AP Physics 1 · Algebra-Based · CED Unit 2

Force & Translational Dynamics

Unit 1 told you how things move. Unit 2 tells you why. Forces, free-body diagrams, Newton's three laws, friction, springs, gravity, and the physics of going in circles. The single heaviest unit on the exam: 18–23% of every paper.

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42 SLIDES · 18–23% OF THE EXAM

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2 · Learning Objectives

By the end of this unit, you will be able to…

Six measurable goals — everything the exam asks in Unit 2 maps back to one of these.

Define a system and locate its center of mass; classify forces as internal or external (2.1.A, 2.1.B)
Draw a correct free-body diagram for any object in any scenario — the single most valuable skill in AP Physics 1 (2.2.A, 2.2.B)
Identify Newton's-third-law pairs and explain why they never cancel on the same object (2.3.A)
Apply ΣF = ma to single objects, multi-block systems, elevators, and inclines (2.4.A, 2.5.A)
Model contact forces quantitatively — universal gravitation, kinetic & static friction, ideal springs (2.6, 2.7, 2.8)
Analyze circular motion: find the net force toward the center, handle vertical circles, and describe orbits with Kepler's third law (2.9.A, 2.9.B)

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Exam weight: 18–23%. More multiple-choice questions come from Unit 2 than any other unit. Master the free-body diagram here and Units 3–6 become applications of it.

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3 · Course & Exam Description

Unit 2 at a glance — every topic, every skill

Directly from the CED "Unit at a Glance" (effective Fall 2024). Nine topics, 22–27 class periods.

Topic	Title	Learning Objectives	Slides
2.1	Systems and Center of Mass	2.1.A, 2.1.B	6–7
2.2	Forces and Free-Body Diagrams	2.2.A, 2.2.B	8–10
2.3	Newton's Third Law	2.3.A	11–12
2.4	Newton's First Law	2.4.A	13–14
2.5	Newton's Second Law	2.5.A	15–18
2.6	Gravitational Force	2.6.A–2.6.D	19–21
2.7	Kinetic and Static Friction	2.7.A, 2.7.B	22–24
2.8	Spring Forces	2.8.A	25–26
2.9	Circular Motion	2.9.A, 2.9.B	27–29

Note: AP Physics 1 does not cover resistive drag forces or terminal velocity — that's Physics C: Mechanics material. If a practice question needs calculus, it's testing the wrong course.

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4 · Topic 2.1–2.2 Vocabulary

Key terms — the words exam questions are built from

System

The object or collection of objects you choose to analyse. Everything outside is the environment. You decide where to draw the line — choosing it well is a skill.

Force F

An interaction between two objects — a push or pull. Unit: newton (N), where 1 N = 1 kg·m/s². A force is a vector and always has an agent: something exerts it on something.

Net force ΣF

The vector sum of all external forces on a system. Not a force itself — it's the result of adding them.

Center of mass

The point that moves as if all the system's mass were concentrated there and all external forces were applied there. For two objects: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂).

Inertia / mass

An object's intrinsic resistance to being accelerated. More mass → same force produces less acceleration. Mass is a scalar in kilograms.

Free-body diagram (FBD)

One object drawn as a dot, with one arrow per external force, pointing away from the dot, labelled, roughly proportional in length. The core tool of this whole unit.

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5 · Notation Cheat Sheet

Force symbols — spoken as, meaning

Symbol	Spoken as	Meaning / Agent
F_g = mg	"gravitational force" / weight	Exerted by Earth (or any planet) on the mass, straight down
F_N or N	"normal force"	Exerted by a surface, perpendicular to the surface, pushing the object away from it
f_s, f_k	"static / kinetic friction"	Exerted by a surface, parallel to it, opposing sliding or its tendency
T	"tension"	Exerted by a rope/string/cable, pulling along the rope away from the object
F_sp = kΔx	"spring force"	Exerted by a spring, along its axis, restoring toward natural length
F_g = Gm₁m₂/r²	"universal gravitation"	Exerted by each mass on the other, attractive, along the line joining them
ΣF	"net force" / "sum of forces"	Vector sum of all external forces — not drawn on an FBD

The classic notation trap

ma is not a force. It's the result of the net force. Never draw an "ma arrow" or a "centripetal force arrow" on a free-body diagram — only actual interaction forces get arrows.

