AP Physics 1 Unit 2 Notes: Understanding Motion Through Forces and Systems

Systems and Center of Mass

What a “system” is (and why you should care)

A system is the collection of objects you choose to analyze together. That choice is not just a formality—it can make problems dramatically easier. In AP Physics 1, you often have multiple objects interacting (blocks connected by a string, a person pushing a cart, two carts colliding, etc.). Instead of writing Newton’s laws separately for each object every time, you can often treat several objects as one system.

The key payoff is this idea:

• Forces between objects inside your chosen system are internal forces.
• Forces exerted on the system by objects outside the system are external forces.

Internal forces often come in Newton’s third-law pairs and can cancel in a system-level equation—so the system approach can reduce the number of unknown forces you must deal with.

Internal vs. external forces (how it works)

Suppose two blocks push on each other while sliding on a table. If your system is “both blocks,” then the contact force between them is internal. If your system is “just the left block,” that same contact force is external (because it’s coming from the other block, which is now outside your system).

This means the same physical force can be internal or external depending on your system choice. A lot of mistakes in Unit 2 come from forgetting that “internal/external” is not an inherent label—it’s defined by your boundary.

Center of mass: the system’s “balance point”

The center of mass is the weighted average position of mass in a system. It’s the point that moves as if all the system’s mass were concentrated there (for translational motion).

For discrete point masses along a line (1D), the center-of-mass position is:

x_{cm} = \frac{\sum m_i x_i}{\sum m_i}

• m_i is the mass of object i
• x_i is its position
• \sum m_i is the total mass

In 2D, you apply the same idea separately for each coordinate:

x_{cm} = \frac{\sum m_i x_i}{\sum m_i}

y_{cm} = \frac{\sum m_i y_i}{\sum m_i}

Why it matters: many AP problems ask about a system’s motion without needing every internal detail. The center of mass lets you describe the system’s translational behavior cleanly.

The center of mass responds only to net external force

The deepest connection between “systems” and Newton’s laws is this system-level version of Newton’s second law:

\sum F_{ext} = M a_{cm}

• \sum F_{ext} is the net external force on the system
• M is the total mass of the system
• a_{cm} is the acceleration of the center of mass

This is powerful because internal forces do not appear in \sum F_{ext}. They can affect how the objects move relative to each other, but they cannot change the motion of the center of mass unless there is an external net force.

Common interpretation that helps

• If \sum F_{ext} = 0, then a_{cm} = 0. The center of mass moves at constant velocity (possibly zero).
• If external force is applied (like a pull or push), the whole system’s center of mass accelerates accordingly.

Worked example 1: center of mass on a line

Two carts on a track:

• m_1 = 2\ \text{kg} at x_1 = 1\ \text{m}
• m_2 = 6\ \text{kg} at x_2 = 5\ \text{m}

Find x_{cm}.

x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

x_{cm} = \frac{(2)(1) + (6)(5)}{2 + 6}

x_{cm} = \frac{2 + 30}{8} = 4\ \text{m}

Notice the center of mass is much closer to the 6 kg cart, as expected.

Worked example 2: using the system equation for connected blocks

Two blocks are connected by a light string on a frictionless surface. A horizontal force F pulls the left block. If you choose the system as “both blocks together,” then the tension in the string is internal.

Let masses be m_1 and m_2 and assume the only external horizontal force on the system is F.

\sum F_{ext,x} = F = (m_1 + m_2)a_{cm}

So the acceleration of the system is:

a = \frac{F}{m_1 + m_2}

You did not need the tension to find the acceleration. Tension matters if you later want the force on each individual block—but for the system’s acceleration, it’s unnecessary.

Exam Focus

• Typical question patterns:
  • “Choose a system boundary” and identify which forces are internal vs. external.
  • Compute x_{cm} or a_{cm} for multiple masses.
  • Use \sum F_{ext} = M a_{cm} to avoid solving for internal forces like tension.
• Common mistakes:
  • Treating an internal force (like tension between two chosen system objects) as if it should appear in \sum F_{ext}.
  • Averaging positions without mass-weighting (using \frac{x_1 + x_2}{2} when masses differ).
  • Forgetting that a_{cm} is controlled only by external net force.

