AP Physics 1  ·  Unit 2: Forces & Translational Dynamics  ·  Lesson 2.1

Deep Dive: Systems and Center of Mass

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
2.1.A.1Concept

What Is a System?

A system is simply the collection of objects you've chosen to study. It could be a single block, two carts connected by a string, or an entire rocket. Everything outside the boundary you've drawn is the environment.

🔑You choose the system. There's no single "correct" system for every problem — the right choice depends on what question you're trying to answer. Choosing wisely is often the hardest and most important step.

A system's properties — its total mass, its momentum, its energy — emerge from how its parts interact. The behavior of the whole system depends both on what's happening inside it and on how it interacts with everything outside its boundary.

2.1.A.2Concept

Internal Forces vs. External Forces

Once you've drawn a system boundary, every force in the problem falls into one of two categories. Internal forces act between objects inside the system. External forces cross the boundary — they come from the environment.

SYSTEM BOUNDARYABinternalexternalcenter of massInternal forces cancel. Only the external force moves the system.
🔑Internal forces always come in equal-and-opposite pairs within the system. When you add up all the forces in the system, those pairs cancel completely. Only external forces can change how the system's center of mass moves.
Internal example

Two skaters pushing off each other. If both skaters are part of your system, the push between them is internal — it cancels and doesn't move the system's overall center of mass.

External example

Gravity pulling down on those same skaters. Gravity comes from outside the skater-skater system, so it's external — and it's the only thing that can move their combined center of mass.

2.1.A.3Concept⚠ Watch Out

When Can You Treat a System as One Object?

This is a judgment call, not a rule. A system can be modeled as a single object when the internal details — how the parts interact, move relative to each other, or change shape — don't matter for the question being asked.

⚠️Watch for this: "Treat the system as one object" is a choice you make, not a law of physics. A rigid car can usually be treated as one object. But a spring-block system, or two carts connected by a stretching string, often can't — the internal motion matters too much.

A useful test: if the parts of your system all move with the same velocity and acceleration at every moment, you can safely collapse it to one point. If parts slide relative to each other, stretch, or behave differently, you need to keep tracking them separately.

2.1.B.1Concept

Finding Center of Mass with Symmetry

The center of mass is the point that represents the average position of all the mass in a system. For objects with symmetrical mass distributions — a uniform ring, a meter stick, a sphere — the center of mass sits right on the line of symmetry.

💡The center of mass doesn't have to be located inside the object. For a ring or a donut shape, the center of mass sits in the empty space at the middle. It's a mathematical point, not necessarily a physical part of the object.
2.1.B.22.1.B.3Math

Calculating Center of Mass

When objects aren't symmetrically arranged, you calculate the center of mass as a weighted average of position — heavier objects pull the center of mass toward themselves.

x_cm = Σmᵢxᵢ / Σmᵢ

In plain terms: multiply each object's mass by its position, add those products together, then divide by the total mass. This is exactly what your infographic calls the "weighted average calculation."

Adjust each mass and position. Watch the center of mass shift toward the heavier object.

Mass A (orange)
mass3 kg
position2 m
Mass B (blue)
mass5 kg
position8 m
0123456789101112A: 3kgB: 5kgx_cm = 5.75 m

Notice the center of mass always sits closer to the heavier object — it's a weighted average, not a midpoint.

ExampleWorked Example — Two-Mass System

A 2 kg mass sits at x = 0 m and a 6 kg mass sits at x = 4 m. Find the center of mass of the system.

ExampleGuided Example — Three-Mass System

Three point masses sit on a line: m₁ = 2 kg at x = 0 m, m₂ = 3 kg at x = 2 m, m₃ = 5 kg at x = 4 m. Find the system's center of mass.

Step 1Set up the formula
x_cm = (m₁x₁ + m₂x₂ + m₃x₃) / (m₁ + m₂ + m₃)
← Back to Lesson 2.1Next: Lesson 2.2 →Free-body diagrams are next — the most-used skill in the rest of the unit.
Built with v0