 = 240 m; decelerating stage = 0.5 x 20 x 5 = 50 m; total = 80 + 240 + 50 = 370 m. (c) Deceleration = 20 / 5 = 4.0 m/s², opposite to the motion. Braking distance is proportional to the square of the speed (s = v² / 2a), so doubling the speed roughly quadruples the braking distance.
2	The car in the previous question has a mass of 1200 kg and experiences a friction force of 800 N. (a) Calculate the net force on the car while it accelerates. (b) Calculate the driving force the engine must provide during that stage. (c) Calculate the kinetic energy at 20 m/s; by what factor does it change if the speed doubles? State the effect on braking distance.	(a) Net force F = ma = 1200 x 2.5 = 3,000 N, in the direction of motion. (b) Driving force = net force + friction = 3,000 + 800 = 3,800 N. (c) Kinetic energy = 0.5mv² = 0.5 x 1200 x 20² = 240,000 J. At double the speed it is 0.5 x 1200 x 40² = 960,000 J, i.e. 4 times as much; with the same braking force the braking distance is also about four times longer.
3	A stone is dropped from rest from the roof of a building 45 m high. Take g = 10 m/s² and ignore air resistance. (a) Calculate the time of fall. (b) Calculate the speed on hitting the ground. (c) If a stone three times as massive were used, would the time of fall change? Explain using Newton’s second law.	(a) From s = 0.5gt², t = sqrt(2s / g) = sqrt(2 x 45 / 10) = 3.0 s. (b) v = gt = 10 x 3.0 = 30 m/s downwards. (c) No change. For free fall a = F / m = mg / m = g, independent of mass; a stone three times as massive has three times the weight but still accelerates at g, so the time is the same.
4	A box of mass 20 kg rests on a frictionless incline of angle 30°. Take g = 10 m/s². (a) Calculate the component of weight along the incline. (b) Calculate the normal reaction from the incline, to three significant figures. (c) Calculate the acceleration down the incline and explain why the normal reaction does not affect it.	(a) Component along the incline = mg sin(theta) = 20 x 10 x sin 30° = 100 N. (b) Normal reaction N = mg cos(theta) = 20 x 10 x cos 30° = 173.2 N. (c) a = g sin(theta) = 10 x 0.5 = 5.0 m/s². The normal reaction is perpendicular to the incline and has no component along it, so without friction it cannot affect the acceleration.

Section B: Momentum, Collisions and Circular Motion (Questions 5-8, 36 marks)

Each question carries 9 marks; state the direction of every vector and quote the conservation law before and after the calculation.

#	Question	Answer and marking notes
5	Trolley A of mass 2 kg moving at 3 m/s collides with a stationary trolley B of mass 3 kg. They stick together and move with a common velocity. (a) Use conservation of momentum to find the common velocity. (b) Calculate the total kinetic energy before and after the collision. (c) Calculate the percentage of kinetic energy lost and state whether the collision is elastic.	(a) Momentum is conserved: 2 x 3 = (2 + 3)v, so v = 6 / 5 = 1.2 m/s, in the original direction of A. (b) Total kinetic energy before = 0.5 x 2 x 3² = 9.0 J; after = 0.5 x 5 x 1.2² = 3.6 J. (c) Loss = 9.0 – 3.6 = 5.4 J, that is 60%. Kinetic energy is not conserved, so the collision is inelastic (perfectly inelastic, since they move together).
6	On a smooth horizontal track, ball A of mass 4 kg moving at 6 m/s hits a stationary ball B of mass 2 kg. The collision is perfectly elastic. (a) Write down the equations for conservation of momentum and of kinetic energy. (b) Calculate the velocities of A and B after the collision, stating their directions. (c) If the two balls had equal masses, explain how the velocities would be exchanged.	(a) Momentum: 4 x 6 = 4v_A + 2v_B, so 24 = 4v_A + 2v_B. Kinetic energy: 0.5 x 4 x 6² = 72.0 J = 0.5 x 4v_A² + 0.5 x 2v_B². (b) Solving gives v_A = 2.0 m/s (original direction) and v_B = 8.0 m/s (original direction). Check momentum: 4 x 2.0 + 2 x 8.0 = 24 kg m/s, unchanged. (c) With equal masses the balls exchange velocities: A stops and B moves off at the original speed of A.
7	A ball of mass 0.5 kg hits a wall perpendicularly at 10 m/s and rebounds at 8 m/s. The contact time is 0.06 s. (a) Calculate the impulse on the ball and state its direction. (b) Calculate the average force exerted by the wall on the ball. (c) Calculate the average acceleration during contact; how would the impulse change if the ball stopped dead on the wall?	(a) Taking the original direction as positive: impulse = m(v – u) = 0.5 x ((8) – (10)) = -9.0 N s. Magnitude 9.0 N s, directed away from the wall, opposite to the original motion. (b) Average force F = impulse / time = 9.0 / 0.06 = 150 N, same direction as the impulse. (c) Average acceleration = change in velocity / time = (8 + 10) / 0.06 = 300 m/s². If the ball stopped dead, the change in velocity would be only 10 m/s and the impulse 5.0 N s, about half as large.
8	An object of mass 0.8 kg moves in a circle of radius 1.2 m at a constant speed of 3 m/s. (a) Calculate the centripetal acceleration. (b) Calculate the centripetal force required and state its direction. (c) Calculate the period of the motion to three significant figures; by what factor does the force change if the speed doubles?	(a) Centripetal acceleration a = v² / r = 3² / 1.2 = 7.5 m/s². (b) F = ma = 0.8 x 7.5 = 6.0 N, directed towards the centre (perpendicular to the velocity). (c) Period T = 2 pi r / v = 2 pi x 1.2 / 3 = 2.51 s. If the speed doubles, v² becomes four times larger, so the force becomes 24.0 N, four times the original.

