Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂): Worked examples

This page is the workout gym for the parent note. We already know the formula: $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{speed of separation}{speed of approach}$ Here we hunt down every kind of collision problem the topic can hand you and solve each fully. Before that, we lay out a map so you never meet a situation we didn't drill.

Sign convention (fix this once, for the whole page)

Every velocity below is signed, and we use ONE rule everywhere.

The two master equations (and the closed form)

Every 1-D restitution problem uses exactly two facts:

Momentum conservation (one equation): $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ .
Restitution law (second equation): $v_{2} - v_{1} = e (u_{1} - u_{2})$ .

Two linear equations, two unknowns $v_{1}, v_{2}$ → there is one canonical closed-form answer. Let us solve the pair once, in general, so every worked example below is just this formula with numbers.

Every worked example either plugs into this closed form or inverts it (solving for $e$ ).

The scenario matrix — and why it is exhaustive

Any collision is fixed by four ingredients: the two initial velocities $u_{1}, u_{2}$ (their signs and order), the value of $e$ (from $0$ to $1$ ), and whether one mass is finite or infinite (a wall). Every possibility falls into one of the branches below — this is a genuine case split, not a wishlist.

Intuition Why these ten cells cover everything

Walk the decision tree once:

Is there a collision at all? A collision needs a positive approach speed $u_{1} - u_{2} > 0$ . If $u_{1} = u_{2}$ (or $u_{1} < u_{2}$ , body 1 falling behind) there is no impact → that is the degenerate branch (Cell H).
If yes, what are the signs of $u_{1}, u_{2}$ ? Either both same sign — a chase (Cell A) — or opposite signs — a head-on (Cell B). (Sign-flips of the whole picture just relabel left/right, so these two exhaust the geometry.)
What is $e$ ? The endpoints $e = 1$ (Cell C) and $e = 0$ (Cell D) are the two physical limits; every $0 < e < 1$ in between is handled by the same closed form (Cells A, B demonstrate it).
Is a mass infinite? Then it is a fixed wall, $u_{2} = v_{2} = 0$ (Cell F).
What are you asked to find? Forward (find $v$ 's: A–D, F), or inverse (find $e$ from velocities (Cell E), from heights (Cell G), or from energy loss (Cell J)).
Is it 1-D or 2-D? A 2-D bounce off a surface reduces to 1-D on the perpendicular axis (Cell I). Every leaf of this tree is a cell — so nothing is left out.

Cell	Case class	What is special	Example
A	Same-direction chase ( $u_{1}, u_{2}$ both $+$ , $u_{1} > u_{2}$ )	both velocities positive	Ex 1
B	Head-on, opposite signs ( $u_{1} > 0$ , $u_{2} < 0$ )	one velocity negative → approach speed adds	Ex 2
C	$e = 1$ elastic limit	KE also conserved; velocities swap for equal mass	Ex 3
D	$e = 0$ perfectly inelastic	bodies stick, one common velocity	Ex 4
E	Find $e$ from data (inverse problem)	solve for $e$ , not for $v$ 's	Ex 5
F	Fixed wall / infinite mass	$u_{2} = v_{2} = 0$ , ball just reverses	Ex 6
G	Bouncing ball, height & multiple bounces	$h \propto v^{2}$ , needs ; $e^{2 n}$	Ex 7
H	Degenerate: equal initial speeds ( $u_{1} = u_{2}$ )	approach $= 0$ , no collision — formula breaks	Ex 8
I	Real-world word problem (2-D bounce off ground)	vertical component obeys $e$ , horizontal untouched	Ex 9
J	Exam twist: back-solve $e$ from KE loss	uses $KE lost \propto (1 - e^{2})$	Ex 10

Prerequisites you may want open: Conservation of Linear Momentum, Elastic Collisions, Perfectly Inelastic Collisions, Kinetic Energy Loss in Collisions, Projectile Motion.

Cell A — same-direction chase

Worked example Ex 1 — a fast ball catching a slow one

$m_{1} = 3$ kg at $u_{1} = + 8$ m/s catches up to $m_{2} = 1$ kg at $u_{2} = + 2$ m/s. Restitution $e = 0.5$ . Find $v_{1}, v_{2}$ .

