AP Physics 1 · Unit 2 · Topic 2.1

Systems and Center of Mass

A system is the object or collection of objects selected for analysis. That choice determines which interactions are internal, which are external, and whether a complex collection can be replaced by a single point located at its center of mass.

Titration response · Systems and Center of MassThe equivalence region produces the steepest change; its location depends on stoichiometry and acid-base strength.

Learning Goals

After this lesson, you should be able to:

draw and justify a system boundary;
classify an interaction as internal or external for that chosen system;
explain when a multi-object system may be modeled as one object;
locate a center of mass by symmetry or mass-weighted coordinates;
calculate a center of mass for a small one- or two-dimensional particle system;
relate individual motion to center-of-mass motion; and
recognize when mass, energy, or interactions cross the system boundary.

1. Choose the System Before the Equation

Draw an imaginary boundary around the object or objects included in the system. An interaction between two included parts is internal. An interaction exerted by something outside the boundary is external.

The same physical force can change classification when the boundary changes. Tension between two connected carts is internal for the two-cart system but external for a system containing only one cart.

2. What Can Cross the Boundary?

System choice	Internal features	Possible exchange with environment
One cart	Cart's constituent structure	Force, energy transfer, added or removed mass
Two linked carts	Cart–connector interactions	Track, Earth, and external pulls
Ball + Earth	Gravitational interaction	Air resistance or an external push
Open container	Contents currently included	Matter and energy may cross

A useful solution explicitly states whether system mass is constant and whether an external interaction is negligible. “Energy is conserved” or “momentum is conserved” is incomplete without this boundary analysis.

3. When Can a System Be One Object?

If internal separations, rotations, and deformations do not affect the question, a macroscopic system may be represented as one object at its center of mass. This simplifies translational motion but does not erase internal behavior.

When internal structure matters—such as a person moving their arms, fuel leaving a rocket, or masses sliding inside a box—the model must retain enough internal information to answer the question.

4. Center of Mass as a Weighted Position

For point masses, the center-of-mass coordinate is a mass-weighted average:

\[x_{cm}=\frac{\sum_i m_i x_i}{\sum_i m_i},\qquad y_{cm}=\frac{\sum_i m_i y_i}{\sum_i m_i}\]

The center lies closer to the larger mass because its coordinate receives greater weight. The result has position units and depends on the chosen origin, but its physical location relative to the system does not depend on how the axes are labeled.

5. Weighted Balance Model

**Systems and Center of Mass**Topic-matched fine pixel-art model · use the lesson's equations and labeled diagrams for exact scientific meaning.

6. Worked Example: Two Masses on a Line

Question: A \(2.0\,\mathrm{kg}\) cart is at \(x_1=0\) and a \(6.0\,\mathrm{kg}\) cart is at \(x_2=4.0\,\mathrm m\). Find the center of mass.

\[x_{cm}=\frac{(2.0)(0)+(6.0)(4.0)}{2.0+6.0}=3.0\,\mathrm m\]

Answer: The center of mass is at \(x=3.0\,\mathrm m\). It lies within the interval and nearer the heavier cart, providing two quick reasonableness checks.

7. Worked Example: Particles in Two Dimensions

Three particles have \((m,x,y)\) values \((1.0\,\mathrm{kg},0,0)\), \((2.0\,\mathrm{kg},3.0\,\mathrm m,0)\), and \((1.0\,\mathrm{kg},0,4.0\,\mathrm m)\).

\[x_{cm}=\frac{(1)(0)+(2)(3)+(1)(0)}{4}=1.5\,\mathrm m\]

\[y_{cm}=\frac{(1)(0)+(2)(0)+(1)(4)}{4}=1.0\,\mathrm m\]

Answer: \((x_{cm},y_{cm})=(1.5\,\mathrm m,1.0\,\mathrm m)\). Calculate each coordinate independently using the same total mass.

8. Symmetry Before Arithmetic

For a uniform object, the center of mass lies on every line or plane of mass symmetry. A uniform rectangle's two symmetry lines intersect at its geometric center. A uniform ring's center of mass is at the empty geometric center even though no material is located there.

