AP Physics C: Mechanics — Cheatsheet
Formulas, exam-day tips, and key terms on one page.
Formulas & relationships
Velocity as a derivative
v(t) = dx/dt = lim (Δt→0) Δx / Δt
The slope of the position–time graph at an instant. Units: metres per second.
Acceleration as a derivative
a(t) = dv/dt = d²x/dt²
First derivative of velocity; second derivative of position. Units: metres per second².
Building motion back up by integration
v(t) = v₀ + ∫ a dt x(t) = x₀ + ∫ v dt
Each indefinite integral introduces a constant; v₀ and x₀ are those constants, set by the initial conditions.
Constant-a kinematics as integral results
v = v₀ + at x = x₀ + v₀t + ½at²
Valid ONLY when a is constant. If a depends on t, integrate a(t) directly instead.
Component form of the motion vectors
r = x î + y ĵ v = (dx/dt) î + (dy/dt) ĵ a = (dvₓ/dt) î + (dv_y/dt) ĵ
A vector of magnitude V at angle θ resolves as Vₓ = V cos θ and V_y = V sin θ; recombine with |V| = √(Vₓ² + V_y²).
Relative-velocity addition
v(A/C) = v(A/B) + v(B/C)
Read the subscripts like a chain: A-relative-to-C equals A-relative-to-B plus B-relative-to-C. The inner label B cancels. Reversing a pair flips the sign: v(B/A) = −v(A/B).
Newton’s second law
ΣF = ma = m dv/dt
The net (vector) force equals mass times acceleration. Applied one axis at a time: ΣFₓ = maₓ and ΣF_y = ma_y.
Friction force
f_k = μ_k N f_s ≤ μ_s N
Kinetic friction has fixed magnitude μ_k N once sliding. Static friction adjusts up to a maximum μ_s N to prevent sliding; it equals whatever is needed below that cap.
Falling object with linear drag
m dv/dt = mg − bv
Down is positive. Weight mg pulls down; drag bv pushes up and grows as v grows. This first-order differential equation governs the whole descent.
Terminal velocity
v_t = mg / b (linear drag) v_t = √(mg / c) (quadratic drag)
Found by setting dv/dt = 0 so the resistive force equals the weight. No calculus needed for v_t itself — only the balance condition.
Atwood machine
a = (m₂ − m₁)g / (m₁ + m₂) T = 2 m₁ m₂ g / (m₁ + m₂)
Two masses hang over an ideal pulley. The heavier mass m₂ falls, the lighter m₁ rises, both with the same |a|. The tension is the same throughout the single string.
Universal gravitation
F = G m₁ m₂ / r²
Attractive, directed along the line between the masses. G = 6.67 × 10⁻¹¹ N·m²/kg². Doubling either mass doubles F; doubling r cuts F to one quarter.
Surface gravity
g = GM / R²
The free-fall acceleration at a spherical planet of mass M and radius R. At a distance r > R from the center, the local acceleration is GM/r².
Work as a line integral
W = ∫ F·dx (constant force: W = Fd cos θ)
The dot product keeps only the force component along the displacement. Units: joules (1 J = 1 N·m). Area under an F-versus-x graph.
Work-energy theorem
W_net = ΔKE = ½mv_f² − ½mv_i²
The net work on an object equals its change in kinetic energy. Positive net work speeds it up; negative net work slows it down.
Potential energy functions
U_grav = mgh U_spring = ½kx²
Each is minus the work done by the conservative force. Potential energy is always measured relative to a chosen reference where U = 0.
Force from potential energy
F = −dU/dx
The force is the negative slope of the potential energy curve. Equilibrium occurs where dU/dx = 0; it is stable at a minimum of U, unstable at a maximum.
Conservation of mechanical energy
KE_i + U_i = KE_f + U_f
Valid when only conservative forces (gravity, springs) do work. Pick a reference level for U and apply it consistently at both instants.
Energy with a nonconservative force
KE_i + U_i + W_nc = KE_f + U_f
W_nc is the (usually negative) work of nonconservative forces. For friction over a distance d, W_nc = −f·d, draining mechanical energy into heat.
Average and instantaneous power
P_avg = W / Δt P_inst = dW/dt = F·v
Instantaneous power is the force dotted with the velocity, so a force perpendicular to the motion delivers zero power. Units: watts (W).
Momentum and Newton's second law
p = mv ΣF = dp/dt
Momentum is a vector (kg·m/s). The net force is the time rate of change of momentum; when m is constant this reduces to ΣF = ma.
Impulse-momentum theorem
J = ∫ F dt = Δp = m v_f − m v_i
Impulse (N·s) equals the change in momentum. Graphically it is the area under a force-versus-time curve; an average force gives J = F_avg Δt.
Conservation of momentum
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Total momentum before equals total momentum after. It is a vector equation — apply it component by component, keeping track of signs.