Section C: Work, Power and Conservation of Energy (Questions 9-12, 32 marks)

Each question carries 8 marks; show the useful energy and the total input energy separately, and support each conclusion with the figures given.

#	Question	Answer and marking notes
9	A worker pushes a crate along horizontal ground with a horizontal force of 250 N through a distance of 12 m in 15 s. (a) Calculate the work done on the crate by the worker. (b) Calculate the average power delivered. (c) If the worker’s efficiency is 80%, calculate the power he actually supplies and explain where the rest of the energy goes.	(a) Work W = Fs = 250 x 12 = 3,000 J. (b) Average power P = W / t = 3,000 / 15 = 200 W. (c) Power supplied = 200 / 0.80 = 250 W; the remaining 50 W (20%) is lost as thermal and sound energy in the worker’s body and in friction with the ground.
10	An object of mass 2.5 kg is raised through 6 m. Take g = 10 m/s². (a) Calculate the gain in gravitational potential energy. (b) If the object is released from rest at that height, calculate its speed on reaching the ground, to three significant figures. (c) Explain why the actual landing speed is usually smaller than the answer to (b).	(a) Gain in potential energy = mgh = 2.5 x 10 x 6 = 150 J. (b) From mgh = 0.5mv², v = sqrt(2gh) = sqrt(2 x 10 x 6) = 11.0 m/s. (c) Air resistance does negative work, so part of the potential energy becomes thermal and sound energy; the kinetic energy and hence the speed on landing are smaller.
11	A crane lifts a load of 400 kg through 15 m in 25 s at constant speed. Take g = 10 m/s² and the input power of the crane is 3000 W. (a) Calculate the useful work done on the load. (b) Calculate the useful power output. (c) Calculate the efficiency of the crane to three significant figures and suggest one way to improve it.	(a) Useful work = mgh = 400 x 10 x 15 = 60,000 J. (b) Useful power = 60,000 / 25 = 2,400 W. (c) Efficiency = 2,400 / 3,000 x 100% = 80.0%. Improve it by reducing the mass of the jib and cable and by lubricating the pulley bearings to cut friction losses.
12	A block of mass 4.0 kg slides from rest down a rough incline of angle 30° from a height of 5 m. Its actual speed at the bottom is 8.0 m/s. Take g = 10 m/s². (a) Calculate the speed at the bottom if the incline were frictionless. (b) Calculate the work done against friction. (c) Calculate the average friction force and explain how a rougher incline changes the final speed.	(a) From mgh = 0.5mv², v = sqrt(2gh) = sqrt(2 x 10 x 5) = 10.0 m/s. (b) Length of incline = 5 / sin 30° = 10.0 m; work done against friction = 0.5 x 4.0 x 10.0² – 0.5 x 4.0 x 8.0² = 72 J. (c) Average friction f = 72 / 10.0 = 7.2 N. A rougher incline means more work is done against friction, so the speed at the bottom is lower.

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