SCIENCE · INVESTIGATE AND EXPLAIN

Preserve two momentum components through a point collision

Momentum is a vector. In a two-dimensional collision with negligible external impulse, conserve its east and north components separately before calculating a magnitude or direction. Adding the incoming speed magnitudes would discard the directions that the conservation statement needs.

Learning outcome: Compare an oblique packet, a head-on partial cancellation and exact zero momentum. Retain both components and kinetic-energy accounts. Explain why the displayed chain is a vector addition diagram rather than the collision trajectory.

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,w=4, common velocity is(2,4/3) m/s. Its speed is√52/3≈2.40370 m/s and direction atan(2/3)≈33.6901° north of east, not the angle obtained by adding speed magnitudes.

Open the experiment

Predict

Exact total momentum cancellation after point sticking means:

Your prediction and reason

Experiment

Compare an oblique packet, a head-on partial cancellation and exact zero momentum. Retain both components and kinetic-energy accounts. Explain why the displayed chain is a vector addition diagram rather than the collision trajectory.

Keep both momentum components through sticking

Worked teaching model · synthetic values · First point has mass 2 kg and velocity (3,0) m/s; second mass 1 kg has selected velocity. They meet and stick at a point with negligible external impulse and no retained rotation. Zero resultant velocity has no direction.

Second point east velocity: -2 m/s Second point north velocity: 4 m/s

Reset keeps your prediction, captures and practice history.

Momentum addition on equal component scales

Spatial axes use the same pixels per physical unit, preserving geometry. 1: component chain (solid).2: total momentum (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Momentum addition on equal component scales

component chain

east momentum (kg m/s)	north momentum (kg m/s)
0	0
6	0
4	4

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.

total momentum

east momentum (kg m/s)	north momentum (kg m/s)
0	0
4	4

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
Total east momentum	4	kg m/s
Total north momentum	4	kg m/s
Common east velocity	1.33333	m/s
Common north velocity	1.33333	m/s
Common speed	1.88562	m/s
Direction counterclockwise from east	45	degrees
Initial kinetic energy	19	J
Final kinetic energy	5.33333	J
Internal energy increase	13.6667	J

Add vector momenta before taking a magnitude; sticking fixes one shared velocity.

The combined mass is 3 kg, so divide each total momentum component by 3. The point-mass approximation excludes retained rotation; extended offset impacts would need an angular-momentum account. Zero final translational kinetic energy does not mean zero initial energy; internal conversion closes the stipulated isolated sticking budget.

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

Add vector momenta before taking a magnitude; sticking fixes one shared velocity.

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 first point has mass 2 kg and velocity(3,0) m/s. The second has mass 1 kg and velocity(u,w). They meet and stick at a point, with no retained rotation in this approximation. Total momentum is(6+u,w) kg m/s, and the common velocity is each component divided by the combined3 kg.

Check an original packet

For u=−2,w=4, total momentum is(4,4), common velocity(4/3,4/3) and common speed 4√2/3≈1.88562 m/s. Initial kinetic energy is 19 J and final kinetic energy 16/3 J, so internal energy increases by 41/3 J. The equal-scale momentum chain depicts components, not separate stages of physical travel.

Test the domain and evidence

At u=−6,w=0, total momentum cancels exactly. The stuck point is at rest and has no velocity direction; assigning an eastward0° direction would invent information. Initial kinetic energy is still27 J, all converted in the stipulated point-sticking account. Offset collisions of extended objects could retain rotation and need angular momentum; they are not represented by this model.

Worked original packet

At u=0,w=4, common velocity is(2,4/3) m/s. Its speed is√52/3≈2.40370 m/s and direction atan(2/3)≈33.6901° north of east, not the angle obtained by adding speed magnitudes.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

A 2 kg point at(3,0) m/s sticks to a 1 kg point at(0,3) m/s. Find the common northward velocity component in m/s.

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.

Apply

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

What changes in the physical model if the sticking objects have finite separation and can rotate?

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: Supply a collision response instead of inventing one from momentum Next: Track an internal energy release in two frames

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