AP Physics C: Mechanics Practice Test 3 – 60 Questions with Answers

Test yourself with 60 original AP Physics C: Mechanics practice questions covering all 7 units of the course: kinematics, force and translational dynamics, work, energy, and power, linear momentum, torque and rotational dynamics, energy and momentum of rotating systems, and oscillations. Every answer includes the full working. All questions are calculus-based. Take g = 9.8 m/s2 unless stated otherwise.

Unit 1: Kinematics

Q1. A ball is thrown horizontally with speed 14 m/s from a cliff 20 m high. How far from the base of the cliff does it land? Take g = 9.8 m/s2.

  • (A) 28.28 m
  • (B) 31.11 m
  • (C) 42.43 m
  • (D) 14.14 m
Show Answer

Answer: A – Time to fall: t = √(2h/g) = √(2×20÷9.8) = 2.02 s. Horizontal distance = v0t = 14×2.02 = 28.28 m.

Q2. A particle moves along a straight line with position x(t) = 4t2 + 7t metres, where t is in seconds. What is the velocity of the particle at t = 3 s?

  • (A) 36 m/s
  • (B) 31 m/s
  • (C) 93 m/s
  • (D) 46.5 m/s
Show Answer

Answer: B – v = dx/dt = 8t + 7. At t = 3 s, v = 8×3 + 7 = 31 m/s.

Q3. A particle has position x(t) = 3t3 – 2t2 metres, where t is in seconds. What is its acceleration at t = 3 s?

  • (A) 12.5 m/s2
  • (B) 25 m/s2
  • (C) 40 m/s2
  • (D) 50 m/s2
Show Answer

Answer: D – v = dx/dt = 9t2 – 4t, so a = dv/dt = 18t – 4. At t = 3 s, a = 18×3 – 4 = 50 m/s2.

Q4. A body moves with velocity v(t) = 6t + 3 m/s, where t is in seconds. What is its displacement from t = 0 to t = 4 s?

  • (A) 48 m
  • (B) 66 m
  • (C) 15 m
  • (D) 60 m
Show Answer

Answer: D – Displacement = ∪v dt from 0 to 4 = (6/2)t2 + 3t evaluated from 0 to 4 = 48 + 12 = 60 m.

Q5. A body starts with velocity 5 m/s and has acceleration a(t) = 2t + 4 m/s2. What is its velocity at t = 3 s?

  • (A) 16 m/s
  • (B) 32.5 m/s
  • (C) 26 m/s
  • (D) 39 m/s
Show Answer

Answer: C – v = v0 + ∪a dt = 5 + (2/2)t2 + 4t from 0 to 3 = 5 + 9 + 12 = 26 m/s.

Q6. A projectile is launched from level ground with speed 31 m/s at 30θ above the horizontal. What is its horizontal range? Take g = 9.8 m/s2.

  • (A) 106.15 m
  • (B) 42.46 m
  • (C) 84.92 m
  • (D) 254.77 m
Show Answer

Answer: C – Range R = v02 sin(2θ)/g = 312 × sin(60θ) ÷ 9.8 = 84.92 m.

Q7. A stone is dropped from rest from a height of 70 m. How long does it take to reach the ground? Take g = 9.8 m/s2.

  • (A) 4.16 s
  • (B) 1.89 s
  • (C) 2.83 s
  • (D) 3.78 s
Show Answer

Answer: D – h = (1/2)gt2, so t = √(2h/g) = √(2×70÷9.8) = 3.78 s.

Unit 2: Force and Translational Dynamics

Q8. A planet has surface gravity 9.8 m/s2. At a height of 3 planet radii above the surface, what is the acceleration due to gravity there?

  • (A) 0.49 m/s2
  • (B) 0.67 m/s2
  • (C) 0.15 m/s2
  • (D) 0.61 m/s2
Show Answer

Answer: D – g’ = g(R/(R + h))2 = 9.8×(1/4)2 = 0.61 m/s2.

Q9. A 5 kg object moves in a circle of radius 4 m with a period of 5 s. What is the magnitude of the centripetal force on it?