Forecast: heavy fast ball hits light slow ball; the light one should get flung forward faster, the heavy one slows a bit but keeps going. Guess $v_{2} > v_{1}$ , both still positive.

Approach speed $= u_{1} - u_{2} = 8 - 2 = 6$ m/s. Why this step? Both move right, so how fast they close in is the difference of their speeds, not the sum.
Restitution equation: $v_{2} - v_{1} = e (u_{1} - u_{2}) = 0.5 \times 6 = 3$ . Why? Newton's law: separation speed is $e$ times approach speed.
Momentum: $3 (8) + 1 (2) = 3 v_{1} + 1 v_{2} \Rightarrow 26 = 3 v_{1} + v_{2}$ . Why? No external push, so total momentum is unchanged — this is our second equation.
Solve the pair. From step 2, $v_{2} = v_{1} + 3$ . Substitute: $26 = 3 v_{1} + (v_{1} + 3) = 4 v_{1} + 3 \Rightarrow v_{1} = \frac{23}{4} = 5.75$ , then $v_{2} = 8.75$ . Why? Two equations, two unknowns → unique answer. (Same result as plugging into the closed form above.)

Verify: $v_{2} - v_{1} = 8.75 - 5.75 = 3 = 0.5 \times 6$ ✓. Momentum after $= 3 (5.75) + 1 (8.75) = 17.25 + 8.75 = 26$ ✓. Both positive, $v_{2} > v_{1}$ ✓ — flung forward as forecast.

Cell B — head-on collision (opposite signs)

Worked example Ex 2 — two balls rushing at each other

$m_{1} = 2$ kg at $u_{1} = + 5$ m/s moving right; $m_{2} = 2$ kg at $u_{2} = - 3$ m/s moving left. $e = 0.6$ . Find final velocities.

Forecast: they smash head-on with equal mass; expect both to reverse or slow strongly. Guess $v_{1}$ becomes negative-ish (bounced back), $v_{2}$ positive.

Approach speed $= u_{1} - u_{2} = 5 - (- 3) = 8$ m/s. Why this step? Here $u_{2}$ is negative because it moves left. Subtracting a negative adds — that's exactly right, since two things rushing toward each other close in at the sum of their speeds.
Restitution: $v_{2} - v_{1} = 0.6 \times 8 = 4.8$ . Why? Same law; the signed subtraction already baked the geometry in.
Momentum: $2 (5) + 2 (- 3) = 2 v_{1} + 2 v_{2} \Rightarrow 10 - 6 = 2 v_{1} + 2 v_{2} \Rightarrow v_{1} + v_{2} = 2$ . Why? Keep signs — leftward momentum is negative.
Solve. $v_{2} = v_{1} + 4.8$ into $v_{1} + v_{2} = 2$ : $2 v_{1} + 4.8 = 2 \Rightarrow v_{1} = - 1.4$ , $v_{2} = 3.4$ . Why? Same elimination as before.

Verify: $v_{2} - v_{1} = 3.4 - (- 1.4) = 4.8$ ✓. Momentum $= 2 (- 1.4) + 2 (3.4) = - 2.8 + 6.8 = 4$ , and before $= 10 - 6 = 4$ ✓. $v_{1} < 0$ (ball 1 bounced back left) as forecast.

Cell C — the elastic limit $e = 1$

Worked example Ex 3 — perfect bounce, equal masses

$m_{1} = m_{2} = 1$ kg, $u_{1} = + 6$ m/s, $u_{2} = 0$ , $e = 1$ . Find final velocities and confirm KE is conserved.

Forecast: the classic Newton's-cradle case — equal masses, perfect bounce. Guess they swap velocities: $v_{1} = 0$ , $v_{2} = 6$ .

