A rebound changes momentum more than a stop
You will be able to: Calculate p_f − p_i with signs and relate contact time to average net force.
Why does reversing direction require a larger impulse than just stopping?
A 0.2 kg ball traveling right at 5 m/s has momentum +1 kg·m/s. Stopping it needs −1 N·s. Sending it back left at 5 m/s needs −2 N·s because its final momentum is −1 kg·m/s.
A useful starting point: Impulse is the signed area under force versus time →
Words and symbols before equations
- Δp
- Final momentum minus initial momentum, including their directions.
- Rebound
- A reversal of velocity direction during an interaction.
- Contact duration Δt
- Time over which the contact interaction occurs.
- Average net force
- Δp/Δt; contact force equals it only when other impulses are negligible or accounted for.
What this picture assumes
A horizontal stop or rebound in a fixed frame. Other horizontal impulses are negligible. Average net force is calculated from the velocity change; no contact pulse shape or peak force is prescribed.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- p_i = 2, p_f = -1 kg·m/s. Impulse -3 N·s; average net force -30 N over 0.1 s. Peak force is not determined by these data.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Choose one positive direction for the entire event. For constant mass, Δp = m(v_f − v_i). A negative final velocity is not a negative speed; it marks reversed motion. Subtracting a positive initial velocity from a negative final velocity adds their magnitudes in Δp.
For a specified Δp, increasing duration reduces the magnitude of average net force: |F_avg| = |Δp|/Δt. This statement concerns average force. Peak force also depends on the pulse shape and cannot be inferred from duration alone.
If gravity matters during vertical contact, the contact force is not the net force. For upward-positive motion, J_contact − mgΔt = Δp, so J_contact = Δp + mgΔt. Our horizontal rebound model neglects other horizontal forces.
A worked example, step by step
A 0.5 kg puck approaches a bumper at +4 m/s and leaves at −2 m/s after 0.10 s. Find impulse and average net horizontal force.
- Set positive direction toward the bumper. p_i = (0.5)(4) = +2 kg·m/s.
- p_f = (0.5)(−2) = −1 kg·m/s.
- Δp = p_f − p_i = −1 − 2 = −3 kg·m/s, so J_net = −3 N·s.
- F_avg = −3/0.10 = −30 N. Its negative sign means force acts away from the bumper on average.
Use velocity signs before subtracting. The change in speed, 2 − 4, is not the change in velocity, −2 − 4.
If the ball rebounds more slowly, is its impulse necessarily smaller than for a stop?
Compare with an explanation
No. For the same incoming velocity, any reversal adds an opposite final momentum and increases impulse magnitude beyond the stopping impulse.
Predict. Change one thing. Explain.
Keep the approach speed and mass fixed. Compare final velocity 0 with a negative final velocity. Then change only contact time and distinguish the unchanged impulse from the changed average force.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
p_i = 2, p_f = -1 kg·m/s. Impulse -3 N·s; average net force -30 N over 0.1 s. Peak force is not determined by these data.
A horizontal stop or rebound in a fixed frame. Other horizontal impulses are negligible. Average net force is calculated from the velocity change; no contact pulse shape or peak force is prescribed.
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 0.1 kg ball arrives horizontally at +10 m/s. Compare the impulse needed to stop it with the impulse needed to return it at −5 m/s. For each, find average net force over 0.05 s.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Stopping impulse = 0.1(0 − 10) = −1 N·s.
- 1 point: Rebound impulse = 0.1(−5 − 10) = −1.5 N·s.
- 1 point: Stopping average force = −1/0.05 = −20 N.
- 1 point: Rebound average force = −1.5/0.05 = −30 N; the reversed final momentum adds to the magnitude of the change.
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 be included in a rebound calculation?
The sign of each velocity in the same coordinate system.
RECALL 2Does a longer collision automatically reduce peak force?
Not without a statement about pulse shape; the fixed-impulse conclusion directly concerns average force.
RECALL 3When does contact impulse equal net impulse?
When other force impulses are negligible, or have already been included correctly.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A rebound changes momentum more than a stop
- J_net = m(v_f − v_i) for fixed mass.
- F_avg,net = m(v_f − v_i)/Δt.
- Same Δp spread over longer time means smaller average force magnitude.
Remember: Use velocity signs before subtracting. The change in speed, 2 − 4, is not the change in velocity, −2 − 4.
Conditions: A horizontal stop or rebound in a fixed frame. Other horizontal impulses are negligible. Average net force is calculated from the velocity change; no contact pulse shape or peak force is prescribed.
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.
Want to work through this with a tutor?
Bring your question about A rebound changes momentum more than a stop. Your explanation and answers remain free to access.