Object vs System in Physics: The Difference

An object is a physical thing whose internal structure is ignored, so it has mass and position but no parts. A system is a collection of objects analyzed together. Neither label is a fact about the thing: both are modeling choices, and the choice sets which forces are external.

AP Physics: Unit 2 (topics 2.1 Systems and Center of Mass, 2.2 Forces and Free-Body Diagrams, 2.3 Newton's Third Law). Object and system are defined by College Board itself, in the appendix Vocabulary and Definitions of Important Ideas in AP Physics printed at the back of all four Course and Exam Descriptions: an object is a physical thing where the internal structure and properties of the thing are ignored, and a system is a collection of objects that are analyzed together. AP Physics 1 EK 1.2.A.1 restates the object model in course language, and Topic 2.1 supplies the system properties, EK 2.1.A.1 through 2.1.A.6, plus EK 2.1.B.3 placing a system's single-object model at its center of mass. EK 2.3.A.2 supplies the consequence that internal forces do not influence the motion of a system's center of mass, and EK 2.2.A.1.ii that an object or system cannot exert a net force on itself. The Topic 2.1 boundary statement limits center-of-mass calculations to five or fewer particles in a two-dimensional configuration or to highly symmetrical systems. The appendix also states that the exam will not directly assess vocabulary and will not ask students to state the difference between a system and an object. Unit 2 carries 18 to 23 percent of the AP Physics 1 multiple-choice section.

Two definitions, both written by College Board

Most physics words are defined in a textbook. These two are defined by the exam board itself, in an appendix printed at the back of all four Course and Exam Descriptions called Vocabulary and Definitions of Important Ideas in AP Physics.

Object. A physical thing where the internal structure and properties of the thing are ignored. AP Physics 1 EK 1.2.A.1 restates it in course language: when using the object model, the size, shape, and internal configuration are ignored, and the object may be treated as a single point with extensive properties such as mass and charge.

System. A collection of objects that are analyzed together.

Read the two definitions next to each other and the difference is not what the thing is. It is what you have agreed to ignore. An object is a thing you have decided has no inside. A system is a group of things you have decided to track as a group, and the appendix is blunt about who is doing the deciding: there is no single right or wrong way to group objects, but often the preferred method is to choose a system that simplifies the analysis.

That is the sentence the whole page hangs on. "Is this an object or a system?" is not a question about the truck, the cart or the block. It is a question about you.

The glossary defines each term on its own at object and system. This page is about the seam between them: when you have to switch models, what happens to the numbers when you do, and which switch costs marks.

Object vs system, side by side

	Object	System
College Board's definition	A physical thing where the internal structure and properties of the thing are ignored	A collection of objects that are analyzed together
Has parts	No, by construction	Yes, and their interactions define its properties (EK 2.1.A.1)
Has a size or a shape	No. Size, shape and internal configuration are ignored (EK 1.2.A.1)	Yes, and it can change: EK 2.1.A.6 says the substructure may change as external variables change
Where a force is applied	Not part of the model. Top, bottom and middle are the same point	Can matter, and for a rigid system it does
Can be compressed, twisted or rotated	No. The appendix says so directly	Yes
Number of free-body diagrams	One	One per part, plus one for the whole thing if you want it
What counts as an internal force	Nothing. There is no inside	Any force between two parts inside the boundary
Who chooses it	You do	You do, and the appendix says there is no single right grouping
The reason to choose it	It removes detail you do not need	It removes forces you do not want to compute
Can become the other	A group of objects becomes one object when their internal behavior does not matter (EK 2.1.A.2)	A system is modeled as a single object located at its center of mass (EK 2.1.B.3)

The last row is the one worth rereading. The two models are not rivals. A system collapses into an object the moment its internals stop mattering, and that collapse is one of the most useful moves available in mechanics.

The case that separates them: two blocks and one push

Put a 3.0 kg block A against a 5.0 kg block B on a frictionless floor and push A with a horizontal 12 N force. Two questions, two models, and each model can only answer one of them.

Question 1: how fast do they accelerate? Take the two blocks as one system. The 12 N is the only external horizontal force. EK 2.5.A.2 gives the acceleration of the system's center of mass, and the total mass is 8.0 kg:

a = \frac{F_{\text{net}}}{m_{\text{sys}}} = \frac{12\ \text{N}}{8.0\ \text{kg}} = 1.5\ \text{m/s}^2

One line. The push A gives B never appeared, because it is internal to the system you chose.

