Read net force from momentum–time graphs
You will be able to: Connect the slope of momentum versus time to net force.
How can a momentum graph tell us the force?
A cart’s momentum rises steadily from +2 to +10 kg·m/s in 4 s. Each second it gains 2 kg·m/s. That steady rate of change corresponds to a +2 N net force.
A useful starting point: Read impulse from a force–time graph →
Words and symbols before equations
- Slope
- Vertical change divided by horizontal change; rise divided by run.
- Momentum–time graph
- Momentum in kg·m/s vertically, time in s horizontally.
- Secant slope
- Slope of a straight line between two points; gives the average net force.
- Tangent slope
- The local steepness of a smooth curve at one instant; gives instantaneous net force.
What this picture assumes
Initial momentum +2 kg·m/s at t=0 s; selected final momentum at t=4 s. Straight-line change means constant net force over this interval.
Connect the picture to the physics
For a straight segment, calculate F_net=Δp/Δt. The units (kg·m/s)/s simplify to newtons. A steeper positive slope means a larger positive net force.
A horizontal momentum graph means zero net force, even if momentum is nonzero. An object can keep moving under balanced forces. A downward slope means negative net force; momentum need not already be negative.
For a curved graph, the whole-interval secant gives average force while local slope gives instantaneous force. At a sharp corner of an idealized piecewise graph, use the slopes immediately before and after; the single-instant derivative is not defined there.
A worked example, step by step
Momentum falls linearly from +6 to −2 kg·m/s in 4 s. Find net force and interpret the moment the graph crosses zero.
- Δp=−2−6=−8 kg·m/s over Δt=4 s.
- F_net=−8/4=−2 N throughout the straight segment.
- p=6−2t becomes zero at t=3 s; for a positive constant mass, velocity is zero at that instant.
- The graph continues downward, so the object reverses after t=3 s. Force is still −2 N at zero momentum.
Graph height is momentum; graph slope is force. Crossing p=0 does not mean the force is zero.
Can momentum be positive while net force is negative?
Compare with an explanation
Yes. A positive graph that slopes downward describes positive momentum decreasing under negative net force.
Predict. Change one thing. Explain.
Initial momentum is +2 kg·m/s and duration is 4 s. Change final momentum to create rising, flat and falling lines. Identify a case with negative force and positive momentum throughout.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Δp=8 kg·m/s over 4 s. Slope = net force = 2 N. The force follows the slope, not the graph’s height.
Initial momentum +2 kg·m/s at t=0 s; selected final momentum at t=4 s. Straight-line change means constant net force over this interval.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a momentum or impulse 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 2 kg cart’s momentum changes linearly from +4 to −4 kg·m/s in 2 s. (a) Find force. (b) Find its initial and final velocities. (c) Find when it is at rest. (d) Explain the force at that instant.
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Compare with the answer and four-point rubric
- 1 point: F_net=(−4−4)/2=−4 N.
- 1 point: v_i=+2 m/s; v_f=−2 m/s.
- 1 point: At t=1 s, p=0.
- 1 point: The line still has slope −4 N, so force is nonzero at the instant of rest.
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 1Slope of p–t?
Net force.
RECALL 2Horizontal p–t means what?
Constant momentum and zero net force.
RECALL 3Curved p–t graph?
Net force varies with the local slope.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Read net force from momentum–time graphs
- F_net,avg=Δp/Δt.
- Instantaneous net force equals local slope of p versus t.
- For constant mass, F_net=mΔv/Δt on a straight segment.
Remember: Graph height is momentum; graph slope is force. Crossing p=0 does not mean the force is zero.
Conditions: Initial momentum +2 kg·m/s at t=0 s; selected final momentum at t=4 s. Straight-line change means constant net force over this interval.
Refresh Kid · 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. Fall-2026 corrections also checked. 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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