Last updated: 2026-07-01

Perfectly inelastic collisions

A collision is perfectly inelastic when the two bodies stick together on impact and move off as a single object. Because they end up joined, they must share one common final velocity. This is the extreme opposite of an elastic collision, in which the bodies bounce apart and keep all of their kinetic energy.

Two principles govern what happens:

• Momentum is always conserved. No outside horizontal force acts during the brief moment of contact, so the total momentum just after impact equals the total just before.
• Kinetic energy is not conserved. Some of it is spent deforming the bodies and is released as heat, sound, and permanent damage. A perfectly inelastic collision loses the largest fraction of kinetic energy that conservation of momentum still allows.

Given the two masses and their initial velocities, the calculator returns the common final velocity, the kinetic energy before and after, and the energy lost.

The final-velocity formula

Write the total momentum before and after the collision and set them equal. Before impact the bodies carry m₁u₁ and m₂u₂; after impact the combined mass (m₁ + m₂) moves at one velocity v:

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

Solving for v gives the result this calculator uses:

$v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

Velocities are signed. Choose one direction as positive; motion the other way is negative. Two bodies approaching head-on therefore have opposite signs, and their momenta partly cancel.

Energy before and after

The kinetic energy before the collision is the sum of ½mv² for each body:

$KE_i = \tfrac{1}{2} m_1 u_1^2 + \tfrac{1}{2} m_2 u_2^2$

After the collision a single mass moves at the common velocity:

$KE_f = \tfrac{1}{2}(m_1 + m_2)\,v^2$

The energy lost is simply the difference:

$\Delta KE = KE_i - KE_f$

ΔKE is never negative: the combined object can never carry more kinetic energy than the bodies brought into the collision.

Worked example

A 2 kg body moving at 5 m/s strikes a stationary 3 kg body and the two stick together.

Final velocity

$v = \frac{2 \times 5 + 3 \times 0}{2 + 3} = \frac{10}{5} = 2\ \text{m/s}$

Kinetic energy before

KE_i = ½ × 2 × 5² + ½ × 3 × 0² = 25 J

Kinetic energy after

KE_f = ½ × (2 + 3) × 2² = ½ × 5 × 4 = 10 J

Kinetic energy lost

ΔKE = 25 − 10 = 15 J

So 60% of the original kinetic energy is converted to heat, sound, and deformation, while all of the momentum (10 kg·m/s) is preserved.

A bullet in a block

The classic ballistic-pendulum setup is perfectly inelastic. A 10 g (0.01 kg) bullet travelling at 400 m/s embeds itself in a 2 kg block at rest:

$v = \frac{0.01 \times 400 + 2 \times 0}{0.01 + 2} = \frac{4}{2.01} \approx 1.99\ \text{m/s}$

The kinetic energy before is ½ × 0.01 × 400² = 800 J, while afterwards it is only ½ × 2.01 × 1.99² ≈ 3.98 J. Almost the entire 800 J is lost — mostly as heat in the bullet and block. This huge energy loss, combined with conserved momentum, is exactly why a ballistic pendulum can measure a bullet's speed from how far the block swings.

Inelastic compared with elastic collisions

Real collisions usually fall between these two ideals. Cars are engineered with crumple zones precisely so that a crash behaves more like an inelastic collision, absorbing kinetic energy in the structure instead of passing it on to the occupants.