Everything below prints as one AP AP Physics C: Mechanics practice paper set: papers A & B with their answer keys, plus the full-length study package exam. Use the Download PDF / Print button (or Cmd/Ctrl+P) to save it.
Paper A
AP Physics C: Mechanics — Practice Paper A
Original unofficial practice questions · paper A · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
The derivative of position gives
A. velocityB. accelerationC. momentumD. forceNewton's 2nd law states
A. F = maB. F = mvC. F = ma²D. F = m/aWork done by a variable force is the
A. integral of F dxB. derivative of FC. product of F and tD. sum of momentaMoment of inertia depends on
A. mass distribution about the axisB. only massC. only shape lengthD. temperatureRotational kinetic energy is
A. ½Iω²B. ½mv²C. IωD. mghGravitational force between masses is proportional to
A. 1/r²B. 1/rC. r²D. rThe momentum of a 2 kg object moving 5 m/s is
A. 10 kg·m/sB. 5 kg·m/sC. 20 kg·m/sD. 2.5Impulse equals
A. F ΔtB. mvC. ½mv²D. mgSimple harmonic motion has acceleration proportional to
A. displacement (opposite)B. velocityC. timeD. energyFor a system with no external torque, angular momentum is
A. conservedB. increasingC. zeroD. doublingSection II — Free Response
A 4 kg block on a frictionless surface is pushed by a force F(t) = 6t. Find the velocity at t = 3 s using F = dp/dt.
7 points · rubric: F=ma/dp 2 pts, integrate 3 pts, evaluate 2 pts.
A uniform rod (mass m, length L, free to pivot) tips from horizontal. Find its angular acceleration using torque about the pivot.
7 points · rubric: Torque 3 pts, I = ml²/3 2 pts, result 2 pts.
Answer Key
1. velocity — dx/dt = v.
2. F = ma — Law.
3. integral of F dx — Line/area integral.
4. mass distribution about the axis — Distribution matters.
5. ½Iω² — Rotational KE.
6. 1/r² — Inverse-square.
7. 10 kg·m/s — p = mv.
8. F Δt — Force × time.
9. displacement (opposite) — a = -ω²x.
10. conserved — Conservation law.
Free response — rubric notes
1. F=ma/dp 2 pts, integrate 3 pts, evaluate 2 pts. · model: a = 6t/4 = 1.5t; v = ∫ 1.5t dt = 0.75t²; at t=3, v = 6.75 m/s.
2. Torque 3 pts, I = ml²/3 2 pts, result 2 pts. · model: τ = mg·L/2; I = mL²/3; α = (3g)/(2L).
Paper B
AP Physics C: Mechanics — Practice Paper B
Original unofficial practice questions · paper B · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
The derivative of position gives
A. forceB. accelerationC. velocityD. momentumNewton's 2nd law states
A. F = mvB. F = ma²C. F = m/aD. F = maWork done by a variable force is the
A. derivative of FB. integral of F dxC. sum of momentaD. product of F and tMoment of inertia depends on
A. temperatureB. only shape lengthC. mass distribution about the axisD. only massRotational kinetic energy is
A. IωB. ½Iω²C. ½mv²D. mghGravitational force between masses is proportional to
A. r²B. rC. 1/r²D. 1/rThe momentum of a 2 kg object moving 5 m/s is
A. 5 kg·m/sB. 2.5C. 20 kg·m/sD. 10 kg·m/sImpulse equals
A. mgB. F ΔtC. ½mv²D. mvSimple harmonic motion has acceleration proportional to
A. velocityB. timeC. displacement (opposite)D. energyFor a system with no external torque, angular momentum is
A. increasingB. doublingC. zeroD. conservedSection II — Free Response
A 4 kg block on a frictionless surface is pushed by a force F(t) = 6t. Find the velocity at t = 3 s using F = dp/dt.
7 points · rubric: F=ma/dp 2 pts, integrate 3 pts, evaluate 2 pts.
A uniform rod (mass m, length L, free to pivot) tips from horizontal. Find its angular acceleration using torque about the pivot.
7 points · rubric: Torque 3 pts, I = ml²/3 2 pts, result 2 pts.
Answer Key
1. velocity — dx/dt = v.
2. F = ma — Law.
3. integral of F dx — Line/area integral.
4. mass distribution about the axis — Distribution matters.
5. ½Iω² — Rotational KE.
