Impulse is the signed area under force versus time
You will be able to: Evaluate impulse from a force–time graph and compare average with peak force.
How can a short force pulse change momentum?
A cart feels a 4 N push for 0.5 s. The impulse is 2 N·s, equal to a 2 kg·m/s momentum change. A triangular pulse with the same duration needs an 8 N peak to deliver that same impulse.
A useful starting point: Zero total momentum does not mean no motion →
Words and symbols before equations
- Impulse J
- Time integral of force, measured in N·s; it is a vector.
- Net impulse
- Integral of the net external force; equals the system’s momentum change.
- Average force F_avg
- A constant force that would give the same impulse over the same interval.
- Peak force
- Largest force magnitude during the interval; generally different from the average.
What this picture assumes
Net horizontal force on a 2 kg cart initially at +1 m/s. Triangle has its peak halfway through the full duration and returns to zero. Rectangle is constant during its interval; endpoint jumps are idealizations. Other horizontal forces are absent.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At t = 0.2 s: impulse 1.2 N·s, momentum 3.2 kg·m/s, velocity 1.6 m/s. Complete pulse impulse 1.2 N·s; average force over the complete pulse 6 N.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
During a short time dt, the momentum increment is dp = F_net dt. Adding gives J_net = ∫F_net(t)dt = p_f − p_i. For a fixed direction, this is signed area under F versus t. If force reverses, opposite areas subtract.
For constant force, J = FΔt. For a triangular pulse starting and ending at zero, J = ½F_peakΔt. Average force is J/Δt, not necessarily the value at the middle or the peak for an arbitrary waveform.
Impulse from one force equals that force’s contribution to momentum change; only net impulse equals the complete Δp. If force direction varies in a plane, integrate the x and y components separately rather than assuming impulse points along the force at every instant.
A worked example, step by step
A net horizontal force increases linearly from 0 to 12 N over 0.10 s and decreases linearly to 0 over another 0.10 s. A 2 kg cart starts at +1 m/s. Find impulse, average force and final velocity.
- The force–time graph is one triangle of total duration 0.20 s and height 12 N.
- Impulse = ½(0.20)(12) = +1.2 N·s.
- F_avg = 1.2/0.20 = 6 N. Initial momentum is 2 kg·m/s, so final momentum is 3.2 kg·m/s.
- v_f = 3.2/2 = +1.6 m/s. Peak force is 12 N; the average is 6 N.
A force–time area gives impulse, not work. Work uses force and displacement.
Can the net impulse be zero while forces acted?
Compare with an explanation
Yes. Positive and negative time areas can cancel even though momentum changed during the interval.
Predict. Change one thing. Explain.
Keep duration fixed and compare rectangular and triangular pulses with the same peak. Then double the triangular peak and check that the impulses match.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At t = 0.2 s: impulse 1.2 N·s, momentum 3.2 kg·m/s, velocity 1.6 m/s. Complete pulse impulse 1.2 N·s; average force over the complete pulse 6 N.
Net horizontal force on a 2 kg cart initially at +1 m/s. Triangle has its peak halfway through the full duration and returns to zero. Rectangle is constant during its interval; endpoint jumps are idealizations. Other horizontal forces are absent.
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 rectangular net force of −6 N acts for 0.25 s on a 0.5 kg cart initially moving at +4 m/s. Find impulse, initial and final momentum, and final velocity.
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Compare with the answer and four-point rubric
- 1 point: J = (−6)(0.25) = −1.5 N·s.
- 1 point: Initial momentum = (0.5)(4) = +2 kg·m/s.
- 1 point: Final momentum = 2 − 1.5 = +0.5 kg·m/s.
- 1 point: Final velocity = 0.5/0.5 = +1 m/s; it slows but does not reverse.
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 impulse an area?
It sums force times each small time interval.
RECALL 2Does equal peak force guarantee equal impulse?
No. Duration and the entire waveform matter.
RECALL 3When can impulse direction be read from force direction?
When the force maintains a fixed direction over the interval.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Impulse is the signed area under force versus time
- J_net = ∫F_net dt = Δp.
- F_avg = J/Δt.
- Triangular fixed-direction pulse: J = ½F_peakΔt.
Remember: A force–time area gives impulse, not work. Work uses force and displacement.
Conditions: Net horizontal force on a 2 kg cart initially at +1 m/s. Triangle has its peak halfway through the full duration and returns to zero. Rectangle is constant during its interval; endpoint jumps are idealizations. Other horizontal forces are absent.
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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