Elastic Collisions — Momentum AND Kinetic Energy Conserved

Spec mapping: OCR H556 Module 3.3 — Distinction between elastic and inelastic collisions; kinetic-energy conservation in elastic collisions; the relative-velocity (approach = separation) rule for 1-D elastic interactions. (Refer to the official OCR H556 specification document for exact wording.)

A collision is any interaction between two (or more) bodies in which they exchange momentum and, possibly, kinetic energy. The classification of collisions according to whether KE is conserved is one of the most important conceptual divisions in mechanics, because it dictates which conservation laws apply and therefore how many equations you have to play with.

Elastic collision — both linear momentum and kinetic energy are conserved.
Inelastic collision — linear momentum is conserved, kinetic energy is not (some has been converted to heat, sound, deformation; see Lesson 8).
Perfectly inelastic collision — a subset of inelastic in which the bodies stick together (Lesson 8). This case loses the maximum possible KE consistent with momentum conservation.

This lesson treats the elastic case — the gold standard of "idealised" collisions. Truly elastic collisions are rare in everyday life (gas molecules in an ideal gas come extremely close; macroscopic steel balls approach but never quite reach it) but the mathematics is essential as a limiting case, a problem-solving template, and the conceptual benchmark against which all real collisions are measured. The PAG 6 air-track investigation gets very close to elastic with magnetic-bumper colliders.

Key Definition: A collision is elastic if and only if the total kinetic energy of the system after the collision equals the total kinetic energy before the collision. Momentum conservation is automatic for any closed system (Lesson 6); the additional KE conservation is the defining feature of elastic interactions.

What Makes a Collision Elastic?

Two conditions must hold simultaneously:

$\sum_i \tfrac{1}{2} m_i u_i^2 = \sum_i \tfrac{1}{2} m_i v_i^2 \qquad \text{(KE conserved)}$

$\sum_i m_i \vec{u}_i = \sum_i m_i \vec{v}_i \qquad \text{(momentum conserved, automatic for closed system)}$

If the total KE is the same before and after, no kinetic energy has been converted into heat, sound, vibration or permanent deformation. The bodies emerge from the collision with exactly the macroscopic mechanical energy they brought in, just redistributed differently between themselves.

Classic examples — listed in increasing order of "how nearly elastic":

Lump of clay onto a wall — perfectly inelastic (clay sticks; nearly all KE → deformation).
Two steel balls in mid-air collision — about 90–95% KE retained; close-but-not-elastic.
Pool or snooker balls (ivory or composite) — about 95–98% KE retained; treated as elastic for A-Level.
Gas molecules in an ideal gas — exactly elastic on the time-average; collisions exchange momentum and direction but not bulk KE.
Rutherford alpha-particle scattering off a gold nucleus — perfectly elastic at the level of individual electromagnetic Coulomb interactions.
Two superconducting magnets repelling head-on — essentially perfectly elastic; no contact, no energy loss.

In OCR exam questions the phrase "the collision is elastic" is your trigger to use both conservation laws. The phrase "the bodies stick together" or "perfectly inelastic" is your trigger to use momentum only and expect a KE deficit.

Diagram — Elastic vs Inelastic Decision Tree

flowchart TD
  A[Collision between two bodies] --> B{Is total KE after = total KE before?}
  B -- Yes --> C[Elastic collision]
  B -- No --> D{Do they stick together?}
  D -- Yes --> E["Perfectly inelastic (KE loss is maximum)"]
  D -- No --> F[Inelastic but not sticky]
  C --> G["Use BOTH: p conservation AND KE conservation"]
  E --> H["Use p only; common v_f; KE loss = ΔKE"]
  F --> I["Use p only; need extra data to fix final state"]

Two Equations, Two Unknowns

For a 1-D elastic collision between masses $m_1$ and $m_2$ with initial velocities $u_1, u_2$ and final velocities $v_1, v_2$ :

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{(1: momentum)}$

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

These are two equations in two unknowns ( $v_1$ and $v_2$ ), so a unique solution exists. Solving them together yields the general result:

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

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

OCR does not require you to memorise these formulae — you can always rederive them from the simultaneous equations — but recognising the structure can save time in exam conditions. They emerge naturally from the relative-velocity trick below.