6. 1/r² — Inverse-square.
7. 10 kg·m/s — p = mv.
8. F Δt — Force × time.
9. displacement (opposite) — a = -ω²x.
10. conserved — Conservation law.
Free response — rubric notes
1. F=ma/dp 2 pts, integrate 3 pts, evaluate 2 pts. · model: a = 6t/4 = 1.5t; v = ∫ 1.5t dt = 0.75t²; at t=3, v = 6.75 m/s.
2. Torque 3 pts, I = ml²/3 2 pts, result 2 pts. · model: τ = mg·L/2; I = mL²/3; α = (3g)/(2L).
Full-length study package exam
AP Physics C: Mechanics — Practice Exam
Section 1: Multiple Choice (35 questions, 45 minutes)
Use g = 10 m/s² unless otherwise stated. Select the best answer.
1. The position of a particle is x(t) = 2t³ − 9t² + 12t − 1. At t = 2 s, the particle's acceleration is: (A) 0 m/s² (B) 3 m/s² (C) 6 m/s² (D) 12 m/s²
2. A ball is thrown straight up with speed v₀. At what height is its speed half of v₀? (A) 3v₀²/(8g) (B) v₀²/(2g) (C) 3v₀²/(4g) (D) v₀²/(4g)
3. A 5 kg block on a frictionless surface is pushed by a force F = 20t N (time in seconds). The velocity at t = 3 s (starting from rest) is: (A) 6 m/s (B) 12 m/s (C) 18 m/s (D) 36 m/s
4. A force F = (3x² + 2) N acts on a 2 kg particle moving along the x-axis. The work done from x = 0 to x = 2 m is: (A) 8 J (B) 10 J (C) 12 J (D) 14 J
5. Two objects undergo a perfectly inelastic collision. The fraction of initial KE lost is: (A) Always 1/2 (B) Always 1 (C) Depends on the masses (D) Depends on the velocities
6. The moment of inertia of a solid sphere of mass M and radius R about a tangent line is: (A) 2MR²/5 (B) 3MR²/5 (C) 7MR²/5 (D) 2MR²/3
7. A disk rolls without slipping down a 30° incline. The acceleration is: (A) g/3 (B) 2g/3 (C) 5g/14 (D) g/2
8. A mass on a spring oscillates with period T. If the spring constant is quadrupled, the new period is: (A) T/4 (B) T/2 (C) 2T (D) 4T
9. A satellite in circular orbit at radius r has orbital energy E. If the radius is doubled, the new energy is: (A) E/2 (B) 2E (C) E/4 (D) 4E
10. The escape velocity from a planet of mass M and radius R is v_e. For a planet of mass 2M and radius 2R: (A) v_e/√2 (B) v_e (C) v_e√2 (D) 2v_e
11. A particle moves with velocity v = (3t i + 2 j) m/s. The magnitude of acceleration is: (A) 0 (B) 2 m/s² (C) 3 m/s² (D) √13 m/s²
12. In projectile motion, the acceleration vector: (A) Is tangent to the trajectory (B) Has constant magnitude and direction
(C) Has constant magnitude, changing direction (D) Changes in both magnitude and direction
13. A block of mass m is on a frictionless incline of angle θ. The acceleration down the incline is: (A) g sin θ (B) g cos θ (C) g tan θ (D) g/sin θ
14. The potential energy function U(x) = 3x² − x³ has stable equilibrium at: (A) x = 0 only (B) x = 2 only (C) x = 0 and x = 2 (D) x = −2 and x = 0
15. A 0.1 kg ball moving at 10 m/s is caught by a 0.9 kg glove initially at rest. The final velocity is: (A) 0.1 m/s (B) 1.0 m/s (C) 5.0 m/s (D) 10 m/s
16. A hoop and a solid cylinder of the same mass and radius are released from rest at the top of an incline. Which reaches the bottom first? (A) Hoop (B) Solid cylinder (C) They arrive together (D) Depends on the angle
17. Angular momentum is conserved when: (A) Net force is zero (B) Net torque is zero (C) Net force and torque are both zero (D) Kinetic energy is conserved
18. A simple pendulum has period 2 s. If its length is quadrupled, the new period is: (A) 1 s (B) 2 s (C) 4 s (D) 8 s
19. Inside a uniform solid sphere at r = R/2, the gravitational field is: (A) GM/(4R²) (B) GM/(2R²) (C) Gm/(R²) (D) 0
20. The total energy of a mass-spring system is E = ½kA². If the amplitude is doubled, the energy: (A) Doubles (B) Triples (C) Quadruples (D) Increases by √2
Section 2: Free Response (3 questions, 45 minutes)
Question 1: Kinematics and Dynamics (15 minutes)
A particle of mass m = 2.0 kg moves along the x-axis under the influence of a force F(x) = −4x + 2 (in N, x in m). At t = 0, the particle is at x = 0 with velocity v₀ = 3.0 m/s.
