SCIENCE · INVESTIGATE AND EXPLAIN

Separate center motion from collision energy

For two masses, kinetic energy splits into center-of-mass translation plus relative motion: K = MVcm²/2 + μurel²/2, where μ = m₁m₂/M. Expanding the two individual kinetic terms verifies this identity and shows why the cross terms cancel.

Learning outcome: Derive the energy split, reconstruct both outgoing velocities from center and relative motion, and check momentum and kinetic loss independently.

Starting recap

Integrate signed force and normalized mass moments. Specify the system before conserving momentum, and independently test energy and observation constraints.

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Try

For equal 2 kg masses and e = 1, outgoing velocities are −1 and 3 m/s and kinetic loss is zero. At e = 0 they both move at 1 m/s and lose the full 8 J of initial relative energy.

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Predict

For equal 2 kg masses and e = 0.5, translational kinetic energy lost is:

Your prediction and reason

Experiment

Derive the energy split, reconstruct both outgoing velocities from center and relative motion, and check momentum and kinetic loss independently.

Decompose collision energy into center and relative motion

Worked teaching model · synthetic values · First mass2 kg initially at+3 m/s, second selected mass initially at−1 m/s. One-dimensional isolated collision with supplied e=(v2−v1)/(u1−u2),0≤e≤1, no stored-energy release. COM and reduced-mass decomposition derived explicitly; e is a response assumption.

Second mass: 2 kg Supplied restitution: 0.5 ratio

Reset keeps your prediction, captures and practice history.

Relative kinetic energy retained across supplied response

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: e² times initial relative energy (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Relative kinetic energy retained across supplied response

e² times initial relative energy

restitution e	relative kinetic energy (J)
0	0
0.0125	0.00125
0.025	0.005
0.0375	0.01125
0.05	0.02
0.0625	0.03125
0.075	0.045
0.0875	0.06125
0.1	0.08
0.1125	0.10125
0.125	0.125
0.1375	0.15125
0.15	0.18
0.1625	0.21125
0.175	0.245
0.1875	0.28125
0.2	0.32
0.2125	0.36125
0.225	0.405
0.2375	0.45125
0.25	0.5
0.2625	0.55125
0.275	0.605
0.2875	0.66125
0.3	0.72
0.3125	0.78125
0.325	0.845
0.3375	0.91125
0.35	0.98
0.3625	1.05125
0.375	1.125
0.3875	1.20125
0.4	1.28
0.4125	1.36125
0.425	1.445
0.4375	1.53125
0.45	1.62
0.4625	1.71125
0.475	1.805
0.4875	1.90125
0.5	2
0.5125	2.10125
0.525	2.205
0.5375	2.31125
0.55	2.42
0.5625	2.53125
0.575	2.645
0.5875	2.76125
0.6	2.88
0.6125	3.00125
0.625	3.125
0.6375	3.25125
0.65	3.38
0.6625	3.51125
0.675	3.645
0.6875	3.78125
0.7	3.92
0.7125	4.06125
0.725	4.205
0.7375	4.35125
0.75	4.5
0.7625	4.65125
0.775	4.805
0.7875	4.96125
0.8	5.12
0.8125	5.28125
0.825	5.445
0.8375	5.61125
0.85	5.78
0.8625	5.95125
0.875	6.125
0.8875	6.30125
0.9	6.48
0.9125	6.66125
0.925	6.845
0.9375	7.03125
0.95	7.22
0.9625	7.41125
0.975	7.605
0.9875	7.80125
1	8

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
Center-of-mass velocity	1	m/s
Reduced mass	1	kg
First final velocity	0	m/s
Second final velocity	2	m/s
Center kinetic energy	2	J
Initial relative kinetic energy	8	J
Final relative kinetic energy	2	J
Internal energy increase	6	J
Final total momentum	4	kg m/s

The COM kinetic part is fixed by momentum; the supplied collision response determines the relative part.

Ktotal=(1/2)M Vcm²+(1/2)mu(u1−u2)², with mu=m1m2/M. Restitution changes relative approach speed4 into separation speed4e, leaving the COM velocity unchanged. Loss is(1/2)mu(1−e²)16. Momentum alone cannot determine e, and e0/e1 respectively give sticking/elastic endpoints under this passive family.

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

The COM kinetic part is fixed by momentum; the supplied collision response determines the relative part.

Observation to attach to a capture

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

Understand

Separate center motion from collision energy

The first particle has mass 2 kg and incoming velocity 3 m/s; the second has mass m₂ and incoming velocity −1 m/s. The approach speed is 4 m/s. A supplied restitution parameter e between 0 and 1 specifies the separation speed 4e; it is a model input, not inferred merely from momentum conservation.

For m₂ = 2 kg, Vcm = 1 m/s and μ = 1 kg. Incoming energy is 2 J of center translation plus 8 J of relative motion. At e = 0.5, outgoing velocities are 0 and 2 m/s, relative energy is 2 J and kinetic loss is 6 J. Total momentum remains unchanged.

In general the relative-energy loss is μ(4²)(1 − e²)/2. The center term is unchanged when external impulse is absent. A shared final velocity at e = 0 removes relative kinetic energy but does not remove the energy of center translation. A kinetic loss is an account of energy leaving these translational degrees of freedom, not evidence of one specified microscopic mechanism.

Changed-case check

Predict before reading the comparison. For equal 2 kg masses and e = 1, outgoing velocities are −1 and 3 m/s and kinetic loss is zero. At e = 0 they both move at 1 m/s and lose the full 8 J of initial relative energy.

Retained investigation

Derive the energy split, reconstruct both outgoing velocities from center and relative motion, and check momentum and kinetic loss independently. Why is the center kinetic term frame dependent while relative kinetic energy is not?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All packets, derivations and prose are original. These bounded mathematical models are not measured experiments, physical procedures, practical certification or institutional credit.

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Public practice checks a specific claim. It does not establish independent mastery.

Why is the center kinetic term frame dependent while relative kinetic energy is not?

Your explanation and limitations

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Previous: Cancel internal impulses only inside the correct system Next: Do not mistake a balanced vector ledger for a passive collision

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