Ten calculus-based lessons across the seven units: kinematics, forces, work and energy, momentum, rotation and oscillations, with derivations and graph reasoning.
A reasoning guide to AP Physics C: Mechanics, not a full course. Unit titles and weightings follow the College Board course page; all numbers in examples are invented for practice, with g = 10 m/s^2 for convenience.
Calculus (derivatives and integrals) at the same time or before, and basic algebra-based physics.
Course outline
Kinematics with calculus: position, velocity and acceleration
Relate position, velocity and acceleration by differentiation and integration, in one dimension.
Projectile motion and vectors
Resolve vectors into components and solve two-dimensional motion with independent axes.
Newton's laws and free-body diagrams
Draw free-body diagrams and apply the second law to single objects and connected systems.
Friction, circular motion and drag forces
Apply friction models, centripetal acceleration and velocity-dependent drag to dynamics problems.
Work, kinetic energy and power
Compute work as an integral and apply the work-energy theorem and power.
Potential energy and conservation of energy
Link force to potential energy, use energy conservation and interpret potential energy graphs.
Linear momentum, impulse and collisions
Apply impulse-momentum, conservation of momentum, center of mass and collision analysis.
Torque and rotational dynamics
Compute torque and moment of inertia and apply the rotational form of Newton's second law.
Rotational energy, angular momentum and rolling
Use rotational kinetic energy, angular momentum conservation and rolling constraints.
Oscillations: simple harmonic motion, springs and pendulums
Derive and use simple harmonic motion for springs and small-angle pendulums.
Sources and curriculum note
Reviewed October 6, 2026. Confirm format on the College Board site for your exam year.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Kinematics with calculus: position, velocity and acceleration
Learning goal: Relate position, velocity and acceleration by differentiation and integration, in one dimension.
Velocity is the derivative of position, v = dx/dt, and acceleration is the derivative of velocity, a = dv/dt. Going the other way, velocity is the integral of acceleration plus an initial velocity, and position is the integral of velocity plus an initial position. AP Physics C expects you to use calculus directly, not only the constant acceleration equations.
On a graph, the slope of a position-time graph is velocity and the slope of a velocity-time graph is acceleration. The area under a velocity-time graph is displacement, and the area under an acceleration-time graph is change in velocity. Negative areas count as negative displacement. Distance traveled adds the absolute areas.
For x(t) = 2t^3, the velocity is 6t^2 and acceleration is 12t. At t = 2 the velocity is 24 m/s and acceleration is 24 m/s^2. If a(t) is given, integrate twice and use two initial conditions.
The constant acceleration equations, v = v0 + at and x = x0 + v0 t + (1/2) a t^2, are special cases when a does not depend on time. They fail when acceleration changes. Always state the sign convention and the positive direction first.
Worked example
x(t) = 2t^3 meters (invented). Find the velocity at t = 2 s.
v = dx/dt = 6t^2.
At t = 2: 6(4).
v = 24.
Velocity is 24 m/s.
Practice problem and solution
v(t) = 4t m/s with x(0) = 0. Find x(3) in meters.
x = 2t^2, so x(3) = 18.
Mental model: Differentiate to go down, integrate to go up. Use calculus when a varies.
Common trap: Using constant acceleration equations when a depends on time.
2. Projectile motion and vectors
Learning goal: Resolve vectors into components and solve two-dimensional motion with independent axes.
Vectors have magnitude and direction. Resolve each into x and y components and handle each axis separately. In projectile motion the horizontal acceleration is 0 and the vertical acceleration is -g, with air resistance ignored unless the problem says otherwise.
Time links the axes. A projectile launched horizontally at 8 m/s from 45 m up falls for t where 45 = (1/2)(10)t^2, so t = 3 s and the horizontal distance is 8 x 3 = 24 m. We use g = 10 m/s^2 for practice; the exam gives its own value.
For an angled launch, vx = v cos(theta) and vy = v sin(theta). At the top of the trajectory the vertical velocity is 0 but the horizontal velocity is unchanged. Maximum range on level ground comes at 45 degrees.
Describe motion with graphs: vx is constant, vy is a straight line with slope -g. Check the final speed by combining components. Free-response answers should show the equation in symbols before substituting numbers.
Worked example
A ball rolls off a 45 m cliff at 8 m/s (invented, g = 10). How far from the base does it land?
