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

WHAT is it?

Why is the numerator $v_2-v_1$ and denominator $u_1-u_2$ (note the swapped order)? Because for the bodies to actually collide, body 1 must be catching up to body 2, so $u_1 > u_2$ → approach $= u_1-u_2 > 0$ . After collision they must move apart, so body 2 ends up faster/ahead, $v_2 > v_1$ → separation $= v_2 - v_1 > 0$ . The swap keeps both quantities positive, so $e \ge 0$ automatically.

WHY does it exist? (The physics behind it)

HOW to derive it from first principles

We model the collision in two phases.

Phase 1 — Deformation: bodies squash together until they momentarily move at a common velocity $v_c$ . The colliding surfaces push on each other; impulse $J_d$ acts.

Phase 2 — Restoration: the squashed material springs back, pushing the bodies apart; impulse $J_r$ acts.

Deriving the velocity formula from impulses. Take body 1 (mass $m_1$ ). Let positive direction be its motion.

Deformation phase (from $u_1$ to common $v_c$ ): $-J_d = m_1(v_c - u_1) \quad\Rightarrow\quad J_d = m_1(u_1 - v_c)$

Restoration phase (from $v_c$ to final $v_1$ ): $-J_r = m_1(v_1 - v_c) \quad\Rightarrow\quad J_r = m_1(v_c - v_1)$

So for body 1: $e = \frac{J_r}{J_d} = \frac{v_c - v_1}{u_1 - v_c}$

Doing the identical bookkeeping for body 2 (forces are reversed, so impulses on it are $+J_d, +J_r$ ): $J_d = m_2(v_c - u_2), \qquad J_r = m_2(v_2 - v_c)$ $\Rightarrow e = \frac{v_2 - v_c}{v_c - u_2}$

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

The special case: ball bouncing off a fixed floor

The floor is "body 2" with infinite mass, so $u_2=v_2=0$ . Drop from height $h_1$ → hits floor at $u_1=\sqrt{2gh_1}$ (downward). Rebounds at $v_1=\sqrt{2gh_2}$ (upward), so in the chosen sign convention separation $=v_2-v_1$ becomes the rebound speed.

Worked examples

Common mistakes (Steel-manned)

Active recall

Recall Cover and answer

Write $e$ in terms of velocities and name each part.
Why do we even need $e$ as a separate equation?
What is conserved when $e<1$ and what is lost?
Derive the bounce-height relation $h_2=e^2 h_1$ .
What does $e=0$ physically mean?

Recall Feynman: explain to a 12-year-old

Imagine two toy cars crashing. Before the crash they were rushing toward each other — that's their "coming-together speed." After the crash they bounce apart — that's their "flying-apart speed." The number $e$ just asks: how much of the speed survived the crash? A super bouncy rubber ball keeps almost all of it ( $e$ near 1). A lump of clay keeps none — it just goes splat and sticks ( $e=0$ ). So $e$ is the "bounciness score" of the crash, and it depends on what stuff the things are made of.

Connections

Conservation of Linear Momentum — the first equation; restitution is the second.
Elastic Collisions — the case $e=1$ where KE is conserved too.
Perfectly Inelastic Collisions — the case $e=0$ , bodies move together.
Impulse and Momentum — the $J_r/J_d$ definition of $e$ .
Kinetic Energy Loss in Collisions — energy lost $\propto (1-e^2)$ .
Projectile Motion — needed for the bouncing-ball height analysis.

Coefficient of restitution formula

e=\dfrac{v_2-v_1}{u_1-u_2}

= separation speed / approach speed

Why are the indices swapped in numerator vs denominator?

So both approach (

u_1>u_2

) and separation (

v_2>v_1

) come out positive, giving

0\le e\le1

What does

e=1

mean physically?

Perfectly elastic collision — speed of separation equals speed of approach, kinetic energy conserved

What does

e=0

mean physically?

Perfectly inelastic — bodies stick and move with common velocity, maximum KE loss

Is momentum conserved when

e<1

Yes, always (no external impulse). Kinetic energy is what is lost

e

as a ratio of impulses

e=J_r/J_d

= restoration impulse / deformation impulse

Bounce height relation for a ball on a fixed floor

e=\sqrt{h_2/h_1}

, so

h_n=e^{2n}h_0

A ball drops 2 m and rebounds 1.28 m; find

e

e=\sqrt{1.28/2}=\sqrt{0.64}=0.8

Why is

e

a ratio of speeds but heights give

e^2

Because

h=v^2/2g

, so height scales with speed squared

Common velocity trick in derivation

Add numerators and denominators of the two impulse ratios;

v_c

cancels, leaving

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab do cheezein takraati hain (collision), toh momentum conservation se sirf ek equation milti hai, lekin final velocities do hoti hain — $v_1$ aur $v_2$ . Ek equation se do unknown solve nahi hote! Isliye humein ek doosri equation chahiye, aur wahi deta hai coefficient of restitution $e$ . Iska matlab simple hai: takraane se pehle dono kitni tezi se paas aa rahe the (approach speed = $u_1-u_2$ ) aur takraane ke baad kitni tezi se door ja rahe hain (separation speed = $v_2-v_1$ ). $e$ in dono ka ratio hai: $e=(v_2-v_1)/(u_1-u_2)$ .

$e$ basically "bounciness score" hai. Agar $e=1$ ho toh perfectly elastic — saari speed wapas mil gayi, kinetic energy bhi conserve. Agar $e=0$ ho toh perfectly inelastic — cheezein chipak jaati hain, splat, jaise geeli mitti. Beech mein jitna $e$ chhota hai utna bouncy. Yaad rakho: momentum har case mein conserve hota hai, chahe $e$ kuch bhi ho — lose sirf kinetic energy hoti hai jab $e<1$ .

Ek important trick: gend (ball) ko fixed floor pe girao. Floor ki mass infinite, toh $u_2=v_2=0$ . Tab $e=\sqrt{h_2/h_1}$ ban jaata hai — yaani rebound height ka aur drop height ka ratio under root. Bahut students galti karte hain $e=h_2/h_1$ likh dete hain — galat! Kyunki height speed ke square pe depend karti hai ( $h=v^2/2g$ ), isliye square root lagana padta hai.

Exam tip (80/20): bas teen cheezein pakad lo — (1) formula with swapped indices, (2) momentum + restitution ko milakar do equations solve karna, (3) bounce-height ka root wala formula. Inhi se 90% questions ban jaate hain.

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

WHAT is it?

WHY does it exist? (The physics behind it)

HOW to derive it from first principles

The special case: ball bouncing off a fixed floor

Worked examples

Common mistakes (Steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Momentum & Collisions

Connections