The steady motion of the center of mass
You will be able to: Calculate center-of-mass velocity as a mass-weighted average.
What keeps moving steadily while two carts bounce?
A heavier cart moving left can balance a lighter cart moving right. Their center of mass may stay still even as both carts move. If the pair receives no net external force, its center of mass keeps the same velocity through the collision.
A useful starting point: Adding a system’s momenta →
Words and symbols before equations
- Center of mass (CM)
- The mass-weighted center of a system.
- Total mass M
- The sum of the included masses, in kg.
- CM velocity v_CM
- Total momentum divided by total mass, in m/s.
- Weighted average
- An average in which a larger mass contributes more.
What this picture assumes
A: 1 kg at +4 m/s; B: selected mass at rest. Velocity bars have a shared scale. This compares separate systems, not a variable-mass process.
Connect the picture to the physics
For two objects, P=m_Av_A+m_Bv_B=Mv_CM. Divide by M=m_A+m_B: v_CM=(m_Av_A+m_Bv_B)/(m_A+m_B). A heavier object contributes more to the numerator and denominator.
For 2 kg at +3 m/s and 3 kg at −2 m/s, P=6−6=0, so v_CM=0. Neither cart must be at rest, and there need not be any object physically located at the center of mass.
With constant total mass and zero net external force, P and v_CM stay constant. Internal collisions can rearrange individual velocities without changing v_CM. A simple arithmetic average of velocities works only when the masses are equal.
A worked example, step by step
A 1 kg cart travels at +4 m/s and a 3 kg cart is at rest. Find v_CM. If they stick with negligible external impulse, predict their shared velocity.
- Total mass M=1+3=4 kg.
- P=(1)(4)+(3)(0)=+4 kg·m/s.
- v_CM=P/M=4/4=+1 m/s, not the unweighted average of 2 m/s.
- After sticking, every part moves together at v_CM=+1 m/s; the internal collision cannot change the pair’s momentum.
The center of mass does not follow the faster object or the ordinary average of unequal-mass velocities.
Can the center of mass have constant velocity while both objects accelerate?
Compare with an explanation
Yes. Equal and opposite internal impulses can change the individual velocities while total P and v_CM stay fixed.
Predict. Change one thing. Explain.
Cart A is 1 kg at +4 m/s; cart B is initially stationary. Increase B’s mass. Compare v_CM with the two cart velocities and explain why it shifts toward B’s velocity.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
P=4 kg·m/s; M=4 kg; v_CM=P/M=1 m/s. Increasing the stationary mass shifts the average toward 0 m/s.
A: 1 kg at +4 m/s; B: selected mass at rest. Velocity bars have a shared scale. This compares separate systems, not a variable-mass process.
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 2 kg cart moves at +5 m/s and a 3 kg cart at −1 m/s. (a) Find P. (b) Find M. (c) Find v_CM. (d) Predict whether an isolated collision changes v_CM.
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Compare with the answer and four-point rubric
- 1 point: P=10−3=+7 kg·m/s.
- 1 point: M=5 kg.
- 1 point: v_CM=7/5=+1.4 m/s.
- 1 point: No. With negligible external impulse and fixed M, total P and v_CM are unchanged.
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 1Center-of-mass velocity formula?
Total momentum divided by total mass.
RECALL 2When is the ordinary velocity average valid?
For equal masses.
RECALL 3Can an internal push change v_CM?
Not for a fixed-mass system with zero net external impulse.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
The steady motion of the center of mass
- v_CM=Σmv/Σm=P/M.
- For fixed total mass, Δv_CM=J_ext/M.
Remember: The center of mass does not follow the faster object or the ordinary average of unequal-mass velocities.
Conditions: A: 1 kg at +4 m/s; B: selected mass at rest. Velocity bars have a shared scale. This compares separate systems, not a variable-mass process.
Refresh Kid · Unit 4 · Objectives 4.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.3, objectives 4.3.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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