Impulse: explain a stop or a rebound
You will be able to: Use impulse–momentum to relate signed velocities, net force and time.
Why does bouncing back change momentum more than stopping?
A 0.5 kg ball approaches a wall at 4 m/s. Stopping it changes its momentum by 2 kg·m/s in magnitude. Sending it back at the same speed changes it by 4 kg·m/s: the wall must undo the incoming momentum and create outgoing momentum.
A useful starting point: Signed momentum →
Words and symbols before equations
- Impulse J
- Force accumulated over time; a vector with units N·s.
- Δ (delta)
- Final value minus initial value; Δp=p_f−p_i.
- Average net force F_avg
- A constant force that would produce the same impulse over the interval.
- Time interval Δt
- Duration in seconds; final time minus initial time.
What this picture assumes
0.5 kg ball initially at +4 m/s. All values refer to the net horizontal impulse; vertical forces are not modeled. Force readout is an average, not a peak.
Connect the picture to the physics
Take motion toward the wall as positive. Initially p_i=(0.5)(4)=+2 kg·m/s. If the ball returns at −4 m/s, p_f=−2 kg·m/s and Δp=−2−(+2)=−4 kg·m/s.
Net impulse equals change in momentum: J_net=Δp=F_net,avg Δt. For constant mass, Δp=m(v_f−v_i). The direction of impulse is the direction of the change, which need not be the initial velocity direction.
If several forces act, include their combined impulse to predict Δp. Using only a contact force is justified in an axis where other impulses cancel or are negligible. Because 1 N=1 kg·m/s², N·s and kg·m/s are equivalent units. These lessons keep mass fixed; continuously changing-mass systems are beyond this quantitative treatment.
A worked example, step by step
A 0.5 kg ball changes from +4 to −2 m/s in 0.10 s. Find its net impulse and average net force.
- p_i=(0.5)(4)=+2 kg·m/s; p_f=(0.5)(−2)=−1 kg·m/s.
- J_net=Δp=−1−2=−3 N·s.
- F_net,avg=J/Δt=−3/0.10=−30 N.
- The average force points away from the wall. It is not necessarily the peak force during contact.
Subtract signed velocities; subtracting speed magnitudes misses the extra change during a rebound.
The ball rebounds with unchanged speed. Is its impulse zero?
Compare with an explanation
No. Its momentum reverses direction; the impulse magnitude is twice its initial momentum magnitude.
Predict. Change one thing. Explain.
The 0.5 kg ball begins at +4 m/s. Compare final velocities 0, −2 and −4 m/s, then change contact time while keeping the same final velocity. Separate impulse from average force.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
p_i=+2; p_f=-1 kg·m/s. Net impulse=-3 N·s over 0.1 s; average net force=-30 N.
0.5 kg ball initially at +4 m/s. All values refer to the net horizontal impulse; vertical forces are not modeled. Force readout is an average, not a 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 0.2 kg ball changes from +5 to −5 m/s in 0.05 s. (a) Find p_i and p_f. (b) Find J_net. (c) Find F_net,avg. (d) Compare with stopping the ball in the same time.
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Compare with the answer and four-point rubric
- 1 point: p_i=+1 and p_f=−1 kg·m/s.
- 1 point: J_net=−2 N·s.
- 1 point: F_net,avg=−40 N.
- 1 point: Stopping gives J=−1 N·s and F_avg=−20 N, half the rebound magnitudes.
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 delta mean?
Final minus initial.
RECALL 2What equals net impulse?
The change in the selected system’s momentum.
RECALL 3Why can a rebound give larger impulse?
Momentum reverses as well as losing its incoming value.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Impulse: explain a stop or a rebound
- J_net=Δp=m(v_f−v_i)=F_net,avg Δt for constant mass.
- Impulse units: N·s = kg·m/s. Average force is not generally peak force.
Remember: Subtract signed velocities; subtracting speed magnitudes misses the extra change during a rebound.
Conditions: 0.5 kg ball initially at +4 m/s. All values refer to the net horizontal impulse; vertical forces are not modeled. Force readout is an average, not a 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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