A Level Physics 3.3 Linear Momentum and Conservation MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 3.3: Linear Momentum and its Conservation. Work through conservation of momentum, elastic and inelastic collisions, and impulse – then open each answer to check the full worked explanation.
Conservation of Momentum
Q1. The principle of conservation of linear momentum states that the total linear momentum of a system of interacting objects remains constant, provided that
- (A) all collisions between the objects are elastic
- (B) the total kinetic energy of the system remains constant
- (C) no external resultant force acts on the system
- (D) no friction acts between the objects in the system
Show Answer
Answer: C – Internal forces between the objects cancel in pairs (Newton’s third law), so only an external resultant force can change the total momentum. Friction between objects inside the system is internal and does not prevent conservation; kinetic energy need not be conserved.
Q2. A trolley of mass 2.0 kg moving at 3.0 m/s collides with a stationary trolley of mass 4.0 kg. The two trolleys stick together. Their common speed is
- (A) 0.50 m/s
- (B) 1.0 m/s
- (C) 1.5 m/s
- (D) 2.0 m/s
Show Answer
Answer: B – Momentum before = 2.0 × 3.0 = 6.0 kg m/s. After the collision the combined mass is 6.0 kg, so v = 6.0 / 6.0 = 1.0 m/s.
Q3. A cannon of mass 400 kg fires a shell of mass 10 kg horizontally at 120 m/s. Assuming the cannon is free to recoil on a frictionless surface, the recoil speed of the cannon is
- (A) 3.0 m/s
- (B) 0.30 m/s
- (C) 12 m/s
- (D) 30 m/s
Show Answer
Answer: A – Total momentum starts at zero, so momentum of shell = momentum of cannon in magnitude: 10 × 120 = 400 × v, giving v = 1200 / 400 = 3.0 m/s in the opposite direction to the shell.
Q4. A stationary object explodes into two fragments of mass 1.0 kg and 3.0 kg. The 1.0 kg fragment flies off at 12 m/s. The speed of the 3.0 kg fragment is
- (A) 36 m/s
- (B) 12 m/s
- (C) 6.0 m/s
- (D) 4.0 m/s
Show Answer
Answer: D – The total momentum was zero, so the fragments carry equal and opposite momenta: 3.0 × v = 1.0 × 12, giving v = 4.0 m/s. In an explosion from rest, the speeds are in the inverse ratio of the masses.
Q5. An object of mass 5.0 kg moving at +4.0 m/s collides head-on with an object of mass 3.0 kg moving at −2.0 m/s. The objects stick together. Their common velocity is
- (A) +3.3 m/s
- (B) +2.6 m/s
- (C) +1.8 m/s
- (D) +0.88 m/s
Show Answer
Answer: C – Taking the +4.0 m/s direction as positive: total momentum = (5.0 × 4.0) + (3.0 × −2.0) = 20 − 6 = 14 kg m/s. Combined mass = 8.0 kg, so v = 14 / 8.0 = 1.75 ≈ +1.8 m/s.
Q6. An ice skater of mass 60 kg, initially at rest on frictionless ice, throws a ball of mass 3.0 kg horizontally at 8.0 m/s. The recoil speed of the skater is
- (A) 0.20 m/s
- (B) 0.40 m/s
- (C) 0.80 m/s
- (D) 2.5 m/s
Show Answer
Answer: B – Momentum gained by the ball = 3.0 × 8.0 = 24 kg m/s, so the skater gains 24 kg m/s in the opposite direction: v = 24 / 60 = 0.40 m/s.
Q7. In a school experiment, two trolleys collide and couple together on a ramp that is tilted slightly so that a component of the trolleys’ weight acts down the slope. The ramp is tilted in order to
- (A) compensate for friction, so that there is approximately no external resultant force along the track and momentum is conserved
- (B) make the trolleys move faster so the collision is easier to time
- (C) increase the friction so that the trolleys stop quickly after colliding
- (D) keep the light gates at the correct height above the track
Show Answer
Answer: A – Friction is an external force that would steadily remove momentum. Tilting the ramp until the trolleys move at constant speed when given a small push balances friction against the component of weight, leaving (approximately) zero external resultant force, so total momentum before and after the collision should match.