  • (A) 31.58 N
  • (B) 39.48 N
  • (C) 41.58 N
  • (D) 23.69 N
Show Answer

Answer: A – Speed v = 2πr/T = 2π×4÷5 = 5.03 m/s. F = mv2/r = 31.58 N.

Q10. A time-varying force F(t) = 3t + 4 N acts on a 2 kg mass. What is the acceleration of the mass at t = 4 s?

  • (A) 10 m/s2
  • (B) 24 m/s2
  • (C) 6 m/s2
  • (D) 8 m/s2
Show Answer

Answer: D – At t = 4 s, F = 3×4 + 4 = 16 N. a = F/m = 16÷2 = 8 m/s2.

Q11. An object of mass 18 kg is on a planet where the acceleration due to gravity is 4 times that on Earth. What is its weight on that planet? Take g = 9.8 m/s2 on Earth.

  • (A) 705.6 N
  • (B) 882 N
  • (C) 700.6 N
  • (D) 695.6 N
Show Answer

Answer: A – Weight = m×gplanet = 18×(4×9.8) = 705.6 N.

Q12. A net force acts on a 5 kg mass and gives it an acceleration of 5 m/s2. What is the magnitude of the net force?

  • (A) 25 N
  • (B) 6.25 N
  • (C) 18.75 N
  • (D) 75 N
Show Answer

Answer: A – F = ma = 5×5 = 25 N.

Q13. Two horizontal forces, 21 N to the right and 9 N to the left, act on a 6 kg block on a frictionless surface. What is the acceleration of the block?

  • (A) 2.5 m/s2
  • (B) 0.5 m/s2
  • (C) 2 m/s2
  • (D) 6 m/s2
Show Answer

Answer: C – Net force = 21 – 9 = 12 N to the right. a = F/m = 12÷6 = 2 m/s2.

Q14. A 12 kg block rests on a horizontal surface with coefficient of static friction μ = 0.3. What is the maximum static friction force that can act on the block? Take g = 9.8 m/s2.

  • (A) 52.92 N
  • (B) 35.28 N
  • (C) 28.22 N
  • (D) 40.28 N
Show Answer

Answer: B – Maximum static friction = μN = μmg = 0.3×12×9.8 = 35.28 N.

Q15. A block slides down a plane inclined at 60θ to the horizontal. The coefficient of kinetic friction is 0.3. What is the acceleration of the block down the plane? Take g = 9.8 m/s2.

  • (A) 7.02 m/s2
  • (B) 4.21 m/s2
  • (C) 2.02 m/s2
  • (D) 17.02 m/s2
Show Answer

Answer: A – a = g(sinθ – μcosθ) = 9.8×(0.87 – 0.3×0.5) = 7.02 m/s2.

Q16. A 20 kg block rests on a plane inclined at 45θ to the horizontal. What is the normal force on the block? Take g = 9.8 m/s2.

  • (A) 103.94 N
  • (B) 173.24 N
  • (C) 138.59 N
  • (D) 277.19 N
Show Answer

Answer: C – Normal force N = mg cosθ = 20×9.8×0.71 = 138.59 N.

Q17. Two masses, 5 kg and 9 kg, are connected by a light string over a frictionless pulley. What is the magnitude of their acceleration? Take g = 9.8 m/s2.

  • (A) 0.7 m/s2
  • (B) 2.8 m/s2
  • (C) 1.4 m/s2
  • (D) 5.6 m/s2
Show Answer

Answer: B – a = (m2 – m1)g/(m1 + m2) = (9 – 5)×9.8÷(5 + 9) = 2.8 m/s2.

Q18. A person of mass 45 kg stands on a scale in a lift accelerating upwards at 1 m/s2. What is the scale reading in newtons? Take g = 9.8 m/s2.

  • (A) 486 N
  • (B) 972 N
  • (C) 607.5 N
  • (D) 388.8 N
Show Answer

Answer: A – Normal force N = m(g + a) = 45×(9.8 + 1) = 486 N.

Q19. A person of mass 60 kg stands on a scale in a lift accelerating downwards at 2 m/s2. What is the scale reading in newtons? Take g = 9.8 m/s2.

