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Conservation of momentum and recoil: Class 9 practice

With no external force, total momentum before = total momentum after. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Momentum p = mv
  • Units (g → kg)

The idea behind conservation of momentum and recoil

With no external force, total momentum before = total momentum after. For a gun and bullet starting at rest, m_b v_b + m_g v_g = 0, so the gun recoils backwards.

Step-by-step method

  1. Write total momentum before.
  2. Write total momentum after with the unknown.
  3. Set equal and solve; the sign shows direction.

Worked example 1 (Basic)

A 10 g bullet is fired at 110 m/s from a 2 kg gun at rest. Find the recoil velocity of the gun.

  1. 0.01 × 110 + 2v = 0
  2. v = -0.55 m/s (backwards)
  3. Answer: 0.55 m/s backwards

Worked example 2 (Standard)

A 50 g bullet is fired at 400 m/s from a 3 kg gun at rest. Find the recoil velocity of the gun.

  1. 0.05 × 400 + 3v = 0
  2. v = -6.667 m/s (backwards)
  3. Answer: 6.667 m/s backwards

Worked example 3 (Challenge)

A 3 kg trolley at 8 m/s collides and sticks to a 5 kg trolley moving at -3 m/s. Find their common velocity.

  1. (3)(8) + (5)(-3) = (8)v
  2. v = 1.13 m/s
  3. Answer: 1.13 m/s

Common mistakes and misconceptions

  • Forgetting to convert grams
  • Dropping the minus sign

Practice questions

  1. (Basic) A 24 g bullet is fired at 440 m/s from a 6 kg gun at rest. Find the recoil velocity of the gun.
  2. (Basic) A 12 g bullet is fired at 490 m/s from a 2 kg gun at rest. Find the recoil velocity of the gun.
  3. (Standard) A 24 g bullet is fired at 410 m/s from a 4 kg gun at rest. Find the recoil velocity of the gun.
  4. (Standard) A 5 g bullet is fired at 130 m/s from a 2 kg gun at rest. Find the recoil velocity of the gun.
  5. (Challenge) A 3 kg trolley at 10 m/s collides and sticks to a 2 kg trolley moving at -3 m/s. Find their common velocity.
  6. (Challenge) A 5 kg trolley at 3 m/s collides and sticks to a 5 kg trolley moving at -1 m/s. Find their common velocity.

Answer key with full working

  • Q1: 0.024 × 440 + 6v = 0 → v = -1.76 m/s (backwards) → Answer: 1.76 m/s backwards
  • Q2: 0.012 × 490 + 2v = 0 → v = -2.94 m/s (backwards) → Answer: 2.94 m/s backwards
  • Q3: 0.024 × 410 + 4v = 0 → v = -2.46 m/s (backwards) → Answer: 2.46 m/s backwards
  • Q4: 0.005 × 130 + 2v = 0 → v = -0.325 m/s (backwards) → Answer: 0.325 m/s backwards
  • Q5: (3)(10) + (2)(-3) = (5)v → v = 4.8 m/s → Answer: 4.8 m/s
  • Q6: (5)(3) + (5)(-1) = (10)v → v = 1 m/s → Answer: 1 m/s

Syllabus note

This skill is listed under "Force and Laws of Motion" for Class 9 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is conservation of momentum and recoil in?

"Force and Laws of Motion", Class 9 Physics (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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