IGCSE Physics 1.6 Momentum MCQs – 30 Practice Questions with Answers (Cambridge 0625)

These 30 original practice questions cover Section 1.6 Momentum of the Cambridge IGCSE Physics (0625) syllabus, from p = mv through conservation of momentum to impulse and safety features. Momentum is a Supplement topic in 0625, so the questions are tagged [Supplement] and are aimed at students taking the extended papers.

Formula recap: momentum p = m × v  |  total momentum before a collision = total momentum after (no external force)  |  impulse = F × t = change in momentum  |  resultant force = rate of change of momentum. Full list: IGCSE Physics Formula Sheet

Momentum Basics: p = mv

Q1. [Supplement] The momentum of a moving object is calculated from

  • (A) mass ÷ velocity
  • (B) mass × velocity
  • (C) mass × acceleration
  • (D) velocity ÷ mass
Show Answer

Answer: B – Momentum p = m × v. It is measured in kg m/s.

Q2. [Supplement] What is the unit of momentum?

  • (A) newton (N)
  • (B) joule (J)
  • (C) kilogram metre per second (kg m/s)
  • (D) metre per second squared (m/s2)
Show Answer

Answer: C – Momentum = mass (kg) × velocity (m/s), so its unit is kg m/s. The newton second (N s) is the same unit, used for impulse.

Q3. [Supplement] A lorry of mass 1500 kg travels at 12 m/s. What is its momentum?

  • (A) 18 000 kg m/s
  • (B) 125 kg m/s
  • (C) 1512 kg m/s
  • (D) 1800 kg m/s
Show Answer

Answer: A – p = m × v = 1500 × 12 = 18 000 kg m/s.

Q4. [Supplement] A cricket ball of mass 0.15 kg is bowled at 20 m/s. Its momentum is

  • (A) 0.0075 kg m/s
  • (B) 133 kg m/s
  • (C) 30 kg m/s
  • (D) 3.0 kg m/s
Show Answer

Answer: D – p = m × v = 0.15 × 20 = 3.0 kg m/s.

Q5. [Supplement] Which object has the greatest momentum?

  • (A) 2 kg moving at 3 m/s
  • (B) 5 kg moving at 2 m/s
  • (C) 1 kg moving at 8 m/s
  • (D) 4 kg moving at 2 m/s
Show Answer

Answer: B – The momenta are A: 6, B: 10, C: 8 and D: 8 kg m/s. The 5 kg object at 2 m/s has the greatest momentum, 10 kg m/s.

Q6. [Supplement] Momentum is a vector quantity. This is because

  • (A) velocity is a vector, so momentum has both magnitude and direction
  • (B) momentum is always a positive quantity
  • (C) mass is a vector quantity
  • (D) momentum depends only on the speed of the object
Show Answer

Answer: A – Momentum = mass (a scalar) × velocity (a vector), so momentum itself is a vector with magnitude and direction.

Q7. [Supplement] Taking east as the positive direction, a 2 kg trolley moves west at 4 m/s. Its momentum is

  • (A) +8 kg m/s
  • (B) −4 kg m/s
  • (C) −8 kg m/s
  • (D) −2 kg m/s
Show Answer

Answer: C – West is the negative direction, so v = −4 m/s and p = 2 × (−4) = −8 kg m/s.

Q8. [Supplement] Trolley A (3 kg) moves east at 2 m/s and trolley B (2 kg) moves west at 3 m/s. Taking east as positive, the total momentum of the two trolleys is

  • (A) 12 kg m/s
  • (B) 6 kg m/s
  • (C) −6 kg m/s
  • (D) 0 kg m/s
Show Answer

Answer: D – pA = 3 × 2 = +6 kg m/s and pB = 2 × (−3) = −6 kg m/s. Total = 6 + (−6) = 0 kg m/s.

Q9. [Supplement] An object has momentum 45 kg m/s and moves at 9 m/s. Its mass is

  • (A) 405 kg
  • (B) 5 kg
  • (C) 0.2 kg
  • (D) 36 kg
Show Answer

Answer: B – m = p ÷ v = 45 ÷ 9 = 5 kg.

Q10. [Supplement] A sprinter of mass 60 kg has momentum 240 kg m/s. At what speed is the sprinter running?

  • (A) 4 m/s
  • (B) 0.25 m/s
  • (C) 14 400 m/s
  • (D) 180 m/s
Show Answer

Answer: A – v = p ÷ m = 240 ÷ 60 = 4 m/s.