  • (A) 514.8 N
  • (B) 473 N
  • (C) 468 N
  • (D) 463 N
Show Answer

Answer: C – Normal force N = m(g – a) = 60×(9.8 – 2) = 468 N.

Unit 3: Work, Energy, and Power

Q20. A 4 kg block attached to a spring of spring constant 200 N/m is released from rest with the spring compressed by 0.3 m on a frictionless surface. What is the speed of the block as it passes the equilibrium position?

  • (A) 1.7 m/s
  • (B) 4.24 m/s
  • (C) 2.12 m/s
  • (D) 1.27 m/s
Show Answer

Answer: C – (1/2)kx2 = (1/2)mv2, so v = √(k/m)×x = √(200÷4)×0.3 = 2.12 m/s.

Q21. A 5 kg object speeds up from 3 m/s to 7 m/s. How much net work is done on the object?

  • (A) 25 J
  • (B) 90 J
  • (C) 100 J
  • (D) 60 J
Show Answer

Answer: C – W = change in KE = (1/2)m(v22 – v12) = 0.5×5×(72 – 32) = 100 J.

Q22. A constant force of 39 N pushes a crate 2 m along a horizontal floor in the direction of the force. How much work is done by the force?

  • (A) 68 J
  • (B) 78 J
  • (C) 85.8 J
  • (D) 58.5 J
Show Answer

Answer: B – W = Fd = 39×2 = 78 J.

Q23. A force of 53 N acts at 30θ to the direction of motion as a block moves 5 m. How much work is done by the force?

  • (A) 183.6 J
  • (B) 234.5 J
  • (C) 458.99 J
  • (D) 229.5 J
Show Answer

Answer: D – W = Fd cosθ = 53×5×0.87 = 229.5 J.

Q24. A variable force F(x) = 40x N stretches a spring from x = 0 to x = 0.5 m. How much work is done? Here x is in metres.

  • (A) 15 J
  • (B) 5 J
  • (C) 7.5 J
  • (D) 3 J
Show Answer

Answer: B – W = ∪F dx from 0 to 0.5 = (1/2)kX2 = 0.5×40×0.52 = 5 J.

Q25. A force F(x) = 6x2 N acts on a particle as it moves from x = 0 to x = 3 m. How much work is done by the force?

  • (A) 40.5 J
  • (B) 44 J
  • (C) 54 J
  • (D) 162 J
Show Answer

Answer: C – W = ∪F dx from 0 to 3 = (6/3)x3 from 0 to 3 = 54 J.

Q26. What is the kinetic energy of a 3 kg object moving at 5 m/s?

  • (A) 27.5 J
  • (B) 56.25 J
  • (C) 46.88 J
  • (D) 37.5 J
Show Answer

Answer: D – KE = (1/2)mv2 = 0.5×3×52 = 37.5 J.

Q27. A spring of spring constant 100 N/m is compressed by 0.1 m. What is the elastic potential energy stored in the spring?

  • (A) 0.5 J
  • (B) 0.3 J
  • (C) 0.4 J
  • (D) 0.38 J
Show Answer

Answer: A – PE = (1/2)kx2 = 0.5×100×0.12 = 0.5 J.

Q28. A 7 kg object is raised vertically by 7 m. By how much does its gravitational potential energy increase? Take g = 9.8 m/s2.

  • (A) 240.1 J
  • (B) 480.2 J
  • (C) 470.2 J
  • (D) 1440.6 J
Show Answer

Answer: B – Change in PE = mgh = 7×9.8×7 = 480.2 J.

Q29. An object is dropped from rest from a height of 40 m. Ignoring air resistance, what is its speed just before it hits the ground? Take g = 9.8 m/s2.

  • (A) 56 m/s
  • (B) 23 m/s
  • (C) 33 m/s
  • (D) 28 m/s
Show Answer

Answer: D – Energy conservation: (1/2)mv2 = mgh, so v = √(2gh) = √(2×9.8×40) = 28 m/s.

Q30. A motor does 123 J of work in 10 s. What is its average power output?