The "Velocity of Approach = Velocity of Separation" Trick

For 1-D elastic collisions, there is a beautifully simple shortcut. Algebraically subtracting equations (1) and (2) and dividing through (a slightly tedious manipulation, but worth doing once) yields:

$u_1 - u_2 = -(v_1 - v_2)$

In words: the relative velocity of approach before the collision equals the relative velocity of separation after it (with the sign reversed because they are now moving apart). The magnitude of the relative velocity is unchanged.

This is a much easier relationship to apply than the full quadratic KE equation. Combined with momentum conservation, it gives a pair of linear simultaneous equations (linear + linear, not linear + quadratic) — vastly faster to solve. OCR explicitly includes this relationship in the specification guidance for the elastic case.

Mathematically, this happens because the KE equation can be factorised: $\frac{1}{2}m_1(u_1^2 - v_1^2) = \frac{1}{2}m_2(v_2^2 - u_2^2)$ becomes $m_1(u_1 - v_1)(u_1 + v_1) = m_2(v_2 - u_2)(v_2 + u_2)$ . Dividing the momentum equation $m_1(u_1 - v_1) = m_2(v_2 - u_2)$ into this gives $u_1 + v_1 = v_2 + u_2$ , i.e. $u_1 - u_2 = -(v_1 - v_2)$ . Neat.

Common Exam Mistake: Forgetting the negative sign in $u_1 - u_2 = -(v_1 - v_2)$ . The relative velocity reverses, not preserves; the bodies were approaching, now they separate.

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.

Momentum (1): $(0.50)(4.0) + (0.50)(0) = (0.50) v_1 + (0.50) v_2 \Rightarrow v_1 + v_2 = 4.0$ … (a)

Relative velocity: $u_1 - u_2 = -(v_1 - v_2) \Rightarrow 4.0 - 0 = -(v_1 - v_2) \Rightarrow v_1 - v_2 = -4.0$ … (b)

Adding (a) + (b): $2 v_1 = 0 \Rightarrow v_1 = 0$ . Substituting back: $v_2 = 4.0$ m s $^{-1}$ .

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

Sanity check (KE): before $= \frac{1}{2}(0.50)(16) = 4.0$ J. After $= \frac{1}{2}(0.50)(0)^2 + \frac{1}{2}(0.50)(16) = 4.0$ J ✓.

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.

Momentum (1): $0.20 \times 5.0 = 0.20 v_1 + 2.0 v_2 \Rightarrow 1.0 = 0.20 v_1 + 2.0 v_2$ … (a)

Relative velocity: $5.0 = -(v_1 - v_2) \Rightarrow v_2 - v_1 = 5.0 \Rightarrow v_2 = v_1 + 5.0$ … (b)

Substitute (b) into (a):

$1.0 = 0.20 v_1 + 2.0(v_1 + 5.0) = 0.20 v_1 + 2.0 v_1 + 10$

$-9.0 = 2.20 v_1 \Rightarrow v_1 = -4.09 \text{ m s}^{-1}$

Then $v_2 = -4.09 + 5.0 = +0.91$ m s $^{-1}$ .

The light ball bounces back at $4.09$ m s $^{-1}$ ; 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 stationary bowling ball — the small projectile reverses, the large target barely moves.

Sanity checks:

Momentum: $0.20(-4.09) + 2.0(0.91) = -0.82 + 1.82 = +1.0$ kg m s $^{-1}$ ✓ (matches $p_{\text{before}} = 1.0$ )
KE before: $\frac{1}{2}(0.20)(25) = 2.5$ J. KE after: $\frac{1}{2}(0.20)(16.7) + \frac{1}{2}(2.0)(0.83) = 1.67 + 0.83 = 2.5$ J ✓

Both quantities are conserved — confirming it is truly an elastic collision.

Worked Example 3 — Heavy Ball Hits Light Ball

A $2.0$ kg ball at $5.0$ m s $^{-1}$ collides elastically with a stationary $0.20$ kg ball. Find the velocities after.

Momentum: $10 = 2.0 v_1 + 0.20 v_2$ … (a)

Relative velocity: $v_2 - v_1 = 5.0 \Rightarrow v_2 = v_1 + 5.0$

Substitute: $10 = 2.0 v_1 + 0.20(v_1 + 5.0) = 2.2 v_1 + 1.0 \Rightarrow v_1 = 9/2.2 \approx 4.09$ m s $^{-1}$ . Then $v_2 \approx 9.09$ m s $^{-1}$ .

