AP Physics C: Mechanics Study Guide 2026

Unit 114–20% of exam

Kinematics

Calculus-based description of motion. Position, velocity, and acceleration are linked by differentiation and integration — the core skill Physics C demands.

AP Physics C: Mechanics is a calculus-based course. Unlike Physics 1, you cannot always use the kinematic equations — they apply only to constant acceleration. When acceleration is a function of time (or position, or velocity), you must integrate or differentiate.

The Calculus Definitions

The three kinematic quantities are related by calculus — this is the defining feature of Physics C:

Velocity is the derivative of position with respect to time: v = dx/dt
Acceleration is the derivative of velocity: a = dv/dt = d²x/dt²
Recovering position from velocity: x(t) = x₀ + ∫v dt
Recovering velocity from acceleration: v(t) = v₀ + ∫a dt

When to Use Integration vs. Kinematic Equations

If a = constant → use the four kinematic equations (v = v₀ + at, etc.)
If a = f(t) → integrate to find v(t), then integrate again to find x(t)
If you're given v(t) and need displacement → integrate: Δx = ∫v dt (area under v-t curve)
If you're given x(t) and need velocity → differentiate: v = dx/dt (slope of x-t curve)

\begin{array}{ll} v(t) = \dfrac{dx}{dt} & a(t) = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2} \\[16pt] x(t) = x_0 + \displaystyle\int_0^t v\,dt' & v(t) = v_0 + \displaystyle\int_0^t a\,dt' \\[16pt] \end{array} \\[4pt] \textbf{Constant acceleration only:} \\[6pt] \begin{array}{ll} v = v_0 + at & x = x_0 + v_0 t + \tfrac{1}{2}at^2 \\[8pt] v^2 = v_0^2 + 2a\Delta x & \bar{v} = \dfrac{v_0 + v}{2} \end{array}

Linked Kinematics Graphs — x(t), v(t), a(t)

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Select a motion scenario and drag the cursor across all three linked graphs simultaneously. The slope of x(t) at any point equals v(t); the slope of v(t) equals a(t); the area under v(t) equals displacement. This is the calculus backbone of all Physics C kinematics problems.

position x(t)

velocity v(t)

acceleration a(t)

← drag to move cursor

t = 2.00 sx = 4.000v = 2.000a = 0.000v = dx/dt | a = dv/dt

Projectile Motion — Parametric

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Projectile Motion — Parametric

Adjust v₀ (initial speed) and θ (launch angle) to see how the parametric trajectory changes. Notice that range is maximized at 45° — and that launch angles supplementary to each other (e.g., 30° and 60°) produce the same range. These are FRQ favorites.

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Key Concepts

Position x(t)Location as a function of time; slope gives instantaneous velocity.

Velocity v(t)Rate of change of position; slope gives acceleration, area gives displacement.

Acceleration a(t)Rate of change of velocity; area gives change in velocity.

Integration (∫)Used to find position from velocity, or velocity from acceleration — reverse differentiation.

Differentiation (d/dt)Used to find velocity from position, or acceleration from velocity.

Instantaneous vs. averageInstantaneous quantities use derivatives; average quantities use total change over total time.

Exam tip: If the FRQ gives you a(t) as a polynomial (e.g., a = 3t² − 4), integrate to get v(t) = t³ − 4t + C and apply the initial condition to find C. Then integrate again for x(t). Always show the integration explicitly — the FRQ rubric awards points for the integral setup, not just the final answer.

Common mistake: Never use v² = v₀² + 2aΔx when acceleration is not constant. If a = f(t), those equations are invalid. The most common Physics C kinematics error is applying constant-acceleration formulas to variable-acceleration problems. Check: is a constant? If not, integrate.

Unit 217–23% of exam

Newton's Laws of Motion

Forces, free body diagrams, and Newton's laws in two dimensions — including circular motion and non-inertial reference frames.

Newton's three laws remain the foundation of classical mechanics. In Physics C, you apply them in 2D with calculus — and in situations (circular motion, springs, drag) where Physics 1 only asked conceptual questions.

Free Body Diagrams — The Non-Negotiable First Step

Every dynamics problem begins with a free body diagram (FBD). An FBD shows:

One dot or box representing the object
Arrows for every force acting ON the object (weight down, normal perpendicular to surface, tension along string, friction opposing motion)
A clearly defined positive direction (or +x/+y axes for 2D)

Then apply ΣF = ma in each direction separately.

Circular Motion

For an object moving in a circle at speed v with radius r, the net force must point toward the center (centripetal direction) with magnitude mv²/r. The centripetal force is not a new force — it is the net result of real forces (gravity, normal, tension, friction).