Forces and Free-Body Diagrams

What a force is

A force is an interaction that can change an object’s motion (cause acceleration) or deform it. In AP Physics 1, forces are vectors: they have magnitude and direction. You do not “have” forces unless there is an interaction—every force comes from another object (even if that object is the Earth).

Why free-body diagrams are essential

A free-body diagram (FBD) is a simplified drawing that shows all external forces acting on a chosen object (or system). It’s the bridge between a physical situation and Newton’s laws.

A good FBD matters because Newton’s second law is not “one equation for the whole picture.” It must be applied to a specific object or system. Your FBD defines what forces belong in your equation.

How to build an FBD correctly (step by step)

1. Choose the object/system you’re analyzing and mentally “isolate” it.
2. List all interactions with other objects: contact forces, gravity, tension, springs, friction, etc.
3. Draw only forces acting on your chosen object/system. Do not draw forces the object exerts on others.
4. Choose axes that simplify components (often along and perpendicular to motion or a surface).
5. Resolve forces into components only if needed for your equations.

A common student trap is drawing what you think should be there (like “a force in the direction of motion”) instead of forces from actual interactions.

Common forces in AP Physics 1 (what they mean)

Weight near Earth

For an object of mass m near Earth:

F_g = mg

• g is the gravitational field magnitude near Earth (often taken as 9.8\ \text{m/s}^2 on AP problems unless stated otherwise)
• Weight is not “mass.” Mass is inertia; weight is a force.

Normal force is not always mg

The normal force adjusts to satisfy Newton’s laws in the direction perpendicular to the surface. On a level surface with no vertical acceleration, it often equals weight, but that’s a special case, not a rule.

Friction: static vs kinetic

• Static friction prevents slipping up to a maximum value:

f_s \le \mu_s N

• Kinetic friction acts during slipping:

f_k = \mu_k N

Key idea: static friction is “whatever it needs to be” (up to a max) to prevent relative motion. Many students incorrectly set f_s = \mu_s N every time.

Spring force (if springs appear)

Hooke’s law (magnitude):

F_s = kx

• k is spring constant
• x is displacement from equilibrium
  Direction is opposite the displacement.

Worked example 1: block on an incline (FBD logic)

A block rests on a frictionless incline at angle \theta. Forces on the block:

• Weight mg downward
• Normal force N perpendicular to incline

Choose axes: x parallel to incline, y perpendicular.

Resolve weight into components:

F_{g,\parallel} = mg\sin\theta

F_{g,\perp} = mg\cos\theta

If the block accelerates down the incline, Newton’s second law along the incline uses only the parallel component.

A frequent error: drawing an extra “force down the ramp” separate from gravity’s component. The component is not a new force—it’s part of the same gravitational force.

Worked example 2: block pulled on a rough surface

A block of mass m is pulled right with force F at constant speed on a horizontal surface with kinetic friction coefficient \mu_k.

Constant speed means acceleration is zero, so net force is zero in each direction.

Vertical forces: N up and mg down. With no vertical acceleration:

N = mg

Kinetic friction magnitude:

f_k = \mu_k N = \mu_k mg

Horizontal equilibrium (right positive):

F - f_k = 0

So:

F = \mu_k mg

This illustrates a big idea: constant speed does not mean no forces—it means forces balance.

Exam Focus

• Typical question patterns:
  • Draw an FBD for an object (or for each object) and write Newton’s 2nd law in components.
  • Determine whether friction is static or kinetic and whether you can use f_s = \mu_s N or must use f_s \le \mu_s N.
  • Identify correct tension/normal directions for connected objects or objects on inclines.
• Common mistakes:
  • Including forces the object exerts on others (wrong object) in the FBD.
  • Setting N = mg automatically even on inclines or when there’s vertical acceleration.
  • Treating “motion direction” as a force (inventing a “force of motion”).

Newton’s Third Law

What Newton’s third law says

Newton’s third law: when two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.

If object A exerts a force on object B, then object B exerts a force on object A:

F_{A\to B} = -F_{B\to A}

This is not an approximation—it’s a fundamental symmetry of interactions.

Why it matters

Newton’s third law is the reason internal forces cancel in a system analysis, and it prevents a common misconception: the more massive object does not exert a bigger force just because it’s more massive. Forces in an interaction pair are always equal and opposite.

So why do objects accelerate differently? Because acceleration depends on mass through Newton’s second law:

a = \frac{F}{m}

Same force, different masses, different accelerations.

