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🎡AP Physics 1 Unit 2 Review

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2.1 Systems and Center of Mass

2.1 Systems and Center of Mass

Written by the Fiveable Content Team • Last updated August 2025
Verified for the 2026 exam
Verified for the 2026 examWritten by the Fiveable Content Team • Last updated August 2025
🎡AP Physics 1
Unit & Topic Study Guides
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A system is an object or a collection of objects that we choose to analyze together. In physics, we often treat objects as having no internal structure to simplify our analysis.

systems

When using a system approach, we can apply Newton's Second Law of Motion to find the acceleration by considering the total mass of the system:

ΣF=ma\Sigma F = ma

Where the forces are the external forces acting on the entire system, m is the total mass, and a is the acceleration of the system.

A system has several important characteristics:

  • It has boundaries (physical or conceptual) that separate what's inside from what's outside
  • It can exchange energy, matter, or both with its surroundings
  • Depending on the interactions acting on it, a system may have zero net external force (equilibrium) or a nonzero net external force (which changes the system's motion)

System Boundaries in Physics Problems

Defining the system boundary is a critical first step in solving physics problems. This choice determines which forces are external and which are internal:

  • External forces act across the system boundary and can change the total momentum
  • Internal forces occur between objects within the system and cannot change the total momentum

A system can interact with its environment across its boundary. These interactions may transfer energy, mass, or both. The key idea is identifying what is inside the system, what is outside it, and which interactions cross the boundary.

For example, if we define a book sliding on a table as our system:

  • The gravitational force, normal force, and friction are all external forces
  • There are no internal forces

If we expand our system to include both the book and the table:

  • The gravitational force remains external
  • The normal force and friction become internal forces (they're interactions between parts of our system)

Properties and Interactions of Systems

System Properties from Object Interactions

Systems have properties that emerge from the interactions of their constituent parts. In AP Physics 1, the properties of a system are determined by the interactions between the objects inside the system. For example, whether a two-block system speeds up together, deforms, or stays at rest depends on how the objects in the system interact with each other and with anything outside the system.

When these individual properties aren't important for understanding the overall behavior, we can simplify and treat the system as a single object. Modeling the behavior of a macroscopic system as a single object is possible when the properties or interactions of its constituent objects are not significant factors.

Here are some key ideas about how systems behave:

  • A car can be treated as one object when analyzing its motion, even though its parts interact internally.
  • A two-block system may be modeled as one object if the blocks move together, but not if they slide relative to each other.
  • A system can exchange energy or mass with the environment, such as fuel leaving a rocket or thermal energy leaving a hot object.
  • Objects within a system can behave differently from one another and from the system as a whole. For example, in a skater-push-off situation, each skater accelerates differently even though the two-skater system has one center of mass. In a block-and-table system, the block may slide relative to the table even though the system as a whole can still be analyzed using external forces.
  • The internal structure of a system affects how we analyze it. For example, a rigid object can often be modeled as a single object, but a spring-block system or two carts connected by a string may require attention to internal forces, stretching, or relative motion.
  • As external conditions change, the system's substructure may change in a way that changes the model. For example, if a force becomes large enough to stretch a spring, compress a bumper, or cause two objects to lose contact, the system can no longer be analyzed the same way as before.

Center of Mass Location

The center of mass is a key concept that represents the average position of mass in a system. It's the point where the entire mass of the system can be considered concentrated.

Center of Mass in Symmetrical Systems

In systems with symmetrical mass distributions, the center of mass lies on the lines of symmetry.

  • The center of mass of a uniform rod is at its midpoint
  • A sphere with uniform density has its center of mass at its geometric center
  • A rectangular plate has its center of mass at the intersection of its diagonals

Center of Mass Calculation

The position of a system's center of mass along an axis can be found using the formula:

Xcm=ΣmixiΣmiX_{cm} = \frac{\Sigma m_{i} x_{i}}{\Sigma m_{i}}

Where:

  • XcmX_{cm} is the center of mass position
  • mim_{i} is the mass of each object
  • xix_{i} is the position of each object

For a two-dimensional system, we calculate both x and y coordinates:

Xcm=ΣmixiΣmiX_{cm} = \frac{\Sigma m_{i} x_{i}}{\Sigma m_{i}}

Ycm=ΣmiyiΣmiY_{cm} = \frac{\Sigma m_{i} y_{i}}{\Sigma m_{i}}

For example, with two objects of masses m1m_1 and m2m_2 at positions x1x_1 and x2x_2, the center of mass is:

Xcm=m1x1+m2x2m1+m2X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

Systems as Singular Objects

When the internal structure and internal interactions of a system are not important for the question being asked, the system can be modeled as a single object located at its center of mass. This model is especially useful for analyzing the translational motion of the whole system, even though different parts of the system may move differently relative to one another.

