Inelastic Collisions in Detail — Lost Energy, Kept Momentum

What makes a collision inelastic

An inelastic collision is one in which the total kinetic energy after impact is less than before. The missing energy has not vanished — it has been converted into heat, sound, and the work of permanently deforming the objects. Squelch a ball of clay against a wall and the dent is the receipt for the lost kinetic energy.

The defining equations:

1. Momentum is conserved (isolated system): $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.
2. Kinetic energy decreases: $KE_{after} < KE_{before}$.

With only one equation for two unknowns, a general inelastic collision is underdetermined — you need extra information (like 'they stick together') to solve it. That extra condition is what makes textbook problems solvable.

The perfectly inelastic limit: sticking together

The perfectly inelastic collision is the extreme: the objects stick together and move off with a single common velocity $v$. Maximum kinetic energy is lost — the most that momentum conservation allows.

$$v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$$

This is just the velocity of the system's center of mass, which no internal collision forces can change. A $2\ \mathrm{kg}$ lump of clay at $3\ \mathrm{m/s}$ striking a stationary $3\ \mathrm{kg}$ lump:

$$v = \frac{(2)(3) + (3)(0)}{2 + 3} = \frac{6}{5} = 1.2\ \mathrm{m/s}$$

Momentum check: before $6\ \mathrm{kg\cdot m/s}$, after $(5)(1.2) = 6$ ✓.

Worked example: counting the lost energy

Continuing the clay example, quantify the loss. Kinetic energy before:

$$KE_{before} = \tfrac{1}{2}(2)(3^2) + 0 = 9\ \mathrm{J}$$

After sticking:

$$KE_{after} = \tfrac{1}{2}(5)(1.2^2) = \tfrac{1}{2}(5)(1.44) = 3.6\ \mathrm{J}$$

Energy lost: $\Delta KE = 9 - 3.6 = 5.4\ \mathrm{J}$ — 60% of the original kinetic energy, converted into deforming the clay and warming it slightly.

The fraction lost depends on the masses: for a light object hitting a massive stationary one and sticking, almost all the kinetic energy disappears (think of a bug on a windshield); for equal masses, exactly half is lost.

Car crashes: engineered inelasticity

Car crashes are the canonical real-world inelastic collisions. Two $1500\ \mathrm{kg}$ cars collide head-on at $10\ \mathrm{m/s}$ each and the wreckage comes to rest (symmetric, so $v = 0$).

Momentum: $(1500)(10) + (1500)(-10) = 0$ before and after ✓. Kinetic energy before: $2 \times \tfrac{1}{2}(1500)(100) = 150{,}000\ \mathrm{J}$; after: zero. All $150\ \mathrm{kJ}$ — roughly the energy of a stick of dynamite — goes into crumpling metal, heat and sound.

This is why modern cars have crumple zones: by extending the collision time, the same momentum change happens with smaller average force ($F = \Delta p/\Delta t$), and the energy is absorbed by sacrificial structure rather than the passenger cabin.

Explosions: inelasticity in reverse

An explosion is an inelastic collision run backward in time. Chemical potential energy converts into kinetic energy, so $KE_{after} > KE_{before}$ — yet momentum conservation still governs.

A $10\ \mathrm{kg}$ firework shell at rest bursts into two fragments of $6\ \mathrm{kg}$ and $4\ \mathrm{kg}$. The $6\ \mathrm{kg}$ piece flies off at $20\ \mathrm{m/s}$. What about the other?

Total momentum must stay zero: $(6)(20) + (4)(v_2) = 0$, so $v_2 = -30\ \mathrm{m/s}$. The lighter fragment recoils faster — the same inverse-mass pattern as collisions, with energy flowing in instead of out.

This is also the physics of rocket propulsion: the rocket and its exhaust are an 'explosion' in slow motion, and momentum conservation is what pushes the rocket forward.

Choosing the right collision model

Choose the right model with this decision tree:

• Do the objects stick together? → Perfectly inelastic: one equation, $v = (m_1u_1 + m_2u_2)/(m_1+m_2)$.
• Is kinetic energy conserved? → Elastic: use both conservation equations (or the closed-form solutions).
• Neither stated? → Momentum conservation always holds for the isolated system; you need one more piece of data (a final velocity, the energy lost, the coefficient of restitution) before the problem is solvable.

And the universal safety checks: verify momentum balances, verify the energy change has the correct sign (loss for inelastic, gain only if there is an energy source like an explosion), and confirm no speed exceeds physical sense.

Variables and units

Common mistakes

Real-world applications

• Car crash safety design manages how kinetic energy converts to crumple-zone deformation.

• Ballistic pendulums historically measured bullet speeds from the swing of a struck block.

• Railway coupling analysis treats wagon joining as perfectly inelastic collisions.

• Forensic accident reconstruction uses post-crash skid and stick-together motion to infer pre-impact speeds.