When objects stick: find the shared velocity
You will be able to: Calculate the shared velocity and kinetic-energy change in a perfectly inelastic collision.
Why does a moving cart slow when it picks up another cart?
A 1 kg cart moving at +4 m/s catches a stationary 3 kg cart and sticks. The same total momentum is now carried by 4 kg, so the pair moves at +1 m/s. Some kinetic energy has become other forms during the impact.
A useful starting point: Classify collisions with an energy check →
Words and symbols before equations
- Shared final velocity V
- The single velocity of objects that remain attached.
- Perfectly inelastic condition
- v_Af=v_Bf=V.
- Kinetic-energy decrease
- K_i−K_f; not destroyed total energy.
What this picture assumes
1 kg cart initially +4 m/s sticks to a stationary target. Negligible external impulse. Energy bars show translational K and its conversion; they do not imply total energy destruction.
Connect the picture to the physics
Select both carts over the short contact interval. With negligible external impulse, m_Av_Ai+m_Bv_Bi=(m_A+m_B)V. Divide by combined mass to find V. Include both signed initial velocities if both carts move.
In this example, 1(4)+3(0)=4V, so V=+1 m/s. K_i=8 J and K_f=½(4)(1²)=2 J: 6 J has become deformation, thermal energy and other forms.
If the target starts at rest, V=[m_A/(m_A+m_B)]v_Ai. Increasing target mass lowers V at fixed incoming mass and velocity. Do not conserve kinetic energy across sticking; use energy conservation only in a later separate stage where its assumptions apply.
A worked example, step by step
A 2 kg cart at +3 m/s sticks to a 1 kg cart at −1 m/s. Find shared velocity and the change in kinetic energy.
- Initial P=2(3)+1(−1)=+5 kg·m/s.
- V=5/(2+1)=+5/3≈1.67 m/s.
- K_i=½(2)(3²)+½(1)(1²)=9.5 J. K_f=½(3)(5/3)²=25/6≈4.17 J.
- K_i−K_f=16/3≈5.33 J changes to other forms. The positive shared velocity is consistent with the positive initial total momentum.
Use momentum to find the immediate shared velocity. Conserving kinetic energy through a sticking impact gives the wrong answer.
Could the shared velocity be zero even when both carts were moving?
Compare with an explanation
Yes. Equal and opposite initial momenta give zero total momentum and therefore zero shared velocity.
Predict. Change one thing. Explain.
A 1 kg cart at +4 m/s sticks to a stationary target. Increase target mass from 1 to 5 kg. Predict V and compare initial and final kinetic energies.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Combined mass=4 kg; shared V=1 m/s. P_i=P_f=4 kg·m/s. K falls from 8 to 2 J; 6 J changes form.
1 kg cart initially +4 m/s sticks to a stationary target. Negligible external impulse. Energy bars show translational K and its conversion; they do not imply total energy destruction.
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 1 kg cart at +6 m/s sticks to a stationary 2 kg cart. (a) Find initial P. (b) Find V. (c) Find K_i and K_f. (d) Explain the difference.
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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=6 J.
- 1 point: The 12 J decrease becomes other forms during impact; total momentum remains +6 kg·m/s.
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 1Final speeds when objects stick?
The same signed velocity.
RECALL 2More stationary target mass does what to V?
Lowers its magnitude for fixed incoming momentum.
RECALL 3Can K be conserved through sticking with relative motion?
No. Some translational kinetic energy changes form.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When objects stick: find the shared velocity
- V=(m_Av_Ai+m_Bv_Bi)/(m_A+m_B), if J_ext≈0.
- K_i−K_f is the decrease in translational kinetic energy.
Remember: Use momentum to find the immediate shared velocity. Conserving kinetic energy through a sticking impact gives the wrong answer.
Conditions: 1 kg cart initially +4 m/s sticks to a stationary target. Negligible external impulse. Energy bars show translational K and its conversion; they do not imply total energy destruction.
Refresh Kid · 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. 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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