Restitution ( $e = 1$ ): $v_{2} - v_{1} = 1 \times (6 - 0) = 6$ . Why? At $e = 1$ separation speed equals approach speed — no speed lost.
Momentum: $6 + 0 = v_{1} + v_{2} \Rightarrow v_{1} + v_{2} = 6$ . Why? Equal masses cancel out of the equation nicely.
Solve: add-and-subtract → $v_{2} = 6$ , $v_{1} = 0$ . Why? $v_{1} + v_{2} = 6$ and $v_{2} - v_{1} = 6$ give $v_{2} = 6, v_{1} = 0$ instantly. (The closed form with $e = 1, m_{1} = m_{2}$ gives exactly this velocity swap.)
KE check: before $= \frac{1}{2} (1) (6^{2}) = 18$ J; after $= \frac{1}{2} (1) (0^{2}) + \frac{1}{2} (1) (6^{2}) = 18$ J. Why this step? $e = 1$ must also conserve kinetic energy — this is the signature of an elastic collision.

Verify: velocities swapped exactly; KE identical (18 J → 18 J) ✓.

Cell D — perfectly inelastic $e = 0$

Worked example Ex 4 — they stick together (with the KE-loss formula derived)

$m_{1} = 4$ kg at $u_{1} = + 5$ m/s hits $m_{2} = 6$ kg at rest ( $u_{2} = 0$ ), $e = 0$ . Find the final velocity and KE lost.

Forecast: $e = 0$ means splat — one common velocity. Total mass 10 kg; guess final speed around $2$ m/s.

Restitution ( $e = 0$ ): $v_{2} - v_{1} = 0 \Rightarrow v_{1} = v_{2} = v_{c}$ . Why? Zero separation speed means they leave together at one common velocity — the definition of perfectly inelastic.
Momentum: $4 (5) + 0 = (4 + 6) v_{c} \Rightarrow 20 = 10 v_{c} \Rightarrow v_{c} = 2$ m/s. Why? With one unknown left, momentum alone finishes the job.
KE lost: before $= \frac{1}{2} (4) (5^{2}) = 50$ J; after $= \frac{1}{2} (10) (2^{2}) = 20$ J; lost $= 30$ J. Why this step? $e = 0$ gives the maximum possible KE loss (all "bounce-back" energy gone to heat/deformation).
Where the general fraction comes from. With $u_{2} = 0$ and $v_{c} = \frac{m _{1} u _{1}}{m _{1} + m _{2}}$ : before $= \frac{1}{2} m_{1} u_{1}^{2}$ ; after $= \frac{1}{2} (m_{1} + m_{2}) v_{c}^{2} = \frac{1}{2} \frac{m _{1}^{2} u _{1}^{2}}{m _{1} + m _{2}}$ . Divide after by before: $\frac{after}{before} = \frac{m _{1}}{m _{1} + m _{2}}$ , so the lost fraction $= 1 - \frac{m _{1}}{m _{1} + m _{2}} = \frac{m _{2}}{m _{1} + m _{2}}$ . Why this step? It shows the " $m_{2} / (m_{1} + m_{2})$ " rule is not magic — it is exactly steps 2–3 done with symbols instead of numbers, linking to Kinetic Energy Loss in Collisions.

Verify: momentum after $= 10 \times 2 = 20 =$ before ✓. Numeric fraction $= 30/50 = 0.6$ ; symbolic rule $\frac{m _{2}}{m _{1} + m _{2}} = \frac{6}{10} = 0.6$ ✓ — the two agree.

Cell E — inverse problem, find $e$

Cell F — ball off a fixed wall

Worked example Ex 6 — bounce off an immovable wall

A ball hits a fixed vertical wall at $u_{1} = + 9$ m/s (toward the wall). Wall is "body 2" with $u_{2} = v_{2} = 0$ . If $e = 0.7$ , find the rebound speed.

Forecast: the wall doesn't move; the ball just comes back slower. Rebound $\approx 0.7 \times 9 \approx 6.3$ m/s, but travelling the other way (negative).

Restitution: $v_{2} - v_{1} = e (u_{1} - u_{2}) \Rightarrow 0 - v_{1} = 0.7 (9 - 0)$ . Why? An "infinitely heavy" wall stays at rest, so $u_{2} = v_{2} = 0$ .
Solve: $- v_{1} = 6.3 \Rightarrow v_{1} = - 6.3$ m/s. Why? The minus sign is the physics telling us the ball now moves in the opposite direction — it bounced back.
Rebound speed (magnitude) $= 6.3$ m/s. Why this step? Speed is the size of velocity; the sign told us direction, the magnitude tells us how fast.