  • (A) 6.15 W
  • (B) 22.3 W
  • (C) 12.3 W
  • (D) 15.38 W
Show Answer

Answer: C – P = W/t = 123÷10 = 12.3 W.


Unit 4: Linear Momentum

Q31. A force F(t) = 2t + 6 N acts on a body from t = 0 to t = 3 s. What is the impulse delivered to the body?

  • (A) 40.5 N·s
  • (B) 21.6 N·s
  • (C) 33.75 N·s
  • (D) 27 N·s
Show Answer

Answer: D – J = ∪F dt from 0 to 3 = (2/2)t2 + 6t from 0 to 3 = 9 + 18 = 27 N·s.

Q32. An impulse of 30 N·s is delivered to a 7 kg object initially at rest. What is its final speed?

  • (A) 3.43 m/s
  • (B) 4.29 m/s
  • (C) 12.86 m/s
  • (D) 1.07 m/s
Show Answer

Answer: B – J = mΔv, so Δv = J/m = 30÷7 = 4.29 m/s.

Q33. A 4 kg trolley moving at 9 m/s collides with a stationary 5 kg trolley and the two stick together. What is their common speed after the collision?

  • (A) 5 m/s
  • (B) 3.2 m/s
  • (C) 6 m/s
  • (D) 4 m/s
Show Answer

Answer: D – Momentum conservation: 4×9 = (4 + 5)v, so v = 36÷9 = 4 m/s.

Q34. A rifle of mass 3 kg fires a bullet of mass 0.01 kg at 372 m/s. What is the recoil speed of the rifle?

  • (A) 0.99 m/s
  • (B) 1.86 m/s
  • (C) 1.36 m/s
  • (D) 1.24 m/s
Show Answer

Answer: D – Momentum conservation: Mvrecoil = mv = 0.013372 = 3.72. So vrecoil = 3.72÷3 = 1.24 m/s.

Q35. A 5 kg object moving at 5 m/s collides head-on with a 3 kg object moving at 2 m/s in the opposite direction, and they stick together. What is their velocity after the collision, taking the first object’s direction as positive?

  • (A) 2.38 m/s
  • (B) 1.9 m/s
  • (C) 2.97 m/s
  • (D) 3.56 m/s
Show Answer

Answer: A – Total momentum = 5×5 – 3×2 = 19 kg·m/s. Dividing by total mass 8 kg gives 2.38 m/s.

Q36. A 3 kg object changes its velocity by 8 m/s in 4 s. What is the average force acting on it?

  • (A) 4.5 N
  • (B) 11 N
  • (C) 4.8 N
  • (D) 6 N
Show Answer

Answer: D – Average force = Δp/Δt = mΔv/Δt = 3×8÷4 = 6 N.

Q37. A force acting on a body rises linearly from zero to 40 N over 4 s and then stops. What is the impulse delivered?

  • (A) 80 N·s
  • (B) 70 N·s
  • (C) 40 N·s
  • (D) 85 N·s
Show Answer

Answer: A – Impulse is the area under the force-time graph: a triangle of area (1/2)×40×4 = 80 N·s.

Q38. A 0.5 kg ball hits a wall at 4 m/s and rebounds at 2 m/s in the opposite direction. What is the magnitude of the change in its momentum?

  • (A) 1.5 kg·m/s
  • (B) 3 kg·m/s
  • (C) 3.3 kg·m/s
  • (D) 9 kg·m/s
Show Answer

Answer: B – Change in momentum = m(vf – vi) in magnitude = 0.5×(2 + 4) = 3 kg·m/s.

Unit 5: Torque and Rotational Dynamics

Q39. A uniform rod of mass 7 kg and length 3 m rotates about an axis through one end, perpendicular to the rod. What is its moment of inertia?

  • (A) 21 kg·m2
  • (B) 11 kg·m2
  • (C) 10.5 kg·m2
  • (D) 16.8 kg·m2
Show Answer

Answer: A – For a rod about one end, I = (1/3)ML2 = (1/3)×7×32 = 21 kg·m2.

Q40. A uniform solid disc of mass 7 kg and radius 0.4 m rotates about its central axis. What is its moment of inertia?

