Elastic Collisions — Momentum AND Kinetic Energy Conserved

A collision is an interaction between two (or more) bodies in which they exchange momentum and, possibly, energy. In physics we classify collisions by what is conserved:

• Elastic collision: both momentum and kinetic energy are conserved.
• Inelastic collision: momentum is conserved, but kinetic energy is not (Lesson 8).
• Perfectly inelastic: a subset of inelastic, in which the bodies stick together (Lesson 8).

This lesson focuses on the elastic case — the gold standard of "idealised" collisions. Truly elastic collisions are rare in everyday life (though gas molecules come extremely close), but the maths is crucial both as a limiting case and as a clean conceptual test. It is examined in OCR Module 3.5.2.

---

What Makes a Collision Elastic?

Two conditions must hold:

Σp(before) = Σp(after) (always true for a closed system) ΣKE(before) = ΣKE(after) (the defining feature of elastic collisions)

If total KE is the same before and after, no energy has been converted to heat, sound, or permanent deformation. Classic examples:

• Two hard steel balls colliding (almost, but not quite elastic).
• Pool or snooker balls to a good approximation.
• Gas molecules in an ideal gas — exactly elastic on average.
• Alpha particles scattering off a heavy nucleus (Rutherford scattering) — perfectly elastic at the level of individual interactions.

---

Diagram — Elastic vs Inelastic Decision Tree

flowchart TD
  A[Collision between two bodies] --> B{Is kinetic energy conserved?}
  B -- Yes --> C[Elastic collision]
  B -- No --> D{Do they stick together?}
  D -- Yes --> E[Perfectly inelastic collision]
  D -- No --> F[Inelastic collision]
  C --> G[Use both p conservation AND KE conservation]
  E --> H[Use p conservation; combined final velocity]
  F --> I[Use p conservation; separate final velocities]

---

Two Equations, Two Unknowns

For a 1-D elastic collision between masses m₁ and m₂ with initial velocities u₁ and u₂ and final velocities v₁ and v₂:

Momentum: m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ ...(1) KE: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² ...(2)

These are two equations in two unknowns (v₁, v₂), so a unique solution exists. Solving them together yields the general result:

v₁ = [(m₁ − m₂) u₁ + 2 m₂ u₂] / (m₁ + m₂) v₂ = [(m₂ − m₁) u₂ + 2 m₁ u₁] / (m₁ + m₂)

OCR does not require you to memorise these formulae — you can always rederive them by solving the simultaneous equations — but recognising them can save time.

---

The "Velocity of Approach = Velocity of Separation" Trick

For 1-D elastic collisions, there is a beautiful short-cut. Subtracting and rearranging equations (1) and (2) yields:

u₁ − u₂ = −(v₁ − v₂)

In words: the relative velocity of approach before the collision equals the relative velocity of separation after it (with a sign flip to indicate reversal). This is much easier to apply than the full KE equation, and OCR explicitly includes it in the specification guidance.

Combined with momentum conservation (equation 1), it gives you a pair of linear equations, which are much easier to solve than a linear + a quadratic.

---

Worked Example 1 — Equal Masses, One Stationary

A 0.50 kg ball moving at 4.0 m s⁻¹ collides elastically with an identical stationary ball. Find the final velocities.

Equation (1): 0.50 × 4.0 + 0.50 × 0 = 0.50 v₁ + 0.50 v₂ ⇒ v₁ + v₂ = 4.0 ...(a)

Relative velocity rule: u₁ − u₂ = −(v₁ − v₂) ⇒ 4.0 − 0 = −(v₁ − v₂) ⇒ v₁ − v₂ = −4.0 ...(b)

Adding (a) + (b): 2 v₁ = 0 ⇒ v₁ = 0 Substituting back: v₂ = 4.0 m s⁻¹

The moving ball stops dead, and the previously stationary ball takes off at the original speed. This is the famous Newton's cradle result — and it only works for equal masses in an elastic collision. Snooker players use it constantly when striking the cue ball at a stationary target ball on a centre-line shot.

---

Worked Example 2 — Light Ball Hits Heavy Ball

A 0.20 kg ball at 5.0 m s⁻¹ collides elastically with a stationary 2.0 kg ball. Find the velocities after.

Equation (1): 0.20 × 5.0 = 0.20 v₁ + 2.0 v₂ ⇒ 1.0 = 0.20 v₁ + 2.0 v₂ ...(a)

Relative velocity: 5.0 = −(v₁ − v₂) ⇒ v₂ − v₁ = 5.0 ...(b)

From (b): v₂ = v₁ + 5.0. Substitute into (a): ⇒ 1.0 = 0.20 v₁ + 2.0 (v₁ + 5.0) ⇒ 1.0 = 0.20 v₁ + 2.0 v₁ + 10 ⇒ −9.0 = 2.20 v₁ ⇒ v₁ = −4.09 m s⁻¹

So v₂ = −4.09 + 5.0 = +0.91 m s⁻¹.

The light ball bounces back at 4.09 m s⁻¹, and the heavy ball moves slowly forward at 0.91 m s⁻¹. This is the intuition you have from bouncing a ping-pong ball off a bowling ball — the small object reverses direction, the large object barely moves.