Use momentum for impact and energy for the rise
You will be able to: Separate a sticking impact from the subsequent conservative swing.
Why does a capture-and-swing problem need two different stages?
A small soft projectile embeds in a hanging catcher. The collision is brief and inelastic; the joined mass then rises. Its maximum height can reveal the speed just after capture, and momentum can then reveal the projectile’s incoming speed.
A useful starting point: A collision can lose kinetic energy without sticking →
Words and symbols before equations
- Capture stage
- The short sticking impact, modeled with negligible horizontal external impulse.
- Swing stage
- The later rise, modeled with negligible drag and a fixed ideal suspension.
- Rise h
- Vertical gain in the joined mass’s center of mass, measured from just after capture.
- Point-mass catcher model
- A small catcher and projectile treated as one translating point on a light string, with rotation neglected.
What this picture assumes
Ideal point-mass capture at the bottom of a sufficiently long light-string suspension. Horizontal external impulse is negligible during capture. The later swing has negligible drag and no rotational energy; g = 10 m/s². It is not a rigid-pivot extended-catcher model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Capture: P = 2 kg·m/s, joined V = 2 m/s; K converted 18 J. Swing: remaining 2 J gives rise h = 0.2 m. Inferring backward from h returns incoming speed 20 m/s.
- 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 capture at the bottom, conserve horizontal momentum for projectile + catcher if the external horizontal impulse is negligible: mu = (M+m)V, for an initially stationary catcher. Do not conserve K during this sticking interaction.
During the rise, use ½(M+m)V² = (M+m)gh if mechanical losses are negligible. Thus V = √(2gh). Momentum of projectile + catcher is not conserved during the swing: gravity and suspension forces change its direction and magnitude.
Combining the stages gives u = [(M+m)/m]√(2gh). This depends on the specified idealization. An extended catcher rotating about a rigid pivot needs rotational dynamics instead. In a classroom, use a low-energy soft projectile and appropriate supervised equipment.
A worked example, step by step
A 0.10 kg soft projectile embeds in a stationary 0.90 kg catcher. The joined point-mass system rises 0.20 m. Use g = 10 m/s² to infer incoming speed and the kinetic energy converted at impact.
- Analyze the rise first: V = √[2(10)(0.20)] = 2 m/s immediately after capture.
- During capture, mu = (M+m)V gives (0.10)u = (1.00)(2), so u = 20 m/s.
- Before impact K = ½(0.10)(20²) = 20 J; just after impact K = ½(1.00)(2²) = 2 J.
- The impact converts 18 J. The remaining 2 J becomes gravitational potential energy during the ideal swing.
Do not use one conservation equation across both stages. Momentum applies to the short horizontal capture; mechanical energy applies to the later ideal rise.
Why can’t you set initial projectile K equal to the final gravitational energy?
Compare with an explanation
The sticking impact converts part of the initial K to internal energy. Only the remaining mechanical energy powers the rise.
Predict. Change one thing. Explain.
Keep projectile mass and speed fixed while increasing catcher mass. Predict the capture velocity, rise height and kinetic energy converted at impact. Compare the two stage ledgers.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Capture: P = 2 kg·m/s, joined V = 2 m/s; K converted 18 J. Swing: remaining 2 J gives rise h = 0.2 m. Inferring backward from h returns incoming speed 20 m/s.
Ideal point-mass capture at the bottom of a sufficiently long light-string suspension. Horizontal external impulse is negligible during capture. The later swing has negligible drag and no rotational energy; g = 10 m/s². It is not a rigid-pivot extended-catcher model.
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 0.05 kg projectile sticks to a stationary 0.45 kg point catcher, then rises 0.05 m. Use g = 10 m/s². Find speed just after capture, initial projectile speed, and the kinetic-energy decrease during impact.
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Compare with the answer and four-point rubric
- 1 point: V = √(2gh) = √(2×10×0.05) = 1 m/s.
- 1 point: Momentum during capture: u = (0.50/0.05)(1) = 10 m/s.
- 1 point: Initial K = ½(0.05)(100) = 2.5 J; post-capture K = ½(0.50)(1) = 0.25 J.
- 1 point: Impact converts 2.25 J; the 0.25 J remaining becomes gravitational energy during the rise.
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 split the process into two stages?
Impact loses mechanical energy, while the later ideal swing exchanges K and U.
RECALL 2Which height is used in √(2gh)?
Vertical center-of-mass rise after capture, not arc length.
RECALL 3What would require a more advanced model?
A catcher with important rotational inertia about a pivot or significant losses during the rise.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Use momentum for impact and energy for the rise
- Capture: mu = (M+m)V with negligible external horizontal impulse.
- Rise: ½(M+m)V² = (M+m)gh, so V = √(2gh).
- Combined inference: u = [(M+m)/m]√(2gh).
Remember: Do not use one conservation equation across both stages. Momentum applies to the short horizontal capture; mechanical energy applies to the later ideal rise.
Conditions: Ideal point-mass capture at the bottom of a sufficiently long light-string suspension. Horizontal external impulse is negligible during capture. The later swing has negligible drag and no rotational energy; g = 10 m/s². It is not a rigid-pivot extended-catcher model.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official Unit 4) · Objectives 4.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.4, objectives 4.4.A. 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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