Recoil trades momentum between separating parts
You will be able to: Use momentum conservation during separation and identify the source of kinetic energy.
How can two objects start moving when their total momentum is zero?
Two carts initially at rest are held together by a compressed spring. When released, a 1 kg cart moves right at 3 m/s while a 3 kg cart moves left at 1 m/s. Their momenta cancel, but spring energy has become 6 J of kinetic energy.
A useful starting point: The center of mass follows the total momentum →
Words and symbols before equations
- Recoil
- Motion of one part opposite the momentum gained by another part.
- Separation or explosion model
- Internal interactions push parts apart; total momentum depends on external impulse.
- Stored internal energy
- Energy available inside the system, such as a compressed spring’s potential energy.
- Common initial velocity V₀
- Velocity of the joined system before separation, in the chosen frame.
What this picture assumes
The numerical model is a one-dimensional separation with zero external impulse. Stored internal energy supplies the K increase. The optional 3D illustration is a separate qualitative six-fragment example: each momentum arrow has an equal opposite partner. It is not a quantitative 3D collision calculation.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- First velocity 3 m/s; second -1 m/s. Total P = 0 kg·m/s before and after. K increases by 6 J, supplied by stored internal energy. Camera affects only the separate qualitative view.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For two parts, (m₁ + m₂)V₀ = m₁v₁ + m₂v₂ when net external impulse is negligible. If the initial system is at rest, m₁v₁ = −m₂v₂. Equal and opposite momenta do not require equal and opposite speeds unless masses match.
Momentum conservation does not forbid a kinetic-energy increase. The internal spring or other stored energy can supply it. Include that energy source in the total-energy account; do not classify a powered separation as an ordinary inelastic collision.
In two or three dimensions, vector momentum still balances. The optional qualitative spatial picture shows opposite momentum pairs whose contributions cancel despite the fragments spreading out. Its arrow lengths are illustrative, and it is not a quantitative three-dimensional collision exercise.
A worked example, step by step
A 6 kg assembly initially moves at +2 m/s and separates into a 2 kg part moving at +5 m/s and a 4 kg part. Find the second velocity and the change in total kinetic energy.
- Initial momentum is (6)(2) = +12 kg·m/s.
- Set 12 = 2(5) + 4v₂, giving v₂ = +0.5 m/s. Recoil relative to the center of mass does not require negative lab velocity.
- Initial K = ½(6)(2²) = 12 J. Final K = ½(2)(5²) + ½(4)(0.5²) = 25.5 J.
- The kinetic increase is 13.5 J, supplied by a decrease in stored internal energy if the system has no external energy input.
Opposite relative recoil does not always mean opposite lab velocities. Keep the initial total momentum instead of assuming it is zero.
If the heavier part moves more slowly, does it receive a smaller impulse?
Compare with an explanation
Not from the mutual interaction. The internal impulses are equal and opposite, but equal momentum changes create smaller velocity changes in a larger mass.
Predict. Change one thing. Explain.
Set initial velocity to zero and increase the second mass. Predict its recoil speed while keeping the first final velocity fixed. Then give the whole assembly an initial velocity and examine the lab-frame signs.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
First velocity 3 m/s; second -1 m/s. Total P = 0 kg·m/s before and after. K increases by 6 J, supplied by stored internal energy. Camera affects only the separate qualitative view.
Optional qualitative 3D view: opposite fragment momenta
This separate qualitative illustration shows cancellation in space. Use the viewing-direction control to rotate the projection. The one-dimensional numerical example above is unchanged.
The numerical model is a one-dimensional separation with zero external impulse. Stored internal energy supplies the K increase. The optional 3D illustration is a separate qualitative six-fragment example: each momentum arrow has an equal opposite partner. It is not a quantitative 3D collision calculation.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant impulse, system boundary, momentum or calculus 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 compressed spring releases two carts from rest: m₁ = 2 kg and m₂ = 4 kg. The first moves at −3 m/s. Find the second velocity, final K and required spring-energy decrease if losses are negligible.
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Compare with the answer and four-point rubric
- 1 point: Initial total momentum is zero.
- 1 point: 0 = 2(−3) + 4v₂ gives v₂ = +1.5 m/s.
- 1 point: Final K = ½(2)(9) + ½(4)(2.25) = 13.5 J.
- 1 point: Spring potential energy must decrease by 13.5 J; momentum conservation alone does not provide this energy.
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 is the heavier part’s recoil speed smaller when starting from rest?
Equal momentum magnitudes divided by a larger mass give a smaller speed.
RECALL 2Where can separation kinetic energy come from?
Stored internal energy, such as an initially compressed spring.
RECALL 3What remains unchanged without external impulse?
The total momentum and center-of-mass velocity of the fixed collection.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Recoil trades momentum between separating parts
- (m₁ + m₂)V₀ = m₁v₁ + m₂v₂ when J_ext = 0.
- For V₀ = 0: v₂ = −(m₁/m₂)v₁.
- An increase in K during separation requires another energy source.
Remember: Opposite relative recoil does not always mean opposite lab velocities. Keep the initial total momentum instead of assuming it is zero.
Conditions: The numerical model is a one-dimensional separation with zero external impulse. Stored internal energy supplies the K increase. The optional 3D illustration is a separate qualitative six-fragment example: each momentum arrow has an equal opposite partner. It is not a quantitative 3D collision calculation.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official 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 alongside the Fall 2026 clarifications. This is Mechanics Unit 4: Linear Momentum. The unit covers Topics 4.1–4.4. Calculus connects force to momentum derivatives and impulse integrals. Collision calculations use one or two dimensions; the optional spatial fragment diagram is qualitative. Changing-mass examples explicitly account for momentum carried across a boundary, so dp/dt is not used blindly for an open system. 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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