Do not mistake a balanced vector ledger for a passive collision
A planar momentum equation determines a missing vector when all masses and the other vector are supplied. It does not by itself prove that the event satisfies an additional energy restriction. Here a 2 kg particle enters with velocity (3, 0) m/s and a 1 kg particle starts at rest, so total momentum is (6, 0) kg m/s and initial kinetic energy is 9 J.
Learning outcome: Derive the admissible velocity disk by completing the square. Test one interior, boundary and exterior point with separate vector and energy ledgers.
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
Change the first outgoing velocity to (2, 1) m/s. The second is (2, −2) m/s and final kinetic energy is 9 J, on the passive elastic boundary. The point (2, 0) gives the minimum 6 J.
Open the experimentPredict
A vector-balanced packet has final kinetic energy 12 J from an initial 9 J. Under the stated passive no-release assumption it is:
Experiment
Derive the admissible velocity disk by completing the square. Test one interior, boundary and exterior point with separate vector and energy ledgers.
Test a momentum-compatible collision against its energy premise
Worked teaching model · synthetic values · First mass2 kg enters at(3,0), second mass1 kg is at rest. Given first outgoing velocity, infer second from zero external impulse. Then test the separate passive loss-only collision assumption; momentum-compatible states can require an internal stored-energy release not included in that passive premise.
Reset keeps your prediction, captures and practice history.
Outgoing momentum chain
Spatial axes use the same pixels per physical unit, preserving geometry. 1: first then second outgoing momentum (solid).2: conserved incoming total (dashed). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Outgoing momentum chain
first then second outgoing momentum
| east momentum (kg m/s) | north momentum (kg m/s) |
|---|---|
| 0 | 0 |
| 4 | 2 |
| 6 | 0 |
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.
conserved incoming total
| east momentum (kg m/s) | north momentum (kg m/s) |
|---|---|
| 0 | 0 |
| 6 | 0 |
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.
| Quantity | Value | Unit |
|---|---|---|
| Inferred second east velocity | 2 | m/s |
| Inferred second north velocity | -2 | m/s |
| Final east momentum | 6 | kg m/s |
| Final north momentum | 0 | kg m/s |
| Initial kinetic energy | 9 | J |
| Final kinetic energy | 9 | J |
| Required internal energy increase | 0 | J |
| Passive loss-only premise feasible | 1 | flag |
| Center kinetic energy | 6 | J |
| Final relative kinetic energy | 3 | J |
This packet conserves momentum and does not require added translational kinetic energy under the passive screen.
Zero external impulse implies v2=(6−2v1x,−2v1y). Check kinetic energy separately: Kf=3(v1x−2)²+3v1y²+6 J. Thus a passive collision requires(v1x−2)²+v1y²≤1. Being inside that disk is a necessary energy condition, not proof that a particular contact material produces the chosen outgoing directions.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
This packet conserves momentum and does not require added translational kinetic energy under the passive screen.
Zero external impulse implies v2=(6−2v1x,−2v1y). Check kinetic energy separately: Kf=3(v1x−2)²+3v1y²+6 J. Thus a passive collision requires(v1x−2)²+v1y²≤1. Being inside that disk is a necessary energy condition, not proof that a particular contact material produces the chosen outgoing directions.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
Understand
Apply
Public practice checks a specific claim. It does not establish independent mastery.
How would permitting a known internal energy release change the admissible set?
Continue
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