Question 2: how hard does A push B? The system model cannot tell you, and not because it is a weak model. It cannot tell you because you deliberately deleted that information when you drew the boundary around both blocks. To get it back you have to redraw: take B alone as an object, and the push from A is now the only horizontal force on it.

F_{\text{A on B}} = m_B a = (5.0\ \text{kg})(1.5\ \text{m/s}^2) = 7.5\ \text{N}

Same situation, same physics, two boundaries, two answers, and both are correct. This is what people mean when they say the choice of system is the first step of a mechanics problem: you are choosing which forces you will have to know about.

Notice which direction the information flows. Going from parts to whole is free: you can always lump. Going from whole to parts is not: once the boundary is drawn, the internal forces are gone from the algebra and the only way back is a second diagram.

A system can be an object, and that is the point

EK 2.1.A.2 licenses the collapse: if the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object. EK 2.1.B.3 says where that single object sits: at the system's center of mass.

The appendix's own example is a box of cereal. In reality it is a cardboard box, a plastic bag and a large number of pieces of cereal, but the complex motion of the pieces inside the box is not important to a person carrying it, so the entire box, bag and cereal can be treated as a single object.

So the two models are a hierarchy, not a pair of alternatives:

Object. No inside at all. One dot, one free-body diagram, mass and position and nothing else.
System modeled as an object. Several things, lumped, with the lump placed at the center of mass. Internal forces vanish from the algebra by EK 2.3.A.2.
System with its parts tracked. Several things, each with its own diagram, and the forces between them now visible and computable.
Rigid system. A system that does not change shape, but different points within the system may move in different directions and with different speeds. This is the appendix's third definition, and it is the model you need the instant rotation is relevant.

Step 4 is where the object model finally runs out, and the appendix picks a wheel to show it. A wheel on an axle keeps its shape, so nothing is bending or breaking, and yet at any instant a point near the rim travels faster and in a different direction than a point nearer the hub. An object has no rim and no hub, so it cannot hold that fact. See rigid system for how the rotational units use the term.

Where the boundary goes decides what is external

This is the payoff, and it is why the choice of system is worth a page rather than a definition.

A force is either exerted by something inside your boundary or by something outside it. Forces from outside are external; forces between two parts inside are internal, and EK 2.3.A.2 disposes of them in one line: interactions between objects within a system, the internal forces, do not influence the motion of a system's center of mass.

Which means the same physical push changes category when you redraw the line:

Situation	System you chose	The push from A on B is	What follows
Two blocks, one shove	Block B alone	External	B accelerates because of it, and you can solve for it
Two blocks, one shove	A and B together	Internal	It cancels with its Newton's third law partner and never enters $\sum F$
Two carts and a compressed spring	One cart	External	That cart's momentum changes
Two carts and a compressed spring	Both carts and the spring	Internal	The total momentum is unchanged, so the center of mass never moves

Nothing in the world changed between rows. Only the boundary moved. The forces page for this is internal vs external forces, and the conservation consequence, that a quantity can be conserved without being constant, is constant vs conserved.

One guard rail from the CED, because it stops the boundary from being drawn absurdly. EK 2.2.A.1.ii says an object or system cannot exert a net force on itself. If your boundary encloses both members of an interaction, that interaction contributes nothing to the net force on the whole thing. You do not get to pull yourself along by your own bootstraps, and you do not get to invent a net force on a system from a shove between its parts.

The object model is a licence to forget, and it has limits

The appendix is unusually candid about what the object model throws away, and each thrown-away feature is a question the model then cannot answer.

Size and shape. The truck looks the same from the top, the bottom or the side. So the object model cannot tell you that a tall truck tips and a low one does not.
Where the force is applied. Pushing the top of a truck has a different effect from pushing the bottom or the middle, but if the truck is modeled as an object, the location of the application of the force is not considered. That is exactly the information torque needs, which is why Unit 5 stops using the object model.
Compression, twisting and rotation. Objects cannot be compressed, twisted or rotated, because the physical dimensions are ignored.
Internal motion. The appendix's caution is a truck carrying unsecured bricks. Sudden accelerations in any direction may have significant effects on how the truck behaves, and none of that is in the object model.

One case looks like a counterexample and is not. Friction depends on the surfaces of two things in contact, and surfaces sound like internal structure. The appendix answers it: two wooden blocks slide across each other differently depending on whether they are covered with sandpaper or with grease, and the surfaces matter when describing their interaction, yet the blocks may still be treated as objects, because the friction force one exerts on the other does not depend on the size or shape of the blocks, and the area in contact does not change the force of friction. The property you need is a coefficient, and a coefficient is a number attached to a pair of surfaces rather than a structure inside either block. See coefficient of friction.

