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AP Physics C: Mechanics mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 35 questions and 4 free-response questions · about 140 minutes

A full-length practice test built from our exam-style questions on every unit of AP Physics C: Mechanics. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page
  1. Before you start
  2. Section I
  3. Section II
  4. After the mock

Before you start

Section I: 35 exam-style questions

Q1Exam-style
A particle moves with v(t) = 4t3 − 3t2 m/s. Find its displacement from t = 0 to t = 2 s.
m

Hint

Displacement is the integral of velocity.

Worked solution
  1. Δx = ∫02 (4t3 − 3t2) dt = [t4 − t3]02.
  2. Δx = 16 − 8 = 8 m.
Q2Exam-style
A ball is launched at 20 m/s at 30° above the horizontal from level ground. Find its maximum height.
m

Hint

Only the vertical component matters for height.

Worked solution
  1. Vertical component: 20 sin 30° = 10 m/s.
  2. At the top vy = 0: h = vy2/(2g) = 100/19.6.
  3. h = 5.10 m.
Q3Exam-style
A particle starts at x = 0 with velocity 1 m/s and has acceleration a(t) = 2t m/s². Find its position at t = 3 s.
m

Hint

Integrate twice, using both initial conditions.

Worked solution
  1. v = t2 + 1 (using v(0) = 1).
  2. x = t3/3 + t (using x(0) = 0).
  3. x(3) = 9 + 3 = 12 m.
Q4Exam-style
A particle moves with x(t) = t3 − 6t2 + 9t. At what times is it momentarily at rest?

Hint

At rest means v = 0.

Worked solution
  1. v = 3t2 − 12t + 9 = 3(t − 1)(t − 3).
  2. v = 0 at t = 1 s and t = 3 s: B.
Q5Challenge
A ball is thrown horizontally at 12 m/s from the top of a cliff 45 m high. How far from the base of the cliff does it land?
m

Hint

Find the time to fall 45 m from the vertical motion; the horizontal velocity stays 12 m/s.

Worked solution
  1. Vertically: 45 = ½(9.8)t2, so t = 3.03 s.
  2. Horizontally: x = 12 × 3.03 = 36.4 m.
Q6Exam-style
Masses of 4 kg and 6 kg hang on either side of a light frictionless pulley. Find the magnitude of their acceleration.
m/s²

Hint

Treat the two masses as one system pulled by the difference in weights.

Worked solution
  1. Net force on the system: (6 − 4)g = 19.6 N; total mass 10 kg.
  2. a = 19.6/10 = 1.96 m/s².
Q7Exam-style
A block slides down a frictionless incline at 30° to the horizontal. Find its acceleration.
m/s²

Hint

Only the component of gravity along the incline accelerates it.

Worked solution
  1. a = g sin 30°.
  2. a = 9.8 × 0.5 = 4.9 m/s².
Q8Exam-style
A 0.20 kg object falls through a fluid that exerts a drag force F = −bv with b = 0.50 kg/s. Find its terminal speed.
m/s

Hint

At terminal speed the acceleration is zero.

Worked solution
  1. At terminal speed: mg = bvT.
  2. vT = 0.20 × 9.8/0.50 = 3.92 m/s.
Q9Exam-style
A 2.0 kg object starts from rest and is pushed by a force F(t) = 6t N along a frictionless surface. Find its speed at t = 2 s.
m/s

Hint

a(t) = F/m. Integrate to find v(t).

Worked solution
  1. a = 6t/2 = 3t, so v = 1.5t2.
  2. v(2) = 1.5 × 4 = 6 m/s.
Q10Challenge
A 2.0 kg puck slides on a horizontal surface with a resistive force F = −bv, b = 0.40 kg/s, and no other horizontal forces. Its initial speed is 10 m/s. Find its speed after 5.0 s.
m/s

Hint

Write m dv/dt = −bv, separate the variables and integrate.

Worked solution
  1. m dv/dt = −bv gives dv/v = −(b/m) dt, so v = v0e−bt/m.
  2. bt/m = 0.40 × 5.0/2.0 = 1.0.
  3. v = 10e−1 = 3.68 m/s.
Q11Exam-style
A spring with k = 200 N/m is compressed 0.10 m and launches a 0.050 kg ball horizontally on a frictionless surface. Find the ball's launch speed.
m/s

Hint

Spring energy becomes kinetic energy.

