Elastic and Inelastic Collisions

In every collision, momentum is conserved (provided no external forces act). But what about kinetic energy? This is where the distinction between elastic and inelastic collisions becomes crucial. Exam questions routinely ask you to classify collisions and calculate energy transfers, so this lesson gives you a rigorous framework.

Kinetic Energy in Collisions

Momentum is always conserved in a collision. Energy is also always conserved — but kinetic energy might not be. During a collision, kinetic energy can be transformed into other forms: heat, sound, deformation of materials, and internal energy.

The amount of kinetic energy retained after a collision determines its classification:

Type	KE After vs Before	Objects After	Real-World Examples
Perfectly elastic	KE fully conserved	Separate	Gas molecule collisions, some atomic collisions
Inelastic	KE partially lost	Separate	Ball bouncing, car crash with rebound
Perfectly inelastic	Maximum KE loss	Stick together	Bullet in block, cars locking together
Superelastic	KE increases	Separate	Explosions, spring-loaded mechanism releases

flowchart TD
    A["Calculate total KE\nBEFORE collision"] --> B["Calculate total KE\nAFTER collision"]
    B --> C{"Compare KE_before\nand KE_after"}
    C -->|"KE_after = KE_before"| D["Perfectly Elastic\ne = 1"]
    C -->|"KE_after < KE_before\nObjects separate"| E["Inelastic\n0 < e < 1"]
    C -->|"KE_after < KE_before\nObjects stick"| F["Perfectly Inelastic\ne = 0"]
    C -->|"KE_after > KE_before"| G["Superelastic\n(explosion/release)"]

Perfectly Elastic Collisions

In a perfectly elastic collision, both momentum and kinetic energy are conserved:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

True perfectly elastic collisions are rare on a macroscopic scale. Even a "super ball" bouncing on a hard floor loses some energy. However, collisions between gas molecules and between subatomic particles are effectively elastic.

Special Case: Equal Masses, One Initially Stationary

When two objects of equal mass collide elastically and one is initially stationary, the moving object stops and the stationary object moves off with the velocity of the incoming object. This is famously demonstrated with Newton's cradle.

For equal masses: if u₂ = 0, then v₁ = 0 and v₂ = u₁.

Worked Example 1

A 2.0 kg ball moving at 5.0 m s⁻¹ collides elastically with a stationary 2.0 kg ball. Confirm the final velocities.

Using the special case: the first ball stops (v₁ = 0) and the second ball moves at 5.0 m s⁻¹ (v₂ = u₁).

Check momentum: 2.0 × 5.0 = 2.0 × 0 + 2.0 × 5.0 → 10 = 10 ✓ Check KE: ½ × 2.0 × 5.0² = ½ × 2.0 × 0² + ½ × 2.0 × 5.0² → 25 = 25 ✓

General Elastic Collision Formulae

For a head-on elastic collision where object 2 is initially stationary:

$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 \qquad v_2 = \frac{2m_1}{m_1 + m_2}u_1$

Mass Ratio	What Happens
m₁ = m₂	Ball 1 stops; ball 2 takes all velocity
m₁ >> m₂	Ball 1 barely affected; ball 2 flies off at ≈2u₁
m₁ << m₂	Ball 1 bounces back at ≈u₁; ball 2 barely moves

Worked Example 2: Unequal Masses

A 1.0 kg ball at 6.0 m s⁻¹ collides elastically and head-on with a stationary 3.0 kg ball.

$v_1 = \frac{1.0 - 3.0}{1.0 + 3.0} \times 6.0 = \frac{-2.0}{4.0} \times 6.0 = -3.0 \text{ m s}^{-1}$

$v_2 = \frac{2 \times 1.0}{1.0 + 3.0} \times 6.0 = \frac{2.0}{4.0} \times 6.0 = 3.0 \text{ m s}^{-1}$

The 1.0 kg ball bounces back at 3.0 m s⁻¹; the 3.0 kg ball moves forward at 3.0 m s⁻¹.

Check momentum: 1.0 × 6.0 = 1.0 × (−3.0) + 3.0 × 3.0 → 6.0 = −3.0 + 9.0 = 6.0 ✓ Check KE: ½(1)(36) = ½(1)(9) + ½(3)(9) → 18 = 4.5 + 13.5 = 18 ✓

Perfectly Inelastic Collisions

In a perfectly inelastic collision, the objects stick together and move with a common velocity after the collision. This is the collision type that loses the maximum amount of kinetic energy (while still conserving momentum).

