SCIENCE · INVESTIGATE AND EXPLAIN

Cancel internal impulses only inside the correct system

For a fixed set of particles, total momentum is M times center-of-mass velocity. Internal forces cancel in equal-and-opposite pairs when both recipients are included. This statement does not mean that every component has constant momentum or that internal energy cannot change.

Learning outcome: Construct each particle ledger before adding it. Repeat after excluding one internal recipient and identify the impulse that has become external to that smaller boundary.

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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With J = −2 N s and K = 0, velocities are (−1, 2, 1) m/s. Total momentum remains 6 kg m/s and the center still moves at 1 m/s despite one component reversing.

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Predict

Changing only internal J while holding K fixed changes which quantity?

Your prediction and reason

Experiment

Construct each particle ledger before adding it. Repeat after excluding one internal recipient and identify the impulse that has become external to that smaller boundary.

Cancel internal impulses only for the complete recipient set

Worked teaching model · synthetic values · Three point masses1,2,3 kg initially share velocity1 m/s. During a short interval first receives internal impulse J and second−J; third receives declared external impulse K. Other impulses are zero. Endpoint momentum accounting, no force-time profile or physical collision procedure.

Internal impulse from second to first: 2 N s External impulse on third particle: 3 N s

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Individual final momenta and their total

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

Equivalent numerical coordinates for Individual final momenta and their total

individual momenta

particle index1,2,3	final momentum (kg m/s)
1	3
2	0
3	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 velocity	3	m/s
Second final velocity	0	m/s
Third final velocity	2	m/s
Initial total momentum	6	kg m/s
Final total momentum	9	kg m/s
Final center-of-mass velocity	1.5	m/s
External impulse	3	N s
Internal pair impulse sum	0	N s

Internal impulses cancel only after both interacting recipients are included in the system.

The first/second momentum changes are+J and−J, while the third receives external K. Total P changes from6 to6+K and Vcm from1 to1+K/6 independently of J. Individual momentum and kinetic energies can change. The endpoint impulse packet does not specify how much energy an internal mechanism supplied or absorbed.

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

Internal impulses cancel only after both interacting recipients are included in the system.

Observation to attach to a capture

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Understand

Cancel internal impulses only inside the correct system

Three supplied particles have masses 1, 2 and 3 kg and initially all move at 1 m/s. An internal interaction gives impulse +J to the first and −J to the second. An external agent gives impulse K to the third. Their final velocities are 1 + J, 1 − J/2 and 1 + K/3 in metres per second when impulses are in N s.

Adding the three momenta gives 6 + K kg m/s: the two J terms cancel exactly. The whole-system center velocity is 1 + K/6 m/s. Changing J redistributes component velocities but cannot change that center motion.

For J = 2 N s and K = 3 N s, velocities are (3, 0, 2) m/s and total momentum is 9 kg m/s. Center velocity is 1.5 m/s. Nothing in the supplied impulses identifies an internal spring, thermal conversion or kinetic-energy conservation, so such a mechanism must not be invented.

Changed-case check

Predict before reading the comparison. With J = −2 N s and K = 0, velocities are (−1, 2, 1) m/s. Total momentum remains 6 kg m/s and the center still moves at 1 m/s despite one component reversing.

Retained investigation

Construct each particle ledger before adding it. Repeat after excluding one internal recipient and identify the impulse that has become external to that smaller boundary. Does momentum conservation alone identify the energy conversion?

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.

Does momentum conservation alone identify the energy conversion?

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