JEE Mains Physics · Work, Energy and Power

Elastic Collisions and Coefficient of Restitution

In an elastic collision both momentum and kinetic energy are conserved; the coefficient of restitution compares the speed of separation with the speed of approach.

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Why this matters

Fifteen PYQs, twelve of them multiple choice, and one from 2026; five carry a figure. Eleven are elastic collisions: velocities after a head-on hit, equal masses exchanging velocities, the share of kinetic energy passed on, a spring squeezed at the common velocity, a pendulum bob striking a block; four use the coefficient of restitution for a ball bouncing on a floor. Two results, the velocities after a hit on a body at rest and the rebound height e²h, answer most of them.

Concept 1 of 2: Head-on elastic collisions

In an elastic collision no kinetic energy is lost, and momentum is conserved as always. For a head-on hit, the two conditions together say the bodies separate as fast as they approached. For a target at rest this gives two short formulas that answer most questions without solving a quadratic.

Definition

Target at rest: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}\,u$ and $v_2 = \dfrac{2m_1}{m_1 + m_2}\,u$ .
Equal masses exchange velocities. A very heavy body hitting a light one at rest sends it off at nearly 2u; a light body hitting a very heavy one bounces back at nearly u.
Fraction of kinetic energy passed to the target: $\dfrac{4m_1m_2}{(m_1 + m_2)^{2}}$ .
General head-on case: momentum plus $u_1 - u_2 = v_2 - v_1$ .
Two blocks with a spring between them: the spring is most compressed when both move at the common velocity, and then $\tfrac12 kx^{2} = \tfrac12\dfrac{m_1m_2}{m_1 + m_2}u^{2}$ .
Glancing elastic hit between equal masses, one at rest: they move off at 90° to each other. For both to leave at equal angles to the original line, $M/m$ can be at most 3.

Elastic collision with a target at rest

v_1 = \frac{m_1 - m_2}{m_1 + m_2}\,u \qquad v_2 = \frac{2m_1}{m_1 + m_2}\,u \qquad \frac{K_2}{K_1} = \frac{4m_1m_2}{(m_1 + m_2)^{2}}

Worked example

A 2 kg ball moving at 6 m/s hits a 4 kg ball at rest head-on, and the collision is elastic. Find both velocities after the collision and the fraction of kinetic energy passed to the 4 kg ball.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q115Moderate

Example 1 · Work, Energy and Power · Elastic Collisions and Coefficient of Restitution

A body of mass

1\text{ }kg

collides head on elastically with a stationary body of mass

3\text{ }kg

. After collision, the smaller body reverses its direction of motion and moves with a speed of

\frac{2\text{ }m}{s}

. The initial speed of the smaller body before collision is

\_\_\_\_

ms^{- 1}

Exchange of velocities happens one collision at a time

With three identical balls in a line, apply the exchanges in the order the collisions happen. Swapping the first and last velocities at once skips the middle ball and gives the wrong final set.

Maximum compression is at the common velocity

Two blocks with a spring between them squeeze it most when they move at the same speed, not when one of them stops. At that instant the kinetic energy missing from the pair is stored in the spring.

A light body bounces back from a heavy one

When m₁ < m₂, v₁ = (m₁ − m₂)u/(m₁ + m₂) is negative: the light body reverses. Taking its speed as positive in the momentum equation gives a wrong starting speed.

Drill 10 more on head-on elastic collisions

Concept 2 of 2: Coefficient of restitution and bouncing balls

The coefficient of restitution e says how springy a collision is: the speed at which the bodies separate divided by the speed at which they approached. For a ball on the floor, it leaves at e times the speed it arrived with. Height goes as speed squared, so each bounce rises to e² of the height before.

Definition

$e = \dfrac{v_2 - v_1}{u_1 - u_2}$ : 1 for elastic, 0 when the bodies stick, in between otherwise.
Ball dropped from h: arrives at $\sqrt{2gh}$ , leaves at $e\sqrt{2gh}$ , rises to $e^{2}h$ .
Fraction of kinetic energy kept per bounce: $e^{2}$ ; fraction lost: $1 - e^{2}$ .
Height after n bounces: $e^{2n}h$ .
Bouncing until it stops: total distance $h\,\dfrac{1 + e^{2}}{1 - e^{2}}$ ; total time $\sqrt{\dfrac{2h}{g}}\,\dfrac{1 + e}{1 - e}$ .
e from two heights: $e = \sqrt{h_2/h_1}$ ; from two speeds: $e = v_2/v_1$ .

Coefficient of restitution

e = \frac{v_2 - v_1}{u_1 - u_2} \qquad h' = e^{2}h \qquad \frac{K'}{K} = e^{2}

Worked example

A ball is dropped from 8 m onto a floor, and the coefficient of restitution is 0.75. Find the height of the first bounce and the percentage of kinetic energy lost in the impact.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q116Moderate

Example 2 · Work, Energy and Power · Elastic Collisions and Coefficient of Restitution

A ball is dropped from a height of

20\text{ }m

. If the coefficient of restitution for the collision between ball and floor is 0.5 , after hitting the floor, the ball rebounds to a height of___

m

Height goes with e², speed with e

A ball that leaves the floor at e times its arrival speed rises to e² times its starting height. Using e for the height ratio gives a rebound that is too high.

Every bounce is travelled twice

After the first fall, each bounce goes up and comes back down. The total distance is h + 2e²h + 2e⁴h + …, which sums to h(1 + e²)/(1 − e²); counting each bounce once gives too little.

Drill 3 more on coefficient of restitution and bouncing balls

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Head-on elastic collisions
Elastic collision with a target at rest
$v_1 = \frac{m_1 - m_2}{m_1 + m_2}\,u \qquad v_2 = \frac{2m_1}{m_1 + m_2}\,u \qquad \frac{K_2}{K_1} = \frac{4m_1m_2}{(m_1 + m_2)^{2}}$
Coefficient of restitution and bouncing balls
Coefficient of restitution
$e = \frac{v_2 - v_1}{u_1 - u_2} \qquad h' = e^{2}h \qquad \frac{K'}{K} = e^{2}$

Watch out for (5)

Exchange of velocities happens one collision at a time
→ Head-on elastic collisions
Maximum compression is at the common velocity
→ Head-on elastic collisions
A light body bounces back from a heavy one
→ Head-on elastic collisions
Height goes with e², speed with e
→ Coefficient of restitution and bouncing balls
Every bounce is travelled twice
→ Coefficient of restitution and bouncing balls

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