Verify: $∣ v_{1} ∣/ u_{1} = 6.3/9 = 0.7 = e$ ✓. For a fixed wall $e$ is simply the ratio of rebound to impact speed.

Cell G — bouncing ball, heights and repeated bounces

Here geometry matters, so we draw it. In the figure below, the horizontal axis is the ball's position as it hops rightward and the vertical axis is its height in metres. The ball is dropped straight down on the left (dotted white line from height $h_{0}$ ), and each successive bounce arc rises to a colour-coded peak — blue for peak 1, yellow for peak 2, pink for peak 3. The dashed guide lines on the left mark those peak heights. The whole point of the figure: each peak sits at exactly $e^{2}$ times the height of the peak before it, so the peaks march downward by a fixed multiplying factor.

Figure — Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂)

Worked example Ex 7 — drop, rebound, and the

$n$ -th bounce A ball is dropped from $h_{0} = 2.0$ m and rebounds to $h_{1} = 1.25$ m. (a) Find $e$ . (b) To what height does it reach after the 3rd bounce?

Forecast: it loses some height each bounce; $e$ noticeably below 1. After 3 bounces it should be quite low, maybe half a metre.

Impact & rebound speeds. Impact speed $= 2 g h_{0}$ , rebound speed $= 2 g h_{1}$ (from the definition box). Why this step? Convert measured heights into the speeds that $e$ actually compares.
Height ratio → $e$ . $e = \frac{2 g h _{1}}{2 g h _{0}} = \frac{h _{1}}{h _{0}} = \frac{1.25}{2.0} = 0.625 = 0.7906 \dots$ . Why the ? Because $h \propto v^{2}$ , so the ratio of heights is $e^{2}$ ; undoing the square needs a square root. In the figure, "peak 1" (blue) sits at $e^{2} h_{0}$ — read the blue marker against the left-hand dashed guide.
After $n$ bounces: $h_{n} = e^{2 n} h_{0}$ . Why? Each bounce multiplies the height by the fixed factor $e^{2}$ (the yellow and pink markers in the figure show the same ratio repeating); three bounces multiply by $(e^{2})^{3} = e^{6}$ .
Plug $n = 3$ : $h_{3} = e^{6} h_{0}$ . Since $e^{2} = h_{1} / h_{0} = 0.625$ , we get $e^{6} = (0.625)^{3} = 0.2441 \dots$ , so $h_{3} = 0.2441 \times 2.0 = 0.488$ m. Why this step? The question asks specifically for the peak after the 3rd bounce, which by the definition of $h_{n}$ is $h_{3}$ ; substituting $n = 3$ into $h_{n} = e^{2 n} h_{0}$ is the direct way to get it, and writing $e^{6} = (e^{2})^{3}$ lets us reuse the $e^{2} = 0.625$ we already know instead of recomputing $e$ .

Verify: $e^{2} = 0.790 6^{2} = 0.625$ ✓; $h_{3} = (0.625)^{3} \times 2.0 = 0.4883$ m ✓ — lower than $h_{1}$ , as forecast.

Cell H — the degenerate case (no collision)

Worked example Ex 8 — what if they move at the same velocity?

$m_{1} = 2$ kg and $m_{2} = 3$ kg both travel at $u_{1} = u_{2} = + 4$ m/s. What does the restitution formula give?

Forecast: they never touch (both moving in step, no closing). Guess: the formula blows up / is undefined.

Approach speed $= u_{1} - u_{2} = 4 - 4 = 0$ . Why this step? The gap between them isn't shrinking — nobody is catching up — so there is nothing to collide.
Formula check: $e = \frac{v _{2} - v _{1}}{0}$ — division by zero, undefined. Why? The restitution law only describes an actual collision; with zero approach speed there is no impact to characterise.
Physical resolution: no collision → velocities are unchanged, $v_{1} = 4$ , $v_{2} = 4$ . Why? No impulse acts on either body.