\sum F = ma \qquad \text{(vector equation — apply in each direction)} \\[14pt] \text{Circular motion: } \sum F_{\text{centripetal}} = \frac{mv^2}{r} = m\omega^2 r \\[14pt] \text{Atwood machine: } a = \frac{(m_1 - m_2)g}{m_1 + m_2} \qquad T = \frac{2m_1 m_2 g}{m_1 + m_2}

Forces and Motion Basics

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Forces and Motion Basics

Apply forces, adjust mass, and observe how net force, acceleration, and friction interact. Build intuition for free body diagram analysis before tackling FRQ problems.

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Key Concepts

Net force (ΣF)Vector sum of all forces on an object; equals ma.

Normal force (N)Contact force perpendicular to the surface — not always equal to mg.

Frictionfₖ = μₖN (kinetic); fₛ ≤ μₛN (static, opposes tendency to slide).

Centripetal accelerationv²/r directed toward the center of circular motion.

TensionForce transmitted through a string or rope — constant throughout massless strings.

Pseudo force (non-inertial frame)Fictitious force appearing when analyzing motion from an accelerating reference frame.

Exam tip: For circular motion at the top of a loop: both weight AND normal force point toward the center. Set N + mg = mv²/r. At minimum speed (N = 0): mg = mv²/r → v_min = √(gr). This derivation appears on the FRQ — show every step. At the bottom of the loop: N − mg = mv²/r → N = mg + mv²/r (you feel heavier). These are the two most-tested cases.

Common mistake: Never write 'centripetal force' on a free body diagram as if it were a real force. Centripetal force is the NET result of real forces toward the center — gravity, normal, tension, or friction. If you draw a separate 'centripetal force' arrow on your FBD, you are double-counting and will get the wrong equation.

Unit 314–17% of exam

Work, Energy & Power

The work-energy theorem with integration, conservative forces and potential energy, and conservation of mechanical energy with non-conservative forces.

Energy methods are often the fastest route to a Physics C answer. When forces vary with position, use integration. When only conservative forces act, use conservation of energy. When non-conservative forces (friction, applied) are present, use the work-energy theorem.

Work with Variable Force

When force varies with position, work is the area under the F-x curve — found by integration:

W = \int_{x_1}^{x_2} F(x)\,dx \qquad \text{(constant force: } W = F \cdot d \cdot \cos\theta\text{)} \\[14pt] \text{Work-Energy Theorem: } W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \\[14pt] W_{\text{non-conservative}} = \Delta KE + \Delta PE \\[10pt] \text{Spring PE: } U_s = \tfrac{1}{2}kx^2 \qquad \text{Grav. PE: } U_g = mgh

Conservative vs. Non-Conservative Forces

A force is conservative if the work it does depends only on the start and end points, not the path taken (gravity, spring force, electric force). The work done by a conservative force equals the negative change in potential energy:

W_conservative = −ΔPE

A force is non-conservative if work depends on path (friction, applied forces, air resistance). When non-conservative forces act:

W_nc = ΔKE + ΔPE = ΔE_mechanical

Power

Power is the rate of doing work: P = dW/dt = F·v (instantaneous power). Average power = W/Δt.

Spring Potential Energy vs. Displacement

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Spring Potential Energy vs. Displacement

Drag the k (spring constant) and E (total energy) sliders. The parabola shows U(x) = ½kx²; the gap between E (total energy, horizontal line) and U(x) is the kinetic energy at that position. The object oscillates between the two intersection points where KE = 0. This is the energy method for finding turning points.

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Key Concepts

Work (W)W = ∫F·dx — energy transferred by a force over a displacement.

Kinetic energy (KE)½mv² — energy of motion; changes equal net work done on the object.

Potential energy (PE)Stored energy due to configuration; defined only for conservative forces.

Work-energy theoremW_net = ΔKE — the net work equals the change in kinetic energy.

Conservation of energyIf only conservative forces act: KE + PE = constant.

Non-conservative workW_nc = ΔKE + ΔPE — friction/applied forces change total mechanical energy.

Power (P)Rate of doing work: P = dW/dt = F·v.

Exam tip: Energy methods sidestep needing to find acceleration at every instant — use them whenever you want speed at a particular position, not as a function of time. For FRQ: if the problem gives a variable force F(x) and asks for speed at position x₂, integrate: W = ∫F dx = ΔKE, then solve for v. You'll often need the fundamental theorem of calculus — write the integral first, then evaluate.