Where it costs a mark

Each of these is a scoring event rather than a philosophical point.

Drawing one free-body diagram for two blocks and then asking for the contact force between them. The force you want is internal to the diagram you drew, so it is not on it. Redraw with one block as the system.
Putting an internal force into $\sum F$ for the whole system. The shove between two carts, the tension in the rope joining them, the friction between a box and the sled carrying it: if both objects are inside your boundary, that force cancels with its third law partner and contributes zero. Including it inflates the acceleration.
Changing the system halfway through a problem without saying so. Free-response scoring follows your stated system. If your first line says "the system is the block and the cart" and your third line uses the tension between them as an external force, the two lines contradict each other and the second one is what is marked.
Assuming momentum conservation without naming the system. EK 4.3.B.2 makes the conservation conditional on the net external force on the selected system being zero. No selected system, no conditional, no credit for the reasoning step.
Using the object model where rotation matters. A wheel, a rod pivoting about one end, a see-saw: the object model has no distance from an axis, so it cannot produce a torque. EK 2.1.A.5 is the general warning: the internal structure of a system affects the analysis of that system.
Treating a person or a car as unable to be an object because it obviously has parts. It can. The appendix says an object can be anything, because what the object is is not important to the analysis. A cow, the Earth, a car and a pencil are all in its list.
Forgetting that the system's own parts may behave differently from the system. EK 2.1.A.4 says individual objects within a chosen system may behave differently from each other as well as from the system as a whole. The center of mass of an exploding firework travels a smooth parabola while not one fragment does.

What the CED asks, and what the exam will not do

AP Physics 1 Topic 2.1, Systems and Center of Mass, carries the object and system content, and its first learning objective is LO 2.1.A, describe the properties and interactions of a system. Its six essential knowledge statements are worth having in one place, because between them they are the whole model:

2.1.A.1 System properties are determined by the interactions between objects within the system.
2.1.A.2 If the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object.
2.1.A.3 Systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass.
2.1.A.4 Individual objects within a chosen system may behave differently from each other as well as from the system as a whole.
2.1.A.5 The internal structure of a system affects the analysis of that system.
2.1.A.6 As variables external to a system are changed, the system's substructure may change.

LO 2.1.B moves to the center of mass, and the Topic 2.1 boundary statement limits the arithmetic: AP Physics 1 only expects students to calculate the center of mass for systems of five or fewer particles arranged in a two-dimensional configuration, or for systems that are highly symmetrical. The equation sheet prints $\vec{x}_{\text{cm}} = \frac{\sum m_i \vec{x}_i}{\sum m_i}$ and, in the momentum group, $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}$ . Neither the word object nor the word system has an equation of its own, which is the tell that they are modeling decisions rather than quantities. The full sheet is here.

And one reassurance, in the appendix's own words: the AP Physics Exam will not directly assess student understanding of physics vocabulary, and students will not be asked to identify the correct definition of acceleration, or the difference between a system and an object. You will not be asked to recite these definitions. You will be asked to pick a system, state it, and stay consistent with it, which is the harder version of the same thing.

Unit 2 is worth 18 to 23 percent of the AP Physics 1 multiple-choice section. Topic 2.1 Systems and Center of Mass has the full framing.

Two blocks, one push: the system answers one question and the object answers the other

Blocks A (3.0 kg) and B (5.0 kg) sit in contact on a frictionless horizontal floor. A constant horizontal force of 12 N is applied to A, pushing it against B. Find the acceleration of the pair, then find the magnitude of the force A exerts on B, and say which model produced each number.

Declare the convention: positive $x$ is the direction of the applied push. Both blocks stay in contact, so both have the same acceleration.
Choose the system first. Take A and B together as one system, modeled as a single object at its center of mass, which EK 2.1.A.2 permits because the internal detail is not needed for this question.
List the external horizontal forces on that system. There is exactly one: the 12 N push. The forces A and B exert on each other are internal, and EK 2.3.A.2 says internal forces do not influence the motion of the system's center of mass.
Apply the second law to the system: $a = F_{\text{net}} / m_{\text{sys}} = 12\ \text{N} / (3.0 + 5.0)\ \text{kg} = 12/8.0 = 1.5\ \text{m/s}^2$ in the $+x$ direction.
Now redraw for the second question. The contact force is invisible in the system model, so take block B alone as the system. Horizontally, the only force on B is the push from A.
Apply the second law to B: $F_{\text{A on B}} = m_B a = (5.0\ \text{kg})(1.5\ \text{m/s}^2) = 7.5\ \text{N}$ in the $+x$ direction.
Check with the other block. On A, $12\ \text{N} - F_{\text{B on A}} = m_A a = (3.0)(1.5) = 4.5\ \text{N}$ , so $F_{\text{B on A}} = 12 - 4.5 = 7.5\ \text{N}$ . Equal in magnitude to $F_{\text{A on B}}$ , as Newton's third law requires, and the two agree to the digit.