The heavy ball barely slows (5.0 → 4.09 m s $^{-1}$ ); the light ball launches at nearly twice the initial speed. In the limit $m_1 \gg m_2$ , $v_2 \to 2 u_1$ — a result that appears in everything from baseball-bat-on-ball physics to gravity-assist manoeuvres in space exploration (where a probe "bounces" elastically off a moving planet and gains roughly twice the planet's orbital speed).

Limiting Cases — Worth Memorising

The general result $v_2 = 2 m_1 u_1/(m_1 + m_2)$ (with $u_2 = 0$ ) has three beautiful limits:

$m_1 = m_2$ : $v_1 = 0$ , $v_2 = u_1$ . The famous "Newton's cradle" swap.
$m_1 \ll m_2$ : $v_1 \approx -u_1$ (bounces straight back); $v_2 \approx 0$ (target barely moves). Like a ping-pong ball off a wall.
$m_1 \gg m_2$ : $v_1 \approx u_1$ (barely changes); $v_2 \approx 2 u_1$ (target launches at double speed). Like a bowling ball pushing a marble.

These limits are instant sanity checks. If a calculation yields $v_2 = 2.5 u_1$ for "large hits small", you have made an arithmetic error somewhere — the maximum is $2 u_1$ exactly.

Worked Example 4 — Head-On Elastic Collision

A $1.0$ kg ball moving east at $6.0$ m s $^{-1}$ collides elastically head-on with a $2.0$ kg ball moving west at $3.0$ m s $^{-1}$ . Find the final velocities.

Take east as positive: $u_1 = +6.0$ , $u_2 = -3.0$ .

Momentum: $1.0(6.0) + 2.0(-3.0) = 1.0 v_1 + 2.0 v_2 \Rightarrow 0 = v_1 + 2 v_2 \Rightarrow v_1 = -2 v_2$ … (a)

Relative velocity: $u_1 - u_2 = 6.0 - (-3.0) = 9.0$ , so $9.0 = -(v_1 - v_2) = v_2 - v_1$ … (b)

Substitute (a) into (b): $v_2 - (-2 v_2) = 9.0 \Rightarrow 3 v_2 = 9.0 \Rightarrow v_2 = +3.0$ m s $^{-1}$ . Then $v_1 = -6.0$ m s $^{-1}$ .

Interpretation: the 1.0 kg ball reverses (now west at 6.0 m s $^{-1}$ ); the 2.0 kg ball reverses (now east at 3.0 m s $^{-1}$ ). Each ball has swapped direction while keeping its original speed. Sanity check (KE): before $= \frac{1}{2}(1.0)(36) + \frac{1}{2}(2.0)(9) = 18 + 9 = 27$ J; after $= 27$ J ✓.

This "reverse the velocities" symmetry is a feature of head-on elastic collisions where the centre-of-mass frame velocity is zero (i.e. $m_1 u_1 + m_2 u_2 = 0$ ). In the CM frame, both balls bounce off each other and reverse direction; in the lab frame, the same is true.

Worked Example 5 — Spec Check: Approach = Separation

A $3.0$ kg trolley moving east at $5.0$ m s $^{-1}$ collides elastically with a $1.0$ kg trolley moving east at $1.0$ m s $^{-1}$ . Find the final velocities by using the approach-equals-separation rule.

Relative velocity of approach $= u_1 - u_2 = 5.0 - 1.0 = 4.0$ m s $^{-1}$ .

Elastic Collisions — Momentum AND Kinetic Energy Conserved

Elastic Collisions — Momentum AND Kinetic Energy Conserved

What Makes a Collision Elastic?

Diagram — Elastic vs Inelastic Decision Tree

Two Equations, Two Unknowns

The "Velocity of Approach = Velocity of Separation" Trick

Worked Example 1 — Equal Masses, One Stationary

Worked Example 2 — Light Ball Hits Heavy Ball

Worked Example 3 — Heavy Ball Hits Light Ball

Limiting Cases — Worth Memorising

Worked Example 4 — Head-On Elastic Collision

Worked Example 5 — Spec Check: Approach = Separation

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