Common mistake: Don't include normal force or perpendicular components in work calculations — they do zero work (perpendicular to displacement). The most common error is adding N·d to the work equation. Only forces with a component ALONG the displacement direction do work.

Unit 414–17% of exam

Systems of Particles & Linear Momentum

Center of mass, impulse-momentum theorem, conservation of momentum, and elastic vs. inelastic collisions.

Momentum is conserved whenever the net external force on a system is zero — regardless of what forces the objects exert on each other internally. This makes it powerful for collision analysis.

Center of Mass

The center of mass (COM) is the average position of all mass in a system, weighted by mass. For a system of particles:

x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \qquad v_{cm} = \frac{\sum m_i v_i}{M_{\text{total}}} = \frac{p_{\text{total}}}{M_{\text{total}}} \\[14pt] \textbf{Impulse-Momentum Theorem:} \quad J = \Delta p = \int F\,dt = F_{\text{avg}} \cdot \Delta t \\[14pt] \textbf{Conservation of momentum (isolated system):} \quad \sum p_i = \sum p_f \\[14pt] \textbf{Elastic collision:} \quad \tfrac{1}{2}m_1 v_{1i}^2 + \tfrac{1}{2}m_2 v_{2i}^2 = \tfrac{1}{2}m_1 v_{1f}^2 + \tfrac{1}{2}m_2 v_{2f}^2

Collision Types

Type	Momentum conserved?	KE conserved?	Objects stick?
Elastic	Yes	Yes	No — bounce
Inelastic	Yes	No — some KE lost	No
Perfectly inelastic	Yes	No — maximum KE lost	Yes — they stick

For perfectly inelastic: m₁v₁ + m₂v₂ = (m₁ + m₂)v_f. For elastic, use both momentum AND kinetic energy conservation to solve for both final velocities.

Key Concepts

Linear momentum (p)p = mv — a vector quantity; conserved when net external force = 0.

Impulse (J)J = ∫F dt = Δp — the area under a force-time curve.

Conservation of momentumTotal momentum is constant for an isolated system (no net external force).

Elastic collisionBoth momentum and kinetic energy are conserved; objects bounce.

Perfectly inelastic collisionMomentum conserved; maximum KE lost; objects stick together.

Center of massPoint where the system's total mass can be considered concentrated; xₓₘ = Σmᵢxᵢ/M.

ExplosionPerfectly inelastic collision in reverse: initial momentum = 0, objects fly apart with equal and opposite momenta.

Exam tip: When a FRQ says 'the collision lasts 0.01 s and the average force is X,' they want impulse: J = FΔt = Δp. When given F(t) during a collision, integrate: J = ∫F dt. The impulse equals the area under the F-t graph — if it's a triangle, area = ½ base × height. Show this explicitly for full credit.

Common mistake: Momentum is ALWAYS conserved in a collision (when external forces are absent), but KE is NOT always conserved. Students often assume elastic collision when they see 'collision' — wrong. Read the problem carefully: 'the objects stick' = perfectly inelastic, 'the collision is elastic' = both conserved. If neither is stated, assume inelastic (most real collisions are).

Unit 514–20% of exam

Rotation

Angular kinematics, torque, moment of inertia, angular momentum, and rolling without slipping — the most calculation-intensive unit.

Rotation is the Physics C unit that most differentiates strong students. Every translational concept has a rotational analog — learn the table of analogs and you learn both simultaneously.

Rotational Analogs

Translational	Symbol	Rotational	Symbol
Displacement	x	Angular displacement	θ
Velocity	v	Angular velocity	ω
Acceleration	a	Angular acceleration	α
Force	F	Torque	τ
Mass	m	Moment of inertia	I
F = ma		τ = Iα
KE = ½mv²		KE = ½Iω²
p = mv		L = Iω

\tau = r \times F = rF\sin\theta \qquad \sum\tau = I\alpha \\[14pt] I = \sum m_i r_i^2 \quad \text{(discrete)} \qquad I = \int r^2\,dm \quad \text{(continuous)} \\[14pt] \textbf{Parallel Axis Theorem:} \quad I = I_{cm} + Md^2 \\[14pt] L = I\omega \qquad \frac{dL}{dt} = \tau_{\text{net}} \\[14pt] \textbf{Rolling without slipping:} \quad v_{cm} = \omega R \quad a_{cm} = \alpha R

Common Moments of Inertia (memorize these)

Object	Axis	I
Solid disk/cylinder	Through center	½MR²
Hollow ring/hoop	Through center	MR²
Solid sphere	Through center	⅖MR²
Hollow sphere	Through center	⅔MR²
Thin rod	Through center	1/12 ML²
Thin rod	Through end	⅓ML²

Rolling Without Slipping

When a wheel rolls without slipping, the contact point has zero velocity. Total KE = ½mv² + ½Iω² (translational + rotational). Substitute ω = v/R to get total KE as a function of v alone.