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6 · Topic 2.1 · Systems

Systems — you choose the boundary

A force analysis is only defined after you decide what's inside the system.

Internal forces act between objects inside the system (e.g. contact forces between two stacked blocks).
External forces come from the environment (gravity from Earth, push from a hand, friction from the table).
Only external forces can change the motion of the system as a whole. Internal forces always cancel in pairs — Newton's third law.
Same physical situation, different systems: analyse block A alone, or blocks A+B together. The FBD changes — forces at the boundary swap between internal and external.

2.1.A — describe properties and interactions of a system

When asked "which forces act on the system?", first state what's in it. A question about "the two-block system" must not include the contact force between the blocks — that's internal.

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Meme break

Choosing a system is like deciding who counts as "in the group project" — draw the boundary wrong and half the forces suddenly become somebody else's problem. (AP readers can tell when you skipped this step.)

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7 · Topic 2.1 · Center of Mass

Center of mass — the system's balance point

2.1.B: locate the center of mass from the constituent parts.

Center of mass (two objects)

x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)

Measured from the same origin as x₁ and x₂ — the cm is always between them, closer to the heavier one.

Worked example

A 2.0 kg block sits at x = 0 m and a 6.0 kg block at x = 4.0 m. Where is the center of mass?

x_cm = (2.0·0 + 6.0·4.0)/(2.0 + 6.0) = 24/8 = 3.0 m — two-thirds of the way toward the heavy block, as expected.

The center of mass of a uniform symmetric object (meterstick, disk, sphere) is at its geometric center.
The net external force determines the acceleration of the center of mass — even if the parts wobble internally.

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8 · Topic 2.2 · Forces

Every force is an interaction between two objects

2.2.A. If you can't name the agent and the recipient, it isn't a force.

Correct naming pattern: "X exerts a ___ force on Y". "The floor exerts a normal force on the box."
No agent, no force. "The force of motion" and "the force of the acceleration" are not things.
Forces come in two families: contact (normal, friction, tension, spring) and field (gravity — acts at a distance).
On the AP exam you will only ever need: weight, normal, friction, tension, spring, and universal gravitation.

Quick check

A ball flies upward after you throw it. Once it leaves your hand, how many forces act on it?

Exactly one: gravity. No "force of the throw" travels with the ball — that was a contact force, and the contact ended. This exact misconception is tested nearly every year.

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9 · Topic 2.2 · Free-Body Diagrams

The five rules of a legal free-body diagram

2.2.B — FRQ graders award a point literally for "a correct FBD". Follow all five.

One object, one dot. Never a sketch of the scene — the object becomes a point particle.
Arrows start at the dot and point away from it, in the direction the force acts.
One arrow per force, each labelled (F_g, F_N, T, f…). No unlabeled arrows.
Arrow lengths ≈ magnitudes. In equilibrium the arrows must visibly balance; sliding down an incline, f + F_N components vs F_g must look right.
No components, no net-force arrow, no ma. Unless the question asks you to resolve, keep it to real forces. ΣF is a sum, not a separate arrow.

Most common FBD errors on the exam

① Drawing F_N = mg on an incline (it's mg·cosθ). ② Adding a "forward motion force" to keep a thrown ball moving. ③ Putting both third-law pairs on the same diagram. ④ Drawing internal forces when the whole system is the object.

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10 · Topic 2.2 · FBD Gallery

Four FBDs you must recognise on sight

Reading check: in ④, why doesn't F_N point straight up? Because "normal" means perpendicular to the surface — and the surface is tilted.

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11 · Topic 2.3 · Newton's Third Law

Newton's Third Law — forces come in pairs

2.3.A: describe the interaction of two objects using paired forces.

Third law statement

F_{A on B} = −F_{B on A}

Equal in magnitude, opposite in direction, same type of force, acting simultaneously on different objects.

The pair test: swap the two objects in the description. "Earth pulls the book down" ⇄ "the book pulls Earth up". If you can't swap them cleanly, it's not a third-law pair.
Same type: a normal force pairs with a normal force; a gravitational force pairs with a gravitational force. F_N and F_g on one book are not a pair.
They never cancel for one object — they act on different objects. Only when you define both objects as one system do internal forces cancel in the sum.
Horse-and-cart paradox resolved: the cart accelerates because of the net force on the cart. The ground-push on the horse's hooves (friction) is what beats the cart's pull on the horse.

