Inelastic Collisions — Momentum Conserved, KE Lost

In the real world, collisions are almost always inelastic: they conserve momentum (because no external force acts during the very brief interaction), but kinetic energy decreases. The "lost" energy is converted into heat, sound, permanent deformation, or vibrations in the colliding bodies. A head-on car crash is inelastic; a lump of putty hitting a wall is perfectly inelastic; even a bouncing ball is inelastic because it never quite reaches the height it was dropped from.

This lesson completes the classification of collision types for OCR Module 3.5.2. We work through standard inelastic and perfectly inelastic examples, calculate the energy lost in each case, and draw up a comparison table so that you can quickly identify which tools apply to which problem.

---

Inelastic vs Perfectly Inelastic

Two types of inelastic collision appear in A-Level questions:

1. Inelastic: the two bodies separate after the collision but total kinetic energy has decreased. Examples: a bouncing ball, most everyday collisions between distinguishable objects.
2. Perfectly inelastic: the two bodies stick together after the collision and move with a common velocity. This type produces the maximum possible KE loss consistent with momentum conservation. Examples: railway trucks coupling, a dart embedding in a board, a bullet lodging in a block.

Both types conserve linear momentum (closed system). Neither conserves kinetic energy.

---

Diagram — Collision Type Decision Tree

flowchart TD
  A[Two bodies collide] --> B{Total KE after = Total KE before?}
  B -- Yes --> C[Elastic]
  B -- No --> D{Do they stick together?}
  D -- Yes --> E[Perfectly inelastic - max KE lost]
  D -- No --> F[Inelastic - partial KE lost]
  C --> G["2 conservation laws: p AND KE"]
  E --> H["1 conservation law: p only; common v_f"]
  F --> I["1 conservation law: p only; need extra info"]

---

Summary Table — Three Collision Types

Feature	Elastic	Inelastic	Perfectly inelastic
Momentum conserved?	Yes	Yes	Yes
Kinetic energy conserved?	Yes	No (some lost)	No (maximum lost)
Do the bodies stick?	No	No	Yes
Equations needed	Both p and KE	p, plus extra data	p only (common v_f)
Real examples	Atomic collisions, snooker balls (approx)	Most everyday collisions, bouncing balls	Car crash + lock, dart in board, coupling trucks

---

Perfectly Inelastic: The One-Equation Case

When two bodies stick together, the problem has a single unknown (the common final velocity), so a single equation — momentum conservation — is enough.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v_f

Solving for v_f:

v_f = (m₁ u₁ + m₂ u₂) / (m₁ + m₂)

The amount of kinetic energy lost is:

ΔKE = KE_before − KE_after

which is always positive (in the direction of loss) for a genuine collision.

---

Worked Example 1 — Coupling Trucks

A 5000 kg railway truck moves at 4.0 m s⁻¹ and collides with a stationary 3000 kg truck on the same track. They couple together. Find the common velocity and the kinetic energy lost.

Momentum: 5000 × 4.0 + 3000 × 0 = 8000 × v_f ⇒ 20 000 = 8000 v_f ⇒ v_f = 2.5 m s⁻¹

KE before = ½ × 5000 × 16 = 40 000 J = 40 kJ KE after = ½ × 8000 × 6.25 = 25 000 J = 25 kJ ΔKE = 15 kJ lost (37.5% of the original KE)

The lost energy goes into sound, heat in the coupling mechanism, and deformation of the linkage. Note that this is the maximum loss consistent with momentum conservation — any other pair of final velocities that conserve momentum (e.g. the first truck continues forward at 3 m s⁻¹ and the second moves at 1.67 m s⁻¹) would have higher final KE. This is a general and important result.

---

Proof — Perfectly Inelastic Collisions Lose the Maximum KE

Consider two masses approaching each other, with any final state (v₁, v₂) satisfying momentum conservation. For a fixed total momentum p_total = (m₁ + m₂) v_cm (where v_cm is the centre-of-mass velocity), the final kinetic energy is:

KE_after = ½ m₁ v₁² + ½ m₂ v₂²

Using the centre-of-mass frame (where total momentum is zero), the minimum KE is achieved when both bodies are at rest relative to the CM — i.e., they move together at v_cm in the lab frame. This is precisely the perfectly inelastic (stuck-together) case, and it has:

KE_min = ½ (m₁ + m₂) v_cm²

Any other outcome must have extra relative motion in the CM frame, and therefore higher total KE. So the perfectly inelastic collision loses the most KE possible while still conserving momentum. This proof is not examined at A-Level, but the result is — OCR may ask "Explain why the collision loses the maximum possible kinetic energy" and the answer is that the relative velocity of separation is zero.

---

Worked Example 2 — Bullet and Block (Ballistic Pendulum Setup)

A 0.020 kg bullet travels at 400 m s⁻¹ horizontally and embeds in a 2.0 kg wooden block resting on a smooth frictionless horizontal surface. Find the common velocity of the bullet-block system and the energy lost.

Momentum: 0.020 × 400 + 2.0 × 0 = 2.02 × v_f ⇒ 8.0 = 2.02 v_f ⇒ v_f = 3.96 m s⁻¹ (to 3 s.f.)

KE before = ½ × 0.020 × 400² = 1600 J KE after = ½ × 2.02 × 3.96² = ½ × 2.02 × 15.7 = 15.8 J ΔKE = 1584 J lost (about 99% of the original KE)

Almost all the kinetic energy is lost, appearing as heat (the bullet and wood are hot), deformation (the bullet is flattened and the wood splintered) and sound. Only about 1% of the bullet's KE becomes macroscopic motion of the block.

This is an extreme but common case: when a light fast object hits a heavy slow (or stationary) object and sticks, the fraction of KE retained is approximately m₁/(m₁ + m₂) ≈ m₁/m₂, which is tiny when one mass is much larger. The exact fraction is:

KE_after / KE_before = m₁ / (m₁ + m₂)

In our example: 0.020 / 2.02 ≈ 0.0099, or 0.99% — agreeing with the direct calculation above.

---

Worked Example 3 — Inelastic but Not Sticking