Coefficient of Restitution

Conservation of momentum alone gives one equation but a collision between two bodies has two unknown final velocities. The missing second equation is Newton's experimental law of restitution, which captures how "bouncy" the collision is through a single number $e$ , the coefficient of restitution. With both equations in hand, every direct-collision problem becomes a pair of simultaneous equations.

1 Where this sits in AQA 7367

This is the heart of the Mechanics option, Paper 3 (7367/3M) — almost every collision question on that paper uses $e$ . Paper 3 is weighted AO1 40% / AO2 25% / AO3 35%, and restitution questions are classic AO3: set up two equations, solve simultaneously, then interpret (does a second collision occur? how much energy was lost?). It builds on conservation of momentum (previous lesson) and on solving simultaneous linear equations from GCSE/A-Level. The "experimental" in Newton's experimental law of restitution is significant: unlike conservation of momentum (which follows from Newton's laws), the restitution relation is an empirical model that fits real collisions well for moderate speeds. AQA expects you to treat it as a given law and apply it, while being aware it is a modelling assumption — a point that can be worth an AO3 comment.

2 Core theory

Newton's law of restitution

For a direct (head-on, one-dimensional) collision,

$e = \frac{\text{speed of separation}}{\text{speed of approach}}, \qquad 0 \le e \le 1.$

$e = 1$ : perfectly elastic — no kinetic energy lost.
$e = 0$ : perfectly inelastic — the bodies coalesce.
$0 < e < 1$ : partially elastic — some KE lost.

Precise signed formulation

Let A have velocity $u_1$ before and $v_1$ after, and B have $u_2$ before and $v_2$ after, all measured in the same positive direction with A approaching B (so $u_1 > u_2$ ). Then

$\text{approach} = u_1 - u_2, \qquad \text{separation} = v_2 - v_1,$

and the law becomes the equation you actually use:

$\boxed{\,v_2 - v_1 = e\,(u_1 - u_2)\,}.$

Pair this with conservation of momentum, $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ , and solve the two simultaneously for $v_1, v_2$ .

Impact with a fixed wall

A wall does not move, so separation speed is the rebound speed and approach speed is the incoming speed: $e = \dfrac{\text{rebound speed}}{\text{approach speed}}$ . The rebound speed is simply $e$ times the incoming speed. Physically the wall is an effectively infinite mass: in the limit $m_2 \to \infty$ the two-body restitution result collapses to "the moving body reverses with its speed scaled by $e$ ", which is why a wall problem needs only the restitution relation, not momentum.

Kinetic energy lost in terms of $e$

Solving the momentum–restitution pair for a collision where $m_2$ is initially at rest gives, after simplification, a clean expression for the kinetic energy lost:

$\Delta KE = \frac{m_1 m_2\,u^2}{2(m_1 + m_2)}\,(1 - e^2),$

where $u$ is the approach speed. The factor $(1 - e^2)$ is the key: when $e = 1$ it vanishes (a perfectly elastic collision loses no KE), and when $e = 0$ it is maximal (coalescence loses the most). Energy loss therefore scales with $(1 - e^2)$ , not with $(1-e)$ — a distinction worth remembering, because a ball with $e = 0.9$ still loses nearly $19\%$ of the relevant KE, not $10\%$ .

For a ball against a wall the energy bookkeeping is even simpler: the wall does no work and gains no KE (it is fixed), so the fraction of the ball's kinetic energy lost is $1 - e^2$ directly. A ball thrown at a wall with $e = 0.7$ returns with only $e^2 = 49\%$ of its kinetic energy — over half is dissipated in a single bounce, which is why a "lively" ball needs $e$ close to $1$ . This wall result is a clean special case of the two-body formula in the limit of an infinite second mass.

3 Worked examples with mark scheme

Worked Example 1 — two particles, one at rest

A $2\ \mathrm{kg}$ ball moving at $8\ \mathrm{m\,s^{-1}}$ strikes a stationary $3\ \mathrm{kg}$ ball; $e = 0.5$ . Find both velocities afterwards.

\begin{aligned} \text{(momentum)}\quad & 2(8) + 3(0) = 2v_1 + 3v_2 &&\Rightarrow\; 16 = 2v_1 + 3v_2 \quad (1)\\ \text{(restitution)}\quad & v_2 - v_1 = 0.5(8-0) = 4 &&\Rightarrow\; v_2 = v_1 + 4 \quad (2) \end{aligned}

Substitute (2) into (1): $16 = 2v_1 + 3(v_1+4) = 5v_1 + 12$ , so $v_1 = 0.8$ and $v_2 = 4.8\ \mathrm{m\,s^{-1}}$ .

Mark scheme: M1 conservation equation; M1 restitution equation; A1 $v_1 = 0.8$ ; A1 $v_2 = 4.8$ . Check: $v_2 > v_1$ , so they separate — physically sensible. (4 marks.)

