Collisions

When two bodies smash into each other, interaction forces act over a very short time. Momentum is always conserved in collisions (no external impulse during the brief impact). Kinetic energy may or may not be conserved, depending on the type of collision.

Concept

Classification:

• Elastic collision: total KE conserved. Idealized — billiard balls approximate this.
• Inelastic collision: KE is not conserved (some converted to heat, sound, deformation). Momentum still conserved.
• Perfectly inelastic collision: bodies stick together after collision. KE loss is maximum (consistent with momentum conservation).

Coefficient of restitution:

$e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = \frac{v_2' - v_1'}{v_1 - v_2}$

(for 1D, primes denote post-collision). $e = 1$ for elastic, $0 \leq e < 1$ for inelastic, $e = 0$ for perfectly inelastic.

1D Elastic Collision (masses $m_1$, $m_2$; initial velocities $u_1$, $u_2$):

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

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

Special cases of 1D elastic collision:

• Equal masses: velocities exchange. $v_1 = u_2$, $v_2 = u_1$.
• Heavy body hits stationary light body: light body rebounds at $\approx 2 u_1$, heavy continues at $\approx u_1$.
• Light body hits stationary heavy body: light body rebounds at $\approx -u_1$, heavy barely moves.

Derivation

1D elastic collision. Momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (1)$

KE:

$\tfrac{1}{2}m_1 u_1^2 + \tfrac{1}{2}m_2 u_2^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 \quad (2)$

Rearranging (1): $m_1(u_1 - v_1) = m_2(v_2 - u_2)$.

Rearranging (2): $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$, factoring: $m_1(u_1-v_1)(u_1+v_1) = m_2(v_2-u_2)(v_2+u_2)$.

Dividing the second by the first:

$u_1 + v_1 = v_2 + u_2 \quad \Longrightarrow \quad u_1 - u_2 = v_2 - v_1$

This says the velocity of approach equals the velocity of separation ($e = 1$). Solving simultaneously with (1) gives the formulas above.

Perfectly inelastic. They stick: $v_1' = v_2' = v$. By momentum conservation,

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

KE loss:

$\Delta K = \tfrac{1}{2}\frac{m_1 m_2}{m_1+m_2}(u_1 - u_2)^2$

Worked Example

A 2 kg ball moving at 4 m/s collides head-on elastically with a stationary 3 kg ball. Find the velocities after collision.

Solution:

Using the elastic-collision formulas with $u_2 = 0$:

$v_1 = \frac{(2-3)(4) + 0}{2+3} = \frac{-4}{5} = -0.8 \, \text{m/s}$

$v_2 = \frac{0 + 2(2)(4)}{5} = \frac{16}{5} = 3.2 \, \text{m/s}$

The 2 kg ball bounces back at 0.8 m/s, and the 3 kg ball moves forward at 3.2 m/s.

Check. Momentum before: $2 \cdot 4 = 8$ kg·m/s. After: $2(-0.8) + 3(3.2) = -1.6 + 9.6 = 8$ kg·m/s. ✓

KE before: $\tfrac{1}{2}(2)(16) = 16 \, \text{J}$. After: $\tfrac{1}{2}(2)(0.64) + \tfrac{1}{2}(3)(10.24) = 0.64 + 15.36 = 16 \, \text{J}$. ✓

Common Confusions

• Momentum is always conserved in any collision (in absence of external impulse), but KE is conserved only in elastic collisions.
• "Sticking together is elastic with zero relative velocity." No — sticking is perfectly inelastic ($e = 0$); much KE is lost.
• Sign of $e$. $e$ is defined as a ratio of magnitudes of separation to approach speeds; some textbooks express it as $(v_2'-v_1')/(u_1-u_2)$, which is positive when the bodies are moving apart after collision.
• In 2D, momentum is conserved in each direction separately. Use components.

Key Takeaways

• Momentum conserved in all collisions (zero net external impulse).
• KE conserved in elastic collisions only.
• Coefficient of restitution $e$: 1 (elastic), 0 (perfectly inelastic), in between for partly inelastic.
• Equal masses in 1D elastic exchange velocities.
• KE loss in perfectly inelastic collision: $\Delta K = \frac{m_1 m_2}{2(m_1+m_2)}(u_1-u_2)^2$.