$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

Worked Example 3

A 10 g bullet travelling at 400 m s⁻¹ embeds itself in a stationary 2.0 kg wooden block on a frictionless surface. Find the velocity of the block after impact and the percentage of kinetic energy lost.

Velocity after impact: $0.010 \times 400 = (0.010 + 2.0) \times v$ $4.0 = 2.01 \times v$ $v = 1.99 \text{ m s}^{-1} \approx 2.0 \text{ m s}^{-1}$

Kinetic energy before: $KE_i = \frac{1}{2} \times 0.010 \times 400^2 = 800 \text{ J}$

Kinetic energy after: $KE_f = \frac{1}{2} \times 2.01 \times 1.99^2 = 3.98 \text{ J}$

Percentage lost: $\frac{800 - 3.98}{800} \times 100 = 99.5\%$

Almost all the kinetic energy is converted into heat, sound, and deformation of the wood. Yet momentum is perfectly conserved.

Why so much KE is lost: Since KE = p²/(2m), the same momentum p spread over a much larger combined mass (2.01 kg vs 0.010 kg) gives much less kinetic energy. This is a general result — the fraction of KE retained equals m₁/(m₁ + m₂) for a bullet-block collision.

Inelastic Collisions (Partially Inelastic)

Most real collisions fall between the two extremes. A bouncing ball, for instance, loses some kinetic energy with each bounce (converted to heat and sound) but does not stick to the floor.

Worked Example 4

A 0.50 kg ball drops from 2.0 m onto a hard floor and bounces to 1.5 m. What percentage of kinetic energy is retained?

KE just before impact (using PE = mgh): $KE_i = mgh_1 = 0.50 \times 9.81 \times 2.0 = 9.81 \text{ J}$

KE just after bounce: $KE_f = mgh_2 = 0.50 \times 9.81 \times 1.5 = 7.36 \text{ J}$

Percentage retained: $\frac{7.36}{9.81} \times 100 = 75\%$

Coefficient of Restitution

The coefficient of restitution, e, is a measure of how "bouncy" a collision is. It is defined as the ratio of the relative speed of separation to the relative speed of approach:

$e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$

e = 1: perfectly elastic collision (no KE lost)
e = 0: perfectly inelastic collision (objects stick together)
0 < e < 1: partially inelastic (some KE lost)

For a ball bouncing off a fixed surface, this simplifies to:

$e = \frac{v_{\text{rebound}}}{v_{\text{impact}}} = \sqrt{\frac{h_{\text{rebound}}}{h_{\text{drop}}}}$

Surface Pair	Typical e
Steel ball on steel plate	0.95
Glass marble on glass	0.90
Tennis ball on hard court	0.75
Football on grass	0.60
Plasticine on any surface	≈0
Bouncy ball (superball)	0.85–0.90

For the ball in Worked Example 4: e = √(1.5/2.0) = √0.75 = 0.866.

Identifying Collision Types in Exam Problems

In exam questions, you determine the collision type by calculating kinetic energy before and after:

Calculate total KE before the collision
Calculate total KE after the collision
Compare:
- If KE_after = KE_before → perfectly elastic
- If KE_after < KE_before and objects stuck → perfectly inelastic
- If KE_after < KE_before and objects separate → inelastic

Important: You cannot determine the collision type from momentum alone — momentum is conserved in all types.

Worked Example 5: 2D Collision Classification

A 0.17 kg ball at 5.0 m s⁻¹ hits an identical stationary ball. After: ball 1 at 3.5 m s⁻¹ at 30°, ball 2 at 3.2 m s⁻¹ at 33° below.

$KE_{\text{before}} = \frac{1}{2}(0.17)(5.0^2) = 2.125 \text{ J}$

$KE_{\text{after}} = \frac{1}{2}(0.17)(3.5^2) + \frac{1}{2}(0.17)(3.2^2) = 1.041 + 0.870 = 1.911 \text{ J}$

$\text{KE lost} = 2.125 - 1.911 = 0.214 \text{ J} \quad (10\%)$

The collision is inelastic (10% of KE lost to deformation and sound).

Common Exam Mistakes

Mistake 1: "Energy is not conserved in an inelastic collision." Correction: Total energy IS always conserved. It is specifically KINETIC energy that is not conserved — it transforms into heat, sound, and deformation.

Mistake 2: "Momentum is not conserved because kinetic energy is lost." Correction: Momentum and kinetic energy are independent. Momentum is ALWAYS conserved (no external forces). KE loss does not affect momentum conservation.