Key Concepts

Torque (τ)τ = r × F = rF sinθ — rotational analog of force; causes angular acceleration.

Moment of inertia (I)Rotational analog of mass: I = Σmr² or ∫r² dm — depends on axis of rotation.

Parallel axis theoremI = I_cm + Md² — shifts the axis from center of mass to a parallel axis at distance d.

Angular momentum (L)L = Iω — conserved when net external torque = 0.

Conservation of LIf Στ_net = 0: L_initial = L_final. Skater pulling in arms → I decreases → ω increases.

Rolling without slippingContact point has v = 0; v_cm = ωR links translational and rotational motion.

Newton's Third Law and Angular Momentum

Veritasium · YouTube

Exam tip: The parallel axis theorem appears on virtually every rotation FRQ. Practice applying it: find I_cm from the formula sheet (given on the AP exam), then add Md² where d is the distance from the center of mass to the new rotation axis. For a rod rotating about its end: I_end = (1/12)ML² + M(L/2)² = (1/3)ML² — you can verify the formula this way.

Common mistake: Students forget the rotational KE term when solving rolling problems. For a sphere rolling down a ramp: mgh = ½mv² + ½Iω². Substitute I = (2/5)mr² and ω = v/r → mgh = ½mv² + (1/5)mv² = (7/10)mv² → v = √(10gh/7). If you forget the rotational term, you get v = √(2gh) — which is always wrong for rolling objects.

Unit 610–15% of exam

Oscillations

Simple harmonic motion — the differential equation, solution, energy analysis, and pendulum physics.

Simple Harmonic Motion (SHM) is the go-to model for anything that oscillates: springs, pendulums, LC circuits (in E&M), and molecular vibrations. Physics C requires you to recognize the SHM differential equation and solve it.

The SHM Condition

A system undergoes SHM when the restoring force is proportional to and opposite to the displacement: F = −kx. This gives the differential equation:

\frac{d^2x}{dt^2} = -\omega^2 x \qquad \omega^2 = \frac{k}{m} \\[14pt] \textbf{Solution:} \quad x(t) = A\cos(\omega t + \phi) \\[10pt] v(t) = -A\omega\sin(\omega t + \phi) \qquad a(t) = -A\omega^2\cos(\omega t + \phi) \\[14pt] T = \frac{2\pi}{\omega} \quad \textbf{Spring:}\; T = 2\pi\sqrt{\frac{m}{k}} \quad \textbf{Pendulum:}\; T = 2\pi\sqrt{\frac{L}{g}} \\[14pt] \textbf{Energy in SHM:} \quad E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2

Masses and Springs Simulation

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Masses and Springs Simulation

Hang different masses, change the spring constant, and observe how period changes. Confirm T = 2π√(m/k): doubling m multiplies T by √2, not 2. Also observe the energy transfer between KE and PE — total stays constant.

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Key Concepts

SHM conditiond²x/dt² = −ω²x — when restoring force is proportional to displacement.

Angular frequency (ω)ω = √(k/m) for spring-mass; ω = √(g/L) for simple pendulum.

Period (T)T = 2π/ω — time for one complete oscillation; independent of amplitude.

Amplitude (A)Maximum displacement from equilibrium; determines total energy E = ½kA².

Phase constant (φ)Sets the initial conditions: if x(0) = A → φ = 0; if x(0) = 0 and moving positive → φ = −π/2.

Energy in SHMTotal E = ½kA² is constant; converts between KE (at equilibrium) and PE (at turning points).

Exam tip: Period is independent of amplitude — this is one of the most tested SHM facts. Doubling the amplitude doubles the max speed and quadruples the total energy, but does NOT change T. The pendulum period is also independent of mass and depends only on length and g. Both of these counterintuitive results appear on the FRQ with 'justify your answer' prompts.

Exam prediction: This topic frequently appears on the AP Physics C: Mechanics exam. See our full AP Physics C: Mechanics predictions →

Common mistake: Don't confuse ω (angular frequency, rad/s) with f (frequency, Hz) or T (period, s). They're related: ω = 2πf = 2π/T. On FRQs, the differential equation d²x/dt² = −ω²x uses ω — if the problem gives you k and m, compute ω = √(k/m) first, then use that to write x(t).

