AP Physics C: Mechanics Practice Test 5 – 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 stone is dropped from rest from a height of 45 m. How long does it take to reach the ground? Take g = 9.8 m/s2.

  • (A) 6.06 s
  • (B) 3.03 s
  • (C) 2.27 s
  • (D) 3.33 s
Show Answer

Answer: B – h = (1/2)gt2, so t = √(2h/g) = √(2×45÷9.8) = 3.03 s.

Q2. A particle moves with velocity v(t) = 6t2 + 6 m/s, where t is in seconds. What is its displacement from t = 0 to t = 4 s?

  • (A) 167.2 m
  • (B) 190 m
  • (C) 142 m
  • (D) 152 m
Show Answer

Answer: D – Displacement = ∪v dt from 0 to 4 = (6/3)t3 + 6t from 0 to 4 = 128 + 24 = 152 m.

Q3. A car moves with velocity v(t) = 12 – 6t m/s, where t is in seconds. At what time does the car momentarily come to rest?

  • (A) 2 s
  • (B) 1.2 s
  • (C) 4 s
  • (D) 6 s
Show Answer

Answer: A – Set v = 0: 12 – 6t = 0, so t = 12÷6 = 2 s.

Q4. A particle has velocity v(t) = 2t2 + 4t m/s, where t is in seconds. What is its acceleration at t = 3 s?

  • (A) 11 m/s2
  • (B) 16 m/s2
  • (C) 12 m/s2
  • (D) 12.8 m/s2
Show Answer

Answer: B – a = dv/dt = 4t + 4. At t = 3 s, a = 4×3 + 4 = 16 m/s2.

Q5. A ball is thrown horizontally with speed 12 m/s from a cliff 45 m high. How far from the base of the cliff does it land? Take g = 9.8 m/s2.

  • (A) 72.73 m
  • (B) 46.37 m
  • (C) 36.37 m
  • (D) 40 m
Show Answer

Answer: C – Time to fall: t = √(2h/g) = √(2×45÷9.8) = 3.03 s. Horizontal distance = v0t = 12×3.03 = 36.37 m.

Q6. A particle moves along a straight line with position x(t) = 3t2 + 4t metres, where t is in seconds. What is the velocity of the particle at t = 5 s?

  • (A) 44 m/s
  • (B) 34 m/s
  • (C) 37.4 m/s
  • (D) 17 m/s
Show Answer

Answer: B – v = dx/dt = 6t + 4. At t = 5 s, v = 6×5 + 4 = 34 m/s.

Q7. A particle has position x(t) = 2t3 – 5t2 metres, where t is in seconds. What is its acceleration at t = 4 s?

  • (A) 38 m/s2
  • (B) 76 m/s2
  • (C) 22.8 m/s2
  • (D) 57 m/s2
Show Answer

Answer: A – v = dx/dt = 6t2 – 10t, so a = dv/dt = 12t – 10. At t = 4 s, a = 12×4 – 10 = 38 m/s2.

Unit 2: Force and Translational Dynamics

Q8. A 9 kg block rests on a horizontal surface with coefficient of static friction μ = 0.4. What is the maximum static friction force that can act on the block? Take g = 9.8 m/s2.

  • (A) 40.28 N
  • (B) 35.28 N
  • (C) 26.46 N
  • (D) 21.17 N
Show Answer

Answer: B – Maximum static friction = μN = μmg = 0.4×9×9.8 = 35.28 N.

Q9. A block slides down a plane inclined at 30θ to the horizontal. The coefficient of kinetic friction is 0.1. What is the acceleration of the block down the plane? Take g = 9.8 m/s2.

  • (A) 2.43 m/s2
  • (B) 4.46 m/s2
  • (C) 4.05 m/s2
  • (D) 3.24 m/s2
Show Answer

Answer: C – a = g(sinθ – μcosθ) = 9.8×(0.5 – 0.1×0.87) = 4.05 m/s2.

Q10. A 5 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) 34.65 N
  • (B) 29.65 N
  • (C) 69.3 N
  • (D) 38.11 N
Show Answer

Answer: A – Normal force N = mg cosθ = 5×9.8×0.71 = 34.65 N.

Q11. Two masses, 5 kg and 10 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) 3.27 m/s2
  • (B) 1.96 m/s2
  • (C) 6.53 m/s2
  • (D) 4.9 m/s2
Show Answer

Answer: A – a = (m2 – m1)g/(m1 + m2) = (10 – 5)×9.8÷(5 + 10) = 3.27 m/s2.

