Derive center and relative energy for an internal release

Name the energy source and boundary

The virtual system begins as two co-moving masses 2 and 3 kg with center velocity V. An unspecified internal store releases energy into their relative motion. No external impulse acts. This is a mathematical event, not a request for an explosive, spring device or physical release. Total energy includes the internal store before the event.

Solve center and relative constraints

Final momentum obeys 2v 1+3v 2=5V. Define the supplied relative separation velocity w=v 1−v 2≥0. Solving the two equations gives v 1=V+3w/5 and v 2=V−2w/5. The more massive fragment has the smaller change relative to the center, because the center-frame momenta are equal and opposite.

Expand the quadratic energy

Insert both velocities into K=v 1²+1.5v 2². Cross terms cancel, leaving K=2.5V²+.6w². The first term is the center translational energy; the second is one-half reduced mass(6/5 kg) times w². This decomposition is an optional derivational extension supporting the core difference between momentum and energy conservation.

Find the minimum at fixed momentum

Since relative energy is nonnegative, the smallest translational K at the given total momentum is P²/(2M), attained when both velocities equal V. Internal release raises K by.6w² while reducing the store by the same amount under the declared ideal accounting. Momentum conservation alone does not prohibit that increase.

Test frame and direction changes

At V=2,w=3, final velocities are 3.8 and.8 m/s; both fragments move right in the laboratory even though they separate. In the center frame their velocities are 1.8 and−1.2. At V=−2 with the same w, total K is unchanged because V² is unchanged, while total momentum changes sign. Relative energy is independent of a uniform frame translation.

Worked original case

V=2,w=3 gives total P 10 kg m/s, center K 10 J and relative K 5.4 J, for total 15.4 J. The increase 5.4 J must come from the stated internal store.

Investigation

Compare w=0,3,5 at V=2, then reverse V. Reconstruct both fragment velocities and the center-frame momenta, retain the internal-energy decrease and explain why the event is not an isolated creation of energy.

Public numerical transfer practice

Try a new practice case

For masses 2 and 3 kg and relative speed 5 m/s, what is the relative kinetic energy in joules?

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.