  • (A) 0.56 kg·m2
  • (B) 1.68 kg·m2
  • (C) 0.28 kg·m2
  • (D) 0.84 kg·m2
Show Answer

Answer: A – For a solid disc, I = (1/2)MR2 = 0.5×7×0.42 = 0.56 kg·m2.

Q41. A thin hoop of mass 3 kg and radius 0.4 m rotates about its central axis. What is its moment of inertia?

  • (A) 0.48 kg·m2
  • (B) 0.24 kg·m2
  • (C) 0.12 kg·m2
  • (D) 0.96 kg·m2
Show Answer

Answer: A – For a thin hoop, I = 1MR2 = 1×3×0.42 = 0.48 kg·m2.

Q42. A net torque of 30 N·m acts on a body with moment of inertia 6 kg·m2. What is its angular acceleration?

  • (A) 4 rad/s2
  • (B) 3.75 rad/s2
  • (C) 7.5 rad/s2
  • (D) 5 rad/s2
Show Answer

Answer: D – α = τnet/I = 30÷6 = 5 rad/s2.

Q43. A wheel starts with angular velocity 5 rad/s and has a constant angular acceleration of 1 rad/s2. What is its angular velocity after 5 s?

  • (A) 15 rad/s
  • (B) 30 rad/s
  • (C) 10 rad/s
  • (D) 20 rad/s
Show Answer

Answer: C – ω = ω0 + αt = 5 + 1×5 = 10 rad/s.

Q44. A disc starts with angular velocity 5 rad/s and angular acceleration 2 rad/s2. Through what angle does it turn in 2 s?

  • (A) 19 rad
  • (B) 9 rad
  • (C) 14 rad
  • (D) 11.2 rad
Show Answer

Answer: C – θ = ω0t + (1/2)αt2 = 5×2 + 0.5×2×22 = 14 rad.

Q45. The angular momentum of a body varies as L(t) = 4t2 kg·m2/s. What is the net torque on the body at t = 4 s?

  • (A) 19.2 N·m
  • (B) 32 N·m
  • (C) 24 N·m
  • (D) 25.6 N·m
Show Answer

Answer: B – τnet = dL/dt = 8t. At t = 4 s, τ = 8×4 = 32 N·m.

Q46. A force of 20 N is applied at a distance of 0.4 m from a pivot, at 90θ to the lever arm. What is the magnitude of the torque about the pivot?

  • (A) 10 N·m
  • (B) 8 N·m
  • (C) 24 N·m
  • (D) 13 N·m
Show Answer

Answer: B – τ = rF sinθ = 0.4×20×1 = 8 N·m.

Unit 6: Energy and Momentum of Rotating Systems

Q47. A solid disc of mass 5 kg rolls without slipping at 5 m/s. What is its total kinetic energy?

  • (A) 93.75 J
  • (B) 83.75 J
  • (C) 88.75 J
  • (D) 75 J
Show Answer

Answer: A – Total KE = translational + rotational = (1/2)mv2 + (1/2)Iω2 = (3/4)mv2 for a disc = 0.75×5×52 = 93.75 J.

Q48. A thin hoop of mass 6 kg rolls without slipping at 4 m/s. What is its total kinetic energy?

  • (A) 96 J
  • (B) 106 J
  • (C) 144 J
  • (D) 105.6 J
Show Answer

Answer: A – For a hoop, I = MR2, so total KE = (1/2)mv2 + (1/2)mv2 = mv2 = 6×42 = 96 J.

Q49. A torque τ(t) = 3t N·m acts on a body from t = 0 to t = 4 s. What is the change in its angular momentum?

  • (A) 12 kg·m2/s
  • (B) 24 kg·m2/s
  • (C) 72 kg·m2/s
  • (D) 19 kg·m2/s
Show Answer

Answer: B – ΔL = ∪τ dt from 0 to 4 = (3/2)t2 from 0 to 4 = 24 kg·m2/s.

Q50. How much work is needed to spin a solid disc of mass 5 kg and radius 0.3 m from rest to an angular velocity of 4 rad/s?

