Physics/Lessons/Further Mechanics/Elastic & Inelastic Collisions
Physics · Further Mechanics · A2 Level

Elastic & Inelastic Collisions

Momentum is conserved in every collision, no exceptions. Kinetic energy is different — sometimes it survives intact, sometimes almost all of it vanishes into heat, sound and deformation. This lesson is about telling these cases apart and calculating what happens in each.

10 min read

Two conservation laws, not one

Every collision in an isolated system conserves momentum — this follows directly from Newton's third law, and holds regardless of what the objects are made of or how they deform. What is not automatically conserved is kinetic energy. Whether it's conserved defines the type of collision:

  • A perfectly elastic collision conserves both momentum and total kinetic energy.
  • An inelastic collision conserves momentum but not kinetic energy — some is transferred to other forms (heat, sound, permanent deformation).
  • A perfectly inelastic collision is the extreme case of inelastic: the colliding objects stick together and move off with a common final velocity, losing the maximum kinetic energy possible while still conserving momentum.
"Elastic" doesn't mean bouncy in the everyday sense

In physics, elastic has a precise meaning: total kinetic energy before equals total kinetic energy after. Truly elastic collisions are rare at everyday scales — a "superball" bounce or snooker-ball collision are close approximations, never perfect, since some sound and heat is always generated. They're essentially exact, however, for collisions between individual atoms, ions or subatomic particles, which have no internal structure to absorb energy as deformation or heat.

Perfectly inelastic collisions

When two objects collide and stick together, conservation of momentum alone is enough to find the common final velocity — no information about kinetic energy needed.

Worked exampleTwo trolleys colliding and sticking
1
A 2.0 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 1.0 kg trolley, and the two stick together. Find their common velocity after the collision.
2
Conserve momentum: $m_1u_1+m_2u_2 = (m_1+m_2)v$.
After sticking together, both trolleys share the same final velocity $v$, so their masses combine on the right.
3
$(2.0\times3.0)+(1.0\times0)=(2.0+1.0)v \implies 6.0=3.0v \implies v=2.0\,\text{m s}^{-1}$.
4
Check the kinetic energy: KE before $=\tfrac12(2.0)(3.0)^2=9.0\,\text{J}$. KE after $=\tfrac12(3.0)(2.0)^2=6.0\,\text{J}$.
3.0 J has been lost — converted to heat and sound in the collision. This confirms the collision was inelastic, as expected whenever objects stick together.
▪ solved

Notice the pattern: whenever objects coalesce, some kinetic energy is always lost. That loss isn't a mistake — it's a genuine physical requirement. A perfectly inelastic collision can never conserve kinetic energy unless the two objects already shared the same velocity before colliding (in which case there's no real "collision").

Elastic collisions

An elastic collision must satisfy both conservation equations simultaneously — momentum and kinetic energy — giving two independent equations, enough in principle to solve for two unknown final velocities.

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$
$\tfrac12 m_1u_1^2+\tfrac12 m_2u_2^2 = \tfrac12 m_1v_1^2+\tfrac12 m_2v_2^2$
A useful shortcut for elastic collisions

Solving the two equations simultaneously for general masses is algebraically heavy, but at A-level it's rarely necessary to derive the general result — most questions either give enough information to use momentum conservation directly (e.g. one object initially at rest) or expect you to combine both conservation equations for the specific numbers given. A useful special case: when a moving object collides elastically with a stationary object of equal mass, the incoming object stops dead and the target moves off with the incoming object's original velocity — a full transfer, only possible because the collision is elastic.

Worked exampleAn elastic collision between unequal masses
1
A 1.0 kg ball moving at 4.0 m s⁻¹ collides elastically head-on with a stationary 3.0 kg ball. Find both final velocities.
2
Momentum: $(1.0)(4.0)+(3.0)(0)=(1.0)v_1+(3.0)v_2 \implies 4.0=v_1+3v_2$.
3
Kinetic energy: $\tfrac12(1.0)(4.0)^2=\tfrac12(1.0)v_1^2+\tfrac12(3.0)v_2^2 \implies 16=v_1^2+3v_2^2$.
4
From the momentum equation, $v_1=4.0-3v_2$. Substituting into the energy equation and simplifying gives $v_2=2.0\,\text{m s}^{-1}$ and $v_1=-2.0\,\text{m s}^{-1}$.
The negative sign on $v_1$ means the lighter ball rebounds backwards — physically sensible, since it struck a heavier stationary target.
▪ solved

The coefficient of restitution Not included at A-level

Real collisions usually fall somewhere between the two extremes — inelastic, but not perfectly so. A single number, the coefficient of restitution $e$, quantifies where on that spectrum a collision sits: the ratio of relative separation speed to relative approach speed.

$e = \dfrac{\text{speed of separation}}{\text{speed of approach}} = \dfrac{v_2-v_1}{u_1-u_2}$
Two balls before and after a collision, showing their velocity directions in each state
Before the collision, ball 1 (velocity $u_1$) approaches ball 2 (velocity $u_2$). After the collision, they separate with velocities $v_1$ and $v_2$ — the coefficient of restitution compares how fast they move apart afterwards to how fast they closed in beforehand.

A perfectly elastic collision has $e=1$ (objects separate exactly as fast as they approached); a perfectly inelastic collision has $e=0$ (no separation — they move off together). Every real collision has $0 \le e < 1$. It isn't part of the core A-level specification, but is a useful extension if you encounter it.

Key takeaways

  • Momentum is conserved in every collision. Kinetic energy is only conserved in a perfectly elastic collision.
  • An inelastic collision loses some kinetic energy to heat, sound or deformation; a perfectly inelastic collision is the extreme case where the objects stick together and lose the maximum possible kinetic energy.
  • For a perfectly inelastic collision, momentum conservation alone gives the common final velocity: $m_1u_1+m_2u_2=(m_1+m_2)v$.
  • For an elastic collision, both momentum and kinetic energy conservation equations must hold simultaneously — two equations for two unknown final velocities.
  • A moving object colliding elastically with an equal stationary mass transfers all its velocity to the target and itself stops.
  • The coefficient of restitution $e=\dfrac{v_2-v_1}{u_1-u_2}$ measures how elastic a real collision is, from $e=0$ (perfectly inelastic) to $e=1$ (perfectly elastic).
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