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12 · Topic 2.3 · Pair or not?

Third-law pairs — real pairs vs impostors

Candidate pair	Third-law pair?	Why
Earth pulls book down / book pulls Earth up	✅ Yes	Swap works, both gravitational, different objects
Table pushes book up / book pushes table down	✅ Yes	Swap works, both normal/contact
Book: F_N up / F_g down	❌ No	Same object, different types (normal vs gravity) — these balance because a = 0, that's first law, not third
Horse pulls cart forward / cart pulls horse back	✅ Yes	Tension pair, different objects
Car tire pushes road back / road pushes tire forward	✅ Yes	Friction pair — this is literally how cars accelerate

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Meme break

Newton's third law is the universe's group chat policy: every message gets an instant equal-and-opposite reply — sent to a different chat. You never see both in your own notifications. (Which is why pairs never cancel on one object.)

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13 · Topic 2.4 · Newton's First Law

Newton's First Law — no net force, no change

2.4.A: describe the conditions under which a system's velocity remains constant.

First law (equilibrium)

ΣF = 0 ⟺ velocity is constant (a = 0)

"Constant velocity" includes v = 0 (at rest) and any nonzero steady speed, in a straight line.

Equilibrium = ΣF = 0. An object at rest and an object cruising at constant velocity are both in equilibrium.
Motion requires no cause — changes in motion do. This kills the Aristotelian "things stop because they run out of force" instinct (they stop because friction acts).
Constant velocity but multiple forces? Fine — they just have to cancel. A car at 30 m/s steady: engine force = drag + friction, exactly.
Exam tell: "moves at constant speed" or "moves with constant velocity" in the stem → write ΣF = 0 first, before touching any equation.

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14 · Topic 2.4 · Don't mix these up

First law vs third law — the balance confusion

The book on the table has F_N = F_g. Why? First law. Not the third. This distinction is a favourite MCQ.

First law explanation (same object)

The book isn't accelerating, so the forces on the book must sum to zero: F_N + F_g = 0. Two forces, one object, opposite directions — balancing, not pairing.

Third law version (two objects)

The third-law partner of F_N (table on book) is the book pressing down on the table. The partner of F_g (Earth on book) is the book pulling up on Earth.

Why the confusion is deadly

"The normal force equals mg because of Newton's third law" is a wrong statement that sounds right. F_N = mg only when vertical acceleration is zero — in a lift accelerating up, F_N = m(g + a) and the first law (with a ≠ 0 → not equilibrium) is the correct tool: ΣF = ma.

Flowchart: Same object? → first law (balance). Different objects, same type? → third law (pair).

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15 · Topic 2.5 · Newton's Second Law

Newton's Second Law — the equation of the unit

2.5.A: describe the conditions under which a system's velocity changes.

Second law

a = ΣF / m

Acceleration is proportional to net force, inversely proportional to mass, and points in the direction of ΣF. Applied per-axis: ΣF_x = ma_x, ΣF_y = ma_y.

a is the effect, ΣF is the cause. Never draw ma as a force; it's the outcome of adding the real forces.
Vector equation, axis by axis. Choose +x along the motion (or along the incline) to make the algebra clean.
Direction check: object slows down while moving right → a points left → ΣF points left. Signs carry meaning.
Units sanity: N = kg·m/s². If your answer's units don't reduce, the setup is wrong.

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16 · Topic 2.5 · Problem-Solving Method

The four-step method (use it every single time)

1 · Draw the FBD for the chosen object. Name every force with an agent.
2 · Choose axes: +x along the acceleration (or along the surface), +y perpendicular. Tilt both for inclines.
3 · Write ΣF = ma per axis, with signs: forces along +axis positive, against negative.
4 · Solve & sanity-check: units, limits (frictionless? a → g on a vertical drop?), magnitude plausibility.

Apply it — 4.0 kg crate, frictionless floor, pulled by 12 N horizontal

Step 1: FBD = F_N up, F_g down, F = 12 N right.

Step 2: a is horizontal → ΣF_y = 0 → F_N = mg. ΣF_x = F.

Step 3–4: a = F/m = 12/4.0 = 3.0 m/s² in the pull direction. ✓ units ✓ reasonable.

Why the method matters: FRQ graders look for the FBD and the axis setup, not just the final number. The four steps are the rubric.

