FULL REVIEW

Full Review: Center of Mass and Systems of Particles

Review the essential center-of-mass relationships, system momentum, and external-force models used in algebra-based mechanics.

TIME

60 minutes

BEST FOR

A complete unit review

FINISH WITH

A readiness check

After this full review, you'll be able to... calculate center of mass in one and two dimensions, connect total momentum to center-of-mass velocity, and analyze how net external force changes center-of-mass motion.

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Unit Review Overview

This Physics Sensei Unit Review reinforces the key ideas and problem-solving skills for Center of Mass and Systems of Particles. Use it for homework support, quiz preparation, exam review, or independent study.

UNIT: MEC-U19

TOPIC: Center of Mass and Systems of Particles

TREATMENT: Algebra-Based

LEVEL: Introductory college physics

BEST USED

✓ To reinforce key concepts

✓ Before starting homework

✓ Before a quiz or exam

Physics Sensei is an independent educational resource organized around core college-physics ideas, problem-solving models, and study workflows.

Your Review Plan

Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.

6 Stages • Approximately 60 minutes.

Warm-Up Check

Activate prior knowledge.

Core Concepts

Review the essential ideas.

Guided Practice

Apply what you learned.

Confidence Check

Confirm your understanding.

Summary

Review the key ideas.

Next Step

Continue your learning.

Warm-Up Check

Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.

ACTIVITY 1

Recall Activity

Use a mass-weighted average to locate the center of mass.

A 2.0 kg mass is at x = 1.0 m and a 3.0 kg mass is at x = 5.0 m. Find xCM.

Reveal Answers

xCM = 3.4 m.

Why it works: xCM = [2.0(1.0) + 3.0(5.0)]/(2.0 + 3.0) = 17/5 = 3.4 m.

ACTIVITY 2

Recall Activity

Relate total momentum to the velocity of the center of mass.

A 2.0 kg cart moves at +4.0 m/s and a 3.0 kg cart moves at −1.0 m/s. Find the total momentum and the center-of-mass velocity.

Reveal Answers

P = +5.0 kg·m/s; VCM = +1.0 m/s.

Why it works: P = 2.0(+4.0) + 3.0(−1.0) = +5.0 kg·m/s. With total mass 5.0 kg, VCM = P/M = +1.0 m/s.

ACTIVITY 3

Recall Activity

Use net external force to determine center-of-mass acceleration.

A 6.0 kg system experiences a net external force of 18 N in the +x direction. Find the acceleration of its center of mass.

Reveal Answers

ACM = +3.0 m/s2.

Why it works: ΣFext = M ACM, so ACM = 18 N / 6.0 kg = 3.0 m/s2 in the +x direction.

Ready to strengthen your understanding?

You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?

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Core Concepts

Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.

KEY CONCEPT 1

Calculating Center of Mass for Discrete Particles

For discrete particles, the center of mass is found by weighting each position by its mass and dividing by the total mass. In two dimensions, apply the same weighted-average rule separately to x and y coordinates. Define the system first and keep units consistent.

xCM = Σ(mixi)/Σmi; yCM = Σ(miyi)/Σmi.

Example: For 2.0 kg at x = 1.0 m and 3.0 kg at x = 5.0 m, xCM = [2(1) + 3(5)]/5 = 3.4 m.

Sensei Note: Do not average positions unless the masses are equal. The center of mass is a weighted average, not generally the geometric midpoint.

KEY CONCEPT 2

Center-of-Mass Velocity and Total Momentum

The total momentum of a system equals the total mass multiplied by the velocity of the center of mass. This relationship lets you replace a complicated collection of particle momenta with one system-level quantity. If net external force is zero, total momentum and center-of-mass velocity remain constant.

P = Σpi = M VCM.

Example: A 2.0 kg cart at +4.0 m/s and a 3.0 kg cart at −1.0 m/s have P = 8 − 3 = 5 kg·m/s. With M = 5.0 kg, VCM = P/M = +1.0 m/s.

Sensei Note: Momentum can be redistributed among particles by internal interactions while the system’s total momentum remains unchanged when the net external force is zero.

KEY CONCEPT 3

External Force and Center-of-Mass Acceleration

For a system of particles, the net external force equals the total mass multiplied by the acceleration of the center of mass. Internal forces cancel in the system force sum under the usual Newtonian pairwise-force model, so they can change individual motions without determining the center-of-mass acceleration.

