Free Momentum questions for IGCSE Physics (0625), Extended tier (Supplement content), part of Motion, forces & energy. Each question has the answer and the marking points an examiner would look for.
Generate a full Momentum worksheet →All Motion, forces & energy questionsA ball of mass 0.5 kg hits a wall at 12 m/s and rebounds along the same line at 10 m/s. It is in contact with the wall for 0.02 s.
What is the average force exerted by the wall on the ball?
Answer: D
What is the unit of impulse?
Answer: A
Car designers use crumple zones and airbags to reduce injuries in crashes.
Explain, using impulse, how a crumple zone reduces the force on a passenger.
A passenger of mass 70 kg is stopped from 15 m/s in 0.50 s. Calculate the average force on the passenger.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | longer time for same change in momentum the change in momentum (impulse) is the same B1 the crumple zone increases the time taken to stop B1 force = change in momentum ÷ time, so the force is smaller B1 | 3 |
| (b) | 2100 N F = Δp ÷ t = 70 × 15 ÷ 0.50 C1 2100 N A1 | 2 |
A trolley of mass 2 kg moving at 6 m/s collides with a stationary trolley of mass 4 kg. They stick together.
State the principle of conservation of momentum.
Calculate the speed of the trolleys after the collision.
Show that kinetic energy is not conserved.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | total momentum unchanged if no external force total momentum before = total momentum after, if no external forces act B1 | 1 |
| (b) | 2 m/s momentum before = 12 kg m/s C1 12 = 6v C1 2 m/s A1 | 3 |
| (c) | compare KE before and after KE before = 36 J B1 KE after = 12 J, which is less B1 | 2 |
Define momentum and impulse.
Define momentum and state its unit.
State the relationship between resultant force and momentum.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | mass × velocity, kg m/s momentum = mass × velocity B1 kg m/s B1 | 2 |
| (b) | F = change in momentum ÷ time resultant force = rate of change of momentum B1 | 1 |
A trolley A of mass 2.00 kg moves at 0.80 m/s towards a stationary trolley B of mass 0.40 kg. The trolleys collide and stick together.
Calculate the momentum of trolley A before the collision.
Use your answer to (a) to calculate the speed of the trolleys immediately after the collision.
Use your answer to (b) to show that kinetic energy is not conserved in the collision.
The collision lasts 0.100 s. Use your answer to (b) to calculate the average force exerted on trolley B.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.60 kg m/s p = mv = 2.00 × 0.80 C1 1.60 kg m/s A1 | 2 |
| (b) | 0.667 m/s 1.60 = (2.00 + 0.40)v C1 0.667 m/s A1 | 2 |
| (c) | KE before 0.640 J, after 0.533 J KE before = ½ × 2.00 × 0.80² = 0.640 J C1 KE after = 0.533 J C1 KE after is less than KE before A1 | 3 |
| (d) | 2.67 N impulse on B = 0.40 × 0.667 = 0.267 N s C1 F = 0.267/0.100 = 2.67 N A1 | 2 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in