Solve a collision with a momentum ledger
You will be able to: Use a before-and-after momentum equation for an approximately isolated pair.
How can we find an unknown velocity without knowing the collision force?
A 2 kg cart approaches a stationary 3 kg cart at +3 m/s. After they interact, the first cart moves at +0.6 m/s. We can find the second cart’s velocity from the momentum transferred between them without reconstructing every instant of contact.
A useful starting point: Choose a system before conserving momentum →
Words and symbols before equations
- Initial/final subscripts i, f
- Values immediately before and after the interaction.
- Momentum ledger
- A signed before-and-after account for every object.
- Isolated approximation
- Net external impulse during the chosen interval is negligible.
What this picture assumes
A: 2 kg initially +3 m/s. B: 3 kg initially at rest. Net external impulse negligible. Selected outcomes conserve momentum; the permitted range has K_f≤K_i and separating or equal final velocities.
Connect the picture to the physics
Use one sign convention and frame for all four velocities. With negligible external impulse, m_Av_Ai+m_Bv_Bi=m_Av_Af+m_Bv_Bf. The force details disappear because the impulses internal to the pair cancel.
Write symbols before numbers and keep the mass labels attached to their own velocities. Here 2(3)+3(0)=2(0.6)+3v_Bf. Solving gives v_Bf=1.6 m/s to the right.
Conservation of momentum alone does not determine both final velocities if neither is known. You need an additional physically justified condition, such as sticking together, known elastic behavior, or a measured final velocity. Also check energy: an arbitrary proposed outcome may require stored energy to be released.
A worked example, step by step
A 1 kg cart at +4 m/s collides with a stationary 3 kg cart. The first cart rebounds at −2 m/s. Neglect external impulse. Find the second cart’s final velocity.
- System: both carts. Right positive, in the track frame.
- Initial P=1(4)+3(0)=+4 kg·m/s.
- Final P=1(−2)+3v_Bf. Set 4=−2+3v_Bf; therefore v_Bf=+2 m/s.
- Check: −2+6=4 kg·m/s. Initial K=8 J; final K=2+6=8 J, consistent with an elastic outcome.
Conservation of momentum supplies a constraint, not permission to assume equal final speeds in every collision.
If both final velocities are unknown, can one momentum equation determine both?
Compare with an explanation
No. An additional condition or measurement is needed.
Predict. Change one thing. Explain.
For a 2 kg cart initially at +3 m/s and a stationary 3 kg cart, vary A’s final velocity between −0.6 and +1.2 m/s. Predict B’s final velocity from the momentum ledger. Check whether kinetic energy stays equal or decreases.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
v_Af=0.6 m/s; required v_Bf=1.6 m/s. P_i=P_f=6 kg·m/s. K_i=9 J and K_f=4.2 J.
A: 2 kg initially +3 m/s. B: 3 kg initially at rest. Net external impulse negligible. Selected outcomes conserve momentum; the permitted range has K_f≤K_i and separating or equal final velocities.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a momentum or impulse relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 2 kg cart at +3 m/s collides with a stationary 1 kg cart. Afterward the first travels at +1 m/s. (a) Specify the system and approximation. (b) Find initial P. (c) Find the second velocity. (d) Verify the total.
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Compare with the answer and four-point rubric
- 1 point: Both carts; negligible net external impulse during collision.
- 1 point: P_i=2(3)=+6 kg·m/s.
- 1 point: 6=2(1)+1v_Bf, giving v_Bf=+4 m/s.
- 1 point: P_f=2+4=6 kg·m/s.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why can collision forces be omitted from the pair’s ledger?
Their internal impulses cancel.
RECALL 2Which velocities belong in the equation?
Those just before and after, all in one frame.
RECALL 3Do colliding objects always share final velocity?
Only when the interaction makes them move together.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Solve a collision with a momentum ledger
- m_Av_Ai+m_Bv_Bi=m_Av_Af+m_Bv_Bf when J_ext≈0.
- If two final velocities are unknown, another condition is needed.
Remember: Conservation of momentum supplies a constraint, not permission to assume equal final speeds in every collision.
Conditions: A: 2 kg initially +3 m/s. B: 3 kg initially at rest. Net external impulse negligible. Selected outcomes conserve momentum; the permitted range has K_f≤K_i and separating or equal final velocities.
Refresh Kid · Unit 4 · Objectives 4.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.3, objectives 4.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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