The center of mass follows the total momentum
You will be able to: Calculate center-of-mass velocity and explain its behavior during internal interactions.
How can interacting objects have one simple overall motion?
A 1 kg cart moves right at 4 m/s while a 3 kg cart is at rest. Their total momentum is 4 kg·m/s and total mass is 4 kg, so their center of mass moves at 1 m/s. Their individual velocities may change in a collision while that value stays fixed.
A useful starting point: Conserve momentum only after checking the boundary →
Words and symbols before equations
- Center of mass R_cm
- Mass-weighted position of a fixed system.
- Center-of-mass velocity V_cm
- Rate of change of the center-of-mass position.
- Total mass M
- Sum of the constituent masses, fixed in this lesson.
- Total momentum P
- Sum of the individual vector momenta; P = MV_cm for fixed total mass.
What this picture assumes
Two noninteracting point objects initially at x = 0 and 8 m, moving with constant velocities. Positions illustrate mass weighting without simulating contact. The same straight center-of-mass motion would survive any internal interaction with zero external force.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Total P = 4 kg·m/s; M = 4 kg; V_cm = 1 m/s. At 2 s, x_cm = 8 m. Unequal masses require a weighted average.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For fixed masses, x_cm = Σmᵢxᵢ/M. Differentiate with respect to time to obtain V_cm = Σmᵢvᵢ/M = P/M. The formula works component by component in a plane; a heavier object has greater weight in the average.
If net external force is zero, dP/dt = 0 and V_cm stays constant, even during a violent internal collision or separation. Constant velocity can be nonzero. The center of mass moves uniformly rather than necessarily remaining at one place.
In a frame moving with V_cm, total momentum is zero. A perfectly inelastic collision can stop all relative motion while preserving center-of-mass motion. An elastic collision can keep relative motion but reverse relative directions. Neither changes V_cm without external impulse.
A worked example, step by step
A 2 kg object at x = 0 moves at +3 m/s. A 1 kg object at x = 6 m moves at −3 m/s. Find initial x_cm and V_cm, then x_cm after 2 s with no net external force.
- Total mass is M = 3 kg and initial x_cm = [2(0) + 1(6)]/3 = 2 m.
- Total momentum is P = 2(3) + 1(−3) = +3 kg·m/s.
- V_cm = P/M = +1 m/s.
- After 2 s, x_cm = 2 + 1(2) = 4 m, regardless of any internal collision during that time.
The center-of-mass velocity is mass weighted. An ordinary average of velocities works only for equal masses.
Does P = 0 locate the center of mass at x = 0?
Compare with an explanation
No. It makes V_cm zero in that frame. Position requires the mass-weighted positions and can be anywhere.
Predict. Change one thing. Explain.
Hold both velocities fixed and change only the second mass. Predict how the weighted velocity shifts. Then choose velocities giving P = 0 and confirm that the parts can still be moving.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Total P = 4 kg·m/s; M = 4 kg; V_cm = 1 m/s. At 2 s, x_cm = 8 m. Unequal masses require a weighted average.
Two noninteracting point objects initially at x = 0 and 8 m, moving with constant velocities. Positions illustrate mass weighting without simulating contact. The same straight center-of-mass motion would survive any internal interaction with zero external force.
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 questionTwo equal 2 kg carts are at x = 0 and x = 8 m, with velocities +3 and −1 m/s. Find x_cm(0), total momentum, V_cm and x_cm at 3 s if external force vanishes.
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Compare with the answer and four-point rubric
- 1 point: Initial x_cm = (2×0 + 2×8)/4 = 4 m.
- 1 point: Total momentum is 2×3 + 2×(−1) = 4 kg·m/s.
- 1 point: V_cm = 4/4 = +1 m/s.
- 1 point: x_cm(3 s) = 4 + 1×3 = 7 m, including through any internal collision.
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 motion does P/M describe?
The center-of-mass velocity for a fixed system.
RECALL 2Can an internal explosion accelerate the center of mass?
Not without a net external force.
RECALL 3How are internal velocities described in the center-of-mass frame?
Their mass-weighted vector sum is zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
The center of mass follows the total momentum
- R_cm = Σmᵢrᵢ/M for fixed masses.
- V_cm = P/M = Σmᵢvᵢ/M.
- Zero net external force implies constant V_cm.
Remember: The center-of-mass velocity is mass weighted. An ordinary average of velocities works only for equal masses.
Conditions: Two noninteracting point objects initially at x = 0 and 8 m, moving with constant velocities. Positions illustrate mass weighting without simulating contact. The same straight center-of-mass motion would survive any internal interaction with zero external force.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official 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 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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