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17 · Topic 2.5 · Elevators

Elevators — the apparent-weight classic

You feel the normal force, not gravity. That's why lifts mess with your stomach.

Worked example — 60 kg person on a scale in a lift

The lift accelerates upward at 2.0 m/s². What does the scale read? (g = 9.8 m/s²)

FBD: F_N up (the scale reading), F_g down. Acceleration is up, so ΣF_y = ma: F_N − mg = ma.

F_N = m(g + a) = 60(9.8 + 2.0) = 708 N — heavier than the true weight of 588 N.

Accelerating down at 2.0 m/s²: F_N = m(g − a) = 468 N — lighter. In free fall (a = −g): F_N = 0, weightless.

Lift motion	Scale reads	Feeling
At rest / constant v	mg	normal
Accelerating up (or braking while going down)	m(g + a) > mg	heavier, pressed down
Accelerating down (or braking while going up)	m(g − a) < mg	lighter, stomach floats
Cable snapped (a = g down)	0	weightless — true free fall

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18 · Topic 2.5 · Systems of Blocks

Two-block systems — solve the system first

The single most common Unit 2 FRQ setup. Strategy: whole system, then the back block.

Worked example

System (3 + 2 = 5 kg): a = F/M = 15/5 = 3.0 m/s² — one equation, done.

Then block 2 alone (only the contact force C pushes it): C = m₂a = 2.0 × 3.0 = 6.0 N.

Check with block 1: F − C = 15 − 6 = 9.0 N = m₁a = 3 × 3.0 ✓ Consistent.

Why this order: going "system → part" needs 2 equations; going block-by-block from the front needs 3 and invites sign errors. The exam's rubric follows the same logic.

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19 · Topic 2.6 · Gravitational Force

Universal gravitation — every mass pulls every other

2.6.A: describe the gravitational interaction between two objects with mass.

Newton's law of universal gravitation

F_g = G·m₁m₂ / r²

G = 6.67 × 10⁻¹¹ N·m²/kg² · r = distance between centers · always attractive, along the line joining the centers.

Inverse-square: double r → force drops to ¼. Triple r → ⅑. The exam loves proportional-reasoning versions of this.
Example: move a satellite from r to 2r from Earth's center → gravitational force × ¼ → and by N2, orbital acceleration × ¼ too.
2.6.B: near Earth's surface, r ≈ R_E barely changes, so F_g = mg is effectively constant — that's the special case, not the rule.
The "r²" is center-to-center: for a satellite at height h, r = R_E + h, not h.

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20 · Topic 2.6 · Gravitational Field

g is just the gravitational field strength

Set F = mg equal to the universal law and g falls out — it's a property of the planet, not the object.

Field strength at distance r from the planet's center

g = GM_planet / r²

No m of the object anywhere — a feather and a hammer share the same g. That's why they fall together (Apollo 15 proved it on the Moon).

Proportional reasoning — exam style

Planet X has 2× Earth's mass and 2× Earth's radius. What is g on its surface?

g_X/g_E = (M_X/M_E)·(R_E/R_X)² = 2 × (1/2)² = 2/4 = ½ → about 4.9 m/s².

Trap

Mass never enters g. If a question implies "heavier objects fall faster", it's testing whether you'll defend the correct model — they don't.

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21 · Topic 2.6 · 2.6.C & 2.6.D

Apparent weight & the two kinds of mass

2.6.C — apparent weight

Apparent weight = what a scale reads = F_N. It differs from mg whenever the object accelerates vertically: orbiting astronauts have F_N = 0 (they and the station fall together), so they're weightless — but gravity on them is nearly 90% of surface value. Weightless ≠ no gravity.

2.6.D — inertial vs gravitational mass

Inertial mass: resistance to acceleration, from a = F/m.
Gravitational mass: strength of gravitational interaction, from F = Gmm/r².
Experiments show they're equal — that's why all objects fall with the same a in vacuum. The equality is an experimental fact, not an obvious one.

Exam-style MCQ

An astronaut floats in the ISS. Which statement is correct?

(A) No gravity acts — wrong; gravity is ~89% of surface g there.
(B) Gravity acts; the astronaut and station are in free fall together, so the floor exerts no normal force. ✅
(C) No forces act at all — never true anywhere.
(D) Centrifugal force cancels gravity — "centrifugal force" is not an interaction; the AP exam rejects it.