Conservation of Momentum and Collisions

Q11. [Supplement] The total momentum of a system of colliding objects stays constant provided that

  • (A) the collision is perfectly elastic
  • (B) the objects stick together after colliding
  • (C) no external resultant force acts on the system
  • (D) kinetic energy is conserved in the collision
Show Answer

Answer: C – Momentum is conserved in every collision (elastic or inelastic) as long as no external resultant force acts. Kinetic energy is only conserved in elastic collisions.

Q12. [Supplement] A 2 kg trolley moving at 3 m/s collides with a stationary 4 kg trolley and they stick together. Their common velocity is

  • (A) 1 m/s
  • (B) 1.5 m/s
  • (C) 2 m/s
  • (D) 3 m/s
Show Answer

Answer: A – Momentum before = 2 × 3 = 6 kg m/s. Combined mass = 6 kg, so v = 6 ÷ 6 = 1 m/s.

Q13. [Supplement] A 5 kg trolley moving at 4 m/s hits a stationary 5 kg trolley and they join together. They move off at

  • (A) 0.5 m/s
  • (B) 1 m/s
  • (C) 4 m/s
  • (D) 2 m/s
Show Answer

Answer: D – Momentum before = 5 × 4 = 20 kg m/s. Combined mass = 10 kg, so v = 20 ÷ 10 = 2 m/s.

Q14. [Supplement] A 3 kg trolley moving at +2 m/s meets a 3 kg trolley moving at −1 m/s (towards it). They stick together. Their velocity after the collision is

  • (A) 0 m/s
  • (B) +0.5 m/s
  • (C) +1.5 m/s
  • (D) +3 m/s
Show Answer

Answer: B – Total momentum = (3 × 2) + (3 × −1) = 6 − 3 = +3 kg m/s. Combined mass = 6 kg, so v = 3 ÷ 6 = +0.5 m/s, in the direction of the first trolley.

Q15. [Supplement] A cannon of mass 500 kg fires a 5 kg shell at 100 m/s. The recoil speed of the cannon is

  • (A) 0.01 m/s
  • (B) 0.1 m/s
  • (C) 1 m/s
  • (D) 10 m/s
Show Answer

Answer: C – Total momentum starts at zero. Shell momentum = 5 × 100 = 500 kg m/s, so the cannon must gain 500 kg m/s backwards: v = 500 ÷ 500 = 1 m/s.

Q16. [Supplement] A 60 kg ice skater standing still throws a 3 kg ball forwards at 8 m/s. The skater recoils at

  • (A) 0.04 m/s
  • (B) 4 m/s
  • (C) 2.5 m/s
  • (D) 0.4 m/s
Show Answer

Answer: D – Ball momentum = 3 × 8 = 24 kg m/s. The skater gains equal momentum backwards: v = 24 ÷ 60 = 0.4 m/s.

Q17. [Supplement] A stationary 6 kg object explodes into two parts: a 2 kg part and a 4 kg part. The 2 kg part flies off at 9 m/s. The speed of the 4 kg part is

  • (A) 4.5 m/s
  • (B) 9 m/s
  • (C) 3 m/s
  • (D) 18 m/s
Show Answer

Answer: A – Total momentum was zero, so the parts have equal and opposite momenta. Momentum of the 2 kg part = 2 × 9 = 18 kg m/s, so v = 18 ÷ 4 = 4.5 m/s in the opposite direction.

Q18. [Supplement] Which quantity is conserved in an elastic collision but is NOT conserved in an inelastic collision?

  • (A) Momentum
  • (B) Kinetic energy
  • (C) Mass
  • (D) Total energy
Show Answer

Answer: B – Momentum is conserved in both types of collision. Kinetic energy is conserved only in elastic collisions; in inelastic collisions some kinetic energy changes into other forms (heat, sound, deformation).

Q19. [Supplement] Two cars collide and lock together, moving off as one. Which statement about the collision is correct?

  • (A) Kinetic energy is conserved but momentum is not
  • (B) Neither momentum nor kinetic energy is conserved
  • (C) Momentum is conserved but kinetic energy is not
  • (D) Both momentum and kinetic energy are conserved
Show Answer

Answer: C – Objects sticking together is the extreme case of an inelastic collision: momentum is conserved (no external force), but a large share of the kinetic energy is lost to deformation, heat and sound.

