SCIENCE · INVESTIGATE AND EXPLAIN

Track an internal energy release in two frames

Two points of masses 1 and 3 kg initially move together at ground velocity U. An internal store separates them with relative final speed u=v1−v2, with no external impulse. Their total momentum remains 4U even though their kinetic energy can increase. The energy source is the internal store, not a violation of momentum conservation.

Learning outcome: Hold relative speed fixed and change U. Compare ground and center-frame velocities, momenta and kinetic energies. Then double u and predict the energy-source requirement before checking.

Starting recap

Choose the system, interval and reference frame before using a conservation statement. Keep vector momentum, scalar energy and transferred power distinct.

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Try

At U=0 and u=4 m/s, the final velocities are3 and−1 m/s. Their momenta cancel, while kinetic energy is 6 J. At U=2 the same relative release instead gives 5 and 1 m/s, adding the same6 J to the initial ground kinetic energy.

Open the experiment

Predict

With negligible external impulse, an internal release can:

Your prediction and reason

Experiment

Hold relative speed fixed and change U. Compare ground and center-frame velocities, momenta and kinetic energies. Then double u and predict the energy-source requirement before checking.

Compare internal release in two inertial frames

Worked teaching model · synthetic values · Two virtual points of masses 1 and 3 kg initially share selected velocity. An internal store releases them with v1−v2 equal to selected nonnegative speed, no external impulse. Endpoint stored-energy loss supplies kinetic increase; no explosive or physical release procedure.

Initial shared ground velocity: 1 m/s Final relative separation speed: 2 m/s

Reset keeps your prediction, captures and practice history.

Internal energy required for relative separation

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: reduced-mass energy (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Internal energy required for relative separation

reduced-mass energy

relative separation speed (m/s)	kinetic energy increase (J)
0	0
0.1	0.00375
0.2	0.015
0.3	0.03375
0.4	0.06
0.5	0.09375
0.6	0.135
0.7	0.18375
0.8	0.24
0.9	0.30375
1	0.375
1.1	0.45375
1.2	0.54
1.3	0.63375
1.4	0.735
1.5	0.84375
1.6	0.96
1.7	1.08375
1.8	1.215
1.9	1.35375
2	1.5
2.1	1.65375
2.2	1.815
2.3	1.98375
2.4	2.16
2.5	2.34375
2.6	2.535
2.7	2.73375
2.8	2.94
2.9	3.15375
3	3.375
3.1	3.60375
3.2	3.84
3.3	4.08375
3.4	4.335
3.5	4.59375
3.6	4.86
3.7	5.13375
3.8	5.415
3.9	5.70375
4	6

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
First final ground velocity	2.5	m/s
Second final ground velocity	0.5	m/s
Initial total momentum	4	kg m/s
Final total momentum	4	kg m/s
Initial ground kinetic energy	2	J
Final ground kinetic energy	3.5	J
Internal stored-energy decrease	1.5	J
Final center-of-mass-frame kinetic energy	1.5	J

Internal energy can become kinetic energy without changing the total momentum.

Momentum and the specified relative speed give v1=U+3u/4 and v2=U−u/4. Subtracting the center-frame speed U leaves opposite component momenta but positive relative kinetic energy 3u²/8. The kinetic-energy increase is the same in these uniformly shifted frames because total momentum is unchanged. The supplied release speed needs an internal energy source; it is not free acceleration.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Internal energy can become kinetic energy without changing the total momentum.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the energy or momentum system

Build the algebraic account

The equations v1+3v2=4U and v1−v2=u give v1=U+3u/4 and v2=U−u/4. The lighter point receives a larger velocity change because the reciprocal internal impulses have equal magnitudes but divide by different masses. Opposite changes need not mean opposite ground velocities.

Check an original packet

In the center-of-mass frame subtract U from both velocities. Momenta become+3u/4 and−3u/4 kg m/s, and total kinetic energy is 3u²/8 J. In the ground frame initial kinetic energy is 2U² and final energy is 2U²+3u²/8. The increase is the same in these uniformly shifted frames because the pair’s total momentum is unchanged.

Test the domain and evidence

At U=1 m/s and u=2 m/s, final velocities are2.5 and 0.5 m/s; both still move right in the ground frame. Momentum is 4 kg m/s before and after. Kinetic energy rises from 2 to 3.5 J, requiring1.5 J from the internal store. This is a virtual endpoint comparison, not an explosive, spring-release or propulsion procedure.

Worked original packet

Bounded investigation

Hold relative speed fixed and change U. Compare ground and center-frame velocities, momenta and kinetic energies. Then double u and predict the energy-source requirement before checking.

Changed-case numerical practice

Try the changed case before feedback

Masses1 and 3 kg initially rest together. After an internal release their relative separation speed is 4 m/s with no external impulse. What kinetic-energy increase is required, in joules?

Numerical result in the requested units

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.

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Why is the kinetic-energy increase invariant here even though kinetic energy itself depends on frame?

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