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22 · Topic 2.7 · Kinetic Friction

Kinetic friction — surfaces already sliding

2.7.A: describe kinetic friction between two surfaces.

Kinetic friction

f_k = μ_k·F_N

Direction: opposite the relative sliding. Magnitude: constant for given surfaces and normal force — it does not depend on speed or contact area.

μ is a property of the pair of surfaces (rubber-on-concrete ≈ 0.8, ice-on-ice ≈ 0.03, Teflon ≈ 0.04). Unitless.
Independent of area — counter-intuitive but tested constantly. Wider tyres don't get more friction from this model.
On an incline: F_N = mg·cosθ, so f_k = μ_kmg·cosθ. Forgetting the cosθ is the #1 algebra slip.
Linearisation lab: measure f_k for several F_N values, plot f_k vs F_N → straight line through origin, slope = μ_k. A favourite experimental-design FRQ.

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23 · Topic 2.7 · Static Friction

Static friction — the self-adjusting force

2.7.B. The single most mis-modeled force on the exam. It has a maximum, not a fixed value.

Static friction

f_s ≤ μ_s·F_N

It matches whatever is needed to prevent sliding, up to the maximum μ_sF_N. Beyond that, slipping starts and kinetic friction takes over.

The classic sequence

You push a 20 kg crate (μ_s = 0.5, μ_k = 0.3) with F = 60 N. It doesn't move. What is friction?

Not μ_smg = 98 N. Equilibrium: f_s = F = 60 N — static friction simply matches your push. It would take 98 N to break it loose.

Push with 120 N: crate slides, and friction drops to f_k = 0.3 × 196 = 59 N. Note μ_s > μ_k — that's why breaking loose feels like a jerk.

Trap

"f_s = μ_sF_N" is only true at the verge of slipping. If the object isn't sliding, find f_s from ΣF = 0 instead.

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24 · Topic 2.7 · The Graph

Friction vs applied force — the picture worth memorising

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Meme break

Friction: turning "ideal physics" into "real physics" since 10,000 BC. (Static friction is the friend who matches your energy until you push too hard — then the friendship drops to a lower, flatter level.)

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25 · Topic 2.8 · Spring Forces

Ideal springs — Hooke's law

2.8.A: describe the force exerted by an ideal spring.

Hooke's law

F_sp = k·Δx (restoring: opposite the stretch/compression)

k = spring constant (N/m) — stiffness. Δx = displacement from the natural (unstretched) length, not from the floor, not from wherever.

Restoring: stretched → pulls back; compressed → pushes out. Always toward natural length. (The minus sign in F = −kΔx encodes this.)
Δx is measured from natural length — if a spring hangs with a mass and sits 0.12 m longer than natural, then Δx = 0.12 m. This trips up more students than the formula itself.
k from a graph: hang masses, plot weight mg vs stretch Δx → straight line, slope = k. Another lab-design FRQ favourite.
"Ideal" means: massless, frictionless, perfectly linear — no limits on Δx.

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26 · Topic 2.8 · Spring Worked Example

Hanging mass on a spring

Reading check

Two forces on the mass only: F_sp up, F_g down. At rest → balance → k·Δx = mg. The spring is stretched below its natural length by Δx — that's the 0.15 m.

Variant: same setup in an elevator accelerating up at a: k·Δx = m(g + a) — the spring stretches further. Links Topics 2.5 + 2.8 in one line, exactly how MCQs combine them.

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27 · Topic 2.9 · Circular Motion

Circular motion — constant speed, changing velocity

2.9.A. Velocity is a vector: turning is accelerating, even at steady speed.

Centripetal acceleration

a_c = v²/r = 4π²r/T²

Direction: toward the center, always. T = period (time per revolution).

Speed constant, direction changing → velocity changing → a ≠ 0.
By N2: ΣF toward center = mv²/r. That net force must be supplied by real forces (tension, gravity, friction, normal).

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28 · Topic 2.9 · Force Inventory

"Centripetal force" is a job description, not a force

Name the real force doing the job in each scenario — a guaranteed MCQ pattern.

