Distinguish elastic and inelastic collisions

Momentum conservation is not enough to classify a collision

With both masses 2 kg, the initial velocities +3 and −1 m/s give total momentum +4 kg·m/s. Initial translational kinetic energy is Kinitial = ½ × 2 × 3² + ½ × 2 × (−1)² = 10 J. Momentum is a signed sum; kinetic energy is a sum of nonnegative terms.

Keep the masses fixed and compare restitution values 1, 0.5 and 0. All three trials conserve the same total momentum because external horizontal impulse is negligible. Their final kinetic energies differ. This comparison tests two separate statements rather than treating every conservation law as interchangeable.

A collision is elastic when total kinetic energy is unchanged between the before and after states. An inelastic collision has a decrease in that kinetic energy, represented here by an increase in internal energy. The model includes no energy release from a spring or chemical source and no export to the surroundings. More general interactions can require additional terms.

The OpenStax collision classification distinguishes elastic, inelastic and sticking outcomes. The widget supplies idealized original examples and does not infer material properties from observations.

Read the restitution ratio as relative speed

The control e is the coefficient of restitution, a dimensionless ratio: e = (vB − vA)/(uA − uB) for this one-dimensional approaching pair. The u values are initial velocities and the v values are final velocities. A begins to the left of B, so uA − uB = 4 m/s is the approach speed; vB − vA is their separation speed afterward.

At e = 1, separation speed equals approach speed. In this two-body, nonrotating model with momentum conservation, that is the elastic limit. At e = 0.5, separation speed is 2 m/s. At e = 0, relative velocity is zero and the model treats the pair as sticking. Real equal outgoing velocities alone do not tell us how an adhesive contact is constructed; sticking is the specified outcome here.

Restitution is not the fraction of total kinetic energy retained. The relationship depends on the center-of-mass motion as well as relative motion. It is also not assumed to be a universal constant for all speeds or contact conditions of a real material.

Work through the three outcomes

At e = 1 with equal masses, A finishes at −1 m/s and B at +3 m/s. Their final kinetic energies are 1 and 9 J. The total remains 10 J, and the internal-energy increase shown is zero. Their velocities exchange in this special equal-mass case; that is not a rule for unequal masses.

At e = 0.5, A finishes at 0 and B at +2 m/s. Total final kinetic energy is 0 + ½ × 2 × 2² = 4 J. Momentum remains 0 + 2 × 2 = 4 kg·m/s. The objects separate, yet the collision is inelastic: 6 J of initial translational kinetic energy becomes combined internal energy. Bouncing or separation does not establish elasticity.

At e = 0, both finish at +1 m/s. Total final kinetic energy is ½ × 4 × 1² = 2 J, with 8 J of combined internal-energy increase. The pair still moves. Perfectly inelastic does not mean that all initial kinetic energy must disappear from translation in the chosen frame.

Capture each outcome. The before energy bar retains the same length; the after bar splits between kinetic energy and an outlined internal-energy increase. Their combined extent matches the initial bar. The individual momentum bars rearrange, but both total-momentum bars stay at +4 kg·m/s.

Why sticking leaves the minimum kinetic energy

For a fixed total momentum and total mass, there is kinetic energy associated with center-of-mass motion. In the equal-mass example, the 4 kg system’s center moves at +1 m/s, contributing 2 J. Relative motion accounts for the remaining initial 8 J.

Sticking removes the relative motion, leaving that 2 J center-of-mass contribution. It is the minimum translational kinetic energy compatible with this conserved momentum. If total momentum is zero, the minimum can be zero instead, as in the 6 kg B sticking case from the preceding lesson.

Explore the relative-energy relationship

Let M = mA + mB be total mass, P total momentum and μ = mA mB/M the reduced mass. Total kinetic energy can be written P²/(2M) + ½μ(vA − vB)². The first term describes center-of-mass motion. Restitution multiplies relative speed by e, so the relative kinetic energy after contact is e² times its initial value. The model's internal-energy increase is therefore ½μ(1 − e²)(uA − uB)². This is why e = 0.5 does not generally retain half of total initial kinetic energy.

Check model limits before interpreting the result

The internal-energy increase is a combined accounting quantity. It does not specify temperature, permanent deformation, or how energy is distributed between the two objects. The before/after rules also do not predict contact duration, peak force or compression. Those would need an interaction model and measurements.

Change B mass to 6 kg at e = 1. Final velocities become −3 and +1 m/s, rather than the −1 and +3 m/s exchange of the equal-mass trial. Verify both total momentum zero and kinetic energy 12 J. Changing mass changes the starting system; comparing results requires keeping that distinction visible.

Transfer with an independent energy calculation

A 1 kg object at +5 m/s and a 3 kg object at −1 m/s stick at +0.5 m/s. Calculate initial and final kinetic energy, the internal-energy increase, and decide whether their nonzero final speed contradicts perfect inelasticity.

Check the collision-energy transfer

Initial kinetic energy is ½ × 1 × 25 + ½ × 3 × 1 = 14 J. Final kinetic energy is ½ × 4 × 0.5² = 0.5 J. The combined internal-energy increase is 13.5 J. The remaining 0.5 J is the center-of-mass energy required by the nonzero total momentum. A common nonzero final velocity is entirely consistent with a perfectly inelastic collision.