Vertical: 45 = 5t^2.
t = 3 s.
x = 8(3).
24 m.
Practice problem and solution
An object is dropped from 20 m with g = 10. Find the fall time in seconds.
20 = 5t^2, so t = 2.
Mental model: Axes are independent. Time links them. Acceleration at the top is g.
Common trap: Setting acceleration to zero at the highest point.
3. Newton's laws and free-body diagrams
Learning goal: Draw free-body diagrams and apply the second law to single objects and connected systems.
Newton's second law says the net force on an object equals its mass times its acceleration: sum of F = ma, as vectors. Start with a free-body diagram showing only forces on the object, not forces it exerts. Choose axes along the acceleration and write one equation per axis.
For connected objects, write the second law for each object and link them with the constraint, such as equal accelerations through a rope. In an Atwood machine, with masses 3 kg and 1 kg and g = 10, the system acceleration is (3 - 1)(10)/(3 + 1) = 5 m/s^2.
Tension in an ideal rope is the same throughout. Normal force is perpendicular to the surface and is not always equal to mg; on an incline it is mg cos(theta). Newton's third law pairs forces on different objects and these pairs never cancel on a single free-body diagram.
Equilibrium means zero net force, not zero velocity. A constant-velocity object is in equilibrium. Check answers for limiting cases, for example masses equal should give zero acceleration.
Worked example
An Atwood machine has 3 kg and 1 kg masses (invented, g = 10). Find the acceleration.
Net driving force: (3 - 1)(10) = 20 N.
Total mass: 4 kg.
a = 20/4.
a = 5 m/s^2.
Practice problem and solution
A net force of 12 N acts on a 3 kg block. Find the acceleration in m/s^2.
a = F/m = 12/3 = 4.
Mental model: Diagram first. One equation per axis. Third law pairs are on different objects.
Common trap: Drawing the reaction force on the same diagram as the action.
4. Friction, circular motion and drag forces
Learning goal: Apply friction models, centripetal acceleration and velocity-dependent drag to dynamics problems.
Kinetic friction has magnitude mu_k N and opposes motion; static friction is at most mu_s N and adjusts to prevent sliding. Friction is not always mu N, so identify whether the object is sliding. On an incline, the block starts to slide when tan(theta) = mu_s.
In uniform circular motion, the acceleration points to the center with magnitude v^2/r. The net inward force is m v^2/r. It is not a new force; label the real forces, such as tension, friction or normal force, that provide it. For a flat turn, the maximum speed is sqrt(mu g r), which is 10 m/s for mu = 0.5, r = 20 m, g = 10.
Drag forces that depend on speed, such as F = -bv, require a differential equation: m dv/dt = mg - bv for a falling object. Separation gives v = (mg/b)(1 - e^(-bt/m)). Terminal velocity occurs when acceleration is 0, so v_t = mg/b.
Check limiting cases: at t = 0, v = 0; at large t, v approaches v_t. The time constant m/b describes how quickly the object approaches terminal speed. AP Physics C asks you to derive these results, not just quote them.
Worked example
Find terminal velocity for m = 2 kg, b = 4 kg/s, g = 10 (invented).
Terminal when a = 0.
mg = bv.
v = 20/4.
5 m/s.
Practice problem and solution
m = 2 kg, b = 4 kg/s, g = 10 m/s^2. Find terminal speed in m/s.
v_t = mg/b = 20/4 = 5.
Mental model: Net inward force is mv^2/r. Terminal speed sets a = 0.
Common trap: Drawing centripetal force as an extra arrow.
5. Work, kinetic energy and power
Learning goal: Compute work as an integral and apply the work-energy theorem and power.
Work is the integral of force along the path: W = integral of F dot dx. For a constant force at angle theta, W = F d cos(theta). Work is a scalar, can be negative, and is zero when the force is perpendicular to the motion. For a variable force F = 4x, work from 0 to 3 m is the integral of 4x dx, which is 18 J.
The work-energy theorem says the net work equals the change in kinetic energy: W_net = delta K, with K = (1/2) m v^2. It applies to any net force, constant or not, and it follows from integrating Newton's second law. The area under a force-position graph is work.
Power is the rate of doing work, P = dW/dt, and P = F dot v for a force acting on a moving object. A 50 N force on an object at 4 m/s delivers 200 W. Average power is total work over time.
