Conservation of Momentum in Collisions

By the end of this lesson, you will be able to state and apply conservation of momentum, classify collisions as elastic/inelastic, and solve 1D collision problems.

Conservation of Momentum

In an isolated system (no external forces), total momentum is conserved:

$$\vec{p}{\text{total, before}} = \vec{p}{\text{total, after}}$$

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

Types of Collisions

Perfectly elastic: both momentum AND kinetic energy are conserved. (Rare in macroscopic world; approximated by billiard balls and atomic/molecular collisions.)

Inelastic: momentum conserved; kinetic energy NOT conserved (some converted to heat, sound, deformation).

Perfectly inelastic: objects stick together after collision; momentum conserved; maximum KE loss.

Example 1 — Perfectly Inelastic Collision

A 3 kg cart at 4 m/s collides with a stationary 2 kg cart and they stick together. Find the final velocity.

$m_1v_1 + m_2v_2 = (m_1+m_2)v_f$

$3(4) + 2(0) = 5v_f \implies v_f = 2.4$ m/s

KE before: $\frac{1}{2}(3)(16) = 24$ J; KE after: $\frac{1}{2}(5)(5.76) = 14.4$ J. Lost: 9.6 J → heat/sound.

Example 2 — Elastic Collision

Two 1 kg balls: ball 1 at 6 m/s collides elastically with stationary ball 2.

For equal masses elastic: they swap velocities.

$v_{1f} = 0$ m/s, $v_{2f} = 6$ m/s.

Check KE: before = $\frac{1}{2}(1)(36) = 18$ J; after = $\frac{1}{2}(1)(36) = 18$ J ✓

Example 3 — Recoil

A 70 kg skater at rest throws a 2 kg ball at 15 m/s. Find the skater's recoil velocity.

Total momentum before = 0 (both at rest):

$0 = 2(15) + 70v_{\text{skater}} \implies v_{\text{skater}} = -30/70 = -0.43$ m/s

The skater moves backward at 0.43 m/s.

Summary

Conservation of momentum: $\Sigma p_i = \Sigma p_f$ in isolated systems.

Elastic: KE and momentum conserved; no permanent deformation.

Inelastic: only momentum conserved; some KE lost.

Perfectly inelastic: objects stick; maximum KE loss.

Recoil: total momentum zero → each object gets equal and opposite momenta.