Q8. A trolley of mass 1.5 kg moving at 2.4 m/s catches up with and sticks to a trolley of mass 2.5 kg moving in the same direction at 0.80 m/s. Their common speed is
- (A) 1.0 m/s
- (B) 1.2 m/s
- (C) 1.6 m/s
- (D) 1.4 m/s
Show Answer
Answer: D – Total momentum = (1.5 × 2.4) + (2.5 × 0.80) = 3.6 + 2.0 = 5.6 kg m/s. Combined mass = 4.0 kg, so v = 5.6 / 4.0 = 1.4 m/s.
Q9. A ball bounces off a fixed wall and rebounds with a smaller speed. Which statement about momentum is correct?
- (A) The momentum of the ball is conserved because the collision is nearly elastic
- (B) The momentum of the ball alone is not conserved, because the wall exerts an external force on the ball
- (C) Momentum is never conserved when one of the objects is stationary
- (D) The total momentum of the ball, wall and Earth is not conserved
Show Answer
Answer: B – The ball’s momentum changes direction, so it is not conserved for the ball alone: the wall (and the Earth attached to it) exerts an external force during the contact. Momentum is conserved only for the complete system of ball + wall + Earth, where the recoil of the Earth is far too small to notice.
Q10. A railway truck of mass 8000 kg moving at 1.5 m/s couples to a stationary truck of mass 12 000 kg. The speed of the coupled trucks is
- (A) 0.50 m/s
- (B) 0.90 m/s
- (C) 0.60 m/s
- (D) 1.0 m/s
Show Answer
Answer: C – Momentum before = 8000 × 1.5 = 12 000 kg m/s. Combined mass = 20 000 kg, so v = 12 000 / 20 000 = 0.60 m/s.
Elastic & Inelastic Collisions
Q11. A collision is described as perfectly elastic when
- (A) total momentum is conserved but total kinetic energy is not
- (B) the objects stick together after the collision
- (C) total kinetic energy is conserved but total momentum is not
- (D) both total momentum and total kinetic energy are conserved
Show Answer
Answer: D – Momentum is conserved in every collision where no external resultant force acts. A collision earns the label “perfectly elastic” only when total kinetic energy is also conserved.
Q12. Two balls move towards each other along the same line, one at 3.0 m/s and the other at 2.0 m/s, and collide perfectly elastically. The speed at which they separate after the collision is
- (A) 5.0 m/s
- (B) 1.0 m/s
- (C) 2.5 m/s
- (D) impossible to find without knowing the masses
Show Answer
Answer: A – In a perfectly elastic collision the relative speed of separation equals the relative speed of approach. The approach speed is 3.0 + 2.0 = 5.0 m/s, so they also separate at 5.0 m/s, whatever their masses.
Q13. A ball of mass 2.0 kg moving at 3.0 m/s collides perfectly elastically with an identical stationary ball. After the collision
- (A) both balls move together at 1.5 m/s
- (B) the first ball stops and the second ball moves off at 3.0 m/s
- (C) the first ball rebounds at 3.0 m/s and the second stays at rest
- (D) both balls move at 1.5 m/s in opposite directions
Show Answer
Answer: B – For an elastic collision between equal masses, the balls exchange velocities. The moving ball stops dead; the stationary ball takes the full 3.0 m/s. Check: momentum 6.0 kg m/s before and after, and kinetic energy 9.0 J before and after.
Q14. An object of mass 1.0 kg moving at 4.0 m/s sticks to a stationary object of mass 3.0 kg. The kinetic energy lost in the collision is
- (A) 2.0 J
- (B) 4.0 J
- (C) 6.0 J
- (D) 8.0 J
Show Answer
Answer: C – Common speed = (1.0 × 4.0) / 4.0 = 1.0 m/s. Kinetic energy before = 0.5 × 1.0 × 4.02 = 8.0 J. Kinetic energy after = 0.5 × 4.0 × 1.02 = 2.0 J. Loss = 8.0 − 2.0 = 6.0 J, transferred to internal energy, sound and deformation.
Q15. Most everyday collisions (for example, two cars colliding) are inelastic, yet momentum is still conserved in them because
- (A) the forces between the objects are internal to the system and no external resultant force of any significance acts during the short collision
- (B) kinetic energy is in fact conserved in all collisions
- (C) the masses of the objects do not change in the collision
- (D) friction between the objects is always negligible
Show Answer
Answer: A – Conservation of momentum needs only the absence of an external resultant force; the impact forces between the two cars are internal to the two-car system and cancel in pairs. Kinetic energy, by contrast, is partly transferred to internal energy, sound and permanent deformation, which is what makes the collision inelastic.