Worked Example 2 — equal masses, perfectly elastic

Two equal-mass particles collide head-on; one is at rest, $e = 1$ . Find the velocities afterwards.

Let each have mass $m$ ; A has speed $u$ , B at rest.

\begin{aligned} mu &= mv_1 + mv_2 &&\Rightarrow\; u = v_1 + v_2 \quad (1)\\ v_2 - v_1 &= u \quad (e=1) &&(2) \end{aligned}

Adding (1) and (2): $2v_2 = 2u \Rightarrow v_2 = u$ , hence $v_1 = 0$ . They exchange velocities — A stops dead, B moves off at $u$ . This famous result (seen in Newton's cradle) holds only for equal masses with $e=1$ .

Mark scheme: M1 both equations; A1 $v_1=0,\ v_2=u$ with the interpretation "they exchange velocities".

Worked Example 3 — successive collisions on a line

Particles A $(4\ \mathrm{kg})$ , B $(2\ \mathrm{kg})$ , C $(1\ \mathrm{kg})$ lie in a line; B and C are at rest. A strikes B at $6\ \mathrm{m\,s^{-1}}$ ; $e = 0.5$ at every impact. Find all final velocities.

Collision 1 (A→B):

\begin{aligned} 4(6) &= 4v_A + 2v_B &&\Rightarrow\; 24 = 4v_A + 2v_B\\ v_B - v_A &= 0.5(6) = 3 &&\Rightarrow\; v_B = v_A + 3 \end{aligned}

So $24 = 4v_A + 2(v_A+3) = 6v_A + 6 \Rightarrow v_A = 3,\ v_B = 6$ .

Collision 2 (B→C), B now moving at $6$ :

\begin{aligned} 2(6) &= 2v_B' + 1v_C &&\Rightarrow\; 12 = 2v_B' + v_C\\ v_C - v_B' &= 0.5(6) = 3 &&\Rightarrow\; v_C = v_B' + 3 \end{aligned}

So $12 = 2v_B' + (v_B'+3) = 3v_B' + 3 \Rightarrow v_B' = 3,\ v_C = 6$ .

Final: $A = 3,\ B = 3,\ C = 6\ \mathrm{m\,s^{-1}}$ . Check for a third collision: after collision 2, A and B both move at $3$ , so A never catches B — no further impact. This "does a further collision occur?" check is an examiner favourite and earns the final A1.

Worked Example 4 — ball against a wall

A ball strikes a wall head-on at $10\ \mathrm{m\,s^{-1}}$ ; the coefficient of restitution between ball and wall is $0.7$ . Find the rebound speed and the fraction of kinetic energy lost.

The wall is fixed, so the rebound speed is $e$ times the approach speed:

$v = e u = 0.7 \times 10 = 7\ \mathrm{m\,s^{-1}}.$

Since KE $\propto v^2$ , the fraction of KE retained is $e^2 = 0.49$ , so the fraction lost is $1 - e^2 = 0.51 = 51\%$ . Mark scheme: M1 $v = eu$ ; A1 $7\ \mathrm{m\,s^{-1}}$ ; M1 fraction lost $= 1 - e^2$ ; A1 $51\%$ . A wall problem needs only restitution — there is no second body to write a momentum equation for.

Worked Example 5 — both bodies moving, same direction

A $3\ \mathrm{kg}$ ball moving at $6\ \mathrm{m\,s^{-1}}$ catches a $2\ \mathrm{kg}$ ball moving at $2\ \mathrm{m\,s^{-1}}$ in the same direction; $e = 0.5$ . Find both final velocities.

\begin{aligned} \text{(momentum)}\quad & 3(6) + 2(2) = 3v_1 + 2v_2 &&\Rightarrow\; 22 = 3v_1 + 2v_2\\ \text{(restitution)}\quad & v_2 - v_1 = 0.5(6 - 2) = 2 &&\Rightarrow\; v_2 = v_1 + 2 \end{aligned}

Substitute: $22 = 3v_1 + 2(v_1 + 2) = 5v_1 + 4 \Rightarrow v_1 = 3.6,\ v_2 = 5.6\ \mathrm{m\,s^{-1}}$ . Mark scheme: M1 momentum with both bodies moving; M1 restitution using the relative approach speed $6-2$ ; A1 $v_1 = 3.6$ ; A1 $v_2 = 5.6$ . The approach speed is the difference of the initial speeds because both move the same way — a classic place to slip.

Worked Example 6 — a general "show that" with restitution

A particle of mass $m$ moving at speed $u$ collides directly with a stationary particle of mass $2m$ ; the coefficient of restitution is $e$ . Show that the first particle's velocity afterwards is $\dfrac{(1 - 2e)u}{3}$ , and find the condition on $e$ for it to rebound.