ΣFext = M ACM.

Example: For a 6.0 kg system under a net external force of 18 N, ACM = 18/6.0 = 3.0 m/s2 in the direction of the net external force.

Sensei Note: Always identify the system boundary before labeling a force internal or external. The same interaction can change category when the chosen system changes.

Ready to apply these ideas?

You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?

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Guided Practice

Now it's time to apply what you've reviewed.

Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.

PRACTICE 1

Worked Example

Compute the center-of-mass coordinate by coordinate.

Three particles have masses 2.0 kg at (0,0), 3.0 kg at (4.0,0), and 5.0 kg at (0,6.0) m. Find xCM and yCM.

Reveal Answers

xCM = 1.2 m; yCM = 3.0 m.

Why it works: The total mass is 10.0 kg. xCM = [2(0)+3(4)+5(0)]/10 = 1.2 m and yCM = [2(0)+3(0)+5(6)]/10 = 3.0 m.

PRACTICE 2

Guided Problem

Find total momentum first, then divide by total mass.

A 4.0 kg cart moves at +3.0 m/s and a 6.0 kg cart moves at −1.0 m/s. Find the total momentum and VCM.

Reveal Answers

P = +6.0 kg·m/s; VCM = +0.60 m/s.

Why it works: P = 4.0(+3.0) + 6.0(−1.0) = +6.0 kg·m/s. With M = 10.0 kg, VCM = P/M = +0.60 m/s.

PRACTICE 3

Independent Problem

Use the net external force on the complete system.

A system of total mass 8.0 kg experiences a constant net external force of 20 N. Find ACM and explain what internal collisions inside the system can and cannot change.

Reveal Answers

ACM = 2.5 m/s2 in the direction of the net external force. Internal collisions can redistribute momentum among particles but cannot change ACM by themselves.

Why it works: ACM = ΣFext/M = 20/8.0 = 2.5 m/s2. Internal forces occur inside the system boundary and cancel from the net system-force sum.

Ready to check your understanding?

You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?

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Confidence Check

You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.

QUICK CHECK 1

Calculate a Two-Dimensional Center of Mass

Apply the mass-weighted average separately to x and y.

Three masses are 1.0 kg at (0,0), 2.0 kg at (3.0,0), and 3.0 kg at (0,4.0) m. Find the center of mass.

Reveal Answers

xCM = 1.0 m; yCM = 2.0 m.

Why it works: The total mass is 6.0 kg. xCM = [1(0)+2(3)+3(0)]/6 = 1.0 m and yCM = [1(0)+2(0)+3(4)]/6 = 2.0 m.

QUICK CHECK 2

Connect Momentum to Center-of-Mass Velocity

Use the system’s total momentum and total mass.

A 3.0 kg cart moves at +2.0 m/s and a 2.0 kg cart moves at −1.0 m/s. Find the total momentum and the center-of-mass velocity.

Reveal Answers

P = +4.0 kg·m/s; VCM = +0.80 m/s.

Why it works: P = 3.0(+2.0) + 2.0(−1.0) = +4.0 kg·m/s. The total mass is 5.0 kg, so VCM = 4.0/5.0 = +0.80 m/s.

QUICK CHECK 3

Analyze External Force on a System

Use only the net external force to determine center-of-mass acceleration.

A 10 kg system experiences a net external force of 25 N in the +x direction. Find the center-of-mass acceleration. What can internal forces change, and what can they not determine?

Reveal Answers

ACM = +2.5 m/s2. Internal forces can change individual particle velocities and redistribute momentum, but they do not determine the acceleration of the center of mass.

Why it works: ΣFext = M ACM gives ACM = 25/10 = +2.5 m/s2. Only the net external force determines the system center-of-mass acceleration.

How did it go?

You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?

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Summary

Before moving on, take one final look at the most important ideas from this review.

KEY TAKEAWAY 1

Use a Mass-Weighted Position

For discrete particles, calculate each coordinate of the center of mass with a mass-weighted average.

KEY TAKEAWAY 2

Connect Total Momentum to Center-of-Mass Velocity

Use P = M VCM to translate between particle-level momentum and system-level motion.

KEY TAKEAWAY 3

Use Only External Force for Center-of-Mass Acceleration

Internal forces redistribute motion inside the system; the net external force determines ACM.

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