Supply a collision response instead of inventing one from momentum

Define the energy or momentum system

A short collision with negligible net external impulse preserves the total momentum of the selected pair. It does not automatically preserve kinetic energy or specify both final velocities. Here a 2 kg point moving at 4 m/s meets an initially resting target of mass M on the same line.

Build the algebraic account

The core endpoints are a perfectly sticking collision and an elastic collision. To connect them, this workbench supplies the empirical recovery factor e=(v2−v1)/(u1−u2) as an optional algebraic model. It is not inferred from momentum conservation. With u1=4 and u2=0, the equations are2v1+Mv2=8 and v2−v1=4e.

Check an original packet

Solving gives v1=4(2−eM)/(2+M) and v2=8(1+e)/(2+M). For M=3 kg and e=0.5, v1=0.4 and v2=2.4 m/s. Final kinetic energy is 8.8 J, compared with 16 J initially, leaving7.2 J as internal energy under the isolated collision account. Both signed velocities and the nonnegative separation speed should be checked.

Test the domain and evidence

At e=0 both points share the center-of-mass velocity and stick. At e=1 total kinetic energy is unchanged; intermediate values convert some translational kinetic energy into internal forms. Actual response depends on materials and contact conditions, which are not measured here. Stored-energy releases could require a different model; the selected0≤e≤1 family deliberately excludes them.

Worked original packet

For equal masses M=2 kg in the elastic endpoint e=1, v1=0 and v2=4 m/s. In the sticking endpoint e=0, both move at 2 m/s and kinetic energy falls from 16 to 8 J.

Bounded investigation

At M=3 kg compare e=0,0.5,1. Retain both conservation and supplied-response equations and distinguish a predicted response from an empirical assumption. Then test the equal-mass elastic endpoint.

Changed-case numerical practice

Try the changed case before feedback

Two equal2 kg points undergo an ideal one-dimensional elastic collision. The first initially moves at 3 m/s and the second is at rest. Find the second point’s final velocity in m/s.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. All packets and diagrams are synthetic teaching evidence. No physical launches, collisions, spring releases, explosive processes or equipment-safety judgments are requested. Retained self-reports and public numerical feedback do not certify supervised laboratory practice or independent mastery.