Q12. A person of mass 75 kg stands on a scale in a lift accelerating upwards at 3 m/s2. What is the scale reading in newtons? Take g = 9.8 m/s2.

  • (A) 960 N
  • (B) 1200 N
  • (C) 576 N
  • (D) 480 N
Show Answer

Answer: A – Normal force N = m(g + a) = 75×(9.8 + 3) = 960 N.

Q13. A person of mass 48 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) 1123.2 N
  • (B) 379.4 N
  • (C) 374.4 N
  • (D) 364.4 N
Show Answer

Answer: C – Normal force N = m(g – a) = 48×(9.8 – 2) = 374.4 N.

Q14. A 7 kg object moves in a horizontal circle of radius 2 m at a constant speed of 9 m/s. What is the magnitude of the centripetal force on it?

  • (A) 283.5 N
  • (B) 293.5 N
  • (C) 278.5 N
  • (D) 288.5 N
Show Answer

Answer: A – F = mv2/r = 7×92÷2 = 283.5 N.

Q15. A spring of spring constant 200 N/m is stretched by 0.1 m from its natural length. What is the magnitude of the spring force?

  • (A) 30 N
  • (B) 15 N
  • (C) 20 N
  • (D) 60 N
Show Answer

Answer: C – F = kx = 200×0.1 = 20 N.

Q16. A planet has surface gravity 15 m/s2. At a height of 3 planet radii above the surface, what is the acceleration due to gravity there?

  • (A) 0.75 m/s2
  • (B) 0.94 m/s2
  • (C) 1.03 m/s2
  • (D) 0.23 m/s2
Show Answer

Answer: B – g’ = g(R/(R + h))2 = 15×(1/4)2 = 0.94 m/s2.

Q17. A 3 kg object moves in a circle of radius 2 m with a period of 2 s. What is the magnitude of the centripetal force on it?

  • (A) 54.22 N
  • (B) 74.02 N
  • (C) 59.22 N
  • (D) 64.22 N
Show Answer

Answer: C – Speed v = 2πr/T = 2π×2÷2 = 6.28 m/s. F = mv2/r = 59.22 N.

Q18. A time-varying force F(t) = 3t + 13 N acts on a 2 kg mass. What is the acceleration of the mass at t = 4 s?

  • (A) 15.62 m/s2
  • (B) 17.5 m/s2
  • (C) 12.5 m/s2
  • (D) 37.5 m/s2
Show Answer

Answer: C – At t = 4 s, F = 3×4 + 13 = 25 N. a = F/m = 25÷2 = 12.5 m/s2.

Q19. An object of mass 6 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) 470.4 N
  • (B) 240.2 N
  • (C) 141.12 N
  • (D) 235.2 N
Show Answer

Answer: D – Weight = m×gplanet = 6×(4×9.8) = 235.2 N.

Unit 3: Work, Energy, and Power

Q20. What is the kinetic energy of a 5 kg object moving at 7 m/s?

  • (A) 122.5 J
  • (B) 30.62 J
  • (C) 117.5 J
  • (D) 367.5 J
Show Answer

Answer: A – KE = (1/2)mv2 = 0.5×5×72 = 122.5 J.

Q21. A spring of spring constant 80 N/m is compressed by 0.2 m. What is the elastic potential energy stored in the spring?

  • (A) 0.8 J
  • (B) 1.6 J
  • (C) 1.76 J
  • (D) 2.4 J
Show Answer

Answer: B – PE = (1/2)kx2 = 0.5×80×0.22 = 1.6 J.

Q22. A 4 kg object is raised vertically by 11 m. By how much does its gravitational potential energy increase? Take g = 9.8 m/s2.

  • (A) 421.2 J
  • (B) 215.6 J
  • (C) 431.2 J
  • (D) 107.8 J
Show Answer

Answer: C – Change in PE = mgh = 4×9.8×11 = 431.2 J.

Q23. An object is dropped from rest from a height of 80 m. Ignoring air resistance, what is its speed just before it hits the ground? Take g = 9.8 m/s2.

  • (A) 31.68 m/s
  • (B) 39.6 m/s
  • (C) 29.7 m/s
  • (D) 79.2 m/s
Show Answer

Answer: B – Energy conservation: (1/2)mv2 = mgh, so v = √(2gh) = √(2×9.8×80) = 39.6 m/s.

