Newton's law of restitution and direct impact

Newton's law of restitution

For two particles moving along the same line:

speed of separation = e × speed of approach.

The coefficient of restitution e satisfies 0 ≤ e ≤ 1 (the spec says you must know this). e = 1 is perfectly elastic; e = 0 is inelastic, and the particles move together (coalesce).

Method

1. Draw before and after diagrams with all velocities in the same positive direction (an unknown that turns out negative means that particle moves the other way).

2. CLM: m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.

3. NEL: v_2 - v_1 = e(u_1 - u_2), where particle 1 is behind particle 2.

4. Solve the simultaneous equations.

Direction conditions

"Show that A changes direction" means show v_A < 0 (in the original positive direction). With letters, you may get conditions on e such as e > 2/3. Pearson's examiners note that an answer with e > 1 or the reciprocal of e signals an error.

Worked example

A particle A of mass 2 kg moving at 6 m/s hits a particle B of mass 3 kg at rest. e = 0.5. Find their velocities after the impact.

1. CLM: 12 = 2v_A + 3v_B.
2. NEL: v_B - v_A = 0.5 × 6 = 3.
3. Substitute v_B = v_A + 3: 12 = 5v_A + 9, so v_A = 0.6 and v_B = 3.6.

Answer: v_A = 0.6 m/s, v_B = 3.6 m/s, both in A's original direction

Check yourself

1. A (1 kg, 8 m/s) catches B (3 kg, 2 m/s) moving in the same direction. e = 0.5. Find the speed of B after the impact.
   Show answer

   4.25 m/s. CLM: 14 = v_A + 3v_B. NEL: v_B - v_A = 3. So 14 = 4v_B - 3, v_B = 4.25 m/s.

2. A (2 kg) moving at 5 m/s and B (1 kg) moving at 4 m/s collide head on. e = 0.6. Find the velocity of A after the impact, in A's original direction of motion.
   Show answer

   0.2 m/s. CLM: 10 - 4 = 2v_A + v_B. NEL: v_B - v_A = 0.6 × 9 = 5.4. So 6 = 3v_A + 5.4, v_A = 0.200 m/s (A keeps moving forward).

3. A (3 kg) moving at 4 m/s hits B (1 kg) at rest. Afterwards B moves at 4.5 m/s. Find e.
   Show answer

   0.5. CLM: 12 = 3v_A + 4.5, so v_A = 2.5. e = (4.5 - 2.5)/4 = 0.5.

4. A particle A of mass 2m moving with speed u hits a particle B of mass 3m at rest. Find the value of e above which A reverses direction. Give your answer as a fraction in its simplest form.
   Show answer

   2/3. CLM: 2u = 2v_A + 3v_B; NEL: v_B - v_A = eu. Then v_A = u(2 - 3e)/5, which is negative when e > 2/3.

5. A 2 kg particle moving at 6 m/s collides with a 4 kg particle at rest, and they coalesce. Find their common speed.
   Show answer

   2 m/s. e = 0: 12 = 6v, v = 2 m/s.

6. What does e = 1 mean for a collision?
   • The particles coalesce
   • All kinetic energy is lost
   • Momentum is not conserved
   • It is perfectly elastic: no kinetic energy is lost
   Show answer

   It is perfectly elastic: no kinetic energy is lost. With e = 1 the separation speed equals the approach speed and kinetic energy is conserved.

7. A student finds e = 1.4 for a collision between two particles. What should they conclude?
   • The collision was super elastic, which M2 allows
   • They have made an error, since 0 ≤ e ≤ 1
   • The particles coalesced
   • Nothing, any positive e is possible
   Show answer

   They have made an error, since 0 ≤ e ≤ 1. The spec requires 0 ≤ e ≤ 1, so e > 1 signals a mistake (often the reciprocal of e).

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy
• Pearson Edexcel IAL Mechanics M2 (WME02) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.