EduNinja

2.1 Systems and Center of Mass

Choose the system

A system is the object or collection of objects chosen for analysis. Draw an imaginary boundary around it: objects inside are the system's constituent parts; everything outside is the environment. The system's properties depend on how its parts interact.

Read the boundary

Ask about the model	Inside the boundary	Across the boundary
What is tracked?	Parts and their internal interactions	Transfers between system and environment
What may change?	Parts may move or behave differently from one another	Energy or mass may enter or leave
Can details be ignored?	Yes, if internal structure does not affect the macroscopic behavior	No, if an external change alters the system's substructure or behavior

Decide how much detail matters

Example: choose a cart plus the blocks fixed on it as one system. If the blocks remain fixed and their arrangement does not matter for the motion being studied, model the whole system as one object. If the blocks slide relative to the cart, the internal structure matters and the parts must be analyzed.

Do not assume isolation

A system is not automatically isolated, rigid, or uniform. The boundary is a modeling choice. State what is inside it, identify relevant interactions, and reconsider the single-object model when external conditions change the system's substructure.

Interpret the model

The center of mass is the position at which a system can be represented as one object for translational analysis. It is a mass-weighted location: larger masses pull the center of mass closer to themselves. In the formula below, $m_i$ is each particle's mass, $\vec{x}_i$ is its signed position from the chosen origin, and $\sum_i m_i$ is the total mass.

Weight each position by mass

$\vec{x}_{cm}=\frac{\sum_i m_i\vec{x}_i}{\sum_i m_i}$

Calculate and interpret

Hypothetical example: a $2.0\,\mathrm{kg}$ particle is at $x=0\,\mathrm{m}$ and a $3.0\,\mathrm{kg}$ particle is at $x=4.0\,\mathrm{m}$ .

$x_{cm}=\dfrac{(2.0\,\mathrm{kg})(0\,\mathrm{m})+(3.0\,\mathrm{kg})(4.0\,\mathrm{m})}{2.0\,\mathrm{kg}+3.0\,\mathrm{kg}}=2.4\,\mathrm{m}.$

The result lies between the particles and closer to the heavier $3.0\,\mathrm{kg}$ particle, as expected.

Use symmetry and respect scope

For a symmetrical mass distribution, the center of mass lies on every applicable line of symmetry; it need not coincide with material. It is not generally the geometric midpoint. AP Physics 1 calculations are limited to five or fewer particles in a two-dimensional arrangement or to highly symmetrical systems.

2.2 Forces and Free-Body Diagrams

Name source and target

A force is a vector that describes an interaction: one object or system exerts the force on another. Write $\vec F_{A\,on\,B}$ to name both the source $A$ and the target $B$ ; its direction tells how $A$ pushes or pulls $B$ .

Classify the interaction

Interaction	Objects involved	What the force means
Contact	Book and table touching	The table exerts an upward contact force on the book
Noncontact	Earth and book	Earth exerts a downward gravitational force on the book
Internal to a chosen system	Two parts inside the boundary	Parts exert forces on each other, but the system does not exert a net force on itself

Understand contact forces

At the macroscopic scale, normal force, friction and tension are contact interactions. Their microscopic origin is electromagnetic interaction between atoms, even when surfaces appear rigid.

Reject force-like inventions

A force is not something an isolated object 'has,' and motion does not require a force in the direction of motion. To validate a claimed force, identify the second interacting object and state who exerts what force on whom.

Isolate one system

A free-body diagram (FBD) isolates one object or system and shows every external force exerted on it. Replace the chosen system by a dot at its center of mass; the completed diagram is the bridge from the physical situation to force equations.

Draw only external forces

Step	AP-compliant action
1. Isolate	Name exactly one object or system
2. Inventory	For each interaction crossing its boundary, identify the force on the system
3. Draw	Use one labeled straight arrow per force, starting at the center-of-mass dot
4. Separate	Draw same-direction forces side by side, never overlapping
5. Choose axes	Align an axis with acceleration when useful; on an incline, choose an axis parallel to the surface

Choose helpful axes

Example: for a block sliding on a rough incline, the FBD contains the gravitational force by Earth, the normal force by the surface, and—when applicable—the friction force by the surface. Choose axes parallel and perpendicular to the incline, then resolve forces algebraically after the FBD is drawn.

Keep components out of the FBD

Do not draw velocity, acceleration, the path, or a force exerted by the chosen system on something else. AP Physics 1 expects individual forces—not their components—on the FBD. Components belong in the later algebra, not as extra arrows on the diagram.

2.3 Newton’s Third Law

Name the reciprocal forces

When objects $A$ and $B$ interact, each exerts a force on the other at the same time. The forces have equal magnitude and opposite direction: $\vec F_{A\,on\,B}=-\vec F_{B\,on\,A}$ . They are a pair because they come from the same interaction.