(a) Write the differential equation of motion and identify the type of motion.
(b) Find the equilibrium position and determine whether it is stable or unstable.
(c) Find the total energy of the system.
(d) Describe the subsequent motion qualitatively, including the amplitude.
Question 2: Rotation and Energy (15 minutes)
A uniform solid cylinder of mass M = 4.0 kg and radius R = 0.2 m has a light string wrapped around it. The string passes over a massless, frictionless pulley and is attached to a hanging block of mass m = 2.0 kg. The cylinder rotates about a fixed horizontal axis through its center. The system is released from rest.
(a) Draw free-body diagrams for both the cylinder and the hanging mass.
(b) Write the equations of motion for the cylinder (rotational) and the block (translational).
(c) Solve for the acceleration of the block and the tension in the string.
(d) After the block falls 1.0 m, find the angular velocity of the cylinder.
Question 3: Gravitation and Energy (15 minutes)
A spacecraft of mass m = 1000 kg is in a circular orbit of radius r₁ = 7.0 × 10⁶ m around Earth (M_E = 6.0 × 10²⁴ kg). It fires its engines to transfer to a circular orbit of radius r₂ = 1.05 × 10⁷ m.
(a) Calculate the orbital speed in each orbit.
(b) Calculate the total mechanical energy in each orbit.
(c) How much work must the engines do to accomplish this transfer? (Assume an idealized Hohmann-like maneuver.)
(d) In which orbit is the gravitational potential energy higher? Explain physically.
End of Practice Exam. Allow 90 minutes total: 45 for MCQ, 45 for FRQ.
Answer Key & Rubric
AP Physics C: Mechanics — Practice Exam Solutions
Section 1: Multiple Choice Solutions
1. (C) a = d²x/dt² = d²(2t³ − 9t² + 12t − 1)/dt² = 12t − 18. At t = 2: a = 24 − 18 = 6 m/s².
2. (A) v² = v₀² − 2gΔh. Set v = v₀/2: v₀²/4 = v₀² − 2gΔh → 2gΔh = 3v₀²/4 → Δh = 3v₀²/(8g).
3. (C) a = F/m = 4t. v = ∫₀³ 4t dt = 2t²|₀³ = 18 m/s.
4. (C) W = ∫₀² (3x² + 2) dx = [x³ + 2x]₀² = 8 + 4 = 12 J.
5. (C) Fraction lost = m₁m₂(m₁ − m₂)²/[(m₁ + m₂)²(m₁v₁ᵢ² + m₂v₂ᵢ²)/2], which depends on the mass ratio.
6. (C) By parallel axis theorem: I_tangent = I_CM + MR² = 2MR²/5 + MR² = 7MR²/5.
7. (B) For a disk: a = g sin θ/(1 + I/mR²) = g sin 30°/(1 + 1/2) = (g/2)/(3/2) = g/3. Wait: sin 30° = 0.5. a = (10)(0.5)/1.5 = 10/3. The answer is (B) 2g/3 = 20/3 ≈ 6.67... No. Let me recalculate: a = g sin θ × 1/(1 + 1/2) = (g sin θ)(2/3). With g = 10, sin 30° = 0.5: a = 10 × 0.5 × 2/3 = 10/3. Hmm, 10/3 = 3.33. That's g/3. (A) g/3.
8. (B) T = 2π√(m/k). If k → 4k: T → T/2.
9. (A) E = −GMm/(2r). If r → 2r: E → E/2.
10. (B) v_e = √(2GM/R). For 2M, 2R: v_e = √(2G·2M/2R) = √(2GM/R) = v_e. Unchanged.
11. (B) a = dv/dt = 3 i. |a| = 3 m/s². (C).
12. (B) Projectile acceleration is g downward: constant magnitude, constant direction.
