How Momentum and Kinetic Energy Classify Collisions

Momentum conservation is the starting point

For an isolated system, the vector sum of momentum is the same before and after every collision:

$$\sum \mathbf p_i=\sum \mathbf p_f.$$

Momentum conservation alone does not tell you whether a collision is elastic. The classification comes from comparing total translational kinetic energy:

• Perfectly elastic: momentum and total kinetic energy are conserved.
• Inelastic: momentum is conserved, but final kinetic energy is lower.
• Perfectly inelastic: the objects stick together and have one final velocity. For the stated initial conditions, this produces the greatest possible kinetic-energy loss consistent with momentum conservation.

A fresh comparison

Suppose a $1.0\,\mathrm{kg}$ cart moving at $+4.0\,\mathrm{m/s}$ hits a stationary $3.0\,\mathrm{kg}$ cart.

If they stick together, momentum conservation gives

$$v_f=\frac{(1.0)(4.0)+(3.0)(0)}{1.0+3.0}=1.0\,\mathrm{m/s}.$$

The kinetic energy falls from $8.0\,\mathrm{J}$ to $2.0\,\mathrm{J}$, so the collision is perfectly inelastic.

If the same idealized carts collide perfectly elastically in one dimension, the final velocities are $-2.0\,\mathrm{m/s}$ and $+2.0\,\mathrm{m/s}$. The final kinetic energy is

$$\tfrac12(1.0)(2.0)^2+\tfrac12(3.0)(2.0)^2=8.0\,\mathrm{J},$$

equal to the initial value.

A reliable classification workflow

1. Define the system and choose a sign convention.
2. Check total momentum using signed velocities.
3. Calculate total kinetic energy using speed squared.
4. Compare $K_i$ and $K_f$.
5. If the bodies share one final velocity, identify the perfectly inelastic case.

In real collisions, missing translational kinetic energy is transformed into internal energy, deformation, heat, sound, or rotation. It is not destroyed.