Verify a true pair

Pair test	Must be true
Interaction	Both forces describe the same interaction
Objects	One force acts on $A$ ; the other acts on $B$
Magnitude	$\|\vec F_{A\,on\,B}\|=\|\vec F_{B\,on\,A}\|$
Direction	The force vectors point oppositely
Timing	Both forces exist simultaneously

Separate object and system views

Example: a book rests on a table. The table's upward force on the book pairs with the book's downward force on the table. The book's weight is not that partner; it pairs with the book's gravitational force on Earth. Third-law partners do not cancel on the book because each pair member acts on a different object. If both objects are inside one chosen system, the internal pair cancels in the system force sum and does not change the system's center-of-mass motion.

Choose the string model

Model	Consequence for tension
Ideal string	Negligible mass, does not stretch, and has the same tension at all points
String or chain with mass	Tension may vary along its length; for a hanging chain it is greater nearer the top
Ideal pulley	Negligible mass and negligible axle friction; it rotates about an axle through its center of mass

Avoid false pairs

Equal-and-opposite arrows on the same object are not a third-law pair. They may instead be balanced forces from different interactions. AP Physics 1 treats tension in a massive string qualitatively, and its noncontact interactions at a distance are limited here to gravity.

2.4 Newton’s First Law

Add forces as vectors

The net force on a system is the vector sum of every external force exerted on it. Direction matters: forces pointing oppositely contribute opposite vector components.

Recognize equilibrium

$\sum_i \vec F_i=0$

Test each direction

Component test	What it tells you
$\sum F_x=0$ and $\sum F_y=0$	The net-force vector is zero, so the system is in translational equilibrium
$\sum F_x\ne0$ , $\sum F_y=0$	Only the $x$ component of velocity changes
$\sum F_x=0$ , $\sum F_y\ne0$	Only the $y$ component of velocity changes

Apply the test

Example: a crate moves right while the floor's upward normal force balances Earth's downward gravitational force. If a rightward pull is larger than the leftward resistive force, the vertical forces are balanced but the horizontal forces are not. The crate's vertical velocity stays constant while its horizontal velocity changes.

Interpret the result

Zero net force does not mean zero velocity or no forces. It means velocity is constant: the system may remain at rest or keep moving at constant speed in a straight line. This statement is verified in an inertial reference frame; a frame that accelerates relative to an inertial frame is not inertial.

2.5 Newton’s Second Law

Recognize unbalanced forces

Forces are unbalanced when their vector sum is not zero. A system's center-of-mass velocity changes only when a nonzero net external force acts on the system.

Connect force, mass, and acceleration

$\vec a_{\text{sys}}=\frac{\sum_i \vec F_i}{m_{\text{sys}}}=\frac{\vec F_{\text{net}}}{m_{\text{sys}}}$

Calculate and interpret

Worked example: choose right as positive. A $4.0\,\text{kg}$ cart has a $12\,\text{N}$ force right and a $4.0\,\text{N}$ force left.

$F_{\text{net}}=+12-4.0=+8.0\,\text{N}$

$a=\dfrac{F_{\text{net}}}{m}=\dfrac{+8.0\,\text{N}}{4.0\,\text{kg}}=+2.0\,\text{m/s}^2$ .

The positive result means the cart's velocity changes by $2.0\,\text{m/s}$ toward the right each second.

Predict proportional changes

Change	Acceleration consequence
Double $\|\vec F_{\text{net}}\|$ at fixed mass	Double $\|\vec a\|$
Double $m_{\text{sys}}$ at fixed net force	Halve $\|\vec a\|$
Reverse $\vec F_{\text{net}}$	Reverse $\vec a$

Separate acceleration from velocity

Acceleration points with the net force, but velocity need not. If the cart is initially moving left while the net force points right, it first slows down. For a multi-object system, sum external forces and use the total system mass to describe the center-of-mass acceleration; internal forces do not enter that external-force sum.

2.6 Gravitational Force

Locate the interaction

Every pair of masses attracts. Each gravitational force acts along the line joining the systems' centers of mass, toward the other system.

Calculate force magnitude

$|\vec F_g|=G\frac{m_1m_2}{r^2}$

Predict proportional changes

Change	New gravitational-force magnitude
Double either mass	$2\|\vec F_g\|$
Double both masses	$4\|\vec F_g\|$
Double center-to-center distance $r$	$\|\vec F_g\|/4$

Move from force to field

A gravitational field describes the force effect at each position without requiring a test object to remain there. For source mass $M$ , $|\vec g|=\dfrac{|\vec F_g|}{m}=G\dfrac{M}{r^2}$ . Field strength has units $\text{N/kg}$ ; if gravity is the only force, its numerical value equals free-fall acceleration in $\text{m/s}^2$ .