Verify: approach $= 0$ makes $e$ undefined ✓ (this is a genuine degenerate input — the formula's denominator vanishes). Momentum trivially conserved since nothing changed: before $= 2 (4) + 3 (4) = 20$ , after $= 2 (4) + 3 (4) = 20$ ✓.

Cell I — real-world 2-D bounce

We now draw the velocity arrows before and after the bounce. The figure uses the 2-D sign convention from the top of the page ( $+ x$ right, $+ y$ up). The blue arrow is the incoming velocity (pointing down-right, into the floor); the pink arrow is the outgoing velocity (up-right); the yellow horizontal arrow is the shared horizontal part, which does not change; the dashed white vertical arrow is the outgoing vertical part; the pink arc marks the rebound angle $θ$ . Watch the yellow horizontal arrow stay full-length while the vertical part shrinks.

Worked example Ex 9 — a ball bounces off the ground at an angle

A ball hits the floor with horizontal component $u_{x} = + 6$ m/s and vertical component $u_{y} = - 8$ m/s (negative because it is moving downward, into the floor — our $+ y$ is up). The floor–ball restitution is $e = 0.75$ . Find the velocity right after the bounce and the rebound angle above the horizontal.

Forecast: the floor can only push up, so only the vertical part is affected; horizontal speed stays the same. Vertical shrinks by factor $e$ , so the ball leaves flatter than it came in.

Split into components. The collision is with a horizontal floor, so restitution applies only to the vertical ( $y$ ) component. Why this step? The floor's push is straight up along $+ y$ ; it exerts no horizontal force, so $u_{x}$ is untouched. In the figure, the yellow horizontal arrow keeps its length while the vertical arrow shrinks.
Horizontal after: $v_{x} = u_{x} = + 6$ m/s. Why? No horizontal impulse → no change (this mirrors the fixed-wall logic of Cell F, applied to the vertical axis instead).
Vertical after: treat the floor as a fixed wall along the $y$ -axis. Its magnitude scales by $e$ : $∣ v_{y} ∣ = e ∣ u_{y} ∣ = 0.75 \times 8 = 6$ m/s. Since it now moves upward in our convention, $v_{y} = + 6$ m/s. Why? Same Cell F reasoning: the perpendicular velocity reverses sign (from $- y$ to $+ y$ ) and scales by $e$ .
Rebound angle above horizontal: $tan θ = \frac{∣ v _{y} ∣}{∣ v _{x} ∣} = \frac{6}{6} = 1 \Rightarrow θ = 4 5^{\circ}$ . Why $tan$ ? On the right triangle formed by the horizontal ( $v_{x}$ ) and vertical ( $v_{y}$ ) arrows in the figure, $tan θ = \frac{opposite (vertical)}{adjacent (horizontal)}$ — the ratio that encodes the launch steepness.

Verify: incoming angle had $tan θ_{in} = 8/6 = 1.333$ ( $θ_{in} = 53.1 3^{\circ}$ ); outgoing $tan θ_{out} = 6/6 = 1$ ( $θ_{out} = 4 5^{\circ}$ ) — the rebound is flatter, exactly as forecast ✓. Speed after $= 6^{2} + 6^{2} = 8.49$ m/s, and $∣ v_{y} ∣/∣ u_{y} ∣ = 6/8 = 0.75 = e$ ✓.

Cell J — exam twist: back-solve $e$ from KE loss

Worked example Ex 10 — find

$e$ from the fraction of KE lost A moving ball $m_{1} = 1$ kg at $u_{1} = + 10$ m/s strikes an identical ball $m_{2} = 1$ kg at rest ( $u_{2} = 0$ ). The collision loses 28% of the total kinetic energy. Find $e$ .

Forecast: modest energy loss → $e$ fairly high, guess around $0.8$ .