Perfectly inelastic collision
m₁v₁ + m₂v₂ = (m₁ + m₂) v_f
The objects stick and share one final velocity v_f. Momentum is conserved; kinetic energy is not.
Center of mass
x_cm = (Σ mᵢ xᵢ) / (Σ mᵢ) x_cm = (1/M) ∫ x dm
Discrete on the left, continuous on the right. For a continuous body, express dm through the density (dm = λ dx for a rod) and integrate.
Thrust and the equation of motion
Thrust = v_ex |dm/dt| m dv/dt = −v_ex dm/dt
v_ex is the exhaust speed relative to the rocket; dm/dt < 0 as fuel is expelled, so the thrust is a forward force of magnitude v_ex times the burn rate.
Ideal rocket equation
Δv = v_ex ln(m_i / m_f)
The speed gained depends on the exhaust speed and the natural log of the initial-to-final mass ratio. Valid with no external forces (deep space).
Angular velocity and acceleration
ω = dθ/dt α = dω/dt = d²θ/dt²
Angular position θ in radians, ω in rad/s, α in rad/s². The direct analogs of v = dx/dt and a = dv/dt.
Linking linear and angular motion
v = rω a_t = rα a_c = ω²r = v²/r
Multiply an angular quantity by the radius to get the corresponding linear one at that point. The centripetal term a_c points toward the center.
Torque
τ = rF sin θ = r_⊥ F
r is the distance from the axis to the point of application; sin θ picks out the perpendicular component. Maximum when the force is applied perpendicular to the radius.
Moment of inertia
I = ∫ r² dm (hoop: MR²; disk: ½MR²; rod about center: 1/12 ML²; sphere: 2/5 MR²)
r is the distance of each mass element from the rotation axis. The same object has different I about different axes.
Newton's second law for rotation
τ_net = Iα
The net torque about an axis equals the moment of inertia about that axis times the angular acceleration. Solve for α = τ_net/I.
Rolling-without-slipping constraint
v_cm = Rω a_cm = Rα
The center-of-mass motion and the rotation are locked together by the wheel radius R. If the wheel slips, this constraint no longer holds.
Acceleration rolling down an incline
a_cm = g sin θ / (1 + I/MR²)
Only the shape factor I/MR² matters. Smaller I/MR² (mass nearer the axis) rolls faster; the plain block with no rotation (ratio 0) accelerates fastest of all at g sin θ.
Rotational and total kinetic energy
KE_rot = ½Iω² KE_total = ½mv_cm² + ½Iω²
The rotational twin of ½mv². For rolling, add the translational and rotational parts, linked by v_cm = Rω.
Rotational work and power
W = ∫ τ dθ P = τω
Torque integrated over angle gives work; torque times angular velocity gives power. Both mirror their translational forms exactly.
Angular momentum
L = Iω (particle: L = rmv sin θ)
Rigid body on the left, single particle on the right. Units: kg·m²/s. A vector directed along the rotation axis.
Torque as the rate of change of angular momentum
τ_net = dL/dt
The rotational twin of F = dp/dt. When τ_net = 0, L is conserved.
Conservation of angular momentum
L = Iω = constant I₁ω₁ = I₂ω₂
Holds when the net external torque is zero. Reducing I (pulling mass inward) speeds up the spin; increasing I slows it.
Circular orbit
GMm/r² = mv²/r → v = √(GM/r) T² = (4π²/GM) r³
Orbital speed falls as 1/√r. The period relation T² ∝ r³ is Kepler's third law, obtained by combining v = √(GM/r) with T = 2πr/v.
Angular momentum at the apses
L = mvr = constant → v_p r_p = v_a r_a
At perihelion and aphelion the velocity is perpendicular to the radius, so L = mvr directly. The closer approach forces the faster speed.
The SHM equation and its solution
d²x/dt² = −ω²x x(t) = A cos(ωt + φ)
A is the amplitude, ω the angular frequency, φ the phase constant set by initial conditions. The period is T = 2π/ω.
Period, frequency, and angular frequency
T = 2π/ω f = 1/T ω = 2πf
ω is in rad/s, f in hertz, T in seconds. Together they describe how fast the oscillation repeats, independent of amplitude.
Mass-spring oscillator
ω = √(k/m) T = 2π√(m/k) f = 1/T
Angular frequency rises with stiffness and falls with mass. The period depends only on m and k — not on how far the spring is pulled.
Simple pendulum
ω = √(g/L) T = 2π√(L/g)
Valid in the small-angle limit sin θ ≈ θ. Remarkably, the period depends on length and g only — not on the bob's mass or (for small swings) the amplitude.
Physical pendulum
T = 2π√(I / (mgd))
I is the moment of inertia about the pivot; d is the pivot-to-center-of-mass distance. Reduces to the simple-pendulum result when the mass is concentrated at distance L.
Energy in SHM
E = ½kA² = ½kx² + ½mv² v_max = Aω = A√(k/m)
Total energy is fixed by the amplitude. The speed is greatest at equilibrium (v_max = Aω) and zero at the turning points.