Mistake 3: Calculating e with wrong sign convention. Correction: e = (speed of separation)/(speed of approach). Both are positive. Use e = (v₂ − v₁)/(u₁ − u₂) with consistent signs for direction.

Summary

Momentum is conserved in all collisions; kinetic energy may or may not be
Perfectly elastic: KE fully conserved (rare in macroscopic collisions)
Perfectly inelastic: objects stick, maximum KE loss
Coefficient of restitution e measures the "bounciness" (0 ≤ e ≤ 1)
Always calculate KE before and after to classify a collision
Energy is always conserved overall — "lost" KE is transformed into heat, sound, and deformation
For elastic collisions with unequal masses: v₁ = (m₁ − m₂)u₁/(m₁ + m₂) and v₂ = 2m₁u₁/(m₁ + m₂)

A-Level Deep Dive: Elastic and Inelastic Collisions

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics covers the conservation of linear momentum applied to one- and two-dimensional collisions and explosions, the distinction between elastic and inelastic collisions through the conservation (or otherwise) of total kinetic energy, and the application of these principles to interactions ranging from macroscopic ball collisions to subatomic scattering (refer to the official Pearson Edexcel specification document for exact wording). The classification of a collision as elastic, inelastic or perfectly inelastic is examined directly in Paper 2 (Topics 5–8), but feeds synoptically into Paper 1 (energy in particle physics, Topic 7) and Paper 3 General and Practical Principles, where unfamiliar contexts (Compton scattering, neutron moderation, alpha particle scattering) require candidates to identify which conservation laws apply. The Edexcel formula booklet supplies the standard kinematic and momentum equations but does not state the special-case results for elastic collisions between equal masses or the formula for the fraction of kinetic energy retained — these must be derived from the simultaneous conservation equations.

Worked example with full mark scheme

Question (8 marks):

A particle A of mass $m_A = 0.20$ kg moves along a straight horizontal line at $u_A = 6.0$ m s $^{-1}$ and collides head-on with a stationary particle B of mass $m_B = 0.30$ kg. The collision is perfectly elastic.

(a) Write down the two conservation equations that govern the collision. (2)

(b) Show by simultaneous solution that the post-collision velocities are $v_A = -1.2$ m s $^{-1}$ and $v_B = +4.8$ m s $^{-1}$ (taking A's initial direction as positive). (4)

Solution with mark scheme:

(a) Step 1 — state the conservation equations.

$m_A u_A + m_B u_B = m_A v_A + m_B v_B \quad \text{(momentum)}$ $\tfrac{1}{2}m_A u_A^2 + \tfrac{1}{2}m_B u_B^2 = \tfrac{1}{2}m_A v_A^2 + \tfrac{1}{2}m_B v_B^2 \quad \text{(kinetic energy)}$

B1 — momentum conservation written with signed velocities. B1 — KE conservation written explicitly (the factor of $\tfrac{1}{2}$ must appear; candidates lose this mark by writing $mv^2$ without the half).

(b) Step 2 — substitute knowns into momentum equation.

With $u_B = 0$ : $0.20 \times 6.0 = 0.20 v_A + 0.30 v_B$ $1.2 = 0.20 v_A + 0.30 v_B \quad (\text{Eq. 1})$

M1 — correct momentum substitution.

Step 3 — substitute knowns into KE equation.

$\tfrac{1}{2}(0.20)(6.0)^2 = \tfrac{1}{2}(0.20) v_A^2 + \tfrac{1}{2}(0.30) v_B^2$ $3.6 = 0.10 v_A^2 + 0.15 v_B^2 \quad (\text{Eq. 2})$

M1 — correct KE substitution.

Step 4 — solve simultaneously.

The slick route uses the standard A-Level result that, for an elastic 1D collision, the relative velocity of separation equals the relative velocity of approach:

$v_B - v_A = u_A - u_B = 6.0 \text{ m s}^{-1} \quad (\text{Eq. 3})$

Combining Eq. 1 and Eq. 3: from Eq. 3, $v_B = v_A + 6.0$ . Substituting into Eq. 1:

$1.2 = 0.20 v_A + 0.30(v_A + 6.0) = 0.50 v_A + 1.8$ $0.50 v_A = -0.6 \implies v_A = -1.2 \text{ m s}^{-1}$ $v_B = -1.2 + 6.0 = +4.8 \text{ m s}^{-1}$

M1 — correct algebraic manipulation (either the relative-velocity short cut or direct quadratic solution of Eq. 1 and Eq. 2).