Q16. A perfectly inelastic collision is one in which
- (A) no kinetic energy at all is lost
- (B) all of the kinetic energy is lost
- (C) momentum is not conserved
- (D) the objects stick together, giving the maximum kinetic energy loss consistent with conservation of momentum
Show Answer
Answer: D – Sticking together leaves only one final velocity, fixed by momentum conservation, so the kinetic energy loss is as large as it can be. The kinetic energy cannot fall to zero unless the total momentum is zero, because a moving combined mass must still carry some kinetic energy.
Q17. A ball of mass 1.0 kg moving at 6.0 m/s collides perfectly elastically with a stationary ball of mass 2.0 kg. The velocities after the collision are
- (A) both balls move at 2.0 m/s
- (B) the 1.0 kg ball rebounds at 2.0 m/s and the 2.0 kg ball moves at 4.0 m/s
- (C) the 1.0 kg ball moves at 1.0 m/s and the 2.0 kg ball at 2.5 m/s
- (D) the balls stick together and move at 2.0 m/s
Show Answer
Answer: B – Momentum check: before = 6.0 kg m/s; after = (1.0 × −2.0) + (2.0 × 4.0) = −2.0 + 8.0 = 6.0 kg m/s. Kinetic energy check: before = 0.5 × 1.0 × 36 = 18 J; after = (0.5 × 1.0 × 4.0) + (0.5 × 2.0 × 16) = 2.0 + 16 = 18 J. Both are conserved, and the relative speed of separation (4.0 + 2.0 = 6.0 m/s) equals the approach speed.
Q18. In an inelastic collision, the total kinetic energy decreases because
- (A) some momentum is destroyed during the impact
- (B) the total mass of the system decreases
- (C) some kinetic energy is transferred to internal energy of the objects, and to sound and deformation
- (D) the missing kinetic energy is destroyed completely
Show Answer
Answer: C – Energy is never destroyed; it is transferred to other forms. Crushed metal, heated tyres and the sound of the crash all come from the “lost” kinetic energy. Momentum, a vector, remains conserved because the impact forces are internal.
Q19. The total kinetic energy of a system can increase during an interaction when
- (A) stored energy, such as chemical energy in an explosion or potential energy in a compressed spring, is released – while momentum is still conserved
- (B) the two objects have different masses
- (C) the objects stick together on impact
- (D) it cannot increase in any interaction
Show Answer
Answer: A – In an explosion, chemical energy is transferred to kinetic energy, so total kinetic energy rises. The fragments still fly apart with equal and opposite momenta, so total momentum stays at its original value (often zero).
Q20. Object A, mass 4.0 kg, moving at 5.0 m/s, collides with stationary object B, mass 2.0 kg. Afterwards A moves at 1.0 m/s and B moves at 8.0 m/s, both in the original direction. Which conclusion is correct?
- (A) The collision is perfectly elastic
- (B) The collision is perfectly inelastic
- (C) Momentum is not conserved in the collision
- (D) Momentum is conserved, but kinetic energy increases from 50 J to 66 J, so energy must have been released (for example by an explosion) – it cannot be an ordinary collision
Show Answer
Answer: D – Momentum: before = 4.0 × 5.0 = 20 kg m/s; after = (4.0 × 1.0) + (2.0 × 8.0) = 4 + 16 = 20 kg m/s, so momentum is conserved. Kinetic energy: before = 0.5 × 4.0 × 25 = 50 J; after = (0.5 × 4.0 × 1.0) + (0.5 × 2.0 × 64) = 2 + 64 = 66 J. Kinetic energy has increased, which is impossible in an ordinary collision – stored energy must have been released.
Impulse & Force-Time Graphs
Q21. A constant force of 50 N acts on a trolley for 0.30 s. The impulse delivered to the trolley is
- (A) 0.60 N s
- (B) 15 N s
- (C) 150 N s
- (D) 1.5 N s
Show Answer
Answer: B – Impulse = FΔt = 50 × 0.30 = 15 N s. The impulse equals the change in momentum of the trolley, 15 kg m/s.
Q22. During a collision, the force exerted on an object rises from zero to a peak of 600 N and falls back to zero, forming a triangle on a force-time graph with a total base of 0.040 s. The impulse is
- (A) 24 N s
- (B) 6.0 N s
- (C) 12 N s
- (D) 24 000 N s
Show Answer
Answer: C – Impulse = area under the force-time graph = 0.5 × 0.040 × 600 = 12 N s.