On the exam
- On the AP exam, given x(t) you should be able to produce v(t) and a(t) instantly by differentiating, and read the reverse too: where the x–t slope is zero the object is momentarily at rest, and where the v–t slope is zero the acceleration is zero.
- Keep the subscript bookkeeping strict: write every velocity as v(object/frame). The chain v(A/C) = v(A/B) + v(B/C) only works when the inner labels match, and swapping any pair introduces a sign flip. Getting the subscripts right turns most relative-motion questions into simple addition.
- On the AP exam, draw the free-body diagram first and tilt your axes to match the motion. Write ΣF = ma along each axis separately. Getting the components of weight right — mg sin θ along a slope, mg cos θ into it — is where most points are won or lost.
- Keep the two formulas straight: F = Gm₁m₂/r² is the force between two masses, while g = GM/R² is the field (acceleration) one mass produces at distance R. Both are inverse-square — whenever a distance changes, square the ratio before scaling.
- Move fluently in both directions: integrate a conservative force to get U (U = −∫F dx), and differentiate U to get the force (F = −dU/dx). On a U-versus-x graph, remember the force points downhill and equilibria sit where the slope is zero.
- Use P = W/Δt when you know a total amount of work over an interval, and P = Fv (or F·v) when you want the power at a specific instant or speed. A force perpendicular to the velocity — like the centripetal force on a satellite — does zero work and delivers zero power.
- For a continuous body, the recipe is always the same: write dm using the density, integrate x dm for the numerator and dm for the total mass M, then divide. Never average the endpoints — that only works for a uniform object.
- The rocket equation comes from momentum conservation with changing mass, not from F = ma with constant mass. Watch the logarithm: the payoff for carrying more fuel diminishes, since Δv grows only as ln(m_i/m_f).
- Two objects with identical mass can have very different moments of inertia. Always ask where the mass sits relative to the axis: the r² weighting in I = ∫r² dm rewards mass at the rim and penalizes it near the center.
- For rolling problems, always bring in the constraint a_cm = Rα to connect the force equation to the torque equation. Remember that the shape factor I/MR² alone decides the race down an incline — mass and radius drop out.
- Keep L = Iω and τ = dL/dt distinct: one is the angular momentum, the other is the rule for how torque changes it. The single most useful consequence is that zero net external torque forces angular momentum to stay constant.
- For circular orbits set gravity equal to the centripetal force (GMm/r² = mv²/r). For elliptical orbits, use conservation of angular momentum (v_p r_p = v_a r_a) at the apses and conservation of energy between them — energy and angular momentum together pin down the whole orbit.
- State the small-angle approximation explicitly on free-response: sin θ ≈ θ is what turns d²θ/dt² = −(g/L)sin θ into SHM. Remember the simple-pendulum period is independent of mass, and use the physical-pendulum formula T = 2π√(I/mgd) whenever the swinging object is an extended body.
How to get a 5
- Always define your coordinate system and direction of positive rotation. Consistency between linear (a) and rotational (α) signs is crucial in rolling problems.
- Don't memorize specific derived formulas; practice deriving them from Newton's Laws and Energy Conservation. You will need to show the steps on the FRQ.
- In variable mass or density problems, always relate dm to dx (or dr) using density (λ, σ, ρ). e.g., dm = λ dx.
- Check your boundary conditions and limits of integration carefully when doing work or center of mass integrals.
Key terms
Kinematic Equations (Calculus) — v(t) = ∫a(t)dt, x(t) = ∫v(t)dt, a = dv/dt, v = dx/dt. Also a = v(dv/dx).
Work (Variable Force) — W = ∫ F(x) dx. The area under the Force vs. Position graph.
Potential Energy from Force — F = -dU/dx. Force is the negative gradient (slope) of potential energy.
Center of Mass (Continuous) — x_cm = (1/M) ∫ x dm. Use a linear mass density λ = dm/dx to integrate.
Moment of Inertia (Continuous) — I = ∫ r² dm. Common results: Solid disk = ½MR², Hoop = MR², Solid sphere = (2/5)MR², Rod (center) = (1/12)ML².
Parallel Axis Theorem — I = I_cm + Md². Used to find I about an axis parallel to the CM axis.
Torque (Cross Product) — τ = r × F. Magnitude is rF sinθ. Direction given by right-hand rule.
Angular Momentum (Particle) — L = r × p. Magnitude is mvr sinθ (or mv_perp r).
Conservation of Angular Momentum — If net external torque is zero, L_initial = L_final (I_i ω_i = I_f ω_f).
Rolling without Slipping — v_cm = Rω and a_cm = Rα. Static friction does no work, so mechanical energy is conserved.
Simple Harmonic Motion Equation — d²x/dt² = -ω²x. The general solution is x(t) = A cos(ωt + φ).
Newton's Law of Universal Gravitation — F_g = G(m₁m₂)/r². Potential energy U_g = -G(m₁m₂)/r.