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Track the momentum.

Six original Cambridge-style questions on the conservation principle, inelastic collisions, recoil and explosions, the elastic relative-speed result, and kinetic energy lost.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Sum the momentum, mind the signs.

01
Analysis
[2 marks]

State the principle of conservation of momentum, including the condition under which it applies.

  • The total momentum of a system stays constant. ✓
  • This holds provided there is no external resultant force acting on the system. ✓
02
Analysis
[3 marks]

A 1.5 kg trolley moving at 4.0 m s⁻¹ collides with and sticks to a stationary 2.5 kg trolley. Find their common velocity afterward.

  • Momentum before = 1.5 × 4.0 + 0 = 6.0 kg m s⁻¹. ✓
  • Combined mass = 1.5 + 2.5 = 4.0 kg. ✓
  • v = 6.0 / 4.0 = 1.5 m s⁻¹ in the original direction. ✓
03
Analysis
[3 marks]

A 800 kg cannon, initially at rest, fires a 5.0 kg shell horizontally at 240 m s⁻¹. Find the recoil velocity of the cannon.

  • Total momentum before = 0, so total after = 0. ✓
  • 800 × V + 5.0 × 240 = 0, so 800V = −1200. ✓
  • V = −1.5 m s⁻¹: the cannon recoils at 1.5 m s⁻¹ in the opposite direction. ✓
04
Analysis
[3 marks]

Two trolleys are held together against a compressed spring and released from rest. A 1.0 kg trolley moves off to the left at 6.0 m s⁻¹. Find the velocity of the 2.0 kg trolley.

  • Total momentum before = 0 (both at rest). ✓
  • Taking right as positive: 1.0 × (−6.0) + 2.0 × v = 0, so 2.0v = 6.0. ✓
  • v = 3.0 m s⁻¹ to the right (opposite to the 1.0 kg trolley). ✓
05
Analysis
[3 marks]

A 2.0 kg ball moving at 6.0 m s⁻¹ makes a head-on elastic collision with a stationary 2.0 kg ball. Use the elastic relative-speed result to find the final velocities.

  • Momentum: 2.0 × 6.0 = 2.0 v₁ + 2.0 v₂, so v₁ + v₂ = 6.0. ✓
  • Elastic: relative speed of separation equals approach, so v₂ − v₁ = 6.0. ✓
  • Solving: v₁ = 0 and v₂ = 6.0 m s⁻¹. For equal masses the balls exchange velocities. ✓
06
Analysis
[3 marks]

A 0.50 kg lump of clay moving at 8.0 m s⁻¹ strikes and sticks to a stationary 1.5 kg block. Find the common velocity and the kinetic energy lost in the collision.

  • Common velocity: v = (0.50 × 8.0) / (0.50 + 1.5) = 4.0 / 2.0 = 2.0 m s⁻¹. ✓
  • KE before = ½ × 0.50 × 8.0² = 16 J; KE after = ½ × 2.0 × 2.0² = 4.0 J. ✓
  • KE lost = 16 − 4.0 = 12 J, transferred to heat, sound and deformation. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.