When objects stick, share momentum—not speed
You will be able to: Conserve total momentum and calculate the kinetic energy converted to other forms.
How do you find a common velocity after a sticking collision?
A 1 kg cart moving at 4 m/s sticks to a stationary 3 kg cart. Their shared final velocity is 1 m/s, not the simple average 2 m/s. Total momentum remains 4 kg·m/s while kinetic energy falls from 8 J to 2 J.
A useful starting point: Test momentum conservation with cart data →
Words and symbols before equations
- Perfectly inelastic collision
- An interaction in which the objects stick and share one final velocity.
- Common final velocity V
- The velocity of both joined objects immediately after impact.
- Converted kinetic energy
- Initial K minus final K; it becomes deformation, thermal energy, sound or other forms.
- Reduced mass μ
- Optional shorthand m₁m₂/(m₁+m₂); it is not a friction coefficient here.
What this picture assumes
An isolated one-dimensional sticking collision. The initially left cart approaches the initially right cart throughout the control range. Vertical external forces balance; horizontal external impulse is negligible. Arrows compare velocities before and after, not positions during deformation.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Final velocities: left 1 m/s, right 1 m/s. Total P = 4 kg·m/s before and after; K_i = 8 J, K_f = 2 J; converted 6 J. Perfectly inelastic model: shared final velocity.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Select both objects and a short interaction interval with negligible external impulse. Momentum gives m₁u₁ + m₂u₂ = (m₁ + m₂)V, where u denotes initial velocity. Solve for V as a mass-weighted average, retaining signs.
Perfectly inelastic does not mean momentum disappears or all kinetic energy vanishes. The final center-of-mass translation remains. K can vanish completely only in the frame where initial total momentum is zero.
Substitute the common V into K_f. The decrease is K_i − K_f = ½[m₁m₂/(m₁+m₂)](u₁ − u₂)². It is nonnegative and depends on relative approach speed, not the arbitrary lab-frame drift. This identity is a useful check, not a substitute for explaining the system.
A worked example, step by step
A 2 kg cart at +3 m/s collides with a 1 kg cart at −3 m/s and they stick. Find final velocity and converted kinetic energy.
- Initial total momentum is 2(3) + 1(−3) = +3 kg·m/s.
- Combined mass is 3 kg, so V = 3/3 = +1 m/s.
- Initial K = ½(2)(9) + ½(1)(9) = 13.5 J; final K = ½(3)(1²) = 1.5 J.
- The collision converts 12 J from translational kinetic energy to other forms while conserving the pair’s total momentum.
Do not conserve kinetic energy through a sticking impact. Use momentum first, then compare energies.
Must joined carts move in the initial direction of the lighter cart?
Compare with an explanation
No. The sign of their total initial momentum sets the final direction.
Predict. Change one thing. Explain.
Keep the first cart at +4 m/s and the second initially at rest. Increase the second mass. Predict the common final speed and the fraction of initial kinetic energy that remains.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Final velocities: left 1 m/s, right 1 m/s. Total P = 4 kg·m/s before and after; K_i = 8 J, K_f = 2 J; converted 6 J. Perfectly inelastic model: shared final velocity.
An isolated one-dimensional sticking collision. The initially left cart approaches the initially right cart throughout the control range. Vertical external forces balance; horizontal external impulse is negligible. Arrows compare velocities before and after, not positions during deformation.
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 1 kg cart at +6 m/s sticks to a 2 kg cart at rest. Find initial momentum, final velocity, initial and final K, and the converted amount.
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Compare with the answer and four-point rubric
- 1 point: P_i = +6 kg·m/s.
- 1 point: V = 6/3 = +2 m/s.
- 1 point: K_i = 18 J and K_f = ½(3)(2²) = 6 J.
- 1 point: Converted kinetic energy is 12 J; total energy remains conserved when internal forms are included.
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 distinguishes a perfectly inelastic collision?
The objects share a common final velocity because they stick.
RECALL 2Can some K remain after sticking?
Yes, as translation of the joined system.
RECALL 3Which speed controls the kinetic-energy decrease?
The initial relative speed u₁ − u₂, together with the masses.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When objects stick, share momentum—not speed
- V = (m₁u₁ + m₂u₂)/(m₁ + m₂), when J_ext ≈ 0.
- K_converted = K_i − K_f.
- For sticking: K_converted = ½[m₁m₂/(m₁+m₂)](u₁ − u₂)².
Remember: Do not conserve kinetic energy through a sticking impact. Use momentum first, then compare energies.
Conditions: An isolated one-dimensional sticking collision. The initially left cart approaches the initially right cart throughout the control range. Vertical external forces balance; horizontal external impulse is negligible. Arrows compare velocities before and after, not positions during deformation.
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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