Q24. A motor does 237 J of work in 3 s. What is its average power output?

  • (A) 69 W
  • (B) 79 W
  • (C) 118.5 W
  • (D) 98.75 W
Show Answer

Answer: B – P = W/t = 237÷3 = 79 W.

Q25. A constant force of 67 N moves an object at a steady speed of 5 m/s in the direction of the force. What power is delivered by the force?

  • (A) 167.5 W
  • (B) 251.25 W
  • (C) 335 W
  • (D) 330 W
Show Answer

Answer: C – P = Fv = 67×5 = 335 W.

Q26. A 6 kg block slides 7 m across a horizontal surface with coefficient of kinetic friction 0.3. What is the magnitude of the work done by friction? Take g = 9.8 m/s2.

  • (A) 118.48 J
  • (B) 74.09 J
  • (C) 370.44 J
  • (D) 123.48 J
Show Answer

Answer: D – Friction force = μmg = 0.3×6×9.8 = 17.64 N. Work magnitude = Fd = 17.64×7 = 123.48 J.

Q27. A 2 kg block attached to a spring of spring constant 100 N/m is released from rest with the spring compressed by 0.1 m on a frictionless surface. What is the speed of the block as it passes the equilibrium position?

  • (A) 0.71 m/s
  • (B) 0.42 m/s
  • (C) 1.06 m/s
  • (D) 0.88 m/s
Show Answer

Answer: A – (1/2)kx2 = (1/2)mv2, so v = √(k/m)×x = √(100÷2)×0.1 = 0.71 m/s.

Q28. A 8 kg object speeds up from 3 m/s to 6 m/s. How much net work is done on the object?

  • (A) 113 J
  • (B) 108 J
  • (C) 216 J
  • (D) 162 J
Show Answer

Answer: B – W = change in KE = (1/2)m(v22 – v12) = 0.5×8×(62 – 32) = 108 J.

Q29. A constant force of 25 N pushes a crate 3 m along a horizontal floor in the direction of the force. How much work is done by the force?

  • (A) 60 J
  • (B) 150 J
  • (C) 225 J
  • (D) 75 J
Show Answer

Answer: D – W = Fd = 25×3 = 75 J.

Q30. A force of 46 N acts at 60θ to the direction of motion as a block moves 7 m. How much work is done by the force?

  • (A) 96.6 J
  • (B) 483 J
  • (C) 128.8 J
  • (D) 161 J
Show Answer

Answer: D – W = Fd cosθ = 46×7×0.5 = 161 J.


Unit 4: Linear Momentum

Q31. A force acting on a body rises linearly from zero to 20 N over 4 s and then stops. What is the impulse delivered?

  • (A) 40 N·s
  • (B) 60 N·s
  • (C) 50 N·s
  • (D) 44 N·s
Show Answer

Answer: A – Impulse is the area under the force-time graph: a triangle of area (1/2)×20×4 = 40 N·s.

Q32. A 2 kg ball hits a wall at 5 m/s and rebounds at 5 m/s in the opposite direction. What is the magnitude of the change in its momentum?

  • (A) 12 kg·m/s
  • (B) 30 kg·m/s
  • (C) 5 kg·m/s
  • (D) 20 kg·m/s
Show Answer

Answer: D – Change in momentum = m(vf – vi) in magnitude = 2×(5 + 5) = 20 kg·m/s.

Q33. What is the momentum of a 11 kg object moving at 11 m/s?

  • (A) 121 kg·m/s
  • (B) 363 kg·m/s
  • (C) 72.6 kg·m/s
  • (D) 96.8 kg·m/s
Show Answer

Answer: A – p = mv = 11×11 = 121 kg·m/s.

Q34. A constant force of 37 N acts on an object for 2 s. What is the magnitude of the impulse delivered?

  • (A) 18.5 N·s
  • (B) 222 N·s
  • (C) 79 N·s
  • (D) 74 N·s
Show Answer

Answer: D – J = FΔt = 37×2 = 74 N·s.

Q35. A force F(t) = 4t + 9 N acts on a body from t = 0 to t = 3 s. What is the impulse delivered to the body?

  • (A) 49.5 N·s
  • (B) 50 N·s
  • (C) 56.25 N·s
  • (D) 45 N·s
Show Answer

Answer: D – J = ∪F dt from 0 to 3 = (4/2)t2 + 9t from 0 to 3 = 18 + 27 = 45 N·s.

