OxfordAQA International AS and A-level Physics: Momentum, impulse and collisions (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow section 3.2.6 (Momentum) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, which is shared AS and A-level content. Take g = 9.81 m s⁻² where needed. Learn the content first from the study guide or the revision notes. The course hub and checklist list the rest of the course.

Questions

1. A cyclist and bike of total mass 58 kg move at 6.5 m s⁻¹. Calculate the momentum, and the speed of a 900 kg car with the same momentum. [3]

2. A 0.60 kg trolley moving at 2.0 m s⁻¹ hits a stationary 0.90 kg trolley. They stick together. Calculate their common speed and the kinetic energy lost. [4]

3. A dodgem car of mass 210 kg moves right at 2.0 m s⁻¹ and hits a 160 kg car moving left at 1.5 m s⁻¹. The first car rebounds at 0.20 m s⁻¹ to the left.

(a) Calculate the velocity of the second car after the collision. [3] (b) Show that the collision is inelastic. [2]

4. A cannon of mass 1200 kg fires an 18 kg ball horizontally at 85 m s⁻¹. The cannon is free to recoil.

(a) Calculate the recoil speed. [2] (b) Compare the kinetic energies of ball and cannon and explain where the energy came from. [2]

5. A footballer kicks a stationary 0.44 kg ball, which leaves at 26 m s⁻¹. The boot touches the ball for 8.0 ms.

(a) Calculate the impulse on the ball. [2] (b) Calculate the average force. [2]

6. A 70 kg crash-test dummy travelling at 16 m s⁻¹ is stopped in 0.080 s by a belt and crumple zone.

(a) Calculate the average force on the dummy. [2] (b) Calculate the force if the stop took 0.020 s, and explain why crumple zones are used. [3]

7. A force acts on a 0.35 kg ball, initially at rest. It is recorded every 1.0 ms.

t / ms	0	1	2	3	4	5	6
F / N	0	300	700	900	650	250	0

(a) Use the trapezium rule to estimate the impulse. [3] (b) Calculate the final speed of the ball. [2] (c) Calculate the average force. [1]

8. A 1.8 kg item is dropped 0.50 m onto a foam base and stops in 25 ms. Calculate the force the base exerts on it. [4]

9. A hose directs 0.45 kg of water per second at 9.0 m s⁻¹ onto a wall. The water stops and runs down. Calculate the force on the wall. [3]

10. A 0.50 kg puck moving at 4.0 m s⁻¹ hits a stationary 1.5 kg puck head-on. After the collision the first puck moves back at 2.0 m s⁻¹ and the second moves forward at 2.0 m s⁻¹.

(a) Show that momentum and kinetic energy are both conserved. [4] (b) State how the speeds of approach and separation confirm that the collision is elastic. [1] (c) The pucks now stick together on impact instead. Calculate their speed and the kinetic energy lost. [3]

11. A 0.020 kg bullet moving at 300 m s⁻¹ embeds in a 1.98 kg block at rest on a smooth surface.

(a) Calculate the speed of the block and bullet together. [2] (b) Calculate the kinetic energy lost. [3] (c) The block is then stopped by a buffer in 0.050 s. Calculate the average force on the buffer. [2] (d) Explain why the force would be greater if the block rebounded. [1]

Answers

1. p = 58 × 6.5 = 377 kg m s⁻¹ [1]. v = p/m = 377 / 900 [1] = 0.42 m s⁻¹ [1] [3] Examiner insight: Give the unit with each answer; a correct number without its unit is incomplete.

2. Before: 0.60 × 2.0 = 1.2 kg m s⁻¹ [1]. After: (0.60 + 0.90)v = 1.2, so v = 0.80 m s⁻¹ [1]. KE before = ½ × 0.60 × 2.0² = 1.2 J, after = ½ × 1.5 × 0.80² = 0.48 J [1]. Lost = 0.72 J [1] [4] Examiner insight: Calculate the two energies separately and subtract; this is easier to credit than a single line of mixed working.

3. (a) Right positive. Before: 210 × 2.0 + 160 × (−1.5) = 420 − 240 = 180 [1]. After: 210 × (−0.20) + 160v = −42 + 160v [1]. v = 222 / 160 = +1.4 m s⁻¹, to the right [1] (b) KE before = ½ × 210 × 2.0² + ½ × 160 × 1.5² = 420 + 180 = 600 J [1]. KE after = 4.2 + ½ × 160 × 1.3875² = 158 J, less than 600 J, so inelastic [1] [5] Examiner insight: A “show that” needs numerical values for both energies and an explicit comparison; the statement alone is not enough.

4. (a) Ball direction positive. 0 = 18 × 85 + 1200v [1]. v = −1.3 m s⁻¹, so the cannon recoils at 1.3 m s⁻¹ [1] (b) KE ball = ½ × 18 × 85² = 65 000 J; KE cannon = ½ × 1200 × 1.275² = 980 J [1]. The energy comes from the chemical energy of the propellant, which is why KE increases although momentum is unchanged [1] [4] Examiner insight: Say which store the energy came from. Writing only “energy is conserved” does not explain why kinetic energy was zero before.