A1 — both numerical answers correct with signs as printed. The negative $v_A$ indicates A rebounds — this is physically reasonable because A is the lighter particle.

$\text{KE}_{\text{before}} = \tfrac{1}{2}(0.20)(6.0)^2 = 3.6 \text{ J}$ $\text{KE}_{\text{after}} = \tfrac{1}{2}(0.20)(1.2)^2 + \tfrac{1}{2}(0.30)(4.8)^2 = 0.144 + 3.456 = 3.60 \text{ J}$

M1 — substitution into the KE expression with both post-collision speeds.

A1 — agreement to within rounding, with units (J).

Total: 8 marks (B2 M4 A2, distributed as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A railway truck of mass 2.0 $\times 10^3$ kg moves along a horizontal track at 3.0 m s $^{-1}$ and collides with a stationary truck of mass 4.0 $\times 10^3$ kg. The trucks couple together on impact.

(a) Determine the common velocity of the coupled trucks immediately after the collision. (2)

(b) Calculate the kinetic energy lost during the collision and express it as a percentage of the initial kinetic energy. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2 — application of momentum conservation): $2.0 \times 10^3 \times 3.0 = (2.0 \times 10^3 + 4.0 \times 10^3) v$ .
A1 (AO1.1 — numerical answer with units): $v = 1.0$ m s $^{-1}$ .

(b)

M1 (AO2): KE before $= \tfrac{1}{2}(2.0 \times 10^3)(3.0)^2 = 9.0 \times 10^3$ J.
M1 (AO2): KE after $= \tfrac{1}{2}(6.0 \times 10^3)(1.0)^2 = 3.0 \times 10^3$ J. Loss $= 6.0 \times 10^3$ J.
A1 (AO1.1): percentage lost $= 6.0/9.0 \times 100\% = 67\%$ (2 s.f.).

(c)

B1 (AO3.1): energy is dissipated as heat in the deformation of the coupling mechanism, sound, and elastic-then-plastic deformation of the buffer springs.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This is a typical Paper 2 Topic 6 question: AO1 for the calculations, AO2 for selecting the perfectly-inelastic momentum equation, AO3 for the qualitative energy-dissipation reasoning. Note that the two-thirds energy loss arises because the mass ratio is 1:2 — a perfectly inelastic collision in which the moving body has mass $m$ and strikes a stationary body of mass $nm$ retains a fraction $1/(1+n)$ of the original KE.

Synoptic links

Connects to:

Conservation of momentum (Topic 6): every collision — elastic, inelastic, or perfectly inelastic — conserves linear momentum provided no external impulse acts. The classification refers only to whether kinetic energy is also conserved. Conflating "momentum conserved" with "elastic" is the single most common Topic 6 error and is examined explicitly through "state and explain" questions.
Neutron moderation in nuclear reactors (Topic 7 — Nuclear Physics): thermal-neutron reactors slow fast fission neutrons by elastic scattering off light nuclei (hydrogen in water, deuterium in heavy water, carbon in graphite). The fraction of energy lost per collision peaks when the moderator nucleus has a mass equal to the neutron — which is why hydrogen is the most efficient moderator per collision. The 1D elastic-collision formulae from this lesson are used directly to derive moderator efficiency.
Particle physics scattering (Topic 7): Rutherford alpha-scattering, electron-proton elastic scattering, and high-energy collider physics all rely on the elastic / inelastic distinction. An "elastic" scattering event preserves the identity and rest mass of the particles; "inelastic" scattering creates new particles or excites internal states, with the missing kinetic energy converted to mass via $E = mc^2$ . The conceptual framework is identical to A-Level collisions, scaled up energetically.
Kinetic theory of gases (Topic 5): the assumption that gas molecules collide elastically with each other and with the container walls is the foundation of the ideal-gas model. If collisions were inelastic, kinetic energy would steadily drain into internal molecular modes and the gas would cool spontaneously — directly contradicting the empirical observation that an isolated gas maintains its temperature.
Coefficient of restitution and energy methods (Mechanics 2 / Edexcel A-Level Maths): the parameter $e = (v_B - v_A)/(u_A - u_B)$ used in this lesson appears in A-Level Mathematics Mechanics modules, where it generalises the elastic / perfectly inelastic dichotomy to a continuous spectrum. $e = 1$ is elastic, $e = 0$ is perfectly inelastic, and $0 < e < 1$ covers everything in between.

Mark-scheme literacy

Elastic / inelastic collision questions on 9PH0 distribute AO marks evenly because the physics demands both calculation and judgement (which equations apply? what fraction of KE is lost?):