AP Physics C: Mechanics FRQ Practice Guide

title: "AP Physics C: Mechanics FRQ Practice Guide" description: "Master the 3 FRQ archetypes: mechanics with calculus, rotation with conservation, gravitation with orbits. Includes integration cheat-sheet and worked examples." date: "2026-01-15" examDate: "May AP Exam" topics:

• FRQ Patterns
• Integration Techniques
• Worked Examples
• Claim-Evidence-Reasoning

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 $a$ to get $v$ and $x$, or differentiate energy to find force).
2. Rotation or angular momentum — compute moment of inertia via integration, apply $\\tau = I\\alpha$, or conserve $L$.
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 $x = 0$ with $v_0 = 0$. It experiences acceleration $a(t) = 4t \\, \\text{m/s}^2$ for $0 \\leq t \\leq 3$ seconds, then $a = 0$ for $t > 3$.

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

Solution

(a) Velocity $v(t)$:

For $0 \\leq t \\leq 3$: $v(t) = v_0 + \\int_0^t a(\\tau) \\, d\\tau = 0 + \\int_0^t 4\\tau \\, d\\tau = 2t^2$

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

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

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

For $0 \\leq t \\leq 3$: $x(t) = 0 + \\int_0^t v(\\tau) \\, d\\tau = \\int_0^t 2\\tau^2 \\, d\\tau = \\frac{2t^3}{3}$

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

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

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

Worked Example 1B: Work and Force from Potential Energy

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

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

Solution

(a) Force from potential:

$F(x) = -\\frac{dU}{dx} = -\\frac{d}{dx}(3x^2 - 2x) = -(6x - 2) = 2 - 6x$

(b) Speed at $x = 2$:

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

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

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

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

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

Archetype 2: Rotation and Angular Momentum

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

Worked Example 2: Moment of Inertia from Integration

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

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

Solution

(a) Moment of inertia:

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

$I = \\int r^2 \\, dm = \\int_0^L x^2 \\cdot \\frac{M}{L} \\, dx = \\frac{M}{L} \\int_0^L x^2 \\, dx = \\frac{M}{L} \\cdot \\frac{L^3}{3} = \\frac{ML^2}{3}$

(b) Angular acceleration:

From $\\tau = I\\alpha$: $\\alpha = \\frac{\\tau_0}{I} = \\frac{\\tau_0}{\\frac{ML^2}{3}} = \\frac{3\\tau_0}{ML^2}$

Archetype 3: Gravitation and Orbital Mechanics

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

Worked Example 3: Escape Velocity and Orbital Speed

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

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

Solution

(a) Escape velocity:

Energy conservation: at the surface, $KE_i + U_i = KE_f + U_f$. Take $U_f = 0$ at infinity.

$\\frac{1}{2}mv_{esc}^2 - \\frac{GMm}{R} = 0$

$v_{esc} = \\sqrt{\\frac{2GM}{R}} = \\sqrt{\\frac{2 \\times 6.67 \\times 10^{-11} \\times 6 \\times 10^{24}}{6.4 \\times 10^6}} \\approx 11.2 \\, \\text{km/s}$

(b) Orbital speed at height $h$:

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

$\\frac{GMm}{r^2} = \\frac{mv^2}{r}$

$v = \\sqrt{\\frac{GM}{r}} = \\sqrt{\\frac{6.67 \\times 10^{-11} \\times 6 \\times 10^{24}}{10^7}} \\approx 6.3 \\, \\text{km/s}$

Integration Cheat-Sheet

| Integral | Result | Common Physics Use | |---|---|---| | $\\int t \\, dt$ | $\\frac{t^2}{2} + C$ | Velocity from constant acceleration | | $\\int t^2 \\, dt$ | $\\frac{t^3}{3} + C$ | Position from $a(t) = kt$ | | $\\int e^{-kt} \\, dt$ | $-\\frac{1}{k}e^{-kt} + C$ | Drag force decay | | $\\int r \\, dr$ | $\\frac{r^2}{2} + C$ | Moment of inertia (cylindrical shell) | | $\\int r^2 \\, dr$ | $\\frac{r^3}{3} + C$ | Moment of inertia (thin spherical shell) | | $\\int \\frac{dr}{r^2}$ | $-\\frac{1}{r} + C$ | Gravitational potential energy (leading to $-GMm/r$) |

Differentiation Cheat-Sheet

| Function | Derivative | Common Physics Use | |---|---|---| | $x(t) = A\\cos(\\omega t + \\phi)$ | $v = -A\\omega\\sin(\\omega t + \\phi)$ | SHM velocity | | $U(x)$ | $F = -\\frac{dU}{dx}$ | Force from potential energy | | $p(t)$ | $F = \\frac{dp}{dt}$ | Force from momentum | | $E(t) = \\frac{1}{2}mv^2 + U(x)$ | $\\frac{dE}{dt} = F_x v_x + F_y v_y$ | Power (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 $v_f = 5 \\, \\text{m/s}$").
• 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 $x = 2$ is $v = 3 \\, \\text{m/s}$.

Evidence: Using energy conservation, $KE_i + U_i = KE_f + U_f$: $0 + 0 = \\frac{1}{2}(2)v^2 + 3$ $v = \\sqrt{3} \\approx 1.73 \\, \\text{m/s}$

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 →.