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AP Physics C Mechanics Formula Sheet

Physics formula sheetsAP Physics CCalculus basedUpdated 1 Nov 2026

∑F⃗=dp⃗dt\sum \vec{F} = \frac{d\vec{p}}{dt}
where
  • ΣF net external force N
  • p linear momentum, mv kg m/s
  • t time s

This AP Physics C mechanics formula sheet is a calculus based study guide to the mechanics course, built on Newton's second law in the form ∑F⃗=dp⃗dt\sum \vec{F} = \frac{d\vec{p}}{dt}. For constant mass it reduces to ∑F=ma\sum F = ma, but the derivative form also handles rockets, conveyor belts and anything else whose momentum changes in less tidy ways.

It is a summary for revision and not the exam board's own sheet, which comes with the exam along with a table of constants. The tables follow the usual order of the course: motion, forces, energy, momentum, rotation, oscillations and gravitation. For the second C course, see the E&M sheet; for how the two fit together, see the AP Physics C overview or the formula sheets hub.

Quick facts

  • What this is

    A study guide, not the official equation table
  • Core law

    ∑F⃗=dp⃗dt\sum \vec{F} = \frac{d\vec{p}}{dt}
  • Energy link

    Fx=−dUdxF_x = -\frac{dU}{dx}
  • Rotation link

    ∑τ=Iα=dLdt\sum \tau = I\alpha = \frac{dL}{dt}
  • Math used

    Derivatives, definite integrals, simple differential equations
  • Value of g

    9.8 m/s² unless told otherwise

Kinematics and forces

Motion and Newton's laws in calculus form
RelationUseSI unitExplained on
v=dxdt, a=dvdt=d2xdt2v = \frac{dx}{dt},\ a = \frac{dv}{dt} = \frac{d^2x}{dt^2}velocity and acceleration from x(t)m/s, m/s²Instantaneous velocity
x=x0+∫v dtx = x_0 + \int v\,dtposition from a velocity functionmDisplacement
∑F⃗=ma⃗\sum \vec{F} = m\vec{a}constant mass dynamicsNNet force
Fdrag=−bvF_{drag} = -bvlinear drag at low speedNDrag force
v(t)=vT(1−e−t/τ), τ=mbv(t) = v_T\left(1 - e^{-t/\tau}\right),\ \tau = \frac{m}{b}falling with linear drag, from restm/sTerminal velocity
vT=mgbv_T = \frac{mg}{b}terminal speed with linear dragm/sTerminal velocity

The drag result comes from solving mdvdt=mg−bvm\frac{dv}{dt} = mg - bv, a first order differential equation. Expect to set it up, separate the variables and sketch the exponential approach to vTv_T.

Work, energy, power and momentum

Energy and momentum relations
RelationUseSI unitExplained on
W=∫x1x2Fx dxW = \int_{x_1}^{x_2} F_x\,dxwork when F changes with xJWork
ΔK=Wnet\Delta K = W_{net}work energy theoremJWork energy theorem
Fx=−dUdxF_x = -\frac{dU}{dx}force from a potential energy curveNPotential energy
Us=12kx2, UG=−Gm1m2rU_s = \frac{1}{2}kx^2,\ U_G = -\frac{Gm_1m_2}{r}spring and gravitational potential energyJSpring energy
P=dWdt=F⃗⋅v⃗P = \frac{dW}{dt} = \vec{F}\cdot\vec{v}instantaneous powerWWork and energy
J⃗=∫F⃗ dt=Δp⃗\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}impulse of a time varying forceN sImpulse
xcm=∫x dm∫dmx_{cm} = \frac{\int x\,dm}{\int dm}center of mass (centre of mass) of a continuous bodymCenter of mass

Example 1: work by a variable force

A force F(x)=6x2F(x) = 6x^2 newtons (x in metres) pushes a 1.5 kg block from rest at x = 0 to x = 1.5 m on a frictionless surface. Find the work done and the final speed.

W=∫01.56x2 dx=[2x3]01.5=2(3.375)=6.75 JW = \int_0^{1.5} 6x^2\,dx = \left[2x^3\right]_0^{1.5} = 2(3.375) = 6.75\ \text{J}.

Work energy theorem: 12(1.5)v2=6.75\frac{1}{2}(1.5)v^2 = 6.75, so v2=9.0v^2 = 9.0 and v=3.0 m/sv = 3.0\ \text{m/s}.