KE-loss formula for equal masses, target at rest. For $m_{1} = m_{2} = m$ , $u_{2} = 0$ : the fraction of KE lost is $\frac{1 - e ^{2}}{2}$ . Why this step? KE lost scales as $(1 - e^{2})$ ; the extra $\frac{1}{2}$ is the equal-mass geometry. We use energy here because energy — not velocities — is what the problem gives us.
Set equal to the data: $\frac{1 - e ^{2}}{2} = 0.28 \Rightarrow 1 - e^{2} = 0.56 \Rightarrow e^{2} = 0.44$ . Why? Solve algebraically for the single unknown $e$ .
Take the root: $e = 0.44 = 0.663$ . Why ? We need $e$ itself, and only $e^{2}$ appeared.

Verify (full recompute). With $e = 0.663$ : restitution $v_{2} - v_{1} = 0.663 \times 10 = 6.63$ ; momentum $v_{1} + v_{2} = 10$ ; so $v_{2} = 8.317$ , $v_{1} = 1.683$ . KE before $= \frac{1}{2} (1 0^{2}) = 50$ J; after $= \frac{1}{2} (1.68 3^{2}) + \frac{1}{2} (8.31 7^{2}) = 1.416 + 34.585 = 36.0$ J. Loss $= 14$ J $= 28%$ of 50 J ✓ — matches the given data.

Active recall

Recall Which cell is this?

A ball is thrown at a wall and comes straight back slower — which cell, and what's special? ::: Cell F: fixed wall, $u_{2} = v_{2} = 0$ , so $e =$ rebound speed ÷ impact speed and the velocity sign flips.

Recall Sign trap

Two balls approach head-on at $+ 4$ and $- 4$ m/s. What is the approach speed? ::: $u_{1} - u_{2} = 4 - (- 4) = 8$ m/s — subtracting the negative adds the speeds.

Recall Why the square root for heights?

A ball drops from $h_{0}$ and rebounds to $h_{1}$ . Why is $e = h_{1} / h_{0}$ and not $h_{1} / h_{0}$ ? ::: Because $e$ is a ratio of speeds and $h \propto v^{2}$ , so $h_{1} / h_{0} = e^{2}$ ; undo the square with a root.

Recall The degenerate input

When does the restitution formula become undefined, and what does it mean? ::: When approach speed $u_{1} - u_{2} = 0$ (equal velocities) — division by zero. It means no collision happens.

Recall The closed form

Write $v_{1}$ in terms of $m_{1}, m_{2}, u_{1}, u_{2}, e$ . ::: $v_{1} = \frac{( m _{1} - e m _{2} ) u _{1} + ( 1 + e ) m _{2} u _{2}}{m _{1} + m _{2}}$ .

Connections

Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂) (index 1.4.8) — the parent this page drills.
Conservation of Linear Momentum — the partner equation in every 2-unknown problem here.
Elastic Collisions — Cell C, $e = 1$ .
Perfectly Inelastic Collisions — Cell D, $e = 0$ .
Kinetic Energy Loss in Collisions — powers Cells D and J.
Projectile Motion — free-fall speeds for Cells G and I.
Impulse and Momentum — the deeper $J_{r} / J_{d}$ view of $e$ .

General final velocity

v_{1}

v_{1} = \frac{( m _{1} - e m _{2} ) u _{1} + ( 1 + e ) m _{2} u _{2}}{m _{1} + m _{2}}

General final velocity

v_{2}

v_{2} = \frac{( 1 + e ) m _{1} u _{1} + ( m _{2} - e m _{1} ) u _{2}}{m _{1} + m _{2}}

Approach speed for head-on balls at

+ 4

and

- 4

m/s

u_{1} - u_{2} = 4 - (- 4) = 8

m/s

Fixed-wall restitution

e =

rebound speed ÷ impact speed, with velocity sign reversed

When is the

e

formula undefined?

When approach speed

u_{1} - u_{2} = 0

(equal velocities, no collision)

Height after

n

bounces

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

2-D floor bounce: which component obeys

e

Only the vertical (perpendicular) component; horizontal is unchanged

Equal-mass, target-at-rest KE loss fraction

(1 - e^{2}) /2

Perfectly inelastic (

u_{2} = 0

) KE loss fraction

m_{2} / (m_{1} + m_{2})