The pair accelerates at $1.5\ \text{m/s}^2$ , and A pushes B with 7.5 N. The first number came from the system model in one line; the second needed the object model and a second diagram. Neither model is more correct. The system model deleted the contact force on purpose, which is exactly why it was faster, and the price of that speed is that you have to redraw the boundary when the deleted quantity is what the question wants.

Walking to the far end of a boat: the system model answers what no object model can

A 60 kg person stands at one end of a 40 kg boat of length 4.0 m, both at rest on water so still that horizontal resistance is negligible. The person walks to the other end of the boat. How far does the boat move relative to the water, and how far does the person move relative to the water?

Declare the convention: positive $x$ points in the direction the person walks, and all displacements are measured relative to the water.
Choose the system: person plus boat. Horizontally there is no external force on that system, because the only horizontal push is the friction between the person's feet and the boat, and both are inside the boundary.
Apply EK 4.3.A.1.ii: the velocity of a system's center of mass is constant in the absence of a net external force. Everything started at rest, so $v_{cm} = 0$ throughout, and the center of mass never moves.
Write that as a displacement condition. If the person moves $+d_p$ and the boat moves $-d_b$ , then $m_p d_p = m_b d_b$ , that is $(60)(d_p) = (40)(d_b)$ .
Add the geometry. Walking from one end to the other means the person moves 4.0 m relative to the boat, so $d_p + d_b = 4.0\ \text{m}$ .
Solve. Substituting $d_b = 4.0 - d_p$ into $60 d_p = 40(4.0 - d_p)$ gives $60 d_p + 40 d_p = 160$ , so $d_p = 160/100 = 1.6\ \text{m}$ and $d_b = 4.0 - 1.6 = 2.4\ \text{m}$ .
Check the balance: $(60\ \text{kg})(1.6\ \text{m}) = 96\ \text{kg}\cdot\text{m}$ and $(40\ \text{kg})(2.4\ \text{m}) = 96\ \text{kg}\cdot\text{m}$ . Equal, so the center of mass held still, as required.
Sanity check the split. The lighter part moves further, and 40 is two thirds of 60, so the boat's 2.4 m should be one and a half times the person's 1.6 m. It is.

The person moves 1.6 m forward and the boat slides 2.4 m backward, relative to the water. Model the whole thing as one object and the correct answer is that nothing happens at all, because the center of mass does not move, and that answer is true and useless. The interesting numbers live in the parts, and EK 2.1.A.4 is the licence to look for them: individual objects within a chosen system may behave differently from the system as a whole.

A rolling wheel: where the object model finally breaks

A wheel of radius 0.35 m rolls without slipping along level ground, and its center moves at a constant 4.2 m/s. Using the object model, how fast is the top of the wheel moving? Using the rigid-system model, how fast are the topmost and bottommost points moving?

The object model first, honestly. An object has no size, so it has no top and no bottom. Every part of it is the single point at the center of mass, and the model's only possible answer is 4.2 m/s for the whole thing.
State why that is not good enough. The question asks about two different locations on the same body, and the appendix's definition of an object is a physical thing where the internal structure and properties of the thing are ignored. Locations are internal structure.
Switch to the rigid-system model, the appendix's third definition: a system that does not change shape, but different points within the system may move in different directions and with different speeds.
Get the angular speed from the rolling condition. Rolling without slipping gives $v_{cm} = r\omega$ , so $\omega = v_{cm}/r = (4.2\ \text{m/s})/(0.35\ \text{m}) = 12\ \text{rad/s}$ .
Rim speed relative to the center: $r\omega = (0.35\ \text{m})(12\ \text{rad/s}) = 4.2\ \text{m/s}$ , which is the same as the center's speed. That equality is the rolling condition, not a coincidence.
Top of the wheel. The rotation carries it forward relative to the center, so the two add: $4.2 + 4.2 = 8.4\ \text{m/s}$ forward.
Bottom of the wheel. The rotation carries it backward relative to the center, so the two subtract: $4.2 - 4.2 = 0\ \text{m/s}$ . The contact point is instantaneously at rest, which is what rolling without slipping means.
Check against the definition of rigid. The distance between the top point and the bottom point is 0.70 m at every instant, so the shape never changed, exactly as a rigid system requires.