  • The motion of a complex system can be analyzed by treating it as a point mass at its center of mass
  • In a uniform gravitational field near Earth's surface, the system's weight can be treated as acting at the center of mass
  • In projectile motion with negligible air resistance, the center of mass follows a parabolic trajectory even if the object rotates

🚫 Boundary Statement

On the exam, students are only required to calculate the center of mass for systems with five or fewer particles in two-dimensional arrangements or for highly symmetrical systems.

Practice Problem 1: System Boundaries

A 2 kg block sits on a 5 kg table. A horizontal force of 10 N is applied to the block, causing it to slide across the table. If the coefficient of kinetic friction between the block and table is 0.2, determine the acceleration of the block when considering: a) the block alone as the system, and b) the block and table together as the system.

Solution

a) When the block alone is our system:

External forces on the block include:

  • Applied force: 10 N (horizontal)
  • Weight: mg = 2 kg × 9.8 m/s² = 19.6 N (down)
  • Normal force: 19.6 N (up)
  • Friction force: Ff = μk × N = 0.2 × 19.6 N = 3.92 N (opposite to motion)

The net horizontal force is: Fnet = 10 N - 3.92 N = 6.08 N

Using Newton's Second Law: a = Fnet/m = 6.08 N/2 kg = 3.04 m/s²

b) This system choice is not appropriate for finding "the acceleration of the block" by using a = F_ext/M for the combined block-table system, because the block is sliding relative to the table. The block and the table do not necessarily share the same acceleration, so the combined system cannot be treated as a single object with one common translational acceleration for this question. If we want the block's acceleration, we must analyze the block itself: $F_f = \mu_k N = 0.2 \times (2)(9.8) = 3.92$ N, so $F_{net, \text{on block}} = 10 - 3.92 = 6.08$ N and $a_{block} = 6.08/2 = 3.04$ m/s². The value $10/7 = 1.43$ m/s² would only apply if the 7 kg system moved together as one object with a common acceleration, which is not the situation described here.

Practice Problem 2: Center of Mass Calculation

Three particles are arranged on a coordinate system. Particle 1 has mass 2 kg and is located at (0,0) m. Particle 2 has mass 4 kg and is located at (3,0) m. Particle 3 has mass 6 kg and is located at (3,4) m. Find the coordinates of the center of mass of this system.

Solution

To find the center of mass, we'll use the formulas:

Xcm=ΣmixiΣmiX_{cm} = \frac{\Sigma m_{i} x_{i}}{\Sigma m_{i}}

Ycm=ΣmiyiΣmiY_{cm} = \frac{\Sigma m_{i} y_{i}}{\Sigma m_{i}}

First, let's calculate the total mass: Total mass = 2 kg + 4 kg + 6 kg = 12 kg

For the x-coordinate: Xcm=(2×0)+(4×3)+(6×3)12=0+12+1812=3012=2.5X_{cm} = \frac{(2 \times 0) + (4 \times 3) + (6 \times 3)}{12} = \frac{0 + 12 + 18}{12} = \frac{30}{12} = 2.5 m

For the y-coordinate: Ycm=(2×0)+(4×0)+(6×4)12=0+0+2412=2412=2Y_{cm} = \frac{(2 \times 0) + (4 \times 0) + (6 \times 4)}{12} = \frac{0 + 0 + 24}{12} = \frac{24}{12} = 2 m

Therefore, the center of mass of the system is located at (2.5, 2) m.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term

Definition

center of mass

The point in a system where all the mass can be considered to be concentrated for the purpose of analyzing motion and forces.

constituent objects

The individual objects that make up a system.

constituent parts

The individual objects or components that make up a larger system.

energy transfer

The movement of energy from one part of a system to another or between a system and its environment.

internal structure

The arrangement and organization of constituent parts within a system that affects how the system behaves and is analyzed.

lines of symmetry

Imaginary lines about which a system's mass is evenly distributed, and where the center of mass is located for symmetrical objects.

macroscopic system

A system large enough to be observed and analyzed at the scale of everyday objects, rather than at the atomic or molecular level.

mass transfer

The movement of matter from one part of a system to another or between a system and its environment.

symmetrical mass distribution

An arrangement of mass in a system where the mass is evenly distributed about one or more lines or planes of symmetry.

system

A collection of objects and their interactions that are studied together as a single unit.

system properties

The characteristics and behaviors of a system that are determined by the interactions between objects within it.

Frequently Asked Questions

What is a system in physics and how do you define it?