Worked solution
  1. ½kx2 = ½mv2, so v = x√(k/m).
  2. v = 0.10 × √(200/0.050) = 0.10 × 63.2.
  3. v = 6.32 m/s.
Q12Exam-style
A particle has potential energy U(x) = 4x2 − x4 joules, with x in metres. Find the force on it at x = 1 m.
N

Hint

F = −dU/dx.

Worked solution
  1. dU/dx = 8x − 4x3, which is 4 at x = 1.
  2. F = −dU/dx = −4 N, in the negative x direction.
Q13Exam-style
A car travels at a constant 20 m/s against a total resistive force of 600 N. Find the power its engine delivers.
W

Hint

At constant speed the driving force equals the resistance.

Worked solution
  1. Driving force = 600 N.
  2. P = Fv = 600 × 20 = 12000 W (12 kW).
Q14Exam-style
On a graph of potential energy U against position x, a particle is in stable equilibrium where:

Hint

F = −dU/dx. What does the force do if the particle moves slightly away?

Worked solution
  1. At a minimum, dU/dx = 0, so F = 0.
  2. Moving either way makes F point back towards the minimum, so the equilibrium is stable: B.
Q15Challenge
A block slides from rest down a frictionless ramp from a height of 2.0 m, then across a rough horizontal floor with μk = 0.25. How far along the floor does it slide before stopping?
m

Hint

All the gravitational energy is removed by friction on the floor.

Worked solution
  1. mgh = μkmg d, and m and g cancel.
  2. d = h/μk = 2.0/0.25 = 8 m.
Q16Exam-style
A 2.0 kg cart moving at 3.0 m/s collides with and sticks to a 1.0 kg cart at rest. Find their common speed.
m/s

Hint

A perfectly inelastic collision: one final velocity.

Worked solution
  1. 2.0 × 3.0 = 3.0v.
  2. v = 2 m/s.
Q17Exam-style
A bat exerts a force F(t) = 2000t(0.20 − t) N on a ball for 0 ≤ t ≤ 0.20 s. Find the impulse.
N·s

Hint

Integrate F(t) over the contact time.

Worked solution
  1. J = ∫00.20 2000(0.20t − t2) dt = 2000[0.10t2 − t3/3]00.20.
  2. = 2000(0.0040 − 0.002667).
  3. J = 2.67 N·s.
Q18Exam-style
A rod lies along the x-axis from x = 0 to x = 3 m, with linear density λ = 2x kg/m. Find the position of its centre of mass.
m

Hint

xcm = ∫ xλ dx / ∫ λ dx.

Worked solution
  1. Mass: ∫03 2x dx = 9 kg.
  2. ∫03 2x2 dx = 18 kg·m.
  3. xcm = 18/9 = 2 m.
Q19Exam-style
A system has no net external force on it. Its centre of mass:

Hint

ΣFext = Macm.

Worked solution
  1. With no external force, acm = 0.
  2. So the centre of mass moves with constant velocity (which may be zero): B.
Q20Challenge
A 1.0 kg ball moving at 6.0 m/s hits a 2.0 kg ball at rest in a head-on, perfectly elastic collision. Find the speed of the 2.0 kg ball afterwards.
m/s

Hint

Use conservation of momentum and of kinetic energy, or the elastic collision result v2 = 2m1u/(m1 + m2).

Worked solution
  1. Momentum: 6 = v1 + 2v2. Elastic, so relative speed reverses: v2 − v1 = 6.
  2. Adding: 3v2 = 12.
  3. v2 = 4 m/s (and the 1.0 kg ball rebounds at 2 m/s).
Q21Exam-style
A uniform rod of mass 3.0 kg and length 2.0 m rotates about one end. Find its rotational inertia.
kg·m²

Hint

Integrate r2 dm with dm = (M/L) dr from 0 to L.

Worked solution
  1. I = ∫0L r2(M/L) dr = ML2/3.
  2. I = 3.0 × 4.0/3 = 4 kg·m².
Q22Exam-style
A net torque of 6.0 N·m acts on a uniform disk of mass 4.0 kg and radius 0.50 m, about its central axis. Find its angular acceleration.
rad/s²

Hint

I = ½MR2 for a disk.

Worked solution
  1. I = ½ × 4.0 × 0.502 = 0.50 kg·m².
  2. α = τ/I = 6.0/0.50 = 12 rad/s².
Q23Exam-style
Find the rotational inertia of a uniform disk of mass 2.0 kg and radius 0.40 m about an axis through its edge, perpendicular to the disk.
kg·m²

Hint

Parallel axis theorem, with d = R.