Springs: the force is F = -kx, and the work done by the spring from 0 to x is -(1/2) k x^2, with stored energy (1/2) k x^2. State which force does the work and the sign convention.
Worked example
F = 4x newtons acts from x = 0 to x = 3 m (invented). Find the work.
W = integral of 4x dx.
= 2x^2.
At 3: 18.
Work is 18 J.
Practice problem and solution
F = 2x newtons from x = 0 to 4 m. Find the work in joules.
The integral of 2x is x^2, and 4^2 = 16.
Mental model: Work is an integral. Net work gives delta K. Power is F dot v.
Common trap: Using one force's work as the net work.
6. Potential energy and conservation of energy
Learning goal: Link force to potential energy, use energy conservation and interpret potential energy graphs.
A conservative force has work that depends only on endpoints, and it has an associated potential energy U with F = -dU/dx. Gravity near Earth gives U = mgh, and a spring gives U = (1/2) k x^2. Friction is non-conservative; it removes mechanical energy as heat.
For U = 3x^2, the force is F = -6x. At x = 2 the force is -12 N, pointing toward smaller x, toward the minimum. Equilibrium points are where F = 0, so dU/dx = 0; stable at minima and unstable at maxima.
Conservation of mechanical energy says K + U is constant when only conservative forces do work. A ball dropped from 5 m reaches v = sqrt(2 g h) = 10 m/s with g = 10. With friction, K_i + U_i + W_nc = K_f + U_f.
On a U versus x graph, total energy is a horizontal line. Kinetic energy is the gap between the line and the curve, and turning points are where the line meets the curve. The particle cannot enter regions where U exceeds the total energy.
Worked example
U = 3x^2 joules (invented). Find the force at x = 2 m.
F = -dU/dx.
dU/dx = 6x.
F = -6(2).
Force is -12 N.
Practice problem and solution
A ball falls from rest from 20 m with g = 10. What is its speed in m/s at the bottom?
v = sqrt(2 x 10 x 20) = sqrt(400) = 20.
Mental model: F = -dU/dx. Energy lines show turning points. Friction removes mechanical energy.
Common trap: Forgetting the minus sign in F = -dU/dx.
7. Linear momentum, impulse and collisions
Learning goal: Apply impulse-momentum, conservation of momentum, center of mass and collision analysis.
Momentum is p = mv. Impulse is the integral of force over time and equals the change in momentum: J = integral of F dt = delta p. The area under a force-time graph is impulse. A long contact time lowers the force for the same delta p.
In an isolated system, with no net external force, total momentum is conserved. A 2 kg object at 6 m/s striking a 4 kg object at rest and sticking gives (2)(6) = (6)v, so v = 2 m/s. Momentum is conserved in all collisions; kinetic energy is conserved only in elastic ones.
In an elastic collision, relative speed of approach equals relative speed of separation. In an inelastic collision, some kinetic energy becomes thermal energy or deformation, and in a perfectly inelastic collision the objects move together. Check energy to classify.
The center of mass is x_cm = (sum m_i x_i)/(sum m_i), and for continuous bodies an integral with density. The center of mass moves as if all external force acted on a point mass. Use it to simplify systems that explode or collide.
Worked example
A 2 kg mass at 6 m/s hits a 4 kg mass at rest and they stick (invented). Find the final speed.
Initial momentum: 12.
Total mass: 6.
v = 12/6.
v = 2 m/s.
Practice problem and solution
A 3 kg mass at 4 m/s hits a 1 kg mass at rest and they stick. Find the speed in m/s.
12/(3 + 1) = 3.
Mental model: Isolated means momentum is conserved. Impulse is force-time area.
Common trap: Applying kinetic energy conservation to a sticking collision.
8. Torque and rotational dynamics
Learning goal: Compute torque and moment of inertia and apply the rotational form of Newton's second law.
Torque is the rotational effect of a force: tau = r x F, with magnitude r F sin(theta), where theta is the angle between r and F. Only the component perpendicular to the lever arm matters. The sign is by direction of rotation, counterclockwise positive by convention.
Moment of inertia measures resistance to angular acceleration, I = integral of r^2 dm. Standard values: a solid disk about its center is (1/2) M R^2, a thin ring is M R^2, a thin rod about its center is (1/12) M L^2 and about an end is (1/3) M L^2. Mass farther from the axis counts more.
