Elastic vs inelastic collision in 1D — derive final velocities

Question

Two bodies of masses $m_1$ and $m_2$ collide head-on in one dimension. Body 1 moves with velocity $u_1$ and body 2 with velocity $u_2$ . Derive expressions for final velocities $v_1$ and $v_2$ for (a) a perfectly elastic collision, and (b) a perfectly inelastic collision.

(NCERT Class 11, Chapter 6)

Solution — Step by Step

In any collision (elastic or inelastic), linear momentum is conserved:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{...(i)}

For an elastic collision, kinetic energy is also conserved:

\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \quad \text{...(ii)}

Instead of solving the KE equation directly (messy algebra), use this equivalent condition for elastic collisions:

u_1 - u_2 = -(v_1 - v_2)

This says: the relative velocity reverses direction after an elastic collision. This is called the coefficient of restitution $e = 1$ condition.

From equations (i) and the relative velocity condition, solve simultaneously:

v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}

v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}

These are the standard NCERT formulae. Memorise the pattern: the “other mass” always appears with coefficient 2.

In a perfectly inelastic collision ( $e = 0$ ), the two bodies move with a common velocity $v$ after collision:

m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

\mathbf{v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}}

KE is not conserved here — some energy converts to heat, sound, and deformation.

Why This Works

The elastic collision formulae come from solving two simultaneous equations (momentum + energy). The “velocity of approach = velocity of separation” shortcut replaces the quadratic KE equation with a linear one, making the algebra much cleaner.

For inelastic collisions, we lose one equation (KE conservation) but gain one constraint (common final velocity), so the system is still solvable.

The coefficient of restitution $e$ bridges both cases: $e = 1$ for perfectly elastic, $e = 0$ for perfectly inelastic, and $0 < e < 1$ for real-world collisions.

Alternative Method — Centre of Mass Frame

In the CM frame, the total momentum is zero. Each body simply reverses its velocity in an elastic collision. Transform back to the lab frame to get the same results. This method is elegant for JEE Advanced problems involving multiple collisions.

Special cases to remember: (1) Equal masses ( $m_1 = m_2$ ): velocities exchange completely in elastic collision. (2) Heavy body hits light body at rest ( $m_1 \gg m_2$ ): light body flies off at $\approx 2u_1$ . (3) Light body hits heavy body at rest: light body bounces back with nearly the same speed.

Common Mistake

Students often apply KE conservation to inelastic collisions. KE is conserved only in perfectly elastic collisions. In NEET/JEE problems, if the question says “bodies stick together” or “embedded”, it is perfectly inelastic — do NOT write the KE conservation equation. Use only momentum conservation.

Question

Solution — Step by Step

Why This Works

Alternative Method — Centre of Mass Frame

Common Mistake

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