Elastic Collisions in Detail — Conserving Momentum and Kinetic Energy

What makes a collision elastic

Collisions split into two kinds by what happens to kinetic energy. In an elastic collision, the total kinetic energy after the impact equals the total before — the objects bounce without permanent deformation, heating or sound losses.

Two conservation laws then apply simultaneously:

1. Momentum is conserved (as in every isolated collision): $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.
2. Kinetic energy is conserved (the elastic condition): $\tfrac{1}{2}m_1u_1^2 + \tfrac{1}{2}m_2u_2^2 = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2$.

Here $u$ denotes velocities before and $v$ velocities after, with signs for direction. Two equations, two unknowns ($v_1$, $v_2$) — a 1D elastic collision is fully determined.

The 1D elastic collision formulas

Solving the two conservation equations together gives closed-form results for the final velocities:

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

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

These look heavy, but three special cases cover most problems:

• Equal masses, one initially at rest ($m_1 = m_2$, $u_2 = 0$): the objects exchange velocities — $v_1 = 0$, $v_2 = u_1$.
• Massive wall ($m_2 \gg m_1$): the light object rebounds at nearly its incoming speed, $v_1 \approx -u_1$.
• Heavy hits light at rest ($m_1 \gg m_2$, $u_2 = 0$): the heavy object barely slows ($v_1 \approx u_1$) while the light one shoots off at $v_2 \approx 2u_1$.

Worked example: billiard balls

A $0.17\ \mathrm{kg}$ billiard ball rolls at $2.0\ \mathrm{m/s}$ into an identical stationary ball, head-on and elastic. Find both final velocities.

Equal masses, one at rest — use the velocity-swap shortcut: $v_1 = 0$, $v_2 = 2.0\ \mathrm{m/s}$.

Check with the books. Momentum before: $(0.17)(2.0) = 0.34\ \mathrm{kg\cdot m/s}$; after: $(0.17)(0) + (0.17)(2.0) = 0.34\ \mathrm{kg\cdot m/s}$ ✓. Kinetic energy before: $\tfrac{1}{2}(0.17)(2.0)^2 = 0.34\ \mathrm{J}$; after: $0 + \tfrac{1}{2}(0.17)(2.0)^2 = 0.34\ \mathrm{J}$ ✓.

In real pool there is a faint click — a whisper of energy lost to sound — so the cue ball retains a trace of motion. The idealization is excellent but not perfect.

Worked example: unequal masses

Now the general case. A $3\ \mathrm{kg}$ cart moving at $4\ \mathrm{m/s}$ hits a $1\ \mathrm{kg}$ cart at rest, head-on and elastic.

$$v_1 = \frac{3-1}{3+1}(4) + 0 = \frac{2}{4}(4) = 2\ \mathrm{m/s}$$

$$v_2 = \frac{2(3)}{3+1}(4) + 0 = \frac{6}{4}(4) = 6\ \mathrm{m/s}$$

Verify momentum: before $12\ \mathrm{kg\cdot m/s}$; after $(3)(2) + (1)(6) = 12$ ✓. Verify energy: before $\tfrac{1}{2}(3)(16) = 24\ \mathrm{J}$; after $\tfrac{1}{2}(3)(4) + \tfrac{1}{2}(1)(36) = 6 + 18 = 24\ \mathrm{J}$ ✓.

The light cart leaves at $6\ \mathrm{m/s}$ — faster than the incoming $4\ \mathrm{m/s}$ — while the heavy cart lumbers on at $2\ \mathrm{m/s}$. Energy conservation permits this because the light cart carries little mass.

How elastic is 'elastic'? The restitution coefficient

Real collisions are never perfectly elastic: some kinetic energy always becomes heat, sound or permanent deformation. The coefficient of restitution $e$ quantifies the bounciness — the ratio of relative separation speed to relative approach speed:

$$e = \frac{v_2 - v_1}{u_1 - u_2}$$

Perfectly elastic means $e = 1$; perfectly inelastic (sticking) means $e = 0$. A basketball has $e \approx 0.8$, a tennis ball around $0.75$, and putty near zero. Sports regulators actually specify $e$ ranges for balls — it is a measurable, standardized property, not just a textbook idealization.

Elastic collisions in 2D: the right-angle rule

In two or three dimensions, elastic collisions need one more ingredient: momentum is a vector, so it is conserved component by component. For smooth spheres, only the velocity components along the line of centers change; perpendicular components pass through untouched.

The famous result: two identical balls colliding elastically with one initially at rest always scatter at 90° to each other. Pool players use this constantly — the cue ball deflects along one leg of a right angle, the object ball along the other. It follows directly from conserving both vector momentum and scalar kinetic energy.

Variables and units

Common mistakes

Real-world applications

• Billiard and pool strategy is applied elastic-collision physics: aim, angles and spin transfer.

• Particle accelerators analyze scattering events assuming elastic collisions between subatomic particles.

• Gas kinetic theory models molecular collisions as elastic to derive pressure and temperature.

• Sports engineers tune ball and racket 'liveliness' by how close impacts come to perfectly elastic.