Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂): Question bank

This is a question bank for hunting down the sneaky misconceptions that hide inside the coefficient of restitution. No heavy arithmetic here — every item is about understanding. Cover the answer, commit to a reason, then reveal.

Everything below builds on the parent note Coefficient of Restitution. If a term feels unfamiliar, re-read that note first — here we assume you already know:

Recall The five facts these traps attack

$e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{separation speed}{approach speed}$ , and $0 \leq e \leq 1$ .
Momentum is conserved for every $e$ ; kinetic energy is only conserved when $e = 1$ .
$e = 1$ = perfect bounce (Elastic Collisions); $e = 0$ = stick together (Perfectly Inelastic Collisions).
$e$ depends on the materials, not on the speeds or masses.
Ball on a fixed floor: $e = h_{2} / h_{1}$ , so $h_{n} = e^{2 n} h_{0}$ .

True or false — justify

Each is a statement. Decide true/false and give the reason — a bare verdict scores zero.

e < 1

, momentum is not conserved during the collision.

False. Momentum is conserved for all

e

because there is no external impulse on the pair; it is kinetic energy that drops when

e < 1

A collision with

e = 1

conserves kinetic energy.

True. Separation speed equals approach speed, and because

KE \propto

speed

^{2}

and momentum is fixed, no kinetic energy is lost — that is exactly the definition of Elastic Collisions.

The value of

e

depends on how fast the two bodies were moving before impact.

Mostly false as an ideal law —

e

is treated as a material constant. (Real materials show a mild speed-dependence, but at this level

e

depends on what the objects are made of, not their speed.)

For two bodies to collide at all, we need

u_{1} > u_{2}

True (with body 1 as the chaser). Body 1 must be closing on body 2, so its velocity along the line exceeds body 2's, making approach speed

u_{1} - u_{2} > 0

After a collision the separation speed can exceed the approach speed.

False for ordinary collisions — that would need

e > 1

, meaning energy was created. It only happens in "super-elastic" events where stored chemical/spring energy is released, not in the mechanical collisions we study.

If both balls end up moving in the same direction, the collision must have been inelastic.

False. Direction of motion afterwards depends on masses and speeds, not on

e

; even a perfectly elastic (

e = 1

) collision can leave both bodies moving forward.

A perfectly inelastic collision (

e = 0

) loses all the kinetic energy.

False. It loses the maximum possible KE consistent with momentum conservation, but the combined mass keeps moving, so it retains the KE of the common-velocity motion.

Doubling the drop height doubles the rebound height for a fixed ball and floor.

True.

h_{2} = e^{2} h_{1}

is linear in

h_{1}

, so with

e

fixed, doubling

h_{1}

doubles

h_{2}

The coefficient of restitution has units of metres per second.

False. It is a ratio of two speeds, so the units cancel —

e

is dimensionless.

For a ball bouncing on the floor,

e = h_{2} / h_{1}

False.

e

is a ratio of speeds and height

\propto

speed

^{2}

, so

e = h_{2} / h_{1}

; the plain height ratio equals

e^{2}

Spot the error

Each line contains a flawed statement or step. Name the mistake and correct it.

"Separation speed

= v_{1} - v_{2}

to match the order of

u_{1} - u_{2}

The indices are deliberately swapped. Since body 2 ends up ahead (

v_{2} > v_{1}

), separation is

v_{2} - v_{1}

; writing

v_{1} - v_{2}

gives a negative, meaningless

e

"Since energy is lost when

e < 1

, momentum must also decrease."

Energy and momentum are independent conservation questions. Momentum is a vector total protected by Newton's third law during the collision; it stays fixed regardless of energy loss.

"A clay lump hitting a wall has

e = 0

, so its final speed is zero."

e = 0

means no separation (no bounce), so it stops relative to the wall — but only because the wall is fixed. Against a free movable body,

e = 0

means they move off together at a common velocity, not that everything stops.

e = J_{r} / J_{d}

, and since restoration always returns more push than deformation,

e > 1

Restoration returns at most as much impulse as deformation stored (

J_{r} \leq J_{d}

), because some energy escapes as heat/sound. Hence

e \leq 1

, never above.