5. (a) Impulse = Δ(mv) = 0.44 × 26 − 0 [1] = 11 N s (11.44) [1] (b) F = impulse / Δt = 11.44 / 0.0080 [1] = 1.4 × 10³ N [1] [4] Examiner insight: The conversion from ms to s is the usual place to lose the answer by a factor of 1000.

6. (a) Δp = 70 × 16 = 1120 kg m s⁻¹ [1]. F = 1120 / 0.080 = 1.4 × 10⁴ N [1] (b) F = 1120 / 0.020 = 5.6 × 10⁴ N [1]. The momentum change is the same, but a longer stopping time [1] means a smaller rate of change of momentum, so a smaller force on the occupant [1] [5] Examiner insight: A strong explanation names both quantities: the change in momentum stays fixed while the time changes.

7. (a) Strips of 0.0010 s. Impulse = 0.0010 × [½(0 + 0) + 300 + 700 + 900 + 650 + 250] [1] = 0.0010 × 2800 [1] = 2.8 N s [1] (b) v = impulse / m = 2.8 / 0.35 [1] = 8.0 m s⁻¹ [1] (c) F = 2.8 / 0.0060 = 470 N [1] [6] Examiner insight: The average force is much lower than the 900 N peak. Check that your answer is plausible against the data.

8. v = √(2gh) = √(2 × 9.81 × 0.50) = 3.13 m s⁻¹ [1]. Δp = 1.8 × 3.13 = 5.64 kg m s⁻¹ [1]. Resultant force = 5.64 / 0.025 = 226 N upwards [1]. The base also supports the weight 1.8 × 9.81 = 17.7 N, so the force from the base = 226 + 17.7 = 240 N [1] [4] Examiner insight: If the question asks for the force from the surface, add the weight; if it asks for the resultant force, do not.

9. Momentum lost per second = 0.45 × 9.0 [1] = 4.05 kg m s⁻¹ per second [1]. F = Δ(mv)/Δt = 4.1 N, acting on the wall away from the hose [1] [3] Examiner insight: Write the mass per second explicitly; it shows you are using the rate of change of momentum rather than F = ma.

10. (a) Before: 0.50 × 4.0 = 2.0 kg m s⁻¹. After: 0.50 × (−2.0) + 1.5 × 2.0 = 2.0 kg m s⁻¹ [1], so momentum is conserved [1]. KE before = ½ × 0.50 × 16 = 4.0 J [1]. KE after = ½ × 0.50 × 4 + ½ × 1.5 × 4 = 1.0 + 3.0 = 4.0 J, so conserved [1] (b) Approach speed = 4.0 m s⁻¹ and separation speed = 2.0 − (−2.0) = 4.0 m s⁻¹, which are equal [1] (c) 2.0 = 2.0v, so v = 1.0 m s⁻¹ [1]. KE after = ½ × 2.0 × 1.0² = 1.0 J [1]. Lost = 4.0 − 1.0 = 3.0 J [1] [8] Examiner insight: Show both totals for each conserved quantity. A sign slip on the rebounding puck is the usual reason the totals fail to match.

11. (a) 0.020 × 300 = 6.0 kg m s⁻¹ [1]. v = 6.0 / 2.0 = 3.0 m s⁻¹ [1] (b) KE before = ½ × 0.020 × 300² = 900 J [1]. KE after = ½ × 2.0 × 3.0² = 9.0 J [1]. Lost = 891 J [1] (c) Δp = 2.0 × 3.0 = 6.0 kg m s⁻¹ [1]. F = 6.0 / 0.050 = 120 N [1] (d) A rebound makes the change in momentum larger (from +mv to −mv), so the same contact time needs a larger force [1] [8] Examiner insight: In (d) compare the changes in momentum, not just the final speeds.

Where marks are usually lost

• Using one sign for velocities in opposite directions.
• Leaving out the direction of a velocity, or quoting only a speed when direction is wanted.
• Forgetting to convert milliseconds to seconds.
• Using the speed rather than the velocity change when a body rebounds.
• Assuming kinetic energy is conserved in an inelastic collision or explosion.
• Stating the answer without comparing the two kinetic energy totals when asked to show a collision is elastic.
• Using the peak force of a graph as the average force.
• Leaving out weight when a vertical impact asks for the force from the surface.
• Explaining crumple zones in terms of “absorbing” force instead of increasing time.

Next steps

• Review the revision notes and study guide.
• Tick off the topic on the printable checklist and visit the course hub.
• Take one of the free 10-minute diagnostics to check what to revise next.
• Book a free trial class if you want help with any of these questions.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Section 3.2.6 Momentum.