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AP Physics C: Mechanics FRQ Practice Guide

How to attack free-response questions and earn easy partial credit.

⏱️ ~3-4 hours of focused work🎯 AP Physics C: Mechanics

The AP Physics C: Mechanics exam has 3 free response questions, each worth 15% of your total score. Every FRQ follows one of three archetypes:

  1. Mechanics with calculus — kinematics, dynamics, or energy (integrate aa to get vv and xx, or differentiate energy to find force).
  2. Rotation or angular momentum — compute moment of inertia via integration, apply τ=Iα\tau = I\alpha, or conserve LL.
  3. Gravitation or orbital mechanics — escape velocity, orbital speed, or Kepler's laws.

This guide walks through each archetype with a worked example, then gives you the integration and differentiation cheat-sheet you'll need.

Archetype 1: Mechanics with Calculus

Setup: You're given acceleration (or force) as a function of time or position. You must integrate to find velocity and position, or differentiate position to find velocity and acceleration.

Worked Example 1A: Particle Motion from Variable Acceleration

Problem: A particle starts at x=0x = 0 with v0=0v_0 = 0. It experiences acceleration a(t)=4t m/s2a(t) = 4t \, \text{m/s}^2 for 0≤t≤30 \leq t \leq 3 seconds, then a=0a = 0 for t>3t > 3.

(a) Find v(t)v(t) for all tt. (b) Find x(3)x(3) and the maximum position. (c) Sketch x(t)x(t), v(t)v(t), a(t)a(t).

Solution

(a) Velocity v(t)v(t):

For 0≤t≤30 \leq t \leq 3: v(t)=v0+∫0ta(τ) dτ=0+∫0t4τ dτ=2t2v(t) = v_0 + \int_0^t a(\tau) \, d\tau = 0 + \int_0^t 4\tau \, d\tau = 2t^2

At t=3t = 3: v(3)=2(9)=18 m/sv(3) = 2(9) = 18 \, \text{m/s}.

For t>3t > 3: a=0a = 0, so v=18 m/sv = 18 \, \text{m/s} (constant).

(b) Position x(t)x(t) and maximum:

For 0≤t≤30 \leq t \leq 3: x(t)=0+∫0tv(τ) dτ=∫0t2τ2 dτ=2t33x(t) = 0 + \int_0^t v(\tau) \, d\tau = \int_0^t 2\tau^2 \, d\tau = \frac{2t^3}{3}

At t=3t = 3: x(3)=2⋅273=18 mx(3) = \frac{2 \cdot 27}{3} = 18 \, \text{m}.

For t>3t > 3: x(t)=18+18(t−3)x(t) = 18 + 18(t - 3). Position increases without bound. There is no maximum (the particle moves at constant velocity forever).

💡 Note: The graph of x(t)x(t) is concave up for t≤3t \leq 3 (because a>0a > 0), then linear for t>3t > 3.


Worked Example 1B: Work and Force from Potential Energy

Problem: A conservative force acts on a 2 kg mass. The potential energy is U(x)=3x2−2xU(x) = 3x^2 - 2x (in joules, for xx in meters).

(a) Find the force F(x)F(x). (b) If the mass starts at x=0x = 0 with v=0v = 0, what is its speed at x=2 mx = 2 \, \text{m}?

Solution

(a) Force from potential:

F(x)=−dUdx=−ddx(3x2−2x)=−(6x−2)=2−6xF(x) = -\frac{dU}{dx} = -\frac{d}{dx}(3x^2 - 2x) = -(6x - 2) = 2 - 6x

(b) Speed at x=2x = 2:

Use energy conservation: E=KE+U=E = KE + U = constant.

At x=0x = 0: E=12(2)(0)2+3(0)2−2(0)=0E = \frac{1}{2}(2)(0)^2 + 3(0)^2 - 2(0) = 0.

At x=2x = 2: 0=12(2)v2+3(4)−2(2)0 = \frac{1}{2}(2)v^2 + 3(4) - 2(2), so 0=v2+12−4=v2+80 = v^2 + 12 - 4 = v^2 + 8.

This gives v2=−8v^2 = -8, which is impossible. The mass cannot reach x=2x = 2 starting from rest at x=0x = 0 with this potential — the potential energy barrier is too high.

⚠️ Trap: Always check that energy conservation makes sense. If v2<0v^2 < 0, the particle is trapped in a potential well.


Archetype 2: Rotation and Angular Momentum

Setup: Compute moment of inertia via integration, apply τ=Iα\tau = I\alpha, or conserve angular momentum.

Worked Example 2: Moment of Inertia from Integration

Problem: A uniform thin rod of mass MM and length LL is rotated about an axis perpendicular to the rod and passing through one end.

(a) Derive the moment of inertia II. (b) The rod is attached to a motor that applies a constant torque τ0\tau_0. Find the angular acceleration α\alpha.

Solution

(a) Moment of inertia:

Set up a coordinate xx along the rod, with x=0x = 0 at the pivot (the end). A small segment dxdx has mass dm=MLdxdm = \frac{M}{L} dx (linear mass density λ=M/L\lambda = M/L).

