Experiment 1
Keep masses fixed and set B velocity to −2 m/s. Predict final velocity and energy loss.
Check the prediction
Equal opposing momenta cancel. Final velocity is zero and 4 J of initial kinetic energy becomes internal energy.
Conservation and symmetry
Account for what crosses the boundary.
Solve a sticking collision and distinguish conserved momentum, kinetic energy loss and total energy conservation.
Start with: Algebra and signed velocity. Momentum is mass times velocity; kinetic energy is one half mass times speed squared.
01 · Commit to a prediction
What do you expect?
At 1 m/s. The same total momentum now belongs to twice the mass. Kinetic energy falls from 2 J to 1 J; the missing joule becomes internal energy.
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02 · Change an assumption
Predict. Change. Explain.
Before changing a control, say what should move and why. Start with the experiments below. Reset restores the starting model; it preserves your written responses.
Experiment 2
Reset. Double only the stationary cart mass.
Check the prediction
Final velocity becomes 2/3 m/s. The larger total mass carries the same initial 2 kg·m/s momentum.
Cart A starts behind cart B. Positive velocity points toward B.
0.5 to 5 · step 0.5
0.5 to 5 · step 0.5
-5 to 5 · step 0.5
-5 to 5 · step 0.5
Calculated model output. The table gives the same values. Displayed values are rounded; calculations keep full precision.
Read the plotted data as a table
03 · Connect the mechanism
From the picture to the quantities.
Choose both carts as the system. During a sticking collision the forces between them exchange momentum internally. If external impulse is negligible, total momentum stays fixed. Kinetic energy need not: deformation and heating store energy in other forms. Conservation constrains the result before you know the complicated collision forces.
P = m₁u₁ + m₂u₂ = (m₁+m₂)v
v = P / (m₁+m₂)
K = ½mu²
ΔKlost = ½[m₁m₂/(m₁+m₂)](u₁−u₂)²
v = P / (m₁+m₂)
K = ½mu²
ΔKlost = ½[m₁m₂/(m₁+m₂)](u₁−u₂)²
m₁,m₂ are positive masses, 0.5 to 5 kg here. u₁,u₂ are signed initial velocities, −5 to 5 m/s. v is the common final velocity. P is momentum in kg·m/s. K is kinetic energy in joules (J). Lost means transferred out of translational kinetic energy, not destroyed.
Worked example
Initially P = 1 × 2 + 1 × 0 = 2 kg·m/s. After sticking, total mass is 2 kg, so v = 1 m/s. Initial K = ½ × 1 × 2² = 2 J. Final K = ½ × 2 × 1² = 1 J. Their difference is 1 J. Independently, the reduced-mass formula gives ½ × (1/2) × (2−0)² = 1 J.
Assumptions and limits
One-dimensional, nonrelativistic carts stick completely, with negligible external impulse. If a wall exerts an impulse, include the wall or add its impulse to the momentum balance. Continuous spatial translation symmetry is associated with momentum conservation and time translation symmetry with energy conservation in an appropriate closed-system formulation. This lab illustrates a balance law; it does not derive Noether’s theorem or claim that kinetic energy alone is the conserved energy.
The annotated sources distinguish established results from this lesson’s original examples.
04 · Follow the structure
Where else does this apply?
Docking
Objects that latch together redistribute momentum across their combined mass.
Boundary: Thrusters, external fields and rotation require a larger model.
Inventory balances
Initial stock plus inflow minus outflow constrains final stock.
Boundary: This is bookkeeping, not a consequence of mechanical translation symmetry; items can also be produced or destroyed.
05 · Retrieve without hints
Close the explanation. Try a new case.
Write an answer before opening its feedback. Later, return directly here without rereading above. Recognition, explanation and transfer are separate outcomes. No page action or answer reveal measures mastery.
Recognition and explanation
Reveal reasoning and rubric
Not if their initial momenta were equal and opposite. Momentum is signed and can sum to zero while kinetic energies are positive.
Self-check: Use vector or signed addition and distinguish energy.
Calculation
Reveal reasoning and rubric
P = 6 kg·m/s, v = 6/3 = 2 m/s. Initial K = 9 J, final K = 6 J, and 3 J becomes internal energy.
Self-check: Show both the momentum balance and the energy destination.
Novel transfer
Reveal reasoning and rubric
No. The wall delivers external impulse to that system. Expand the system to include Earth or account for that impulse. Energy may also flow into deformation and sound.
Self-check: Choose the boundary before applying the conservation claim.
On a later day, try again and record actual evidence in the curriculum. A later unaided explanation and a fresh transfer problem give stronger evidence than immediate familiarity. No reminder is scheduled.
Sources & scope
What supports the lesson?
- Feynman Lectures I, chapter 10. Momentum conservation and its system boundary.
- Feynman Lectures III, chapter 17. Connections between spatial/time symmetries and conserved quantities.
Original teaching examples. Reference links need a connection; the lesson itself does not. Built 2026-10-11. Learner understanding is not assessed.