Same momentum change, different stopping time
You will be able to: Predict how average force changes when the same momentum change takes longer.
Why does moving your hands backward soften a catch?
When catching a ball, your hands can move with it briefly before bringing it to rest. If the incoming and outgoing velocities stay the same, the momentum change is the same. Spreading that change over more time lowers average net force.
A useful starting point: Impulse and momentum change →
Words and symbols before equations
- Stopping interval
- Time from the start of slowing to rest.
- Average net force
- Total impulse divided by duration; not peak contact force.
- Fixed momentum change
- The same mass and the same initial and final velocities in each comparison.
What this picture assumes
0.5 kg ball stops from +4 m/s horizontally. The rectangular graph is an equivalent constant net force giving the same impulse. It does not predict the real contact-force peak.
Connect the picture to the physics
For a horizontal catch, take the incoming direction positive. A 0.5 kg ball moving at +4 m/s stops, so J_net=0−(0.5)(4)=−2 N·s.
Stopping in 0.10 s gives F_avg=−20 N. Stopping in 0.40 s gives −5 N. The impulse did not shrink; the time increased by four, so the average force magnitude fell by four.
This comparison holds when Δp is fixed. A rebound changes the final velocity and therefore the impulse. Peak force cannot be inferred from average force without knowing the force’s time dependence. In a vertical catch, include gravity to distinguish net force from hand force.
A worked example, step by step
A 1 kg object moving at +6 m/s is stopped horizontally in 0.20 s or 0.60 s. Find the impulse and average net force for each stop.
- In both cases Δp=1(0−6)=−6 kg·m/s, so J_net=−6 N·s.
- Short stop: F_avg=−6/0.20=−30 N.
- Long stop: F_avg=−6/0.60=−10 N.
- Tripling the duration divides the average force magnitude by three, provided both stops begin and end at the stated velocities.
A longer stop lowers average force for a fixed momentum change; it does not make that impulse smaller.
Does a longer stopping time guarantee a lower peak force?
Compare with an explanation
No. It lowers the magnitude of the average net force for fixed Δp. Peak force also depends on the pulse shape.
Predict. Change one thing. Explain.
Change the stopping time of the same 0.5 kg ball from +4 m/s to rest. Compare the widths and heights of equal-area rectangular force pulses. These rectangles represent equivalent average forces.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Stopping impulse stays −2 N·s. Duration=0.1 s; average net force=-20 N. The signed rectangular area stays −2 N·s even though height and width change.
0.5 kg ball stops from +4 m/s horizontally. The rectangular graph is an equivalent constant net force giving the same impulse. It does not predict the real contact-force peak.
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 0.4 kg ball moves horizontally at +5 m/s and stops. (a) Find J_net. (b) Find F_avg for a 0.1 s stop. (c) Repeat for 0.25 s. (d) Explain what cannot be concluded about peak force.
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Compare with the answer and four-point rubric
- 1 point: J_net=0−0.4(5)=−2 N·s.
- 1 point: F_avg=−20 N.
- 1 point: F_avg=−8 N.
- 1 point: Peak force requires the pulse shape; average values alone do not specify it.
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 must stay fixed in the softer-catch comparison?
Mass and initial/final velocities, hence Δp.
RECALL 2Triple stopping time at fixed Δp?
Average force magnitude becomes one third.
RECALL 3Does the hand’s force always equal net force?
No. Other forces, such as gravity, may contribute.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Same momentum change, different stopping time
- F_net,avg=Δp/Δt.
- At fixed Δp, force magnitude is inversely proportional to stopping time.
Remember: A longer stop lowers average force for a fixed momentum change; it does not make that impulse smaller.
Conditions: 0.5 kg ball stops from +4 m/s horizontally. The rectangular graph is an equivalent constant net force giving the same impulse. It does not predict the real contact-force peak.
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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