Conserve momentum only after checking the boundary
You will be able to: Separate internal interaction impulses from the net external impulse.
When is a system’s total momentum constant?
Two carts push on each other. Cart A can gain +3 N·s while cart B gains −3 N·s. Each cart’s momentum changes, but the pair’s total does not—unless something outside the pair also delivers an impulse.
A useful starting point: Track momentum when matter enters a system →
Words and symbols before equations
- Internal force
- An interaction between two objects inside the selected system.
- External force
- A force from an object outside that boundary.
- Isolated along an axis
- Net external impulse along that axis is zero or negligible over the interval.
- Momentum conservation
- Total momentum has the same before and after value when net external impulse is zero.
What this picture assumes
Initial momenta are p_A = +1 and p_B = −1 kg·m/s. The interaction gives A an impulse J and B −J. Any additional external impulse in this illustration acts on A only. No collision energy model is assumed.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Internal impulses are 3 and -3 N·s. External impulse 0 N·s acts on A. Final p_A = 4, p_B = -4; total P = 0 kg·m/s, equal to the total external impulse from the initial zero total.
- 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 a fixed collection of objects, add their momentum-change equations. Internal forces come in equal and opposite Newton’s-third-law pairs. Integrating each pair over the same time interval gives equal and opposite internal impulses, which cancel from the system total.
The remaining equation is P_f − P_i = J_ext. If J_ext = 0, total P is unchanged. External forces need not each be zero: weight and support can balance vertically. If only the horizontal external impulse is negligible, conserve only the horizontal component.
During a short collision, a small external force may have little time to accumulate impulse compared with the contact impulse. Estimate F_extΔt against the momentum scale. A longer observation interval may include significant friction or gravity, so “momentum is conserved in a collision” is a model decision, not a universal rule for every boundary.
| Question | Cart A alone | Carts A + B |
|---|---|---|
| Interaction from B on A | External | Internal |
| Momentum change | Includes B’s impulse | Only net external impulse changes total |
| Conservation test | Check all external impulses | Internal impulses cancel |
A worked example, step by step
During a 0.02 s collision, cart A receives +4 N·s from cart B. The two-cart system also feels a net external force of +2 N throughout. Find B’s interaction impulse and the pair’s total momentum change.
- Newton’s third law gives B an interaction impulse of −4 N·s.
- Internal impulses sum to +4 − 4 = 0 N·s.
- External impulse is (2 N)(0.02 s) = +0.04 N·s.
- The pair’s total momentum changes by +0.04 kg·m/s. This is 1% of the 4 kg·m/s interaction scale, so neglecting it may be a useful approximation if that accuracy is acceptable.
Internal forces cancel from the total, not from the force balance on each individual object.
Can total momentum have the same endpoints even if net external force was not always zero?
Compare with an explanation
Yes. Opposite external impulses can cancel over an interval. Zero net external force at every instant is a sufficient condition for constant momentum throughout.
Predict. Change one thing. Explain.
Change the internal impulse while holding external impulse zero. Watch the object momenta change in opposite ways. Then add an external impulse and identify the exact change in total momentum.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Internal impulses are 3 and -3 N·s. External impulse 0 N·s acts on A. Final p_A = 4, p_B = -4; total P = 0 kg·m/s, equal to the total external impulse from the initial zero total.
Initial momenta are p_A = +1 and p_B = −1 kg·m/s. The interaction gives A an impulse J and B −J. Any additional external impulse in this illustration acts on A only. No collision energy model is assumed.
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 two-cart system starts with P_i = +6 kg·m/s. During interaction, internal impulses are +2 and −2 N·s, while a −3 N external force acts for 0.10 s. Find the final total momentum and justify whether exact conservation applies.
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Compare with the answer and four-point rubric
- 1 point: Internal impulses cancel in the system total.
- 1 point: External impulse is −3×0.10 = −0.30 N·s.
- 1 point: P_f = 6 − 0.30 = +5.70 kg·m/s.
- 1 point: Exact conservation does not apply to this boundary and interval because net external impulse is nonzero.
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 1What changes total momentum?
Net external impulse for the fixed collection being analyzed.
RECALL 2Why do internal impulses cancel?
They come from equal and opposite forces integrated over the same interval.
RECALL 3Why specify the interval?
A small external force may be negligible during a brief collision but significant over a longer time.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Conserve momentum only after checking the boundary
- P_f − P_i = J_ext for a fixed system.
- J_internal on A = −J_internal on B.
- Conserve a component only when external impulse in that component is negligible.
Remember: Internal forces cancel from the total, not from the force balance on each individual object.
Conditions: Initial momenta are p_A = +1 and p_B = −1 kg·m/s. The interaction gives A an impulse J and B −J. Any additional external impulse in this illustration acts on A only. No collision energy model is assumed.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official Unit 4) · Objectives 4.3.A; 4.3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.3, objectives 4.3.A; 4.3.B. 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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