Unit 710–14% of exam

Gravitation

Newton's law of universal gravitation, gravitational field and potential energy, and orbital mechanics derived from calculus.

Gravitation unifies terrestrial and celestial mechanics under one law. In Physics C, you must derive orbital speed, period, and energy from first principles — the AP exam FRQ frequently asks you to show the derivation step by step.

Newton's Law of Universal Gravitation

Every mass attracts every other mass with a force directed along the line connecting them:

F_g = \frac{Gm_1m_2}{r^2} \qquad G = 6.674\times10^{-11}\text{ N·m}^2/\text{kg}^2 \\[14pt] \textbf{Gravitational field:} \quad g(r) = \frac{GM}{r^2} \quad \text{(outside a sphere)} \\[14pt] \textbf{Gravitational PE:} \quad U_g = -\frac{Gm_1m_2}{r} \quad \text{(zero at } r\to\infty\text{)} \\[14pt] \textbf{Circular orbit derivation:} \quad F_g = \frac{mv^2}{r} \;\Rightarrow\; v_{orb} = \sqrt{\frac{GM}{r}} \\[14pt] \textbf{Kepler's Third Law:} \quad T^2 = \frac{4\pi^2}{GM}r^3

Gravitational Potential Energy — The Negative Convention

Uₘ = −Gm₁m₂/r is always negative (bound system). At r = ∞, U = 0 (the reference point). Bringing masses together (decreasing r) releases energy — U becomes more negative while KE increases. This is why satellites speed up as they fall inward.

Escape Velocity

The minimum speed to escape a planet's gravitational field (reach r = ∞ with KE = 0):

½mv² − GMm/R = 0 → v_escape = √(2GM/R)

Orbital Energy

Total mechanical energy of a circular orbit: E = KE + U = ½mv² − GMm/r = −GMm/(2r). This is always negative (bound orbit). A more negative total energy = tighter orbit = smaller radius.

Gravitational Field Strength vs. Distance

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Gravitational Field Strength vs. Distance

Shows g(r) = GM/r² for Earth outside its surface. Notice the 1/r² dependence — doubling your distance from Earth's center reduces g by a factor of 4. The vertical line marks Earth's surface radius. Gravity inside a uniform sphere decreases linearly, but this simulation shows the exterior.

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Key Concepts

Gravitational constant GG = 6.674 × 10⁻¹¹ N·m²/kg² — universal constant in Newton's gravity law.

Gravitational field gForce per unit mass: g = GM/r² (outside a uniform sphere); direction toward mass.

Gravitational PEU = −GMm/r; always negative; zero reference at r = ∞.

Orbital speedv = √(GM/r) for circular orbit — derived by setting F_g = mv²/r.

Escape velocityv_esc = √(2GM/R) — minimum speed to reach r = ∞ from planet surface.

Kepler's Third LawT² ∝ r³ — derived from F_g = mv²/r and T = 2πr/v.

Orbital energyE_total = −GMm/(2r) for circular orbit; negative = bound.

Exam tip: The orbital mechanics derivation is a guaranteed FRQ question type. Practice this from scratch: (1) set F_g = centripetal force: GMm/r² = mv²/r → v = √(GM/r); (2) find period: T = 2πr/v = 2πr/√(GM/r) = 2πr^(3/2)/√(GM); (3) square both sides → T² = 4π²r³/(GM). Always start from Newton's second law — never just state the final result.

Exam prediction: This topic frequently appears on the AP Physics C: Mechanics exam. See our full AP Physics C: Mechanics predictions →

Common mistake: Don't use Uₘ = mgh for gravitational PE in orbital mechanics — that approximation only works near Earth's surface where g is approximately constant. For satellite problems, rockets, or anything involving large changes in r, use U = −GMm/r. The sign matters critically: satellites in lower orbits have lower (more negative) total energy but higher speed.

AP Physics C: Mechanics Study Guide

Kinematics

The Calculus Definitions

When to Use Integration vs. Kinematic Equations

Newton's Laws of Motion

Free Body Diagrams — The Non-Negotiable First Step

Circular Motion

Work, Energy & Power

Work with Variable Force

Conservative vs. Non-Conservative Forces

Power

Systems of Particles & Linear Momentum

Center of Mass

Collision Types

Rotation

Rotational Analogs

Common Moments of Inertia (memorize these)

Rolling Without Slipping

Oscillations

The SHM Condition

Gravitation

Newton's Law of Universal Gravitation

Gravitational Potential Energy — The Negative Convention

Escape Velocity

Orbital Energy

Know exactly what to study for AP Physics C: Mechanics