How to recognize third-law pairs

A third-law pair must satisfy all of these:

• Same interaction (contact, gravity, tension, etc.)
• Same magnitude, opposite direction
• Act on different objects

A very common exam trap is confusing “balanced forces” on one object with a third-law pair. Balanced forces are two forces on the same object that sum to zero. Third-law pairs are on different objects.

Examples of third-law pairs (conceptual)

1. Book on a table

   • Earth pulls book down (weight on the book).
   • Book pulls Earth up (gravitational force on Earth).
     These are a third-law pair.

   The normal force on the book and the weight on the book are not a third-law pair because they act on the same object.

2. Pushing a wall

   • You push the wall.
   • The wall pushes you back with equal magnitude.

3. Tension in a rope

   • Rope pulls on block.
   • Block pulls on rope.

Worked example: two skaters pushing off

Two skaters push off each other on frictionless ice. Skater 1 has mass m_1 = 50\ \text{kg} and skater 2 has mass m_2 = 75\ \text{kg}. During the push, the interaction force magnitudes are equal.

Let the magnitude of the force each exerts be F. Then accelerations during the push are:

a_1 = \frac{F}{m_1}

a_2 = \frac{F}{m_2}

Since m_1 < m_2, skater 1 has the larger acceleration. That does not mean skater 1 experienced a larger force—it means the same force produced a larger acceleration due to smaller mass.

Exam Focus

• Typical question patterns:
  • Identify the correct third-law pair for a given force (especially weight, normal, tension, friction).
  • Concept questions asking whether forces are “equal” during interactions (trucks vs cars, etc.).
  • Use third-law reasoning to explain system internal-force cancellation.
• Common mistakes:
  • Claiming the heavier object exerts a greater force in an interaction.
  • Pairing forces acting on the same object (like N and mg) as third-law pairs.
  • Mixing up “equal and opposite” with “canceling” (they only cancel if they act on the same object, which third-law pairs do not).

Newton’s First Law

What Newton’s first law says

Newton’s first law: if the net external force on an object is zero, the object maintains constant velocity (which includes staying at rest).

In equation form, the condition is:

\sum F = 0 \Rightarrow a = 0

This law is really the definition of inertia: the tendency of an object to resist changes in its velocity. Mass is the quantitative measure of inertia.

Why it matters

Newton’s first law helps you interpret situations where nothing seems to be “making” an object move. If a hockey puck glides at near-constant speed, it’s not because a force keeps it moving; it’s because there’s very little net force opposing its motion.

It also teaches the key mindset shift away from Aristotle’s idea (“motion requires force”) toward Newtonian thinking (“change in motion requires net force”).

Equilibrium: translational (and how you use it)

When \sum F = 0, the object is in translational equilibrium and acceleration is zero. You treat this with Newton’s second law set to zero in each direction:

\sum F_x = 0

\sum F_y = 0

Even though this section focuses on Newton’s first law conceptually, exam problems often use it through these component equations.

Worked example 1: hanging mass at rest

A mass m hangs from a vertical rope and is at rest.

Forces on the mass:

• Tension T upward
• Weight mg downward

Net force is zero:

\sum F_y = T - mg = 0

So:

T = mg

This is equilibrium, not a third-law statement. The third-law pair of the rope’s tension on the mass is the mass’s pull on the rope, not the weight.

Worked example 2: elevator moving at constant velocity

An elevator moves upward at constant speed. Acceleration is zero, so net force is zero.

If the cable tension is T upward and weight is mg downward:

T - mg = 0

T = mg

Many students incorrectly think “moving up” implies T > mg. That would be true only if the elevator were speeding up upward.

Common misconception: “no motion means no forces”

If something is at rest, it’s tempting to say there are no forces. In reality, it often means forces are present but balanced. A book at rest has both mg and N acting.

Exam Focus

• Typical question patterns:
  • Determine unknown forces in equilibrium (objects at rest or moving at constant velocity).
  • Concept questions contrasting constant velocity with constant acceleration.
  • Use equilibrium to relate forces (like T = mg in a static hang).
• Common mistakes:
  • Assuming motion implies net force in the direction of motion (confusing velocity with acceleration).
  • Forgetting to set up separate component equations when net force is zero.
  • Treating equilibrium as “no forces” rather than “net force zero.”

