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Conservation of momentum in collisions

When two particles collide directly, the impulses they give each other are equal and opposite, so the total momentum is unchanged. M1 uses this for one-dimensional collisions only (Newton's law of restitution is not in M1).

The principle

Total momentum before = total momentum after:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2.

If they coalesce (stick together): m_1u_1 + m_2u_2 = (m_1 + m_2)v.

Signs and directions

Draw a before and after diagram with all velocities pointing in the positive direction. Any velocity found to be negative is moving the other way. "The direction of motion of P is reversed" means v_P has the opposite sign to u_P.

Impulse in a collision

The impulse on B is m_B(v_B - u_B); the impulse on A is equal and opposite. Exam questions often ask for this after you find the velocities.

Worked example

A 3 kg particle at 4 m/s hits a 2 kg particle at rest. Afterwards the 3 kg particle moves at 1.5 m/s in the same direction. Find the speed of the 2 kg particle and the impulse on it.

  1. Before: 3(4) + 0 = 12.
  2. After: 3(1.5) + 2v = 4.5 + 2v.
  3. So v = 3.75 m/s; impulse = 2 × 3.75 = 7.5 N s.

Answer: 3.75 m/s and 7.5 N s

Key idea: Total momentum is conserved. Use a positive direction and draw before and after. Impulses in a collision are equal and opposite.

Check yourself

  1. A 4 kg particle moving at 6 m/s collides with a 2 kg particle at rest. After the collision the 4 kg particle moves at 2 m/s in the same direction. Find the speed of the 2 kg particle.
    Show answer

    8 m/s. 24 = 8 + 2v, so v = 8 m/s.

  2. A 4 kg particle moving at 5 m/s collides with a 1 kg particle moving at 3 m/s in the opposite direction. They coalesce. Find their common speed.
    Show answer

    3.4 m/s. 4(5) + 1(-3) = 5v, so v = 17/5 = 3.4 m/s.

  3. P (2 kg) moves at 6 m/s towards Q (m kg) moving at 2 m/s towards P. After the collision both directions are reversed: P moves at 1 m/s and Q at 3 m/s. Find m.
    Show answer

    2.8 kg. Positive in P's original direction: 12 - 2m = -2 + 3m, so 5m = 14 and m = 2.8 kg.

  4. For the collision in the first question (4 kg at 6 m/s hits 2 kg at rest; 4 kg continues at 2 m/s), find the magnitude of the impulse on the 4 kg particle.
    Show answer

    16 N s. 4(2 - 6) = -16, so 16 N s (equal and opposite to the impulse on the other particle: 2 × 8 = 16).

  5. A gun of mass 4 kg fires a 0.02 kg bullet horizontally at 400 m/s. Find the recoil speed of the gun.
    Show answer

    2 m/s. Total momentum before is 0, so 4v = 0.02 × 400 = 8, v = 2 m/s.

  6. A 3 kg particle at 2 m/s collides with a 1 kg particle at 6 m/s moving towards it. After the collision the 1 kg particle moves at 1.5 m/s, having reversed its direction. Find the velocity of the 3 kg particle, taking its original direction as positive.
    Show answer

    -0.5 m/s. 3(2) + 1(-6) = 3v + 1(1.5), so 0 = 3v + 1.5 and v = -0.5 m/s: the 3 kg particle also reverses, at 0.5 m/s.

  7. Why is total momentum conserved when two particles collide?
    • They exert equal and opposite impulses on each other
    • Kinetic energy is always conserved
    • The particles have equal masses
    • No forces act on either particle
    Show answer

    They exert equal and opposite impulses on each other. Newton's third law gives equal and opposite forces for the same time, so the momentum gained by one equals that lost by the other.

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Last updated 7 November 2026. Spotted a mistake? Tell us.