Distinguish mass and weight

The gravitational force from an astronomical body on a relatively small nearby object is its weight: $\text{Weight}=F_g=mg$ . Mass is measured in kilograms; weight is a force measured in newtons. Here $r$ is center-to-center distance, not the gap between surfaces.

Check whether change is negligible

Gravitational force can be treated as constant over a motion when the change in center-to-center distance is small enough that the resulting change in gravitational force is negligible for the analysis.

Use the near-Earth approximation

$g\approx 10\ \text{N/kg}\approx 10\ \text{m/s}^2\quad\text{near Earth's surface}$

Apply it locally

Example: for an object moving through a classroom-scale height near Earth's surface, its distance from Earth's center changes by a negligible fraction. Use one nearly constant $g$ , so a fixed mass has nearly constant weight $F_g=mg$ throughout that motion.

Know when it fails

This is an approximation, not a universal rule. If the change in center-to-center distance is large enough that $G M/r^2$ changes appreciably, calculate the changing field or force instead of using one constant $g$ .

Name the measured force

Your apparent weight is the magnitude of the normal force exerted on you by a supporting surface. Your gravitational force is $F_g=mg$ ; these are different forces and need not have equal magnitudes.

Write the vertical force sum

$\sum F_y=N-mg=ma_y\quad(\text{upward positive})$

Compare acceleration cases

Vertical contact case	Apparent weight $N$
$a_y>0$	$N>mg$ : you feel heavier
$a_y=0$	$N=mg$
$a_y<0$ while contact remains	$0<N<mg$ : you feel lighter
Free fall	$N=0$ : weightless

Explain weightlessness

Example: you and a freely falling elevator accelerate together under gravity alone. The floor no longer needs to push on you, so $N=0$ and your apparent weight is zero—even though gravity still acts and $F_g=mg$ is not zero.

Apply the boundary

Weightlessness means zero support force, not zero gravity. The equivalence principle says that, using only local observations, an observer in a noninertial frame cannot distinguish apparent-weight effects from those produced by a gravitational field.

Measure resistance to change

Inertial mass measures how strongly an object's motion resists changing during an interaction. For the same net force, a larger inertial mass produces a smaller acceleration.

Measure gravitational interaction

Gravitational mass determines how strongly a system participates in gravitational attraction: it appears in the gravitational-force relationship between masses.

Separate the two roles

Role	Revealed by	Relationship
Inertial mass $m_i$	Response to net force	$\vec F_{\text{net}}=m_i\vec a$
Gravitational mass $m_g$	Strength of gravitational interaction	$\|\vec F_g\|\propto m_g$

Interpret equivalence

Experiments verify that inertial and gravitational mass are equivalent: for a given object, their measured values are equal. The roles remain conceptually distinct—one describes response to force, the other gravitational interaction—so equivalence is an empirical result, not merely a definition.

2.7 Kinetic and Static Friction

Recognize sliding

Kinetic friction acts when two surfaces in contact slide relative to each other. On each surface, it points opposite that surface's motion relative to the other surface.

Calculate the magnitude

$|\vec F_{f,k}|=\mu_k|\vec F_n|$

Interpret each quantity

Quantity	Meaning
$F_n$	Perpendicular contact-force component, directed away from the surface
$\mu_k$	Dimensionless coefficient set by the material pair
$F_{f,k}$	Friction magnitude; direction is chosen separately from relative sliding

Apply the model

Worked example: a $5.0\,\text{kg}$ block slides on a horizontal surface with $\mu_k=0.20$ . With no vertical acceleration, $F_n=mg=(5.0\,\text{kg})(10\,\text{m/s}^2)=50\,\text{N}$ . Therefore $F_{f,k}=\mu_kF_n=(0.20)(50\,\text{N})=10\,\text{N}$ , opposite the block's motion relative to the surface.

Respect the boundary

In this friction model, the force does not depend on the apparent contact area. Also, $F_n$ is not automatically $mg$ ; determine it from forces perpendicular to the surface and the motion in that direction.

Recognize no sliding

Static friction may act when contacting surfaces are not sliding relative to each other. It points in the direction needed to prevent the impending relative motion.

Use the allowed range

$|\vec F_{f,s}|\leq \mu_s|\vec F_n|$

Separate actual from maximum

Static friction self-adjusts from zero up to a maximum: $F_{f,s,\max}=\mu_sF_n$ . Below the threshold, solve for the friction actually required to prevent sliding; use the maximum only at impending slip.

