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SCIENCE · INVESTIGATE AND EXPLAIN

Choose a conservation statement separately for each stage

A sequence of events does not inherit one conservation shortcut for every stage. In this virtual packet, an incoming point of mass m and speed u sticks to a stationary2 kg target during a short horizontal collision. The combined point then rises on an ideal frictionless guide. Rotation and external horizontal impulse during the collision are neglected.

Learning outcome: Compare incoming masses 0.5,1,2 kg at one speed and retain collision momentum, collision energy conversion and later rise separately. Identify which external interactions are neglected in each stage and which are not.

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

For m=2 kg,u=4 m/s, common speed is 2 m/s, collision kinetic energy falls from 16 to 8 J, and subsequent maximum rise is 0.2 m. The two energy portions sum to the original16 J.

Open the experiment

Predict

Using initial incoming kinetic energy directly for the final rise would:

Experiment

Compare incoming masses 0.5,1,2 kg at one speed and retain collision momentum, collision energy conversion and later rise separately. Identify which external interactions are neglected in each stage and which are not.

Use different conservation statements in consecutive stages

Worked teaching model · synthetic values · Incoming point sticks to a stationary 2 kg point in a short horizontal collision with negligible external horizontal impulse. The combined point then rises on an ideal frictionless guide under g=10. No rotation; collision kinetic loss is retained separately. Virtual only, not a ballistic device.

Reset keeps your prediction, captures and practice history.

Later rise versus incoming speed

maximum vertical rise (m)incoming speed (m/s)060.20

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

Equivalent numerical coordinates for Later rise versus incoming speed

two-stage prediction

incoming speed (m/s)maximum vertical rise (m)
20.0222222
2.10.0245
2.20.0268889
2.30.0293889
2.40.032
2.50.0347222
2.60.0375556
2.70.0405
2.80.0435556
2.90.0467222
30.05
3.10.0533889
3.20.0568889
3.30.0605
3.40.0642222
3.50.0680556
3.60.072
3.70.0760556
3.80.0802222
3.90.0845
40.0888889
4.10.0933889
4.20.098
4.30.102722
4.40.107556
4.50.1125
4.60.117556
4.70.122722
4.80.128
4.90.133389
50.138889
5.10.1445
5.20.150222
5.30.156056
5.40.162
5.50.168056
5.60.174222
5.70.1805
5.80.186889
5.90.193389
60.2

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
QuantityValueUnit
Common velocity immediately after collision1.33333m/s
Initial incoming kinetic energy8J
Kinetic energy immediately after collision2.66667J
Collision internal energy increase5.33333J
Later maximum rise0.0888889m
Gravitational gain during rise2.66667J
Initial horizontal momentum4kg m/s
Immediate post-collision horizontal momentum4kg m/s

Momentum closes the short collision; mechanical energy closes the subsequent ideal rise.

During sticking, horizontal impulse is neglected and mv=(m+2)V, but kinetic energy decreases. During the later frictionless climb, V²/2=gh while gravity and guide forces change point momentum. Applying initial kinetic energy directly to final gravitational gain skips the collision loss; applying collision momentum conservation through the climb ignores external impulse.

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

Momentum closes the short collision; mechanical energy closes the subsequent ideal rise.

During sticking, horizontal impulse is neglected and mv=(m+2)V, but kinetic energy decreases. During the later frictionless climb, V²/2=gh while gravity and guide forces change point momentum. Applying initial kinetic energy directly to final gravitational gain skips the collision loss; applying collision momentum conservation through the climb ignores external impulse.

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

Understand

Define the energy or momentum system

A sequence of events does not inherit one conservation shortcut for every stage. In this virtual packet, an incoming point of mass m and speed u sticks to a stationary2 kg target during a short horizontal collision. The combined point then rises on an ideal frictionless guide. Rotation and external horizontal impulse during the collision are neglected.

Build the algebraic account

During the short sticking stage, horizontal momentum gives mu=(m+2)V. Kinetic energy is not conserved: compare initial mu²/2 with immediate final(m+2)V²/2 and retain the difference as internal conversion. The common speed V, not the incoming speed u, is the initial condition for the next stage.

Check an original packet

During the later ideal rise, mechanical energy is conserved and V²/2=gh, so h=V²/(2g). Gravity and guide forces transfer momentum during this longer stage. Extending the collision’s momentum conservation through the rise would ignore those external impulses. Conversely, carrying the incoming kinetic energy directly into gravitational gain would ignore sticking loss.

Test the domain and evidence

For m=1 kg,u=4 m/s and g=10, common speed is 4/3 m/s. Initial kinetic energy is 8 J, immediate final energy 8/3 J and collision conversion16/3 J. The rise is 4/45≈0.088889 m, whose gravitational gain is 8/3 J. This mathematical two-stage model does not describe or request any ballistic apparatus, projectile trap or physical collision.

Worked original packet

For m=2 kg,u=4 m/s, common speed is 2 m/s, collision kinetic energy falls from 16 to 8 J, and subsequent maximum rise is 0.2 m. The two energy portions sum to the original16 J.

Bounded investigation

Compare incoming masses 0.5,1,2 kg at one speed and retain collision momentum, collision energy conversion and later rise separately. Identify which external interactions are neglected in each stage and which are not.

Changed-case numerical practice

Try the changed case before feedback

A 1 kg point at 6 m/s sticks to a stationary2 kg point, then rises without loss under g=10. Find maximum rise in metres.

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

Using initial incoming kinetic energy directly for the final rise would:

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

Why does an approximate negligible-impulse claim require a stated interval rather than a permanent system label?

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.

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