The rotational second law is the sum of tau = I alpha. A 10 N force at 0.5 m, perpendicular, on an object with I = 2.5 kg m^2 gives tau = 5 N m and alpha = 2 rad/s^2. For rolling objects pair this with the translational second law and a no-slip constraint a = alpha R.
The parallel-axis theorem, I = I_cm + M d^2, shifts the axis by distance d. Use it for rods about an end and for composite objects. Static equilibrium needs both zero net force and zero net torque about any chosen axis.
Worked example
A 10 N force acts 0.5 m from the axis, perpendicular; I = 2.5 kg m^2 (invented). Find alpha.
tau = 0.5(10) = 5 N m.
alpha = tau/I.
5/2.5.
alpha = 2 rad/s^2.
Practice problem and solution
A torque of 12 N m acts on I = 4 kg m^2. Find alpha in rad/s^2.
alpha = 12/4 = 3.
Mental model: Torque needs the perpendicular component. tau = I alpha. Pick the axis wisely.
Common trap: Using the full force instead of the component perpendicular to r.
9. Rotational energy, angular momentum and rolling
Learning goal: Use rotational kinetic energy, angular momentum conservation and rolling constraints.
Rotational kinetic energy is K = (1/2) I omega^2. Angular momentum of a rigid body is L = I omega, and for a point mass L = r m v sin(theta). The rate of change of L equals net torque, which mirrors F = dp/dt.
If net external torque on a system is 0, angular momentum is conserved. An ice skater with I = 4 and omega = 3 has L = 12; pulling the arms in to I = 2 gives omega = 6. Kinetic energy rises because the skater does work pulling arms inward.
For rolling without slipping, v = omega R and the total kinetic energy is (1/2) m v^2 + (1/2) I omega^2. Objects with smaller I/(m R^2) get more translational speed, so a solid disk beats a ring down the same incline. Static friction acts but does no work on the rolling object.
Rotating collisions, such as a clay ball hitting a rod, conserve angular momentum about the pivot but not linear momentum since the pivot exerts force. Always say which axis you use for L and why torque from external forces about that axis is zero.
Worked example
A skater has I = 4 kg m^2 and omega = 3 rad/s, then I becomes 2 (invented). Find the new omega.
L = 12.
L is conserved.
omega = 12/2.
6 rad/s.
Practice problem and solution
I = 6 kg m^2, omega = 2 rad/s; I decreases to 3 with no external torque. Find the new omega in rad/s.
L = 12, so omega = 12/3 = 4.
Mental model: Conserve L when torque is zero. Rolling splits energy between translation and rotation.
Common trap: Forgetting the rotational term in the energy of rolling.
10. Oscillations: simple harmonic motion, springs and pendulums
Learning goal: Derive and use simple harmonic motion for springs and small-angle pendulums.
Simple harmonic motion arises when the restoring force is proportional to displacement: F = -kx. Newton's second law gives d^2x/dt^2 = -(k/m) x, with solution x = A cos(omega t + phi), where omega = sqrt(k/m). The period is T = 2 pi/omega, independent of amplitude.
For a spring with m = 1 kg and k = 4 N/m, omega = 2 rad/s and T = pi s. With amplitude 0.5 m the maximum speed is A omega = 1 m/s at the center, and the maximum acceleration is A omega^2 = 2 m/s^2 at the ends. Velocity is a quarter period out of phase with position.
A simple pendulum approximates SHM for small angles, because sin(theta) is close to theta. Its period is T = 2 pi sqrt(L/g), independent of mass. A physical pendulum uses T = 2 pi sqrt(I/(m g d)), where d is the distance from the pivot to the center of mass.
Energy in SHM swaps between kinetic and potential: total E = (1/2) k A^2. At the center K is maximum and U is zero; at the ends K is zero. Vertical springs shift the equilibrium point but not the period.
Worked example
m = 1 kg, k = 4 N/m, A = 0.5 m (invented). Find the maximum speed.
omega = sqrt(4/1) = 2.
v_max = A omega.
0.5 x 2.
1 m/s.
Practice problem and solution
A spring has k = 100 N/m and m = 4 kg. Find omega in rad/s.
sqrt(100/4) = sqrt(25) = 5.
Mental model: F = -kx gives SHM. Period is independent of amplitude. Energy is 1/2 k A^2.
Common trap: Thinking a larger amplitude means a longer period.