Scenario	Real force(s) pointing toward center
Ball on a string, horizontal circle	Tension T (= mv²/r)
Car rounding a flat curve	Static friction from the road (yes, friction points sideways here)
Planet orbiting the Sun	Gravity (= GmM/r²)
Rider at the top of a vertical loop	F_N + mg, both downward: F_N + mg = mv²/r
Rider at the bottom of a vertical loop	F_N up, mg down: F_N − mg = mv²/r → F_N = mg + mv²/r (heaviest here)
Conical pendulum	Horizontal component of tension: T·sinθ = mv²/r, with T·cosθ = mg

Top-of-loop minimum speed

At the top, if v is too small, F_N would need to be negative (impossible) → contact lost. Minimum contact case: F_N = 0 → v_min = √(gr). A rollercoaster just barely maintaining contact at the top is moving at exactly this speed.

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29 · Topic 2.9 · Orbits

Circular orbits & Kepler's third law

2.9.B: describe circular orbits using T² ∝ r³.

Derivation in two lines

GmM/r² = m·4π²r/T²

⇒ T² = (4π²/GM)·r³ — T² ∝ r³ for any object orbiting the same central mass M. The orbiting object's own mass cancels.

Farther orbit → longer period, slower orbital speed (v = √(GM/r)). Outer planets crawl; the Moon takes 27 days to lap Earth.
Proportional reasoning: if r ×9, then T ×√(9³) = ×27.

Exam-style

Satellite A orbits at r; satellite B at 4r, around the same planet. Compare periods.

T_B/T_A = √((4r)³/r³) = √64 = 8. B takes 8× longer per orbit.

And speeds: v_B/v_A = √(r/4r) = ½. Farther = slower.

Why the mass cancels: gravity is proportional to the orbiter's mass, and so is the ma it must supply. Equal-×-equal → mass drops out → astronauts and stations share one orbit.

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30 · Practice · MCQ

Click your answer — the deck will grade it

A 60 kg student stands on a bathroom scale in an elevator. The scale reads 468 N. Take g = 9.8 m/s². What can you conclude about the elevator's motion?

AMoving upward at constant speed

BAccelerating downward at about 2.0 m/s²

CAccelerating upward at about 2.0 m/s²

DMoving downward at constant speed

✅ Answer: B. True weight = 60 × 9.8 = 588 N. The scale reads F_N = 468 N, which is less than the weight. Take up as positive: ΣF_y = F_N − mg = m·a_y → a_y = (468 − 588)/60 = −2.0 m/s², i.e. accelerating downward at 2.0 m/s².

The lesson that matters more than the arithmetic: a scale reading tells you about acceleration, never about velocity. Options A and D ("constant speed") would both give F_N = 588 N, so they're ruled out instantly — but which one you're actually doing (up or down, speeding or slowing) is unknowable from the reading alone. That "cannot be determined" move is a favourite AP trap.

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31 · Practice · MCQ

Click your answer

A car accelerates forward along a level road. What force directly accelerates the car?

AThe engine's force on the car

BThe normal force from the road

CStatic friction from the road on the tyres, pointing forward

DThe reaction to the tyres' force on the road

✅ Answer: C. Tyres push road backward (action); road pushes tyres forward (reaction) — that external friction from the road is the only horizontal force on the car, so it is the ma. Option A fails because the engine is internal to the car — internal forces can't accelerate the center of mass. Option D's reaction acts on the road, not the car. This is the third law + second law working together, in one MCQ.

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32 · Practice · Free Response

FRQ walkthrough — two hanging masses (Atwood machine)

(a) Derive the acceleration — system method

External forces on the two-mass system: weights 3g down on the left, 2g down on the right. Tension is internal — gone from the math.

(3g − 2g) = (3 + 2)a → a = g/5 ≈ 1.96 m/s², heavy side descending.

(b) Find the tension — single-object method

2 kg mass, accelerating up at g/5: T − 2g = 2(g/5) → T = 2g(1 + 1/5) = 2.4g ≈ 23.5 N.

Check: 3 kg side → 3g − T = 29.4 − 23.5 = 5.9 = 3 × 1.96 ✓. And note g/5 < a < g, T between 2g and 3g — always sanity-range your Atwood answers.

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33 · Common Mistakes

Five ways students lose Unit 2 points

1 · Inventing forces

"Force of the throw", "centrifugal force", "force of motion" — none are interactions. If you can't name agent + recipient, it doesn't go on the FBD.

2 · Cancelling third-law pairs on one object

Action–reaction pairs act on different objects. They can never appear on the same FBD, let alone cancel.

3 · Normal force = mg, always

Only when vertical a = 0 (and no vertical force components). Inclines: F_N = mg·cosθ. Elevators: m(g ± a). Always solve for F_N, never assume.