Worked solution
  1. I = ½MR2 + MR2 = 1.5MR2.
  2. I = 1.5 × 2.0 × 0.16 = 0.48 kg·m².
Q24Exam-style
A solid sphere and a hoop are released from rest at the top of the same incline and roll without slipping. Which reaches the bottom first?

Hint

The one with less of its energy going into rotation moves faster.

Worked solution
  1. The sphere has I = (2/5)MR2, the hoop MR2.
  2. The sphere puts a smaller share of its energy into rotation, so it moves faster and arrives first: A.
Q25Challenge
A 1.0 kg block hangs from a light string wrapped around a pulley, a uniform disk of mass 2.0 kg and radius 0.10 m on a frictionless axle. Find the block's acceleration.
m/s²

Hint

Write F = ma for the block and τ = Iα for the pulley, with a = Rα.

Worked solution
  1. Block: mg − T = ma. Pulley: TR = (½MR2)(a/R), so T = ½Ma.
  2. mg = (m + ½M)a, so a = 9.8/(1.0 + 1.0).
  3. a = 4.9 m/s².
Q26Exam-style
A skater spins at 2.0 rev/s with rotational inertia 4.0 kg·m², then pulls in her arms to 1.6 kg·m². Find her new rate of spin in rev/s.
rev/s

Hint

Conserve angular momentum. Units of rev/s work on both sides.

Worked solution
  1. I1ω1 = I2ω2: 4.0 × 2.0 = 1.6ω2.
  2. ω2 = 5 rev/s.
Q27Exam-style
A solid cylinder rolls without slipping from rest down a ramp, dropping 1.5 m in height. Find its speed at the bottom.
m/s

Hint

Share the energy between translation and rotation, with I = ½MR2 and v = Rω.

Worked solution
  1. Mgh = ½Mv2 + ½(½MR2)(v/R)2 = ¾Mv2.
  2. v = √(4gh/3) = √(4 × 9.8 × 1.5/3) = √19.6.
  3. v = 4.43 m/s.
Q28Exam-style
A uniform rod of length 1.2 m is pivoted at one end and released from rest in a horizontal position. Find its angular speed as it swings through the vertical.
rad/s

Hint

The centre of mass falls L/2. Use I = ML2/3 about the pivot.

Worked solution
  1. Mg(L/2) = ½(ML2/3)ω2, so ω = √(3g/L).
  2. ω = √(3 × 9.8/1.2) = √24.5.
  3. ω = 4.95 rad/s.
Q29Exam-style
A particle moves in a straight line at constant velocity, not through point P. Its angular momentum about P:

Hint

τ = dL/dt. Is there a torque about P?

Worked solution
  1. No net force acts, so there is no torque about P.
  2. So L = mvr⊥ is constant, with r⊥ the fixed perpendicular distance from P to the line: C.
Q30Challenge
A uniform rod of mass 1.2 kg and length 1.0 m can rotate freely about its centre on a horizontal frictionless surface. A 0.20 kg ball of clay moving at 5.0 m/s, perpendicular to the rod, hits one end and sticks. Find the angular speed afterwards.
rad/s

Hint

Conserve angular momentum about the centre; the clay adds m(L/2)2 to I.

Worked solution
  1. Before: L = mv(L/2) = 0.20 × 5.0 × 0.50 = 0.50 kg·m²/s.
  2. After: I = ML2/12 + m(L/2)2 = 0.10 + 0.05 = 0.15 kg·m².
  3. ω = 0.50/0.15 = 3.33 rad/s.
Q31Exam-style
An oscillator has position x(t) = 0.050 cos(10t), in metres. Find its maximum speed.
m/s

Hint

Differentiate, or use vmax = Aω.

Worked solution
  1. v = −0.50 sin(10t), so vmax = Aω = 0.050 × 10.
  2. vmax = 0.5 m/s.
Q32Exam-style
For the same oscillator, find the maximum acceleration.
m/s²

Hint

amax = Aω2.

Worked solution
  1. amax = 0.050 × 102 = 5 m/s².
Q33Exam-style
A mass on a spring with k = 200 N/m oscillates with amplitude 0.10 m. Find the total mechanical energy.
J

Hint

At maximum displacement all the energy is in the spring.