I=∫r2 dm=∫0Lx2⋅ML dx=ML∫0Lx2 dx=ML⋅L33=ML23I = \int r^2 \, dm = \int_0^L x^2 \cdot \frac{M}{L} \, dx = \frac{M}{L} \int_0^L x^2 \, dx = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3}

(b) Angular acceleration:

From τ=Iα\tau = I\alpha: α=τ0I=τ0ML23=3τ0ML2\alpha = \frac{\tau_0}{I} = \frac{\tau_0}{\frac{ML^2}{3}} = \frac{3\tau_0}{ML^2}


Archetype 3: Gravitation and Orbital Mechanics

Setup: Apply Newton's gravitational law, energy conservation with gravitational PE, or orbital kinematics (a=v2/ra = v^2/r).

Worked Example 3: Escape Velocity and Orbital Speed

Problem: Earth has mass M=6×1024 kgM = 6 \times 10^{24} \, \text{kg} and radius R=6.4×106 mR = 6.4 \times 10^6 \, \text{m}.

(a) Derive the escape velocity from Earth's surface. (b) A satellite orbits at height h=3.6×106 mh = 3.6 \times 10^6 \, \text{m} above the surface. Find its orbital speed.

Solution

(a) Escape velocity:

Energy conservation: at the surface, KEi+Ui=KEf+UfKE_i + U_i = KE_f + U_f. Take Uf=0U_f = 0 at infinity.

12mvesc2−GMmR=0\frac{1}{2}mv_{esc}^2 - \frac{GMm}{R} = 0

vesc=2GMR=2×6.67×10−11×6×10246.4×106≈11.2 km/sv_{esc} = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^6}} \approx 11.2 \, \text{km/s}

(b) Orbital speed at height hh:

At orbital radius r=R+h=6.4×106+3.6×106=107 mr = R + h = 6.4 \times 10^6 + 3.6 \times 10^6 = 10^7 \, \text{m}, gravity provides centripetal force:

GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}

v=GMr=6.67×10−11×6×1024107≈6.3 km/sv = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{10^7}} \approx 6.3 \, \text{km/s}


Integration Cheat-Sheet

IntegralResultCommon Physics Use
∫t dt\int t \, dtt22+C\frac{t^2}{2} + CVelocity from constant acceleration
∫t2 dt\int t^2 \, dtt33+C\frac{t^3}{3} + CPosition from a(t)=kta(t) = kt
∫e−kt dt\int e^{-kt} \, dt−1ke−kt+C-\frac{1}{k}e^{-kt} + CDrag force decay
∫r dr\int r \, drr22+C\frac{r^2}{2} + CMoment of inertia (cylindrical shell)
∫r2 dr\int r^2 \, drr33+C\frac{r^3}{3} + CMoment of inertia (thin spherical shell)
∫drr2\int \frac{dr}{r^2}−1r+C-\frac{1}{r} + CGravitational potential energy (leading to −GMm/r-GMm/r)

Differentiation Cheat-Sheet

FunctionDerivativeCommon Physics Use
x(t)=Acos⁡(ωt+ϕ)x(t) = A\cos(\omega t + \phi)v=−Aωsin⁡(ωt+ϕ)v = -A\omega\sin(\omega t + \phi)SHM velocity
U(x)U(x)F=−dUdxF = -\frac{dU}{dx}Force from potential energy
p(t)p(t)F=dpdtF = \frac{dp}{dt}Force from momentum
E(t)=12mv2+U(x)E(t) = \frac{1}{2}mv^2 + U(x)dEdt=Fxvx+Fyvy\frac{dE}{dt} = F_x v_x + F_y v_yPower (rate of energy change)

FRQ Scoring: Claim–Evidence–Reasoning

College Board rubrics follow a CER structure:

  • Claim: State what you are finding (e.g., "The final velocity is vf=5 m/sv_f = 5 \, \text{m/s}").
  • Evidence: Show your work — the integral, the conservation law you cited, the calculation.
  • Reasoning: Explain why your approach is correct (e.g., "Mechanical energy is conserved because only the conservative gravitational force does work").

Example FRQ response structure:

Claim: The speed of the particle at x=2x = 2 is v=3 m/sv = 3 \, \text{m/s}.

Evidence: Using energy conservation, KEi+Ui=KEf+UfKE_i + U_i = KE_f + U_f: 0+0=12(2)v2+30 + 0 = \frac{1}{2}(2)v^2 + 3 v=3≈1.73 m/sv = \sqrt{3} \approx 1.73 \, \text{m/s}

Reasoning: Mechanical energy is conserved because gravity is a conservative force and no friction acts.


Practice Routine

  1. Pick one archetype per day. Solve the worked example from memory (no looking).
  2. Solve the problem yourself, then check the solution.
  3. Identify the integral or conservation law. What made this problem tick?
  4. Write a brief CER response. 2–3 sentences on why your approach is correct.
  5. Repeat with a different problem in the same archetype until you feel confident.

After practicing all three archetypes, mix them: on a single mock exam, you'll see one of each, randomly ordered.


Ready to practice? Return to the 3-day cram → or check the last-minute formulas →.

Other AP Physics C: Mechanics study plans