Answer: 6.75 J (about 6.8 J) and 3.0 m/s.

Rotation

Rotational dynamics and angular momentum
RelationUseSI unitExplained on
ω=dθdt, α=dωdt\omega = \frac{d\theta}{dt},\ \alpha = \frac{d\omega}{dt}angular velocity and accelerationrad/s, rad/s²Angular velocity
τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F}torque as a cross productN mTorque
I=∫r2 dmI = \int r^2\,dmrotational inertia by integrationkg m²Moment of inertia
I=Icm+Md2I = I_{cm} + Md^2inertia about a shifted, parallel axiskg m²Moment of inertia
∑τ=Iα\sum\tau = I\alphaNewton's second law for rotationN mAngular acceleration
L⃗=r⃗×p⃗, L=Iω\vec{L} = \vec{r}\times\vec{p},\ L = I\omegaangular momentum of a particle and of a rigid bodykg m²/sAngular momentum
Krot=12Iω2K_{rot} = \frac{1}{2}I\omega^2kinetic energy of spinningJKinetic energy

Example 2: rotational inertia of a rod about one end

Find I for a uniform rod of mass 0.60 kg and length 1.2 m, pivoted at one end.

Mass per length λ=M/L\lambda = M/L, so I=∫0LMLx2 dx=13ML2I = \int_0^L \frac{M}{L}x^2\,dx = \frac{1}{3}ML^2.

I=13(0.60)(1.2)2=0.8643=0.288 kg m2I = \frac{1}{3}(0.60)(1.2)^2 = \frac{0.864}{3} = 0.288\ \text{kg m}^2. Answer: about 0.29 kg m². Check with the parallel axis theorem: 112ML2+M(L2)2=13ML2\frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2.

Oscillations and gravitation

Simple harmonic motion and orbits
RelationUseSI unitExplained on
x(t)=Acos⁡(ωt+ϕ)x(t) = A\cos(\omega t + \phi)position in simple harmonic motionmAmplitude
ω=km, T=2πmk\omega = \sqrt{\frac{k}{m}},\ T = 2\pi\sqrt{\frac{m}{k}}mass on a springrad/s, sPeriod
T=2πℓgT = 2\pi\sqrt{\frac{\ell}{g}}simple pendulum, small anglessPeriod
T=2πImgdT = 2\pi\sqrt{\frac{I}{mgd}}physical pendulum, d from pivot to center of masssFrequency
FG=Gm1m2r2F_G = \frac{Gm_1m_2}{r^2}universal gravitationNGravitation
v=GMrv = \sqrt{\frac{GM}{r}}speed in a circular orbitm/sOrbital velocity

Any system whose equation of motion has the form d2xdt2=−ω2x\frac{d^2x}{dt^2} = -\omega^2 x is simple harmonic, and ω can be read straight off. That is the quickest route to the period of an unfamiliar oscillator.

Exam habits for AP Physics C mechanics formula sheet questions

  • State the law, then the calculus. Write ΣF = ma or ΣF = dp/dt before you integrate.
  • Choose dm carefully. For a rod use λ dx, for a disk use σ 2πr dr.
  • Keep signs in F = −dU/dx. Force points downhill on a U against x graph.
  • Check the limits. A drag answer should give v = 0 at t = 0 and v → v_T at large t.
If a question says a quantity is proportional to time or position, it is telling you to integrate rather than use the constant acceleration equations.

Frequently asked questions

Is this the exam board's physics C mechanics formula sheet?

No. It is an independent study guide. The official equation tables are provided with the exam, so practise past questions with that version too.

What topics are on the AP Physics C mechanics exam?

In general terms: kinematics, Newton's laws, work, energy and power, linear momentum, rotation and angular momentum, oscillations, and gravitation. Check the current course outline for details.

Where do I get the AP Physics C: Mechanics formula sheet itself?

From the exam board, which publishes the current equation tables in its course materials and provides them in the exam. This page explains the same physics in plain words.

Do I need integrals for AP Physics C mechanics?

Yes. Typical uses are work from a varying force, impulse from a varying force, center of mass and rotational inertia of continuous bodies, and motion with drag.

How does AP Physics C mechanics compare with AP Physics 1?

They cover much of the same mechanics, but AP Physics C uses calculus throughout and goes into more depth on topics such as drag, variable forces and rotational inertia.