Q20. [Supplement] A 1 kg trolley moving at 6 m/s catches up with a 3 kg trolley moving at 2 m/s in the same direction. They stick together and move at

  • (A) 2 m/s
  • (B) 4 m/s
  • (C) 6 m/s
  • (D) 3 m/s
Show Answer

Answer: D – Total momentum = (1 × 6) + (3 × 2) = 6 + 6 = 12 kg m/s. Combined mass = 4 kg, so v = 12 ÷ 4 = 3 m/s.


Impulse and Force in Collisions

Q21. [Supplement] The impulse of a force is equal to

  • (A) the change in momentum of the object
  • (B) mass × acceleration of the object
  • (C) force ÷ time
  • (D) momentum ÷ mass
Show Answer

Answer: A – Impulse = F × t, and this equals the change in momentum, Δp = mv − mu.

Q22. [Supplement] A force of 200 N acts on an object for 0.5 s. The impulse of the force is

  • (A) 400 N s
  • (B) 100 N s
  • (C) 40 N s
  • (D) 1000 N s
Show Answer

Answer: B – Impulse = F × t = 200 × 0.5 = 100 N s. This is also the change in momentum, 100 kg m/s.

Q23. [Supplement] A ball of mass 0.2 kg hits a wall at 10 m/s and rebounds at 8 m/s. The magnitude of its change in momentum is

  • (A) 0.4 kg m/s
  • (B) 2.0 kg m/s
  • (C) 3.6 kg m/s
  • (D) 18 kg m/s
Show Answer

Answer: C – The direction reverses, so the velocities have opposite signs: Δp = 0.2 × (8 − (−10)) = 0.2 × 18 = 3.6 kg m/s.

Q24. [Supplement] A car of mass 1000 kg travelling at 20 m/s is brought to rest in 5 s by its brakes. The average resultant force on the car is

  • (A) 200 N
  • (B) 100 N
  • (C) 20 000 N
  • (D) 4000 N
Show Answer

Answer: D – Change in momentum = 1000 × 20 = 20 000 kg m/s. Force = rate of change of momentum = 20 000 ÷ 5 = 4000 N.

Q25. [Supplement] Cars are built with crumple zones at the front. In a crash, a crumple zone protects the passengers because it

  • (A) increases the change in momentum of the car
  • (B) increases the time taken for the car to stop, so the force is smaller
  • (C) decreases the mass of the car during the crash
  • (D) increases the speed at which the passengers stop
Show Answer

Answer: B – The change in momentum is fixed, but the crumple zone lengthens the stopping time. Since F = Δp ÷ t, a longer time means a smaller force.

Q26. [Supplement] An airbag inflates during a car crash. It reduces injury to the driver because

  • (A) it spreads the stopping of the driver over a longer time, reducing the force on the driver
  • (B) it removes the momentum of the driver completely before the crash
  • (C) it makes the driver stop in a shorter time
  • (D) it increases the change in momentum so the force is spread out
Show Answer

Answer: A – The airbag cushions the driver and increases the time (and distance) over which the driver stops. For the same change in momentum, a longer time means a smaller force, applied over a larger area.

Q27. [Supplement] A goalkeeper catches a 0.4 kg ball moving at 12 m/s, stopping it in 0.05 s. The average force exerted by the goalkeeper’s hands on the ball is

  • (A) 4.8 N
  • (B) 240 N
  • (C) 96 N
  • (D) 0.6 N
Show Answer

Answer: C – Change in momentum = 0.4 × 12 = 4.8 kg m/s. Force = 4.8 ÷ 0.05 = 96 N.

Q28. [Supplement] The momentum of an object changes by 30 kg m/s in a time of 2 s. The resultant force acting on the object is

  • (A) 60 N
  • (B) 28 N
  • (C) 32 N
  • (D) 15 N
Show Answer

Answer: D – Resultant force = rate of change of momentum = 30 ÷ 2 = 15 N.

Q29. [Supplement] A runner of mass 70 kg runs at 8 m/s and then stops. The impulse needed to bring the runner to rest is

  • (A) 8.75 N s
  • (B) 560 N s
  • (C) 70 N s
  • (D) 0.11 N s
Show Answer

Answer: B – Impulse = change in momentum = 70 × 8 = 560 N s (directed opposite to the motion).

Q30. [Supplement] A bat strikes a ball and the ball’s momentum changes by 9 kg m/s during a contact time of 0.003 s. The average force exerted on the ball is

  • (A) 3000 N
  • (B) 27 N
  • (C) 0.027 N
  • (D) 450 N
Show Answer

Answer: A – F = Δp ÷ t = 9 ÷ 0.003 = 3000 N. The very short contact time is why collision forces are so large.

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