Coefficient of Restitution

Equivalent forms

e = \sqrt{h_1/h_0}

e = v_{\text{out}}/v_{\text{in}} \;(\text{wall})

A single empirical number that interpolates between billiard balls and putty.

Symbol	Name	SI unit	Dimension	Range
$e$	Coefficient of restitutionoutput Ratio of relative speed after to before, with sign reversed		1	0 – 1
$v1$	Initial velocity 1 Pre-collision velocity of body 1	m/s	LT^-1	-20 – 20
$v2$	Initial velocity 2 Pre-collision velocity of body 2	m/s	LT^-1	-20 – 20
$v1_prime$	Final velocity 1 Post-collision velocity of body 1	m/s	LT^-1	-20 – 20
$v2_prime$	Final velocity 2 Post-collision velocity of body 2	m/s	LT^-1	-20 – 20

Unit systems

Symbol	SI	CGS	Imperial
$e$	—	—	—
$v1$	m/s	cm/s	ft/s
$v2$	m/s	cm/s	ft/s
$v1_prime$	m/s	cm/s	ft/s
$v2_prime$	m/s	cm/s	ft/s

Where it holds

1D head-on collisions; for oblique impacts use the component along the line of impact. e depends on material AND impact speed at high speeds.

Dimensional analysis

e \to [1]

[LT^{-1}]/[LT^{-1}] \to [1]

Dimensionless.

\checkmark

Discovery

Isaac Newton · 1687

Newton's Principia tabulated 'rebound ratios' for materials — the empirical seed of what we now call e.

Try this

A tennis ball bounces to 60% of its drop height — what does that tell you about the collision?

Compute the coefficient of restitution from drop height h_0 = 1 m and rebound height h_1 = 0.56 m.

Research status: stable

Real-world applications

Sports equipment certification (tennis, golf, basketball)
Automotive crash testing
Industrial machine design (ball mills, hammer mills)
Granular flow modeling

Common misconceptions

e is a property of one object — it's a property of the colliding pair (and speed)
e > 1 is possible in everyday collisions — it isn't (passive)
e is constant for all speeds — at high impact speeds e drops noticeably

Experimental verification

Drop a ball from height h_0, measure rebound h_1:

e = \sqrt{h_1/h_0}

. Used in ASTM standards for sports-ball quality (tennis: 0.53–0.58, basketball: 0.76).

Derivation

Newton's empirical rule of impact: relative speed of separation

= e \times

relative speed of approach.

Solving simultaneously with momentum conservation gives the post-collision velocities for any e.

Limiting cases

e = 1

⟶ Perfectly elasticRelative velocity simply reverses; KE conserved.

e = 0

⟶ Perfectly inelasticBodies stick;

v_{1}' = v_{2}

'; maximum KE lost.

e > 1⟶ Unphysical for passive collisionsEnergy added — would require an explosive or spring release.

What if…

What if e changes with speed?

Use empirical fit e(v); at high v, polymers and metals lose more energy.

What if impact is oblique?

Apply e only to the velocity component along the line of impact.

How do I use e in a simulation?

$v_{1}' - v_{2}' = -e(v_{1} - v_{2})$ , combined with momentum conservation.

1

Bouncing ball

Given ·

v1:: -4.43
v2:: 0
v1 prime:: 3.32
v2 prime:: 0

Find · e

Steps

Drop from $h_0 = 1\,\mathrm{m}$ → impact speed $\sqrt{2gh} \approx 4.43\,\mathrm{m/s}$ downward $(v_{1} = -4.43)$
Rebound to $h_1 = 0.56\,\mathrm{m} \to v_{1}' \approx +3.32\,\mathrm{m/s}$ upward
$e = -(3.32)/(-4.43) = 0.75$

Result ·

e = -(3.32 - 0)/(-4.43 - 0) = 0.75

2

Bumper cars

Given ·

v1:: 5
v2:: -5
v1 prime:: -3
v2 prime:: 3

Find · e

Steps

Relative approach: $v_{1} - v_{2} = 10$
Relative separation: $v_{1}' - v_{2}' = -6$
$e = -(-6)/10 = 0.6$

Result ·

e = 0.6

1D Elastic Collision (Final Velocities)→Perfectly Inelastic Collision (1D)→Linear Momentum→