Derive collision endpoints from momentum and relative motion

Start with an impulse boundary

During a sufficiently short collision, net external impulse may be negligible compared with the exchanged internal impulses. That is an approximation to justify, not a rule that gravity or supports vanish. Under it, total momentum before and after is equal component by component. This one equation alone cannot determine two unknown final velocities.

Use the elastic or sticking condition

An elastic collision also conserves total kinetic energy. Together with momentum, this implies reversal of relative velocity: v 2−v 1=u 1−u 2 for an approaching one-dimensional pair. In a completely inelastic event the objects share a final velocity P/M. Inelastic does not always mean sticking; separate objects can lose translational kinetic energy.

Parameterize an optional intermediate family

For comparison, the workbench uses v 2−v 1=e(u 1−u 2) with 0≤e≤1. This coefficient-of-restitution family is optional phenomenological enrichment; it does not identify the collision microphysics. With m 1=2,u 1=4,u 2=−1 and second mass m 2, let P=8−m 2 and M=2+m 2. Solving gives v 1=(P−5m 2e)/M and v 2=(P+10e)/M.

Audit kinetic energy separately

The decrease is one-half reduced mass times(1−e²)(u 1−u 2)², where reduced mass=m 1m 2/M. At m 2=3,e=1, velocities are−2 and 3 m/s and K is unchanged. At e=0 both velocities are 1 m/s, leaving K=2.5 J from 17.5 J initially. The 15 J difference becomes other internal channels in the enlarged account, not destroyed total energy.

Challenge a proposed result

Check momentum first, then the supplied collision condition and nonnegative energy loss for this bounded family. A pair of velocities can conserve momentum but violate elasticity. A result with e>1 would require an internal energy source not included in this passive family. A real event also needs uncertainty and external-impulse evidence.

Worked original case

For masses 2 and 3 with velocities 4 and−1, elastic endpoints−2 and 3 conserve momentum 5 and K 17.5. Sticking endpoints 1 and 1 retain momentum 5 but only K 2.5.

Investigation

At second mass 3 compare e=0,.5,1. Independently solve both equations, calculate K before/after and retain the changed constraint. Identify an external impulse that would invalidate the isolated-event approximation.

Public numerical transfer practice

Try a new practice case

Masses 2 and 3 kg with initial velocities 4 and−1 m/s stick. What is their common final velocity in m/s?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original content checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax University Physics momentum. Packets are supplied/generated, not performed collisions or instrument measurements. No physical impact, release, explosive materials or moving machinery is requested. Public solutions and virtual captures do not certify independent mastery or supervised laboratory work.