FULL REVIEW
Full Review: Center of Mass and Systems of Particles
Review center-of-mass models for discrete and continuous systems, system momentum, and the external-force dynamics of the center of mass.
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 for discrete and continuous distributions, connect total momentum to center-of-mass velocity, and derive how net external force governs center-of-mass acceleration.
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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: Calculus-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
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
Choose the correct center-of-mass model for a discrete or continuous mass distribution.
State the discrete-particle expression for rCM and the corresponding integral expression for a continuous distribution. Explain what dm represents.
Reveal Answers
Discrete: rCM = (1/M)Σmiri. Continuous: rCM = (1/M)∫r dm, with M = ∫dm.
Why it works: A discrete set contributes separate masses mi. A continuous body is partitioned into differential mass elements dm, which are expressed from the relevant density model.
ACTIVITY 2
Recall Activity
Connect the time derivative of center-of-mass position to total momentum.
Starting from rCM = (1/M)Σmiri for constant total mass, state the result obtained by differentiating with respect to time and identify its connection to total momentum.
Reveal Answers
VCM = (1/M)Σmivi, so P = MVCM.
Why it works: For constant M, differentiate the center-of-mass definition to obtain the weighted average of particle velocities; multiplying by M gives total momentum.
ACTIVITY 3
Recall Activity
Relate the time derivative of total momentum to net external force.
For a constant-mass system, show the chain of relationships connecting dP/dt, ΣFext, and MACM.
Reveal Answers
dP/dt = ΣFext = MACM.
Why it works: Newton’s second law for the complete system gives dP/dt = ΣFext. With P = MVCM and constant M, differentiating gives dP/dt = MACM.
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?
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
Center of Mass for Discrete and Continuous Systems
For discrete particles, center of mass is a mass-weighted average of position vectors. For a continuous body, replace the sum with an integral over differential mass elements. Express dm using the appropriate density model, set physically correct limits, and use symmetry before integrating whenever possible.
rCM = (1/M)Σmiri; rCM = (1/M)∫r dm; M = ∫dm.
Example: For a rod on 0 ≤ x ≤ L with linear density λ(x) proportional to x, write dm = λ(x)dx and evaluate xCM = [∫x dm]/[∫dm]. Because more mass lies near x = L, the result must lie to the right of L/2.
Sensei Note: Set up dm before integrating. A correct integral with the wrong mass element or limits is still the wrong physical model.
KEY CONCEPT 2
Center-of-Mass Velocity and Total Momentum
For constant total mass, differentiating the center-of-mass position gives the center-of-mass velocity. Multiplying by total mass produces the system’s total momentum. This provides the bridge from particle-level motion to a single system-level velocity.
VCM = drCM/dt; P = Σmivi = MVCM.
Example: For particles of constant masses, differentiate rCM = (1/M)Σmiri to obtain VCM = (1/M)Σmivi, so MVCM = Σmivi = P.
Sensei Note: The simple relation P = M VCM assumes the system mass is treated consistently. For the introductory mechanics systems here, M is constant.
KEY CONCEPT 3
External Force, Momentum, and Center-of-Mass Acceleration
Newton’s second law for the system follows from the rate of change of total momentum. Internal force pairs cancel from the system force sum under the standard Newtonian model, leaving the net external force. For constant total mass, differentiating P = M VCM gives ΣFext = M ACM.
dP/dt = ΣFext = MACM.
Example: If ΣFext = 0, then dP/dt = 0, so P is constant. For constant M, VCM is constant as well, even if particles collide, explode apart, or exchange momentum internally.
Sensei Note: Internal forces may be large, but they do not appear in the net external-force equation for the complete system. Define the system boundary before summing forces.
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?
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
Build the differential mass element, then integrate the weighted position.
A thin rod extends from x = 0 to x = L with linear density λ(x) = kx. Find the center of mass in terms of L.
Reveal Answers
xCM = 2L/3.
Why it works: dm = kx dx. Then M = ∫₀ᴸ kx dx = kL2/2 and ∫₀ᴸ x dm = ∫₀ᴸ kx2 dx = kL3/3. Therefore xCM = 2L/3.
PRACTICE 2
Guided Problem
Use the discrete definition, then differentiate to connect center-of-mass motion to momentum.
For a constant-mass particle system, start from rCM = (1/M)Σmiri and derive P = MVCM.
Reveal Answers
P = MVCM.
Why it works: Differentiate the center-of-mass definition for constant masses: VCM = (1/M)Σmivi. Multiplying by M gives total momentum.
PRACTICE 3
Independent Problem
Differentiate total momentum and identify which force terms survive at the system level.
Show that for a constant-mass system dP/dt = ΣFext leads to ACM = ΣFext/M, and explain why internal force pairs do not determine ACM.
Reveal Answers
ACM = ΣFext/M for constant total mass; internal force pairs do not determine ACM.
Why it works: dP/dt = ΣFext. Since P = MVCM and M is constant, dP/dt = MACM. Internal Newtonian force pairs cancel from the complete 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?
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
Set Up a Continuous Center-of-Mass Integral
Choose the correct differential mass element and limits.
A thin rod occupies 0 ≤ x ≤ L with linear density λ(x) = kx2. Write the integral expression for xCM. You do not need to evaluate it.
Reveal Answers
xCM = [∫₀ᴸ x(kx2)dx]/[∫₀ᴸ kx2dx].
Why it works: The mass element is dm = λ(x)dx = kx2dx. Substitute dm into xCM = (1/M)∫x dm, with M = ∫dm and both integrals evaluated from 0 to L.
QUICK CHECK 2
Derive the Momentum Relation
Differentiate the center-of-mass definition for constant total mass.
Starting from rCM = (1/M)Σmiri, derive the relation between total momentum P and VCM.
Reveal Answers
P = MVCM.
Why it works: For constant total mass, differentiating gives VCM = (1/M)Σmivi. The numerator is total momentum.
QUICK CHECK 3
Connect Momentum Rate to External Force
State the system-level chain of relationships for constant total mass.
Show how dP/dt = ΣFext leads to ΣFext = MACM, and explain why internal force pairs do not determine ACM for the complete system.
Reveal Answers
dP/dt = ΣFext = MACM. Internal force pairs can redistribute momentum among particles but do not determine ACM for the complete system.
Why it works: Newton’s second law for the system gives dP/dt = ΣFext. With constant M and P = MVCM, differentiation gives ΣFext = MACM; internal pair forces cancel in the system sum.
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?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Choose Σ or ∫ from the Mass Model
Use a discrete sum for particles and an integral with the correct dm for continuous mass distributions.
KEY TAKEAWAY 2
Differentiate rCM to Reach System Momentum
For constant total mass, differentiating center-of-mass position gives P = M VCM.
KEY TAKEAWAY 3
Differentiate P to Reach External-Force Dynamics
The system relation dP/dt = ΣFext gives ΣFext = M ACM for constant total mass.
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