The object model says 4.2 m/s and can say nothing else. The rigid-system model says the top moves at 8.4 m/s, twice the speed of the center, and the bottom is momentarily at 0 m/s. Same wheel, same instant, and a factor of two between two answers depending on which model you brought. This is the appendix's own criterion in action: if the rotation of the wheel is relevant to the analysis of the situation, it cannot be modeled as an object. Rolling is the topic that owns it.

Frequently asked questions

What is the difference between an object and a system in AP Physics?

An object is a physical thing whose internal structure and properties are ignored, so it has mass and position but no size, no shape and no parts. A system is a collection of objects that are analyzed together. Both definitions come from College Board's own appendix, Vocabulary and Definitions of Important Ideas in AP Physics, printed at the back of all four Course and Exam Descriptions. The key point is that neither label is a property of the thing itself. A truck can be an object when you are driving it and a system when you are asking how the load inside it shifts. You choose, and the appendix says there is no single right or wrong way to group objects, only groupings that make the analysis simpler or harder.

Can a system be treated as a single object?

Yes, and it is the most useful move in mechanics. AP Physics 1 EK 2.1.A.2 says that if the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object. EK 2.1.B.3 places that single object at the system's center of mass. College Board's own example is a box of cereal: a cardboard box, a plastic bag and hundreds of pieces, but since the motion of the individual pieces does not matter to someone carrying it, the whole box, bag and cereal is one object. The moment the internal behavior does start to matter, the collapse is no longer allowed.

How do I choose the system in a physics problem?

Choose the boundary that turns the forces you do not want to compute into internal forces. If the question asks for the acceleration of two connected blocks, enclose both, and the tension between them disappears from your equations. If the question asks for that tension, enclose one block, and the tension becomes an external force you can solve for. The rule of thumb is that a force between two things inside your boundary cancels with its Newton's third law partner and contributes nothing to the net force on the whole, per EK 2.3.A.2, while a force crossing the boundary drives the system. Write your choice down before your first equation, because free-response scoring follows the system you stated.

What is a rigid system, and how is it different from an object?

College Board defines a rigid system as a system that does not change shape, but where different points within the system may move in different directions and with different speeds. An object has no size at all, so it has no different points, which is why it cannot describe a rotating wheel: a spot near the rim moves faster than a spot near the hub at the same instant. The appendix uses exactly that wheel to make the point, and adds that for a rigid system the location at which a force is exerted matters, so pushing the top of a wheel does something different from pushing at the axle. That distinction is what makes torque possible, and it is why the AP rotational units talk about rigid systems rather than objects.

Why do internal forces cancel but external forces do not?

Because an internal force has both of its Newton's third law partners inside the boundary you drew. Newton's third law says the two are equal in magnitude and opposite in direction, so when you add up every force on the system they sum to zero, no matter how large each one is. AP Physics 1 EK 2.3.A.2 states the consequence: interactions between objects within a system, the internal forces, do not influence the motion of a system's center of mass. An external force has only one of its partners inside; the other acts on something outside the boundary and is simply not in your sum, so nothing cancels it. This is also why EK 2.2.A.1.ii can say that an object or system cannot exert a net force on itself.

Is a person an object or a system?

Either, depending on the question. If you are working out how fast a runner accelerates from a push off the ground, model the runner as an object: mass and position are all you need. If you are asking how a diver changes rotation speed by tucking, the object model cannot help, because it has no shape to change, and you need a system whose parts can move relative to one another. College Board's appendix is explicit that an object can be anything, listing a cow, the Earth, a car and a pencil, because what the object is is not important to the analysis. The properties that matter, such as mass, a coefficient of friction and a speed, are all the object model keeps.

Does the object model ignore friction, since friction depends on surfaces?

No, and College Board's appendix addresses this directly because it looks like a contradiction. Friction is the interaction of two objects in physical contact, and the amount of friction is tied to their surfaces: two wooden blocks slide differently if they are covered with sandpaper rather than grease. But the blocks may still be modeled as objects, because the friction force one exerts on the other does not depend on the size or shape of the blocks, and the contact area does not change the force of friction. The surface information is carried by a coefficient attached to the pair of surfaces, not by any structure inside either block, so nothing the object model deleted was needed.