A system in physics is whatever group of objects you choose to analyze—you define it by drawing a boundary that includes some masses and excludes others. Pick the boundary so that the interactions you care about are either internal (between parts) or external (with the environment). If internal structure and individual interactions don’t matter for the behavior you need, treat the whole system as a single object located at its center of mass. Systems can be isolated (no external forces) or open (external forces or mass/energy transfer), and internal forces cancel pairwise while external forces change the system’s momentum. For center of mass use the mass-weighted average x_cm = (Σ m_i x_i)/(Σ m_i) or the integral form for continuous bodies (r_cm = ∫ r dm / ∫ dm). This is exactly what Topic 2.1 and the CED expect—see the Topic 2.1 study guide on Fiveable for examples and practice (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO). For extra practice problems across Unit 2, check Fiveable’s practice page (https://library.fiveable.me/practice/ap-physics-1-revised).

How do I find the center of mass of two objects?

Think of the center of mass (CM) as the mass-weighted average position of the system. For two point objects on a line, use x_cm = (m1 x1 + m2 x2) / (m1 + m2). Here x1 and x2 are their positions (vectors if in 2D/3D) and m1, m2 are their masses. If m1 = m2 the CM is the midpoint; if one mass is much larger, the CM is closer to that mass. For two objects in 2D or 3D apply the same formula component-wise: r_cm = (m1 r1 + m2 r2)/(m1 + m2). This is exactly the AP idea of a system being treated as a single object at its center of mass (CED 2.1.B.2–B.4). Practice setting coordinates so calculations are simple (e.g., put one object at x = 0). For more review and examples, see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and try problems on the unit practice page (https://library.fiveable.me/practice/ap-physics-1-revised).

Why does the center of mass matter in physics problems?

The center of mass (CM) matters because it’s the single point you can treat a system as if all its mass were concentrated for analyzing translational motion. For many AP problems you don’t need the full internal structure—just the CM—so you can use ΣF_ext = M a_cm (mass-weighted average position x_cm = Σ m_i x_i / Σ m_i) to get acceleration, predict motion, or apply conservation of momentum in collisions. The CM also tells you where gravity effectively acts (center of gravity) and is the natural point to compute torques and rotational behavior about the body. Practically: it simplifies multi-particle systems, helps decide when internal forces cancel, and makes conservation laws easier to apply on the exam (Topic 2.1.B in the CED). For a quick refresher, see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO); review the whole unit (https://library.fiveable.me/ap-physics-1-revised/unit-2) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

What's the difference between center of mass and center of gravity?

Center of mass (CM) is the mass-weighted average position of a system: r_cm = (Σ m_i r_i)/(Σ m_i). It’s a purely geometric/mass property that tells you where you can treat the whole system as a point mass for translational motion (CED 2.1.B). The center of gravity (CG) is the point where the total gravitational force on the object can be considered to act. In a uniform gravitational field (g constant over the object) CM = CG, so you don’t have to worry about a difference on most AP problems. They differ when g varies across the object (very large bodies or strong tidal fields): then the CG shifts relative to the CM because weights are mass×local g, not just mass. For AP Physics 1, you’ll usually treat systems at their center of mass (CED 2.1.B.2–B.4) and assume CM = CG unless a problem explicitly varies gravity. For a quick review, see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO). For more practice, check the unit practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

I'm confused about when to treat a system as one object vs multiple objects - can someone explain?

Think of a “system” as whatever set of parts you choose to simplify the problem. You treat a system as one object when the internal details (how parts interact with each other) don’t affect the macroscopic behavior you care about—then you can model the whole thing located at its center of mass and use ∑Fext = M a_cm. If internal forces, different motions of parts, or energy transfers between parts matter, treat parts separately. Quick rules: - Use one-object model when only translational motion of the whole matters (e.g., a rigid wagon being pushed). - Split into multiple objects when parts move differently, stick/slide, rotate differently, or internal forces do work (e.g., two blocks colliding and sticking). - Always check external vs internal forces: internal forces cancel in ∑Fext but can change internal energy/rotation (CED 2.1.A.1–5). For locating the one-object approximation, use x_cm = (Σ m_i x_i)/(Σ m_i). If you want guided examples and AP-style practice, see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and try practice problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

How do you calculate center of mass using that formula with the summation?

