Types of collisions — elastic, inelastic, perfectly inelastic with conservation laws

Question

How do elastic, inelastic, and perfectly inelastic collisions differ? Which conservation laws apply in each case?

Solution — Step by Step

In ANY collision (elastic, inelastic, or perfectly inelastic), momentum is conserved as long as no external force acts:

m_1\vec{v}_1 + m_2\vec{v}_2 = m_1\vec{v}_1' + m_2\vec{v}_2'

This is because internal forces (between the colliding objects) are equal and opposite (Newton’s third law), so they cancel out.

Both momentum and kinetic energy are conserved.

\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2

For 1D elastic collision (object 2 initially at rest):

v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 \quad \text{and} \quad v_2' = \frac{2m_1}{m_1 + m_2}v_1

Special cases:

Equal masses ( $m_1 = m_2$ ): velocities exchange ( $v_1' = 0, v_2' = v_1$ )
Heavy hitting light ( $m_1 \gg m_2$ ): heavy barely slows, light flies off at $\approx 2v_1$
Light hitting heavy ( $m_1 \ll m_2$ ): light bounces back at $\approx -v_1$ , heavy barely moves

Maximum kinetic energy is lost. The objects stick together and move with a common velocity:

v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}

KE lost:

\Delta KE = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(v_1 - v_2)^2

This lost KE converts to heat, sound, and deformation.

graph TD
    A[Collision] --> B{KE conserved?}
    B -->|Yes| C[Elastic: objects bounce apart]
    B -->|No| D{Objects stick?}
    D -->|Yes| E[Perfectly inelastic: maximum KE loss]
    D -->|No| F[Inelastic: some KE lost]
    C --> G[Momentum conserved + KE conserved]
    E --> H[Momentum conserved, KE NOT conserved]
    F --> H

Why This Works

Property	Elastic	Inelastic	Perfectly Inelastic
Momentum conserved?	Yes	Yes	Yes
KE conserved?	Yes	No	No (maximum loss)
Objects after collision	Separate	Separate	Stick together
Coefficient of restitution ( $e$ )	1	0 < $e$ < 1	0
Examples	Billiard balls (approx.), atomic collisions	Most real collisions	Bullet embedding in block

The coefficient of restitution $e = \frac{v_2' - v_1'}{v_1 - v_2}$ quantifies the “bounciness.” $e = 1$ means perfectly elastic; $e = 0$ means perfectly inelastic.

Alternative Method

For JEE problems involving oblique (2D) collisions, apply momentum conservation component-wise:

Along the line of impact: use coefficient of restitution
Perpendicular to the line of impact: velocities remain unchanged (no force in that direction)

This is commonly tested with two spheres colliding at an angle or a ball hitting a wall obliquely.

Common Mistake

Students assume “inelastic collision means momentum is not conserved.” This is wrong. Momentum is ALWAYS conserved in all collisions (assuming no external force). It is kinetic energy that may or may not be conserved. The word “inelastic” refers to KE loss, not momentum loss. NEET and JEE both test this as a conceptual true/false question.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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