"After many bounces

h_{n} = e^{n} h_{0}

Each bounce multiplies the speed by

e

, and height goes as speed

^{2}

, so each bounce multiplies height by

e^{2}

. After

n

bounces

h_{n} = e^{2 n} h_{0}

e

tells you the fraction of kinetic energy that survives the collision."

e

is the fraction of relative speed that survives, not energy. The fraction of KE tied up in the relative motion scales as

e^{2}

, and the total KE-loss law involves

(1 - e^{2})

— see Kinetic Energy Loss in Collisions.

"Because the floor doesn't move, momentum is not conserved in a bounce."

Momentum is conserved — the Earth (the floor) takes an equal and opposite momentum. It's just enormous in mass, so its velocity change is undetectable; that's why we model

u_{2} = v_{2} = 0

Why questions

Answer with the underlying reason, not just a restatement.

Why do we even need

e

as a separate equation?

Conservation of Linear Momentum gives one equation but a collision has two unknown final velocities, so a second relation is required, and energy isn't generally conserved — Newton's restitution law supplies it.

Why does swapping the indices (

v_{2} - v_{1}

over

u_{1} - u_{2}

) keep

e

positive?

Because physically the chaser is faster before (

u_{1} > u_{2}

) and the chased is faster after (

v_{2} > v_{1}

), so both differences are positive and their ratio is

\geq 0

automatically.

Why does the common velocity

v_{c}

vanish from the final formula?

The "add numerators, add denominators" trick applied to the two impulse ratios cancels every

v_{c}

term, so we recover

e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

without ever knowing

v_{c}

Why is the bounce-height relation a square root, not a straight ratio?

Because the ball's speed at each surface obeys

v = 2 g h

, so height depends on speed squared; taking the speed ratio

e

from a height ratio therefore needs a square root (Projectile Motion supplies the

v

–

h

link).

Why does

e

depend on the materials and not on the masses?

e

measures how completely the deformation springs back (

J_{r} / J_{d}

), which is a property of the material's elasticity; masses affect the final velocities but not the bounce-back fraction itself.

Why can't a normal collision have

e > 1

That would mean the separation speed exceeds the approach speed, i.e. mechanical energy was created out of nothing, violating energy conservation for a passive collision.

Why is

e = 0

the case of maximum kinetic energy loss?

With zero separation speed the two bodies share one velocity, which is the only outcome (given fixed momentum) that leaves the least KE — any bounce would return some relative motion and hence more KE.

Edge cases

The scenarios people forget. Reason through each boundary.

What is

e

when a ball is dropped and never bounces (stays on the floor)?

e = 0

: rebound height

h_{2} = 0

gives

e = 0/ h_{1} = 0

, a perfectly inelastic contact.

What does

e = 1

predict for a bouncing ball?

The ball returns to its original drop height forever, since

h_{2} = e^{2} h_{1} = h_{1}

— an idealised perpetual bounce with no energy loss.

Two bodies of equal mass,

e = 1

, one at rest — what happens?

They exchange velocities: the moving one stops and the stationary one moves off with the incoming speed, the classic elastic equal-mass swap.

What if the two bodies start with the same velocity (

u_{1} = u_{2}

Approach speed is zero, so there is no collision to speak of —

e

is undefined (division by zero) because nothing is closing in.

In the impulse picture, what does

J_{r} = 0

correspond to?

No restoration push at all, so the material never springs back:

e = J_{r} / J_{d} = 0

, a perfectly plastic (splat) collision.

What happens to

e

if we reverse the chosen positive direction for the whole problem?

Nothing — all velocities flip sign together, so both the numerator and denominator flip and the ratio

e

is unchanged;

e

is convention-independent.

If the "floor" is actually a free block of large but finite mass, is

e = h_{2} / h_{1}

still exact?

Not exactly — that formula assumes the floor's mass is infinite so it stays still. With finite mass the block recoils, taking some speed, so the rebound height slightly under-reads the true material

e

Connections

Conservation of Linear Momentum — the always-true equation these traps repeatedly test.
Elastic Collisions — the $e = 1$ boundary explored above.
Perfectly Inelastic Collisions — the $e = 0$ boundary.
Impulse and Momentum — source of the $J_{r} / J_{d}$ view of $e$ .
Kinetic Energy Loss in Collisions — the $(1 - e^{2})$ energy story behind several traps.
Projectile Motion — the $v = 2 g h$ link the height traps rely on.