Differentiate momentum; integrate force
You will be able to: Use F_net = dp/dt and p(t) = p₀ + ∫F_net dt with correct initial conditions.
How do a momentum function and force history describe the same motion?
A 2 kg cart begins at 1 m/s, so p₀ = 2 kg·m/s. If its net force grows as F(t) = (4 N/s)t, then at 2 s the impulse is 8 N·s and momentum is 10 kg·m/s. Its speed is 5 m/s.
A useful starting point: A rebound changes momentum more than a stop →
Words and symbols before equations
- Derivative dp/dt
- The slope of a momentum–time graph; units kg·m/s² = N.
- Definite integral
- Accumulated signed area between specified time limits.
- Initial condition p₀
- The momentum at the chosen starting time; integration alone does not supply it.
- Force coefficient c
- For F = ct, c has units N/s so the product has units N.
What this picture assumes
A fixed-mass 2 kg particle with prescribed net force F = A + bt, starting at t = 0 with momentum p₀. A signed force can slow or reverse motion; this is not a friction-only model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At 4 s: F_net = -4 N, impulse = 0 N·s, p = 3 kg·m/s and v = 1.5 m/s. The initial momentum shifts p without changing its slope.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Newton’s second law for a fixed collection of matter is F_ext = dP/dt. For one constant-mass object it reduces to F_net = m dv/dt = ma. On a p-versus-t graph, the tangent slope is force; graph height is momentum.
Integrate to recover momentum: p(t) = p₀ + ∫₀ᵗF_net(τ)dτ. The symbol τ is a temporary integration variable, letting t denote the final time. For F = ct, p = p₀ + ½ct², and v = p/m if mass is constant.
An interval of negative force gives a decreasing p, but not necessarily negative velocity. The object reverses only when p crosses zero. If a force changes direction over time, the accumulated impulse need not point in its final direction.
A worked example, step by step
A 2 kg cart starts with p₀ = +3 kg·m/s. Its net force is F(t) = 4 N − (2 N/s)t for 0 ≤ t ≤ 4 s. Find p(t), p(4) and when momentum is largest.
- Integrate from zero: p(t) = 3 kg·m/s + (4 N)t − (1 N/s)t².
- At t = 4 s, the signed impulse is 16 − 16 = 0 N·s, so p(4) = 3 kg·m/s.
- The derivative is F = 4 − 2t in SI units. It changes from positive to negative at t = 2 s.
- Momentum is largest at 2 s: p = 3 + 8 − 4 = 7 kg·m/s. Final speed is p(4)/m = 1.5 m/s, equal to its initial speed despite intermediate changes.
Zero final impulse does not mean zero force throughout the interval. Keep the integration constant p₀.
A momentum graph is flat at a positive value. What is the net force?
Compare with an explanation
Zero. The object has positive momentum but its rate of change is zero.
Predict. Change one thing. Explain.
Set force to 4 − 2t in SI units and p₀ to 3 kg·m/s. Move through t = 2 s and 4 s. Identify the momentum maximum and the interval where the negative force removes the earlier impulse.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At 4 s: F_net = -4 N, impulse = 0 N·s, p = 3 kg·m/s and v = 1.5 m/s. The initial momentum shifts p without changing its slope.
A fixed-mass 2 kg particle with prescribed net force F = A + bt, starting at t = 0 with momentum p₀. A signed force can slow or reverse motion; this is not a friction-only 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 constant-mass 2 kg cart has p(t) = 2 kg·m/s + (3 N/s)t² for 0 ≤ t ≤ 2 s. Derive its net force, find impulse over the interval, and calculate final velocity.
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Compare with the answer and four-point rubric
- 1 point: Differentiate: F(t) = (6 N/s)t.
- 1 point: Impulse = p(2) − p(0) = 12 N·s.
- 1 point: Final momentum is 14 kg·m/s.
- 1 point: Final velocity is p/m = 14/2 = 7 m/s; the initial momentum must not be omitted.
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 does a force–time area give?
The change in momentum over the interval.
RECALL 2What does a momentum–time slope give?
The net external force on a fixed collection of matter.
RECALL 3When does decreasing p mean motion has reversed?
Only after the signed momentum crosses zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Differentiate momentum; integrate force
- F_ext = dP/dt for a fixed collection of matter.
- p(t) = p₀ + ∫₀ᵗ F_net(τ)dτ.
- For constant mass, v(t) = p(t)/m.
Remember: Zero final impulse does not mean zero force throughout the interval. Keep the integration constant p₀.
Conditions: A fixed-mass 2 kg particle with prescribed net force F = A + bt, starting at t = 0 with momentum p₀. A signed force can slow or reverse motion; this is not a friction-only model.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official Unit 4) · Objectives 4.2.A; 4.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.2, objectives 4.2.A; 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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