SCIENCE · INVESTIGATE AND EXPLAIN

Conserve planar momentum and test an inferred external impulse

Keep vector components through a sticking collision before interpreting speed or energy.

Learning outcome: Compare b=2 and−2 at a=3. Reconstruct vector and energy results, then suppose the measured final velocity is(1,1): calculate the required residual impulse and list two non-force alternatives that could explain it.

Starting recap

Integrate net external force, preserve vector and system boundaries, and test collision and evidence assumptions independently.

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Try

a=3,b=2 gives initial P=(6,6), common V=(1.2,1.2), Kf 7.2 J and K decrease 7.8 J.

Open the experiment

Predict

Can adding initial momentum magnitudes replace separate component conservation?

Your prediction and reason

Experiment

Compare b=2 and−2 at a=3. Reconstruct vector and energy results, then suppose the measured final velocity is(1,1): calculate the required residual impulse and list two non-force alternatives that could explain it.

Reconcile perpendicular momenta in a sticking event

Worked teaching model · synthetic values · Mass2 kg initially(a,0), mass3 kg initially(0,b), then common velocity after an ideal short sticking event. External impulse negligible; internal-energy increase may include deformation, thermal and other channels, not solely heat.

First x velocity: 3 m/s Second y velocity: 2 m/s

Reset keeps your prediction, captures and practice history.

Momentum component vectors

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Momentum component vectors: numerical coordinates

momentum x (kg m/s)	momentum y (kg m/s)
0	0
6	0
6	6

Calculated values under the declared model
Quantity	Value	Unit
Final common velocity x	1.2	m/s
Final common velocity y	1.2	m/s
Final speed	1.69706	m/s
Direction from positive x	45	degrees
Initial translational kinetic energy	15	J
Final translational kinetic energy	7.2	J
Decrease of translational kinetic energy	7.8	J

Conserve each momentum component, then reconstruct the common velocity vector.

The plotted head-to-tail component chain has independently scaled axes; calculate angles from components rather than screen geometry. It is not a spatial trajectory. A measured final velocity different from this prediction requires an external impulse or a failed mass/event/measurement assumption. The scalar speed sum is not a momentum sum.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Conserve each momentum component, then reconstruct the common velocity vector.

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 vector initial state

A 2 kg object initially moves with velocity(a,0), while a 3 kg object moves with(0,b). Their total momentum is(2a,3b), not the sum of momentum magnitudes. Negative b reverses the vertical component without changing the corresponding kinetic-energy contribution.

Apply the common-velocity constraint

After an ideal short sticking event, the 5 kg combined object moves with V=(2a/5,3b/5) if external impulse is negligible. Its speed is the magnitude of that vector and its direction follows the ratio of signed components. The head-to-tail momentum plot is a component chain, not the spatial route taken by either object. Its axes rescale independently, so do not infer an angle such as 45 degrees from the screen geometry; calculate direction from the labeled momentum components.

Check the energy difference

Initial translational K=a²+1.5b², while final K=2.5(Vx²+Vy²). For a=3,b=2, V=(1.2,1.2) m/s, speed 1.2sqrt(2), angle 45 degrees, and K drops from 15 to 7.2 J. The missing 7.8 J belongs to other channels such as deformation or thermal energy, subject to a larger-system accounting.

Invert a discrepant observation

If a measured common velocity differs from the prediction, compute external impulse candidate Jext=5Vmeasured−Pinitial. That discrepancy could also signal a mass error, incomplete sticking, unmeasured fragments or velocity calibration fault. It is evidence of a failed assumed account, not automatic proof of a new external interaction.

Compare a changed direction

For the same a=3 with b=−2, final velocity(1.2,−1.2) has the same speed and kinetic loss but a different direction. Scalar energy alone cannot distinguish these endpoints. A practical investigation would require independent component tracking and a justified short-event boundary; no physical collision is requested here.

Worked original case

a=3,b=2 gives initial P=(6,6), common V=(1.2,1.2), Kf 7.2 J and K decrease 7.8 J.

Investigation

Public numerical transfer practice

Try a new practice case

For a=3,b=2, suppose final common velocity is measured as(1,1) m/s. What is the x component of the external impulse required by the stated masses, in N s?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

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.

Apply

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

What information does a momentum vector diagram preserve that a speed-only graph loses?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Derive collision endpoints from momentum and relative motion Next: Derive center and relative energy for an internal release

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