United StatesPhysics C: MechanicsSyllabus dot point

How do we define a system and locate its center of mass, and why does only the net external force govern the motion of the center of mass?

Topic 2.1 Systems and Center of Mass: define a system, locate the center of mass by a mass-weighted average (including by integration for continuous bodies), and apply that only external forces accelerate the center of mass.

A focused answer to AP Physics C: Mechanics Topic 2.1, covering the idea of a system, the center of mass as a mass-weighted average for discrete particles and by integration for continuous bodies, the velocity and acceleration of the center of mass, and why only external forces change the center-of-mass motion, with calculus-based worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this topic is asking
Systems and internal versus external forces
Locating the center of mass
Motion of the center of mass
Try this

What this topic is asking

The College Board (Topic 2.1) wants you to define a system of objects, to locate its center of mass as a mass-weighted average (a sum for particles, an integral for continuous bodies), and to apply the principle that only external forces accelerate the center of mass. This idea threads through the whole course: momentum conservation, collisions and rotational dynamics all rest on the center-of-mass concept established here.

Systems and internal versus external forces

Choosing the system is the first strategic decision in many problems. Treat two colliding carts as one system and the collision forces become internal (and cancel), so the system's momentum is conserved. Treat a single cart as the system and that same collision force is external, changing its momentum. The physics is the same; the bookkeeping changes with the boundary you draw.

Locating the center of mass

For a set of discrete particles, the center of mass is the mass-weighted average of their positions:

x_{cm} = \frac{\sum_i m_i x_i}{\sum_i m_i}, \qquad \vec{r}_{cm} = \frac{\sum_i m_i \vec{r}_i}{M},

where $M = \sum m_i$ is the total mass. For a continuous body the sum becomes an integral over the mass elements $dm$ :

x_{cm} = \frac{1}{M}\int x\,dm.

For a one-dimensional rod with linear density $\lambda(x)$ , write $dm = \lambda(x)\,dx$ ; for a sheet use a surface density $\sigma$ , and for a solid use a volume density $\rho$ . This integral form is distinctive to AP Physics C and appears regularly on the exam for rods and other shapes of varying density. A uniform symmetric body has its center of mass at its geometric center.

Motion of the center of mass

Differentiating the center-of-mass position gives its velocity and acceleration:

\vec{v}_{cm} = \frac{\sum_i m_i \vec{v}_i}{M}, \qquad \vec{a}_{cm} = \frac{\sum_i m_i \vec{a}_i}{M}.

Multiplying the acceleration by $M$ and using Newton's second law on each particle, the internal forces cancel in third-law pairs, leaving

\sum \vec{F}_{ext} = M\,\vec{a}_{cm}.

This is profound: no matter how complicated the internal interactions, the center of mass moves as if all the mass were there and the net external force acted on it alone. If the net external force is zero, the center of mass moves at constant velocity (or stays at rest), which is the basis of momentum conservation in the later topics.

Center of mass of two particles, then the recoil

(a) Particles of mass $2.0$ kg at $x = 1.0$ m and $4.0$ kg at $x = 4.0$ m. Find $x_{cm}$ . (b) The particles, initially at rest, repel and fly apart on a frictionless surface. Describe the motion of the center of mass.

step 1 Apply the discrete formula

$x_{cm} = \dfrac{(2.0)(1.0) + (4.0)(4.0)}{2.0 + 4.0} = \dfrac{2.0 + 16}{6.0} = \dfrac{18}{6.0} = 3.0$ m.

step 2 Check it lies toward the heavier mass

$3.0$ m is closer to the $4.0$ kg particle at $x = 4.0$ m than to the $2.0$ kg particle, as expected for a mass-weighted average.

step 3 Identify the forces in part (b)

The repulsion is internal to the two-particle system, and the surface is frictionless, so the net external horizontal force is zero.

step 4 Conclude the center-of-mass motion

With $\sum \vec{F}_{ext} = 0$ , $\vec{a}_{cm} = 0$ . The center of mass stays at $x = 3.0$ m even as the particles move apart in opposite directions.

Try this

Q1. Particles of $3.0$ kg at the origin and $1.0$ kg at $x = 8.0$ m. Calculate the center of mass. [2 points]

Cue. $x_{cm} = \dfrac{(3.0)(0) + (1.0)(8.0)}{4.0} = 2.0$ m.

Q2. A uniform rod has its center of mass at $L/2$ . Explain how the center of mass shifts if one end is made denser. [2 points]

Cue. The mass-weighted average moves toward the denser end, past $L/2$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)5 marksSection II (FRQ, calculus). A thin rod of length

L

lies along the

x

-axis from

x = 0

x = L

. Its linear mass density increases as

\lambda(x) = \lambda_0\,\dfrac{x}{L}

. (a) Derive the total mass of the rod. (b) Derive the position of the center of mass. (c) Explain qualitatively why the center of mass lies past the midpoint.

Show worked answer →

A 5-point calculus FRQ on the center of mass of a continuous body.

(a) Total mass (2 points): $M = \int_0^L \lambda(x)\,dx = \int_0^L \lambda_0\dfrac{x}{L}\,dx = \dfrac{\lambda_0}{L}\cdot\dfrac{L^2}{2} = \dfrac{\lambda_0 L}{2}$ .
(b) Center of mass (2 points): $x_{cm} = \dfrac{1}{M}\int_0^L x\,\lambda(x)\,dx = \dfrac{1}{M}\int_0^L \lambda_0\dfrac{x^2}{L}\,dx = \dfrac{1}{M}\cdot\dfrac{\lambda_0}{L}\cdot\dfrac{L^3}{3} = \dfrac{\lambda_0 L^2/3}{\lambda_0 L/2} = \dfrac{2L}{3}$ .
(c) Reasoning (1 point): more mass is concentrated toward the heavy end ( $x = L$ ), so the mass-weighted average is pulled past the geometric midpoint $L/2$ to $2L/3$ .

Markers reward setting up the density-weighted integrals with correct limits and simplifying to $2L/3$ .

AP 2021 (style)1 marksSection I (multiple choice). Two skaters of mass

40

kg and

60

kg stand at rest on frictionless ice and push off each other. After they separate, the center of mass of the two-skater system... (A) moves toward the lighter skater (B) moves toward the heavier skater (C) remains at rest (D) accelerates. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (C).

The push is an internal force, and there is no net external horizontal force (frictionless ice), so the center of mass has zero acceleration. It started at rest, so it stays at rest even as the skaters move apart. Internal forces cannot move the center of mass. The trap is to think the heavier or lighter skater "wins"; the center of mass is unmoved.

Related dot points

Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)