Test the threshold

Threshold example: a surface has $F_n=50\,\text{N}$ and $\mu_s=0.40$ , so $F_{f,s,\max}=(0.40)(50\,\text{N})=20\,\text{N}$ . A $12\,\text{N}$ horizontal push produces $12\,\text{N}$ of opposing static friction and no slip. A requested $25\,\text{N}$ exceeds the $20\,\text{N}$ limit, so static friction cannot prevent sliding.

Switch models after slip

Static friction is not always $\mu_sF_n$ . Once the surfaces slide, switch to kinetic friction. For a given surface pair, $\mu_s$ is typically greater than $\mu_k$ , so the maximum static friction is typically greater than the kinetic-friction magnitude for the same normal force.

2.8 Spring Forces

Define the ideal model

An ideal spring has negligible mass. Its force is proportional to the change in length $\Delta\vec x$ measured from its relaxed length.

Use Hooke's law

$\vec F_s=-k\,\Delta\vec x$

Read the restoring direction

Spring change	Displacement sign	Spring-force direction
Stretched to the right	$\Delta x>0$	$F_s<0$ : left, back toward equilibrium
Compressed to the left	$\Delta x<0$	$F_s>0$ : right, back toward equilibrium
Relaxed	$\Delta x=0$	$F_s=0$

Calculate and interpret

Worked example: take right as positive. A spring with $k=200\,\text{N/m}$ is stretched $\Delta x=+0.030\,\text{m}$ .

$F_s=-k\Delta x=-(200\,\text{N/m})(+0.030\,\text{m})=-6.0\,\text{N}$ .

The spring exerts a $6.0\,\text{N}$ force to the left, toward equilibrium.

Separate sign and magnitude

The minus sign gives direction; force magnitude is $|F_s|=k|\Delta x|$ . The spring constant $k$ has units $\text{N/m}$ and measures stiffness. Hooke's law is the ideal linear model: doubling the displacement doubles the force magnitude within that model.

2.9 Circular Motion

Separate the directions

At any point on the circle:

Velocity $\vec v$ : tangent to the path; gives the instantaneous motion direction.
Centripetal acceleration $\vec a_c$ : toward the center; changes velocity direction.
Tangential acceleration $\vec a_t$ : tangent to the path; changes speed.
Net acceleration: $\vec a=\vec a_c+\vec a_t$ ; changes direction, speed, or both.

Connect speed, radius, and timing

For radius $r$ and tangential speed $v$ , $a_c=\dfrac{v^2}{r}$ . In uniform circular motion, $a_t=0$ even though $a_c\ne0$ . One revolution takes period $T$ and frequency is revolutions per second: $T=\dfrac{1}{f}$ and $T=\dfrac{2\pi r}{v}$ .

Identify the real inward forces

There is no new force called 'centripetal force.' The inward net component of real forces satisfies $\sum F_{\text{inward}}=ma_c$ . Gravity can supply it alone; normal force and static friction components can supply it on a banked path; a tension component supplies it for a conical pendulum.

Apply the loop threshold

Top-of-loop threshold: at the minimum speed that maintains contact, the contact force has just fallen to zero, so gravity alone supplies the inward force: $mg=m\dfrac{v^2}{r}$ . Therefore $v_{\min}=\sqrt{gr}$ . Below this speed, the required inward force would exceed what gravity provides at that point, so contact cannot be maintained.

Respect the course boundary

Constant speed does not mean zero acceleration in a circle because velocity direction keeps changing. For banked curves, AP Physics 1 expects quantitative analysis only when no friction is required for uniform circular motion; friction-required banked curves are analyzed qualitatively.

Choose the inward force

For a satellite of mass $m$ in a circular orbit of radius $R$ around a central mass $M$ , gravitational attraction supplies the entire inward force. Orbit radius is measured from the central body's center of mass.

Derive the period relation

$G\dfrac{Mm}{R^2}=m\dfrac{v^2}{R}$ and $v=\dfrac{2\pi R}{T}$ . Substitution cancels the satellite mass and gives Kepler's third-law form for a circular orbit: $T^2=\dfrac{4\pi^2}{GM}R^3$ .

Predict proportional changes

Comparison	Consequence
Same $M$ : $R$ increases by factor $q$	$T$ increases by $q^{3/2}$
Same $R$ : $M$ increases by factor $q$	$T$ decreases by $\sqrt q$
Satellite mass $m$ changes	Circular-orbit period at the same $R$ is unchanged

Apply the ratio and boundary

Ratio example: around the same central body, if orbit B has $R_B=4R_A$ , then $\dfrac{T_B}{T_A}=\left(\dfrac{R_B}{R_A}\right)^{3/2}=4^{3/2}=8$ . The farther circular orbit takes eight times as long. AP Physics 1 does not require Kepler's first or second laws; keep this derivation to circular orbits.

Exam analysis

Most tested topics