Newton’s Second Law

What Newton’s second law says

Newton’s second law connects forces to acceleration. The net force on an object equals its mass times its acceleration:

\sum F = ma

• \sum F is the vector sum of all external forces on the object
• m is the object’s inertial mass
• a is the acceleration vector

This is the central tool in Unit 2 because it turns a physical scenario (forces) into a quantitative prediction (acceleration or a required force).

Why it matters

Newton’s second law explains essentially all non-circular translational dynamics problems you’ll see in AP Physics 1:

• Why friction reduces acceleration
• Why heavier objects accelerate less for the same applied force
• How tension relates to acceleration in connected systems
• How to find a required force to produce a desired motion

It also creates a consistent problem-solving method: FBD → component equations → solve.

How to use it correctly (the mechanism)

1. Choose the object (or system) and draw the FBD.
2. Pick axes. If there is an incline, aligning an axis with the incline often simplifies the math.
3. Write Newton’s second law in components:

\sum F_x = ma_x

\sum F_y = ma_y

4. Substitute force models (like F_g = mg, f_k = \mu_k N).
5. Solve algebraically, checking sign conventions and directions.

A crucial mindset: \sum F is the net force, not “the biggest force” or “the applied force.” If multiple forces act, you must add them vectorially.

Units and interpretation

From \sum F = ma, force units are:

1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2

If the net force is zero, acceleration is zero (consistent with Newton’s first law).

Example 1: pushing a crate with friction (acceleration)

A crate of mass m = 10\ \text{kg} is pushed horizontally with F = 50\ \text{N} on a level surface. The kinetic friction coefficient is \mu_k = 0.20. Find the acceleration.

Step 1: Forces

• Horizontal: applied force F right, kinetic friction f_k left
• Vertical: normal N up, weight mg down

Step 2: Vertical equation (no vertical acceleration)

\sum F_y = N - mg = 0

N = mg

Step 3: Friction

f_k = \mu_k N = \mu_k mg

Using g = 9.8\ \text{m/s}^2:

f_k = (0.20)(10)(9.8) = 19.6\ \text{N}

Step 4: Horizontal equation

\sum F_x = F - f_k = ma

50 - 19.6 = 10a

30.4 = 10a

a = 3.04\ \text{m/s}^2

A common mistake here is using f_k = \mu_k mg without first confirming N = mg. On a level surface with no vertical acceleration, it’s valid; on an incline or with vertical pulling components, it may not be.

Example 2: two-block system (why “system” choice changes difficulty)

Two blocks on a frictionless table are connected by a light string. m_1 = 2\ \text{kg} and m_2 = 3\ \text{kg}. A horizontal force F = 20\ \text{N} pulls on m_1 to the right. Find the acceleration and the tension.

Part A: Acceleration using the system approach
Choose the system as both blocks together. The tension is internal, so it does not appear in the net external force on the system.

\sum F_{ext,x} = F = (m_1 + m_2)a

20 = (2 + 3)a

a = 4\ \text{m/s}^2

Part B: Tension by isolating one block
Now isolate m_2. The only horizontal force on m_2 is the tension T to the right.

\sum F_x = T = m_2 a

T = (3)(4) = 12\ \text{N}

This two-step method is common on AP: use a system equation to get a, then use one-object Newton’s second law to get internal forces.

Static friction nuance (a frequent AP trap)

If an object does not slip, friction is static, and its value is whatever is needed (up to a maximum) to satisfy Newton’s second law.

For instance, a box on a rough floor is pulled gently and does not move. If the pull is 10\ \text{N} and the box remains at rest, then static friction is 10\ \text{N} (opposite direction), provided the maximum static friction is at least that large. You should not jump to f_s = \mu_s N unless you are at the threshold of slipping.

Exam Focus

• Typical question patterns:
  • Write Newton’s 2nd law in components from an FBD and solve for a, N, T, or friction.
  • Use the “system first” trick for connected objects to avoid internal forces when finding acceleration.
  • Determine whether static friction is at maximum or just balancing another force.
• Common mistakes:
  • Using \sum F = ma with a single force instead of the net force (forgetting to subtract opposing forces).
  • Mixing up action-reaction pairs as if they act on the same object in the same equation.
  • Automatically using f_s = \mu_s N even when the object is not on the verge of slipping.