  • (A) 2.25 J
  • (B) 0.45 J
  • (C) 1.8 J
  • (D) 1.08 J
Show Answer

Answer: C – Work = change in rotational KE = (1/2)Iω2. I = (1/2)MR2 = 0.22 kg·m2, so W = 0.5×0.22×42 = 1.8 J.

Q51. A solid disc of mass 4 kg and radius 0.3 m rotates at 3 rad/s about its central axis. What is its rotational kinetic energy?

  • (A) 1.62 J
  • (B) 0.81 J
  • (C) 2.43 J
  • (D) 0.49 J
Show Answer

Answer: B – I = (1/2)MR2 = 0.18 kg·m2. KE = (1/2)Iω2 = 0.5×0.18×32 = 0.81 J.

Q52. A body with moment of inertia 4 kg·m2 rotates at 5 rad/s. What is its angular momentum?

  • (A) 16 kg·m2/s
  • (B) 20 kg·m2/s
  • (C) 5 kg·m2/s
  • (D) 30 kg·m2/s
Show Answer

Answer: B – L = Iω = 4×5 = 20 kg·m2/s.

Q53. A spinning skater with moment of inertia 2 kg·m2 rotates at 7 rad/s, then pulls in her arms so her moment of inertia becomes 1 kg·m2. What is her new angular velocity, assuming no external torque?

  • (A) 14 rad/s
  • (B) 15.4 rad/s
  • (C) 10.5 rad/s
  • (D) 8.4 rad/s
Show Answer

Answer: A – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = 2×7÷1 = 14 rad/s.

Unit 7: Oscillations

Q54. A spring of spring constant 80 N/m oscillates with amplitude 0.3 m. What is the total mechanical energy of the oscillator?

  • (A) 1.8 J
  • (B) 2.16 J
  • (C) 3.6 J
  • (D) 5.4 J
Show Answer

Answer: C – E = (1/2)kA2 = 0.5×80×0.32 = 3.6 J.

Q55. An oscillator completes one full oscillation in 1 s. What is its frequency?

  • (A) 0.75 Hz
  • (B) 0.6 Hz
  • (C) 0.8 Hz
  • (D) 1 Hz
Show Answer

Answer: D – Frequency f = 1/T = 1÷1 = 1 Hz.

Q56. A simple pendulum has a period of 0.5 s for small oscillations. What is its length? Take g = 9.8 m/s2.

  • (A) 0.05 m
  • (B) 0.06 m
  • (C) 0.19 m
  • (D) 0.03 m
Show Answer

Answer: B – T = 2π√(L/g), so L = g(T/2π)2 = 9.8×(0.5÷2π)2 = 0.06 m.

Q57. A 4 kg mass on a spring of spring constant 200 N/m oscillates on a frictionless surface. What is the period of oscillation?

  • (A) 1.78 s
  • (B) 1.11 s
  • (C) 0.89 s
  • (D) 0.98 s
Show Answer

Answer: C – T = 2π√(m/k) = 2π√(4÷200) = 0.89 s.

Q58. A simple pendulum of length 0.4 m oscillates with small amplitude. What is its period? Take g = 9.8 m/s2.

  • (A) 1.02 s
  • (B) 0.63 s
  • (C) 1.27 s
  • (D) 1.4 s
Show Answer

Answer: C – T = 2π√(L/g) = 2π√(0.4÷9.8) = 1.27 s.

Q59. A mass of 1 kg oscillates on a spring of spring constant 100 N/m. What is the angular frequency of the oscillation?

  • (A) 20 rad/s
  • (B) 12.5 rad/s
  • (C) 8 rad/s
  • (D) 10 rad/s
Show Answer

Answer: D – ω = √(k/m) = √(100÷1) = 10 rad/s.

Q60. A 2 kg mass oscillates on a spring of spring constant 36 N/m with amplitude 0.3 m. What is its maximum speed?

  • (A) 1.02 m/s
  • (B) 1.27 m/s
  • (C) 1.4 m/s
  • (D) 1.91 m/s
Show Answer

Answer: B – ω = √(k/m) = 4.24 rad/s. Maximum speed vmax = ωA = 4.24×0.3 = 1.27 m/s.

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