Q23. A ball of mass 0.15 kg hits a wall at 20 m/s and rebounds at 15 m/s. The ball is in contact with the wall for 0.050 s. The average force exerted on the ball during the contact is
- (A) 15 N
- (B) 45 N
- (C) 75 N
- (D) 105 N
Show Answer
Answer: D – Momentum is a vector, so the rebound counts as a change from +20 m/s to −15 m/s: Δp = 0.15 × (15 − (−20)) = 0.15 × 35 = 5.25 kg m/s. Average force = Δp / Δt = 5.25 / 0.050 = 105 N. (Using only the change in speed, 5 m/s, wrongly gives 15 N.)
Q24. Airbags and crumple zones reduce injuries in a car crash because they
- (A) allow the same change of momentum to happen over a longer time, so the average force on the passengers is smaller
- (B) reduce the change of momentum of the passengers to almost zero
- (C) increase the contact time so that the force on the passengers increases
- (D) absorb the momentum of the car so that no impulse acts on the passengers
Show Answer
Answer: A – The passengers must lose the same momentum whether they stop suddenly or gradually, so the impulse is fixed. Spreading it over a longer time (F = Δp / Δt) reduces the average force, and it is the force that causes injury.
Q25. A car of mass 1200 kg travelling at 30 m/s is brought to rest in 6.0 s by a constant braking force (take this as the only horizontal force). The magnitude of the braking force is
- (A) 200 N
- (B) 3600 N
- (C) 6000 N
- (D) 36 000 N
Show Answer
Answer: C – Newton’s second law in momentum form: F = Δp / Δt = (1200 × 30) / 6.0 = 36 000 / 6.0 = 6000 N.
Q26. A force-time graph for a collision is a trapezium: the force rises uniformly from 0 to 400 N in 0.020 s, stays at 400 N for a further 0.040 s, then falls uniformly to zero in 0.020 s. The impulse is
- (A) 16 N s
- (B) 24 N s
- (C) 32 N s
- (D) 32 000 N s
Show Answer
Answer: B – Impulse = area of the trapezium. The parallel sides are the top (0.040 s) and the base (0.020 + 0.040 + 0.020 = 0.080 s), and the height is 400 N: area = 0.5 × (0.040 + 0.080) × 400 = 0.5 × 0.120 × 400 = 24 N s.
Q27. A ball of mass 0.40 kg, initially at rest, receives an impulse of 8.0 N s from a bat. The speed of the ball immediately afterwards is
- (A) 3.2 m/s
- (B) 0.050 m/s
- (C) 8.0 m/s
- (D) 20 m/s
Show Answer
Answer: D – Impulse = change of momentum: 8.0 = 0.40 × v, so v = 8.0 / 0.40 = 20 m/s.
Q28. A gymnast landing from a jump is advised to bend her knees as she lands. This reduces the force on her legs because bending her knees
- (A) increases the time taken for her momentum to fall to zero, so the average force is smaller for the same change of momentum
- (B) reduces the change in her momentum
- (C) increases the impulse exerted by the floor
- (D) reduces her mass while she is landing
Show Answer
Answer: A – Her momentum change is fixed by her landing speed and mass. Bending the knees lengthens the stopping time Δt, and since F = Δp / Δt, the average force on her legs falls.
Q29. Two collisions, X and Y, act on identical objects. The force-time graphs for X and Y have exactly the same area, but the graph for Y is spread over twice the time of X with a lower peak. Which statement is correct?
- (A) Collision Y produces the larger change of momentum
- (B) Collision X produces the larger change of momentum
- (C) Both collisions produce the same change of momentum, but Y exerts the smaller peak force
- (D) The peak force is the same in both collisions because the areas are equal
Show Answer
Answer: C – Impulse = area under the force-time graph = change of momentum, so equal areas mean equal changes of momentum. Spreading the same area over a longer time forces the peak (and average) force down.
Q30. A bullet of mass 0.020 kg accelerates from rest to 400 m/s while travelling along the barrel of a rifle. It spends 0.0040 s in the barrel. The average force exerted on the bullet is
- (A) 200 N
- (B) 2000 N
- (C) 8.0 N
- (D) 20 000 N
Show Answer
Answer: B – Change of momentum = 0.020 × 400 = 8.0 kg m/s. Average force = Δp / Δt = 8.0 / 0.0040 = 2000 N.
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