If density is not uniform or a piece is removed, geometric symmetry may be lost. The center shifts toward the region containing more mass. AP Physics 1 expects calculations for no more than five particles in two dimensions or for highly symmetric systems, not calculus-based density integrals.

9. Center-of-Mass Velocity

For a constant-mass particle system, the center-of-mass velocity is the same mass-weighted average applied to velocity components:

\[\vec v_{cm}=\frac{\sum_i m_i\vec v_i}{\sum_i m_i}\]

Individual parts may move in different directions while the center of mass remains at rest. Two equal masses moving with equal and opposite velocities have \(\vec v_{cm}=0\), even though neither mass is stationary.

10. External Force Controls Center-of-Mass Motion

For a constant-mass system, internal interaction forces cancel in the system total. The net external force determines the center-of-mass acceleration:

\[\sum\vec F_{ext}=M\vec a_{cm}\]

Internal forces can dramatically change the motion of individual pieces without changing the center-of-mass trajectory. During an explosion far from significant external forces, fragments scatter, but the center of mass continues with its prior constant velocity.

11. AP Reasoning: Two Carts and a Compressed Spring

Two carts begin at rest with a compressed spring between them on a nearly frictionless track. After release, the carts move apart.

Claim: the center of mass remains at rest. Evidence: spring forces are internal to the two-cart-plus-spring system, while the horizontal external force is negligible. Reasoning: internal forces change the carts' individual velocities but cannot accelerate the center of mass when the net external force is zero.

12. System-Choice Comparison

For the same spring-cart event:

System = one cart: the spring force crosses the boundary and is external.
System = both carts, spring excluded: contact forces from the spring are external to the selected carts.
System = both carts + spring: spring–cart forces are internal, while track and Earth interactions remain external.

The physics has not changed; only the accounting has. A correct argument must remain consistent with its stated boundary.

13. Investigation: Locate a Meterstick System's Center of Mass

Attach up to four known clip masses at measured positions on a meterstick. Include the meterstick's own mass at its independently measured balance point.

Choose an origin and calculate the predicted \(x_{cm}\).
Support the assembled system at the predicted point and observe whether it balances.
Move one mass and predict the direction and size of the center shift before testing.
Repeat with the same total added mass arranged symmetrically.
Compare prediction and measurement with position uncertainty.

Use a low support height so the apparatus cannot fall far.

14. Common Traps and Fast Checks

Trap: averaging coordinates without mass weights. Check: equal weighting is valid only for equal masses.
Trap: forgetting a mass located at the origin. Check: its numerator contribution is zero, but it remains in total mass.
Trap: insisting the center of mass must contain material. Check: rings and hollow objects provide counterexamples.
Trap: calling a force internal without stating the system. Check: draw the boundary first.
Trap: assuming every constituent moves like the center of mass. Check: compare individual and system-wide motion separately.
Trap: placing \(x_{cm}\) outside the interval between two positive point masses. Check: a weighted average must lie between them.

15. Key Takeaways

The system boundary determines internal and external interactions.
A complex system can be represented at its center of mass when internal details are irrelevant.
Center of mass is a mass-weighted position and may lie in empty space.
Symmetry can locate the center without calculation.
Individual parts may move differently from the center of mass.
Net external force, not internal interaction forces, controls center-of-mass acceleration.

Checkpoint

Masses \(m\), \(2m\), and \(m\) are located at \((0,0)\), \((3\,\mathrm m,0)\), and \((0,2\,\mathrm m)\). Calculate the center of mass. Then explain how its motion would respond if the particles exerted forces only on one another.

Concept scope cross-checked with the College Board AP Physics 1 course framework and the open educational references OpenStax University Physics: Center of Mass and OpenStax Physics: Systems, Internal, and External Forces. Explanations, diagram, examples, and investigation are independently written for Insomnia Studyroom.

Official curriculum reference: College Board AP Physics 1 course page. The explanation and worked example are independently written for this study site.