Track momentum when matter enters a system
You will be able to: Apply the product rule and account for momentum carried by incoming matter.
Can momentum increase even while velocity stays constant?
Sand falls vertically into a cart moving horizontally at 2 m/s. To keep that speed, each new kilogram of sand must acquire 2 kg·m/s of horizontal momentum. With 0.5 kg arriving each second and zero incoming horizontal speed, the required forward force is 1 N.
A useful starting point: Differentiate momentum; integrate force →
Words and symbols before equations
- Mass rate dm/dt
- Mass added per second, in kg/s; positive for the accretion example.
- Product rule
- d(mv)/dt = m dv/dt + v dm/dt in one dimension.
- Open system
- A boundary through which matter can pass; that matter may carry momentum.
- Incoming velocity u
- Velocity of the arriving matter in the same inertial frame as the cart.
What this picture assumes
A cart initially contains 2 kg, collects mass at constant rate α, and is externally maintained at velocity v. Incoming matter has ground-frame horizontal velocity u. The balance explicitly includes incoming momentum; vertical forces and energy loss are outside this horizontal model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Mass 3 kg; stored p 6 kg·m/s. Stored rate 1 N = incoming rate 0 N + external force 1 N. Cart velocity stays 2 m/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
A changing product mv can increase because v changes, m changes, or both. At constant v, d(mv)/dt = v dm/dt. The derivative describes the rate of momentum stored in the chosen contents, but identifying it with external force requires checking the boundary.
For a cart collecting mass at rate α with incoming horizontal velocity u, momentum balance is F_ext = d(mv)/dt − αu = m dv/dt + α(v − u). Derive it by including the small incoming parcel in a fixed collection for a short time: the parcel already has momentum u dm before joining.
At constant cart speed, F_ext = α(v − u). If incoming sand has u = 0, this becomes v dm/dt, the constant-velocity special case. If u = v, no horizontal force is needed to speed up the incoming material. This controlled accretion example is not a general rocket equation.
A worked example, step by step
A cart is maintained at +3 m/s while collecting material at 2 kg/s. The incoming material has horizontal velocity +1 m/s. Find the external horizontal force and compare it with the rate of change of stored cart momentum.
- Use the same ground frame and right-positive axis for both velocities.
- Stored momentum grows at vα = 3×2 = 6 kg·m/s².
- Incoming material brings uα = 1×2 = 2 kg·m/s² across the boundary.
- F_ext = 6 − 2 = +4 N. Equating 6 N directly to the external force would ignore incoming momentum.
F_ext = dP/dt directly describes a fixed collection of matter. An open boundary also needs momentum-flow terms; do not use v dm/dt blindly.
If material arrives already moving with the cart, why can stored momentum still grow without a force?
Compare with an explanation
The incoming matter brings exactly that momentum across the boundary. A fixed collection including the incoming matter has no unexplained momentum gain.
Predict. Change one thing. Explain.
Hold cart speed and mass rate fixed. Change incoming horizontal velocity from 0 to the cart speed, then above it. Explain when the maintaining force is forward, zero or backward.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Mass 3 kg; stored p 6 kg·m/s. Stored rate 1 N = incoming rate 0 N + external force 1 N. Cart velocity stays 2 m/s.
A cart initially contains 2 kg, collects mass at constant rate α, and is externally maintained at velocity v. Incoming matter has ground-frame horizontal velocity u. The balance explicitly includes incoming momentum; vertical forces and energy loss are outside this horizontal model.
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 cart stays at +5 m/s while mass arrives at 1 kg/s. Compare incoming speeds 0, +5 and +7 m/s. Compute the required external force for each and explain the last sign.
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Compare with the answer and four-point rubric
- 1 point: For u = 0, F_ext = 1(5 − 0) = +5 N.
- 1 point: For u = 5 m/s, F_ext = 0 N.
- 1 point: For u = 7 m/s, F_ext = 1(5 − 7) = −2 N.
- 1 point: The faster incoming mass supplies excess forward momentum, so a backward force is required to maintain the cart’s speed.
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 1Can momentum change at constant velocity?
Yes, when the mass of the chosen contents changes.
RECALL 2What does an open-system momentum balance add?
Momentum carried in or out by matter crossing the boundary.
RECALL 3When is the special result F = v dm/dt applicable here?
For constant cart speed and zero incoming momentum per unit mass along the analyzed axis.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Track momentum when matter enters a system
- Product rule: d(mv)/dt = m dv/dt + v dm/dt.
- Accretion at rate α: F_ext = m dv/dt + α(v − u).
- Constant v and incoming u = 0: F_ext = v dm/dt.
Remember: F_ext = dP/dt directly describes a fixed collection of matter. An open boundary also needs momentum-flow terms; do not use v dm/dt blindly.
Conditions: A cart initially contains 2 kg, collects mass at constant rate α, and is externally maintained at velocity v. Incoming matter has ground-frame horizontal velocity u. The balance explicitly includes incoming momentum; vertical forces and energy loss are outside this horizontal model.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official Unit 4) · Objectives 4.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.2, objectives 4.2.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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