13. (A) N = mg cos θ, net force along incline = mg sin θ. a = g sin θ.
14. (B) F = −dU/dx = −6x + 3x². Equilibrium: F = 0 → x(3x − 6) = 0 → x = 0, x = 2. dF/dx = −6 + 6x. At x = 0: dF/dx = −6 < 0 → stable. At x = 2: dF/dx = 6 > 0 → unstable. Only x = 0 is stable. (A).
15. (B) Conservation of momentum: (0.1)(10) = (0.1 + 0.9)v → v = 1.0 m/s.
16. (B) Solid cylinder has I = ½MR² < I_hoop = MR², so less KE goes to rotation, more to translation. The cylinder wins.
17. (B) Angular momentum is conserved when net external torque is zero.
18. (C) T = 2π√(L/g). If L → 4L: T → 2T.
19. (B) g = GMr/R³ = GM(R/2)/R³ = GM/(2R²).
20. (C) E = ½kA². If A → 2A: E → 4E.
Section 2: Free Response Solutions
Question 1
(a) m(d²x/dt²) = −4x + 2 = −4(x − 0.5)
d²x/dt² = −2(x − 0.5)
Let x' = x − 0.5: d²x'/dt² = −2x'. This is SHM with ω² = 2, ω = √2 rad/s.
(b) Equilibrium at F = 0: −4x + 2 = 0 → x_eq = 0.5 m.
dF/dx = −4 < 0, so this is stable equilibrium (force is restoring).
(c) At t = 0: x = 0, v = 3.0 m/s.
E = ½mv² + ½k_eff(x − x_eq)² = ½(2)(9) + ½(4)(0.25) = 9 + 0.5 = 9.5 J
(where k_eff = 4 from the linearized force F = −k_eff(x − x_eq))
(d) The motion is SHM about x_eq = 0.5 m with angular frequency ω = √2 rad/s.
Amplitude: E = ½k_eff A² → 9.5 = ½(4)A² → A = √4.75 ≈ 2.18 m.
The particle oscillates between x ≈ −1.68 m and x ≈ 2.68 m about the equilibrium at 0.5 m.
Question 2
(a) FBDs:
- Cylinder: Tension T pulling tangent at top (provides torque). Weight Mg down at center. Normal force N up at axle.
- Block: Weight mg down. Tension T up.
(b) For the block: mg − T = ma ... (1)
For the cylinder: TR = Iα = (½MR²)(a/R) → T = ½Ma ... (2)
(The string constraint gives a = Rα.)
(c) Substitute (2) into (1):
mg − ½Ma = ma → a = mg/(m + M/2) = (2)(10)/(2 + 2) = 20/4 = 5.0 m/s²
T = ½Ma = ½(4)(5) = 10 N
(d) Energy conservation: mgΔy = ½mv² + ½Iω²
(2)(10)(1) = ½(2)v² + ½(½)(4)(0.2)²(v/0.2)²
20 = v² + ½(2)(0.04)(v²/0.04) = v² + v² = 2v²
v² = 10, v = √10 m/s
ω = v/R = √10/0.2 = 5√10 = 15.8 rad/s
Question 3
(a) v = √(GM/r)
v₁ = √(6.67 × 10⁻¹¹ × 6 × 10²⁴/7 × 10⁶) = √(5.72 × 10⁷) = 7,563 m/s
v₂ = √(6.67 × 10⁻¹¹ × 6 × 10²⁴/1.05 × 10⁷) = √(3.81 × 10⁷) = 6,175 m/s
(b) E = −GMm/(2r)
E₁ = −(6.67 × 10⁻¹¹)(6 × 10²⁴)(1000)/(2 × 7 × 10⁶) = −2.86 × 10¹⁰ J
E₂ = −(6.67 × 10⁻¹¹)(6 × 10²⁴)(1000)/(2 × 1.05 × 10⁷) = −1.91 × 10¹⁰ J
(c) Work = ΔE = E₂ − E₁ = (−1.91 + 2.86) × 10¹⁰ = 9.5 × 10⁹ J
(d) The higher orbit (r₂) has higher (less negative) potential energy. Physically, moving to a higher orbit means doing work against gravity, increasing the gravitational PE. The total energy also increases (is less negative), even though the kinetic energy decreases.
Score yourself: each MCQ is worth about 1.4 points, each FRQ part is worth 3–4 points. A raw score of ~85–95 typically earns a 5.