Coefficient of Restitution

Newton's law of restitution introduces the coefficient of restitution e, which measures the "bounciness" of a collision. This is a central concept in AQA Further Mechanics 1.

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1 Newton's Law of Restitution

For a direct (head-on) collision between two particles:

e = speed of separation / speed of approach

where 0 <= e <= 1.

• e = 1: perfectly elastic (no KE lost)
• e = 0: perfectly inelastic (particles coalesce)
• 0 < e < 1: partially elastic

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2 Precise Formulation

If particle A has velocity u_1 before and v_1 after, and particle B has velocity u_2 before and v_2 after (all measured in the same positive direction):

Speed of approach = u_1 - u_2 (assuming A approaches B, so u_1 > u_2)

Speed of separation = v_2 - v_1 (B moves faster than A after collision)

e = (v_2 - v_1) / (u_1 - u_2)

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3 Worked Example 1 -- Ball Hitting a Wall

A ball hits a wall at 10 m s^(-1). The coefficient of restitution is 0.7. Find the rebound speed.

Solution:

The wall does not move, so speed of approach = 10, speed of separation = rebound speed.

e = rebound speed / 10

0.7 = v / 10, so v = 7 m s^(-1)

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4 Worked Example 2 -- Two Particles

A 2 kg ball at 8 m s^(-1) collides with a 3 kg ball at rest. If e = 0.5, find the velocities after collision.

Solution:

Conservation of momentum: 2(8) + 3(0) = 2v_1 + 3v_2

16 = 2v_1 + 3v_2 ... (1)

Newton's law of restitution: v_2 - v_1 = e(u_1 - u_2) = 0.5(8 - 0) = 4

v_2 - v_1 = 4 ... (2)

From (2): v_2 = v_1 + 4. Substituting into (1):

16 = 2v_1 + 3(v_1 + 4) = 5v_1 + 12

v_1 = 4/5 = 0.8 m s^(-1)

v_2 = 4.8 m s^(-1)

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5 Worked Example 3 -- Equal Masses, Perfectly Elastic

Two equal-mass particles have a perfectly elastic head-on collision. One is initially at rest. Find the velocities after.

Solution:

Let each have mass m. Particle A at speed u, B at rest.

Momentum: mu = mv_1 + mv_2, so u = v_1 + v_2 ... (1)

Restitution (e = 1): v_2 - v_1 = u ... (2)

Adding: 2v_2 = 2u, v_2 = u. Then v_1 = 0.

They exchange velocities. A stops and B moves at speed u.