Worked solution
  1. E = ½kA2 = ½ × 200 × 0.010.
  2. E = 1 J.
Q34Exam-style
The amplitude of a mass-spring oscillator is doubled. What happens to its period?

Hint

Look at T = 2π√(m/k). Does A appear?

Worked solution
  1. The period depends only on m and k.
  2. So it stays the same: C. The maximum speed doubles instead.
Q35Challenge
A uniform rod 1.0 m long swings with small oscillations about a pivot at one end. Find its period.
s

Hint

Physical pendulum: T = 2π√(I/(mgd)), with I = mL2/3 and d = L/2.

Worked solution
  1. T = 2π√((mL2/3)/(mgL/2)) = 2π√(2L/(3g)).
  2. T = 2π√(2.0/29.4) = 2π × 0.261.
  3. T = 1.64 s.

Section II: 4 free-response questions

Free-response question 1: Kinematics

Mathematical routines style · 6 points · about 15 minutes

A particle moves along the x-axis with position x(t) = t3 − 6t2 + 9t, in metres, for 0 ≤ t ≤ 4 s.

  1. (a) Find expressions for the velocity v(t) and acceleration a(t).
  2. (b) Find the times when the particle is at rest, and its position at each.
  3. (c) Find the total distance travelled from t = 0 to t = 4 s.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

v = 3t2 − 12t + 9; a = 6t − 12.

Part (b)

v = 3(t − 1)(t − 3) = 0 at t = 1 s (x = 4 m) and t = 3 s (x = 0).

Part (c)

x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4. Distance = 4 + 4 + 4 = 12 m.

Your mark: 0 of 6

Free-response question 2: Work, energy and power

Mathematical routines style · 6 points · about 20 minutes

A 0.50 kg block is attached to a nonlinear spring that exerts a force F(x) = −kx − βx3, with k = 50 N/m and β = 200 N/m3. The block moves on a frictionless horizontal surface.

  1. (a) Derive an expression for the potential energy U(x), taking U(0) = 0.
  2. (b) The block is released from rest at x = 0.20 m. Find its speed at x = 0.
  3. (c) Is x = 0 a position of stable equilibrium? Justify your answer.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

U = −∫0x F dx = ½kx2 + ¼βx4.

Part (b)

U(0.20) = 25(0.04) + 50(0.0016) = 1.08 J. ½(0.50)v2 = 1.08, so v = 2.08 m/s.

Part (c)

Yes: U has a minimum at x = 0, and for any small displacement F points back towards x = 0.

Your mark: 0 of 6

Free-response question 3: Torque and rotational dynamics

Mathematical routines style · 7 points · about 20 minutes

A uniform rod of mass M and length L is pivoted at one end on a frictionless axle. It is held horizontal and then released from rest.

  1. (a) Use integration to show that the rod's rotational inertia about the pivot is ML2/3.
  2. (b) Find the angular acceleration just after release, for L = 1.2 m.
  3. (c) Show that the free end initially accelerates faster than g, and explain how that is possible.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

I = ∫0L r2(M/L) dr = (M/L)(L3/3) = ML2/3.

Part (b)

τ = MgL/2, so α = (MgL/2)/(ML2/3) = 3g/(2L) = 12.25 rad/s2.

Part (c)

a = αL = 3g/2 > g. The rod is rigid, so the pivot and the rest of the rod push the end down faster than free fall.

Your mark: 0 of 7

Free-response question 4: Oscillations

Mathematical routines style · 6 points · about 15 minutes

A block of mass m on a frictionless horizontal surface is attached to a spring of spring constant k. It is pulled a distance A from equilibrium and released from rest at t = 0.

  1. (a) Write the differential equation for the block's position x(t).
  2. (b) Show that x(t) = A cos(ωt) is a solution, and find ω.
  3. (c) For m = 0.20 kg, k = 80 N/m and A = 0.050 m, find the maximum speed.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

m d2x/dt2 = −kx.

Part (b)

d2x/dt2 = −Aω2 cos(ωt) = −ω2x. Matching gives ω = √(k/m). x(0) = A, as required.

Part (c)

ω = √(80/0.20) = 20 rad/s. vmax = Aω = 1.0 m/s.

Your mark: 0 of 6

After the mock

Note every question you missed and its unit, then go back to that unit's practice page: all AP Physics C: Mechanics units. Then try the real thing with official AP Physics C: Mechanics past papers.