Use the mass-weighted average formula: pick an origin, list each particle’s position vector xi and mass mi, then compute x_cm = (Σ mi xi) / (Σ mi). Do the same for y and z components so r_cm = (Σ mi ri)/(Σ mi). Steps: (1) choose coordinates and origin, (2) write each mi and its coordinate(s), (3) multiply each mass by its coordinate, (4) sum those products and divide by total mass. Example (1D): masses 2 kg at x=0 m and 3 kg at x=4 m → x_cm = (2·0 + 3·4)/(2+3) = 12/5 = 2.4 m. For continuous bodies replace Σmi with integrals: r_cm = ∫ r dm / ∫ dm (use λ = dm/dℓ, ρ, etc.). This is exactly the AP formula in the CED (2.1.B.2–3); practice both discrete and continuous problems to match exam tasks. Review this topic on Fiveable’s study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and try extra practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

What does it mean when they say a system has symmetrical mass distribution?

If a system has a symmetrical mass distribution, its mass is arranged so one or more symmetry lines or planes make mass on one side mirror the mass on the other side. For AP Physics 1 (CED 2.1.B.1) that means the center of mass must lie on those lines/planes of symmetry. Examples: a uniform rod has its COM at the midpoint (symmetry along the rod); a uniform disk’s COM is at the geometric center (radial symmetry). Practically, symmetry lets you place the system’s single “effective” object (CED 2.1.A.2 & 2.1.B.4) at that line without summing masses. Use symmetry first on free-response problems to locate the COM quickly; if symmetry’s broken, compute x_cm = Σ m_i x_i or the integral form (CED 2.1.B.2–3). For more examples and practice on Topic 2.1, see the Topic Study Guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and the Unit 2 overview (https://library.fiveable.me/ap-physics-1-revised/unit-2). For more problems, check the practice bank (https://library.fiveable.me/practice/ap-physics-1-revised).

Why do we need to know about linear mass density for center of mass problems?

You need linear mass density because many COM problems deal with objects whose mass isn’t spread evenly (nonuniform rods, bars, or thin wires). The CED says for those you model the object as lots of tiny masses dm and use r_cm = (∫ r dm)/(∫ dm). Linear mass density λ(ℓ) = dm/dℓ gives you dm in terms of position: dm = λ(ℓ) dℓ. That lets you convert the integral into something you can evaluate (for example, x_cm = [∫ x λ(x) dx]/[∫ λ(x) dx]). If λ is constant you recover the uniform case; if λ varies (say λ = kx), the weighting shifts the COM toward heavier regions. This skill (using dm and λ in integrals) is explicitly in the CED (2.1.B.3 and 2.1.B.3.i) and shows up on free-response questions where you must set up and evaluate the integral. For extra practice and worked examples see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and more problems at the Unit 2 page (https://library.fiveable.me/ap-physics-1-revised/unit-2) or practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

Can someone explain the integral formula for center of mass in simple terms?

Think of the integral formula as the continuous version of a mass-weighted average. For discrete masses you use x_cm = (Σ m_i x_i)/(Σ m_i). If mass is spread out, break the object into tiny bits dm and add their contributions: r_cm = (∫ r dm) / (∫ dm). What that means practically: - Pick coordinates and express the position of each tiny piece as r (or x, y, z). - Write dm in terms of a density: for a rod dm = λ(x) dx (linear), for a plate dm = σ dA (areal), for a solid dm = ρ dV (volume). - Integrate r·dm over the whole object for the numerator and integrate dm (total mass M) for the denominator. Quick example: uniform rod length L, λ = M/L, so x_cm = (1/M)∫_0^L x λ dx = (1/L)∫_0^L x dx = L/2. This matches the CED idea that center of mass is a mass-weighted average (2.1.B.2–3) and lets you treat a system as a point mass for translational motion on the AP exam. For more practice and the topic study guide, see Fiveable’s Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO).

How does the internal structure of a system affect its behavior?

The internal structure matters because it determines which parts interact and how those interactions affect the whole system. If internal forces (between parts) are much smaller or don’t change the macroscopic motion, you can model the system as a single object located at its center of mass (x_cm = Σ m_i x_i / Σ m_i). But when internal structure matters—nonuniform mass distribution, parts that move relative to one another, or internal energy transfers—you must analyze constituent parts, internal forces, torques about the center of mass, and possible changes in substructure as external variables change (CED EK 2.1.A.1–6, 2.1.B). Internal forces always cancel for the system’s net external momentum, but they can change internal energy or rotational motion, so the chosen system boundary decides whether simpler “single-object” models apply. For more examples and exam-style guidance on choosing systems and locating centers of mass, see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and the Unit 2 overview (https://library.fiveable.me/ap-physics-1-revised/unit-2). For practice, try problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

What happens to a system's center of mass when external forces are applied?