4 · Static friction = μ_sF_N, always

That's the ceiling. Until slipping, f_s is whatever ΣF = 0 demands. Write "f_s ≤ μ_sF_N" and use equilibrium.

5 · Drawing "centripetal force" as an extra arrow

It's a role filled by real forces. Draw the real forces; the net inward sum is the centripetal force. Adding a fifth arrow = automatic point loss on the FBD.

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34 · AP Exam · MCQ Patterns

How Unit 2 appears in the multiple-choice section

FBD identification: "Which diagram is correct for the object in the situation?" — expect 2–3 of these. Check arrow count, direction (⊥ surface for F_N), and relative lengths.
Proportional reasoning: "If the mass is doubled and the net force halved, the acceleration…" — a = F/m, so ×2/×2⁻¹… answer: ¼. No numbers needed.
Third-law discrimination: paired-text questions asking you to justify why F_N and F_g are not a third-law pair.
Graphs: f vs F applied (slide 24's graph), F_N vs time in an elevator ride.
Circular reasoning-lite: "At the top of the loop, which expression gives F_N?" — derive from mv²/r with signs.

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18–23% of the exam means roughly 11–14 of the 60 MCQs and usually 1 of the 4 FRQs. No other unit is worth more.

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35 · AP Exam · FRQ Patterns

The Unit 2 FRQ blueprint

Typical 12-point dynamics question — usually parts (a)–(d), building upward.

Part	What's asked	How to bank the point
(a)	Draw/complete the FBD	Dot + labelled arrows, lengths sensible, no extras
(b)	Derive an expression (symbols!)	Start at ΣF = ma in writing; end boxed; no numbers until asked
(c)	Calculate with numbers	Substitute at the very end; keep units; 2–3 sig figs
(d)	Justify / compare ("if μ doubles, does T increase?")	Cite a principle (N1/N2/N3), reference your (b) equation, answer in a full sentence

Task verbs matter: "derive" wants algebra from a law; "calculate" wants numbers; "justify" wants physics reasoning in words. Answering the wrong verb is the cheapest way to lose a point you knew.

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36 · Real-World Applications

Unit 2 in the wild

🚗 Seatbelts & crash physics

In a sudden stop the car decelerates but you keep going (N1). The belt supplies the external force your body needs to decelerate with the car. Airbags stretch the stopping time — same Δv, smaller F (impulse logic, arriving in Unit 4).

🏀 Basketball grip

You dribble because of static friction between fingers and ball — μ_s of leather-on-skin is high. Dust kills the μ, and the whole game changes. Ask any point guard.

🛰️ GPS satellites

They orbit at r ≈ 26,600 km from Earth's center — Kepler's third law sets their 12-hour period, and inverse-square gravity sets their orbital speed. Your location fix is Unit 2 running 24/7.

🎢 Rollercoaster top-of-loop

Designers keep v_top ≥ √(gr) so F_N ≥ 0 and riders stay pressed to seats. The "weightless" flutter you feel is F_N dipping toward zero — Topic 2.9 made physical.

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37 · YouTube & Video

Watch these alongside the deck

Full unit · interactive

Khan Academy — Force and Translational Dynamics

Khan Academy · Official AP/college-aligned lessons for every Topic 2.1–2.9, with practice sets

Free-body diagrams · playlist

Flipping Physics — AP Physics 1 lessons

Flipping Physics · Bite-sized FBD and Newton's-law videos made specifically for the AP 1 exam

Problem drills · video series

The Organic Chemistry Tutor — Newton's Laws problem marathons

Back-to-back worked problems: inclines, Atwood machines, elevator cases

Links verified against the official Khan Academy AP Physics 1 course structure. Prefer the Khan lessons for concepts and Flipping Physics for FBD craft; use OCT for volume practice the night before a test.

