Choose a system and locate its center of mass

1. Orient: Choose a system and locate its center of mass

A system is the collection of objects included in an analysis. A force between two members is internal; a force exerted by something outside the boundary is external. These words depend on your chosen boundary, not on the kind of force. String tension can be external for one cart but internal for a two-cart system that includes the string.

2. Explain and derive

For point masses along one axis, the center-of-mass coordinate is xCM=Σmᵢxᵢ/Σmᵢ. The numerator weights each coordinate by its mass and the denominator normalizes by total mass. Equal masses give the midpoint, but unequal masses pull the weighted location toward the heavier object. The result has units of length.

Rewrite the definition as Σmᵢ(xᵢ−xCM)=0. The signed mass-weighted offsets about the center cancel. Shifting every coordinate by the same amount shifts xCM by that amount without changing the relative positions. A hollow ring can have its center of mass in empty space; the center is a useful representative coordinate, not necessarily a material point.

For a fixed-mass system in an inertial frame, the net external force determines center-of-mass acceleration through Fext=M aCM. Internal forces can rearrange members without accelerating the system center by themselves. This does not imply each member has the same acceleration. The present static-position model locates the center; it does not simulate the internal motion.

3. Work an example

For 2 kg at x=0 m and 4 kg at x=6 m, xCM=(2×0+4×6)/6=4 m. Weighted offsets are 2(0−4)=−8 kg m and 4(6−4)=+8 kg m. The center is closer to the 4 kg member, while the geometric midpoint remains 3 m.

Extend the system to two dimensions

Compute each center coordinate separately: xCM=Σmᵢxᵢ/M and yCM=Σmᵢyᵢ/M. For five point masses, take 1 kg at (0,0) m, 1 kg at (4,0) m, 2 kg at (0,3) m, 2 kg at (4,3) m and 4 kg at (2,1) m. Total mass is 10 kg. The weighted x sum is 20 kg m and weighted y sum is 16 kg m, giving (xCM,yCM)=(2,1.6) m. Symmetry explains xCM=2 m; adding the 4 kg member at y=1 m pulls yCM down from 2 m for the four other masses alone to 1.6 m; it remains above the geometric mid-height of 1.5 m.

Changed-case action: Add 1 kg to the member at (2,1) m without moving any point. The new weighted sums are 22 and 17 kg m and mass is 11 kg, so the center is (2,17/11) m. Explain why xCM stays fixed while yCM moves down. The interactive model is deliberately one-dimensional; this written evidence task supplies the additional two-dimensional construction.

Transfer: For 1 kg at (0,0), 2 kg at (6,0) and 3 kg at (0,4), calculate both coordinates before checking: weighted sums are 12 and 12 kg m, total mass 6 kg, so the center is (2,2) m. Retain the two sums, not only the final pair.

4. Predict and investigate

Predict first: If both masses double without moving, will the center shift? Predict from the ratio before changing controls.

Compare equal masses, a heavier left mass and a heavier right mass. Calculate the weighted offsets for each case. Find two different mass pairs with the same center and explain the shared mass ratio. State which body-to-body forces become internal when both objects enter the system. Use the lesson controls and record the settings with your evidence.

5. Transfer and explain

Try a new practice case

Find the center-of-mass coordinate of 2 kg at x=−2 m and 3 kg at x=8 m.

Your numerical result (m) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

How could the center of mass move while one member is momentarily stationary? Distinguish a system statement from a statement about each member.

Retain your first answer, force diagram, equations, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Supplied synthetic outputs do not certify actual laboratory skill.

When internal structure and the boundary matter

An open system can exchange both energy and mass with its environment; the fixed-mass center-of-mass equation used here is not a complete equation for a rocket that ejects material. Even a closed system can change internal structure when external conditions change. Two equal masses connected by a deformable spring can move apart under equal outward external pulls while their center stays fixed. A one-point model captures the center but loses their separation and stored energy.

Predict and check: Put equal masses at x=−1 and +1 m. Symmetric outward movement to −1.5 and +1.5 m leaves xCM=0 but increases separation from 2 to 3 m. Moving only the right member to +1.5 m instead gives xCM=0.25 m. Name which internal coordinate a single point would omit and whether external work or transferred mass is involved in each stipulated change. No physical stretching experiment is requested.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 2 topic mapping: 2.1. Checked 30 October 2026. This is a content cross-reference, not AP authorization.