Take $u$ 's direction as positive.

\begin{aligned} \text{(momentum)}\quad & m u = m v_1 + 2m v_2 &&\Rightarrow\; u = v_1 + 2v_2\\ \text{(restitution)}\quad & v_2 - v_1 = e u &&\Rightarrow\; v_2 = v_1 + eu \end{aligned}

Substitute $v_2$ : $u = v_1 + 2(v_1 + eu) = 3v_1 + 2eu$ , so

$v_1 = \frac{u - 2eu}{3} = \frac{(1 - 2e)u}{3}.$

The first particle rebounds (moves in the negative direction) when $v_1 < 0$ , i.e. $1 - 2e < 0$ , so $e > \tfrac12$ . Mark scheme: M1 both equations; M1 eliminate $v_2$ ; A1 $v_1 = \tfrac{(1-2e)u}{3}$ ; A1 condition $e > \tfrac12$ . Keeping the algebra symbolic is essential — the question rewards the general relationship, and the rebound condition falls straight out of the sign of the answer.

Worked Example 7 — successive bounces with a time-of-flight twist

A ball is dropped from $1.8\ \mathrm{m}$ onto a floor with $e = 0.5$ (take $g = 9.8\ \mathrm{m\,s^{-2}}$ ). Find (a) the speed just before the first impact, (b) the height of the first bounce, (c) the height of the second bounce.

(a) Free fall through $1.8\ \mathrm{m}$ : $v^2 = 2gh = 2(9.8)(1.8) = 35.28$ , so $v = \sqrt{35.28} \approx 5.94\ \mathrm{m\,s^{-1}}$ .

(b) Rebound speed $= e v = 0.5(5.94) = 2.97\ \mathrm{m\,s^{-1}}$ ; first-bounce height $= \dfrac{(ev)^2}{2g} = e^2 h = 0.25(1.8) = 0.45\ \mathrm{m}$ .

(c) Each bounce scales the height by $e^2 = 0.25$ , so the second-bounce height $= e^4 h = 0.0625(1.8) = 0.1125\ \mathrm{m}$ .

Mark scheme: M1 $v = \sqrt{2gh}$ ; A1 $5.94\ \mathrm{m\,s^{-1}}$ ; M1 rebound height $= e^2 h$ ; A1 $0.45\ \mathrm{m}$ ; A1 $0.1125\ \mathrm{m}$ . Note you can answer (b) and (c) entirely through $e^{2n}h$ without ever computing a speed — but the "show that" version would demand the $\sqrt{2gh}$ route.

4 Specimen-style exam question

(Specimen-style.) A ball is dropped from rest at height $h$ onto a horizontal floor with coefficient of restitution $e$ . (a) Show that the height of the first bounce is $e^2 h$ . (b) Hence find the height after $n$ bounces. (c) Evaluate for $h = 5\ \mathrm{m},\ e = 0.6$ after one bounce.

(a) Falling through $h$ , the impact speed is $u = \sqrt{2gh}$ . The rebound speed is $v = eu = e\sqrt{2gh}$ . Rising freely, it reaches height $h_1$ where $0 = v^2 - 2gh_1$ :

$h_1 = \frac{v^2}{2g} = \frac{e^2(2gh)}{2g} = e^2 h.$

(b) Each bounce multiplies the height by $e^2$ , so after $n$ bounces $h_n = e^{2n} h$ (a geometric sequence with ratio $e^2$ ).

(c) $h_1 = (0.6)^2 (5) = 0.36 \times 5 = 1.8\ \mathrm{m}.$

Coefficient of Restitution

Coefficient of Restitution

1 Where this sits in AQA 7367

2 Core theory

Newton's law of restitution

Precise signed formulation

Impact with a fixed wall

Kinetic energy lost in terms of $e$

3 Worked examples with mark scheme

Worked Example 1 — two particles, one at rest

Worked Example 2 — equal masses, perfectly elastic

Worked Example 3 — successive collisions on a line

Worked Example 4 — ball against a wall

Worked Example 5 — both bodies moving, same direction

Worked Example 6 — a general "show that" with restitution

Worked Example 7 — successive bounces with a time-of-flight twist

4 Specimen-style exam question

5 Synoptic links

More in Mathematics

Coefficient of Restitution

Coefficient of Restitution

1 Where this sits in AQA 7367

2 Core theory

Newton's law of restitution

Precise signed formulation

Impact with a fixed wall

Kinetic energy lost in terms of eee

3 Worked examples with mark scheme

Worked Example 1 — two particles, one at rest

Worked Example 2 — equal masses, perfectly elastic

Worked Example 3 — successive collisions on a line

Worked Example 4 — ball against a wall

Worked Example 5 — both bodies moving, same direction

Worked Example 6 — a general "show that" with restitution

Worked Example 7 — successive bounces with a time-of-flight twist

4 Specimen-style exam question

5 Synoptic links

More in Mathematics

Kinetic energy lost in terms of $e$