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38 · Formula Summary

Every equation this unit can ask you to use

Equation	Use it when	Watch out for
ΣF = ma	Any non-equilibrium dynamics problem	a is the effect, not a force
F_g = mg	Near a planet's surface	g = 9.8 m/s² on the AP sheet (10 OK for estimates)
F_g = Gm₁m₂/r²	Any two masses; orbits	r is center-to-center; inverse-square
g = GM/r²	Field strength at distance r	Independent of the object's mass
f_k = μ_kF_N	Surfaces sliding	Opposes relative motion; area-independent
f_s ≤ μ_sF_N	No sliding yet	Find it from ΣF = 0 unless at the verge
F_sp = kΔx	Ideal springs	Δx from natural length; restoring direction
a_c = v²/r = 4π²r/T²	Uniform circular motion	Points to the center; not a new force
T² = (4π²/GM)r³	Circular orbits (Kepler III)	Depends only on central mass M
x_cm = (m₁x₁+m₂x₂)/(m₁+m₂)	Two-object systems	Closer to the heavier mass

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39 · Review · Concept Map

The whole unit on one page

How to read it: every box feeds ΣF = ma. Left side = how you set problems up; right side = the force models you plug in; bottom = the special-motion case.

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40 · Practice · Drill

Ten-minute drill — answers at the bottom (no peeking)

A 5.0 kg box slides on a floor with μ_k = 0.20. Find the friction force and the box's acceleration if pushed then released.
A 1200 kg car rounds a flat 50 m-radius curve at 15 m/s. What friction force does the road supply?
In an elevator a 70 kg person feels 700 N from the floor (g = 9.8). Find a_y.
A spring (k = 250 N/m) stretches 8.0 cm holding a mass at rest. Find the mass.
Two masses 4 kg and 6 kg hang on an ideal Atwood setup. Find a and T.
A planet has half Earth's radius and half Earth's mass. Surface g?

Answers: ① f_k = 9.8 N, a = 1.96 ≈ 2.0 m/s² opposite motion ② F = mv²/r = 1200×225/50 = 5.4×10³ N ③ a_y = 700/70 − 9.8 = 0.20 m/s² up ④ m = kΔx/g = 250×0.08/9.8 ≈ 2.0 kg ⑤ a = g/5 ≈ 1.96 m/s², T = 2.4g ≈ 23.5 N (slide 32, scaled) ⑥ g = GM/r² → (½)/(½)² = ×2 → ≈ 19.6 m/s²

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41 · References

References & where to go deeper

College Board, AP Physics 1 Course and Exam Description (effective Fall 2024) — Unit 2: Force and Translational Dynamics, Topics 2.1–2.9, exam weighting 18–23%.
Khan Academy — AP/College Physics 1, unit "Force and translational dynamics" (aligned topic-by-topic to the 2024 CED).
Flipping Physics — AP Physics 1 video series (free-body diagrams taught exceptionally well).
The Organic Chemistry Tutor — YouTube problem drills for Newton's laws, inclines, and Atwood machines.
AP Physics 1 equation sheet — you get this in the exam; know where every Unit 2 equation lives on it.

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Final nudge

The reference list is ordered by how much it will actually move your score: CED first (it's the contract), Khan for structure, OCT for reps. Reading order is a free mark.

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42 · Coverage Checklist

Content coverage checklist — CED Unit 2, fully mapped

Every topic and learning objective in the Unit at a Glance, and where it lives in this deck.

2.1 Systems: internal vs external forces; choosing the boundary (slide 6)
2.1 Center of mass location and x_cm calculation (slide 7)
2.2 Force as an interaction; agents and recipients (slide 8)
2.2 Free-body diagrams: rules, gallery of four classics (slides 9–10)
2.3 Third-law pairs: definition, pair test, impostors table (slides 11–12)
2.4 First law & equilibrium; first-vs-third distinction (slides 13–14)
2.5 ΣF = ma per axis; four-step method; elevator apparent weight; multi-block systems (slides 15–18)
2.6 Universal gravitation, inverse-square reasoning, g = GM/r², apparent weight, inertial vs gravitational mass (slides 19–21)
2.7 Kinetic friction model, static friction's inequality, the f-vs-F graph, linearisation lab (slides 22–24)
2.8 Ideal spring force, Δx from natural length, k from data (slides 25–26)
2.9 a_c = v²/r, force inventory for circles, vertical circles & conical pendulum, Kepler's third law (slides 27–29)
Skills MCQ drills ×2, full FRQ walkthrough, exam-pattern guides, formula sheet, drill set (slides 30–35, 38, 40)

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Unit 2 complete. Next: Unit 3 — Work, Energy, and Power, where ΣF = ma gets a rival: conservation of energy. You now own the heaviest unit on the exam.

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Unit 2 1 / 42 Force & Translational Dynamics