When external forces act on a system, they change the motion of the system’s center of mass (CM). The key AP relation is ∑F_ext = M a_cm—the vector sum of external forces equals the total mass times the CM acceleration. So: - If no net external force (isolated system), the CM moves at constant velocity (Newton’s first law for the CM). - A nonzero net external force produces acceleration of the CM in the direction of that net force. - An impulse from an external force changes the CM momentum (Δp_cm = ∫∑F_ext dt). Note: Internal forces can change how mass is distributed (so the CM location can shift relative to the parts), but internal forces alone cannot change the CM’s motion. If mass is added/removed or redistributed because of interaction with the environment, you must account for that in the system model (CED 2.1.A.3, 2.1.B.2). For a quick review see the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO). For broader unit review and practice problems, check Unit 2 (https://library.fiveable.me/ap-physics-1-revised/unit-2) and the 1000+ practice questions (https://library.fiveable.me/practice/ap-physics-1-revised).

I don't understand how individual objects in a system can behave differently from the whole system - what does this mean?

Think of a system like a team: each player can run, pass, or fall, but what matters for some questions is the team’s overall position. In physics, individual parts can have different motions (speeds, directions, rotations) while the system as a whole has a single center-of-mass (CM) motion. The CM moves as if all the system’s mass were concentrated there and only external forces act on it (useful when internal details don’t affect the big picture—CED 2.1.A.2, 2.1.B.4). Examples: - Two people on skates push off: each moves opposite ways, but if no external force acts, the CM stays at constant velocity (internal forces cancel). - A wobbling dumbbell: its ends move differently, but the CM follows a smooth path determined by external forces. Why this matters for AP Physics 1: problems often let you model a composite object as a single mass at its CM to simplify F = ma or momentum questions (see Topic 2.1 study guide: https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO). For more practice, check the Unit 2 overview (https://library.fiveable.me/ap-physics-1-revised/unit-2) and the practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

How do you solve center of mass problems for objects that aren't uniform?

Use the mass-weighted average idea from the CED: x_cm = (∑ m_i x_i)/(∑ m_i) for particles, and for nonuniform solids use r_cm = (∫ r dm)/(∫ dm) (CED 2.1.B.2–3). Steps you can follow: 1. Pick coordinates and axis (use symmetry if any). 2. Break the object into small pieces: dm = ρ(x) dV (3D), dm = σ(x) dA (2D), or dm = λ(x) dℓ (1D rod). CED notes λ = d m/dℓ. 3. Write r (or x) of each piece and express dm in terms of the given density function. 4. Compute M = ∫ dm and numerator ∫ r dm, then divide: r_cm = (∫ r dm)/M. 5. For composite objects, treat each piece as a point mass at its own center of mass and use the discrete sum. On the AP exam you may be asked to set up and evaluate these integrals (CED 2.1.B.3). If you want worked examples and extra practice, check the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and more practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

What's the point of modeling a whole system as just one object at the center of mass?

Because external forces determine the motion of the whole system’s translation, you can often ignore the messy internal details and treat the system as one object located at its center of mass (COM). Per the CED, if the behavior of constituent parts isn’t important for the macroscopic motion, the system can be modeled as a single object (2.1.A.2 and 2.1.B.4). Practically: the net external force equals total mass times the COM acceleration ( ΣFext = M a_cm ), and internal forces cancel (they don’t change the COM motion). That makes problems with collisions, gravity, or overall motion much simpler. Limits: don’t do this if you care about rotation, internal motion, changing mass distribution, or torques about a point—then the system’s substructure matters (2.1.A.4–5). For how to find the COM, use the mass-weighted average formula in the CED (2.1.B.2–3). More examples and practice are in the Topic 2.1 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO) and extra problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

Why does the center of mass location change when you add or remove objects from a system?

The center of mass (COM) is a mass-weighted average of where all the pieces of a system sit: x_cm = (Σ m_i x_i)/(Σ m_i) (CED 2.1.B.2). When you add or remove an object you change the terms in that average—either a new m_i and its position x_i are included or an existing one is dropped—so the weighted balance point moves. For example, two equal masses at x = 0 and x = 2 have x_cm = 1. If you add a heavy mass at x = 4, the denominator and numerator both grow, pulling x_cm closer to 4. If you remove the mass at x = 0, x_cm shifts toward the remaining mass. This is just bookkeeping: heavier pieces pull the COM more. On the AP exam you may be asked to compute or reason about COM shifts for particle systems or composite bodies (see Topic 2.1 study guide for worked examples: https://library.fiveable.me/ap-physics-1-revised/unit-2/1-systems-and-center-of-mass/study-guide/nielAWaOcpzSSLLO). For extra practice problems, try the unit practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

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