Q36. An impulse of 40 N·s is delivered to a 2 kg object initially at rest. What is its final speed?

  • (A) 20 m/s
  • (B) 60 m/s
  • (C) 30 m/s
  • (D) 25 m/s
Show Answer

Answer: A – J = mΔv, so Δv = J/m = 40÷2 = 20 m/s.

Q37. A 3 kg trolley moving at 10 m/s collides with a stationary 3 kg trolley and the two stick together. What is their common speed after the collision?

  • (A) 2.5 m/s
  • (B) 3.75 m/s
  • (C) 5 m/s
  • (D) 3 m/s
Show Answer

Answer: C – Momentum conservation: 3×10 = (3 + 3)v, so v = 30÷6 = 5 m/s.

Q38. A rifle of mass 3 kg fires a bullet of mass 0.02 kg at 333 m/s. What is the recoil speed of the rifle?

  • (A) 1.33 m/s
  • (B) 1.67 m/s
  • (C) 2.22 m/s
  • (D) 3.33 m/s
Show Answer

Answer: C – Momentum conservation: Mvrecoil = mv = 0.023333 = 6.66. So vrecoil = 6.66÷3 = 2.22 m/s.

Unit 5: Torque and Rotational Dynamics

Q39. The angular momentum of a body varies as L(t) = 5t2 kg·m2/s. What is the net torque on the body at t = 4 s?

  • (A) 20 N·m
  • (B) 60 N·m
  • (C) 40 N·m
  • (D) 50 N·m
Show Answer

Answer: C – τnet = dL/dt = 10t. At t = 4 s, τ = 10×4 = 40 N·m.

Q40. A force of 43 N is applied at a distance of 0.5 m from a pivot, at 90θ to the lever arm. What is the magnitude of the torque about the pivot?

  • (A) 26.88 N·m
  • (B) 16.5 N·m
  • (C) 32.25 N·m
  • (D) 21.5 N·m
Show Answer

Answer: D – τ = rF sinθ = 0.5×43×1 = 21.5 N·m.

Q41. Two forces act on a beam pivoted at its centre: 28 N at 0.5 m on one side and 30 N at 0.2 m on the other side, in opposite rotational senses. What is the magnitude of the net torque?

  • (A) 8 N·m
  • (B) 12 N·m
  • (C) 10 N·m
  • (D) 6 N·m
Show Answer

Answer: A – Net torque = |r1F1 – r2F2| = |0.5×28 – 0.2×30| = 8 N·m.

Q42. Two point masses, 1 kg at 3 m and 3 kg at 2 m from an axis, rotate about that axis. What is their total moment of inertia about the axis?

  • (A) 16.8 kg·m2
  • (B) 26.25 kg·m2
  • (C) 21 kg·m2
  • (D) 5.25 kg·m2
Show Answer

Answer: C – I = Σmr2 = 1×32 + 3×22 = 21 kg·m2.

Q43. A uniform rod of mass 5 kg and length 1 m rotates about an axis through one end, perpendicular to the rod. What is its moment of inertia?

  • (A) 0.42 kg·m2
  • (B) 1.25 kg·m2
  • (C) 5 kg·m2
  • (D) 1.67 kg·m2
Show Answer

Answer: D – For a rod about one end, I = (1/3)ML2 = (1/3)×5×12 = 1.67 kg·m2.

Q44. A uniform solid disc of mass 2 kg and radius 0.4 m rotates about its central axis. What is its moment of inertia?

  • (A) 0.24 kg·m2
  • (B) 0.16 kg·m2
  • (C) 0.08 kg·m2
  • (D) 0.1 kg·m2
Show Answer

Answer: B – For a solid disc, I = (1/2)MR2 = 0.5×2×0.42 = 0.16 kg·m2.

Q45. A solid sphere of mass 5 kg and radius 0.4 m rotates about its central axis. What is its moment of inertia?

  • (A) 0.96 kg·m2
  • (B) 0.32 kg·m2
  • (C) 0.24 kg·m2
  • (D) 0.4 kg·m2
Show Answer

Answer: B – For a solid sphere, I = 0.4MR2 = 0.4×5×0.42 = 0.32 kg·m2.

Q46. A net torque of 15 N·m acts on a body with moment of inertia 2 kg·m2. What is its angular acceleration?

  • (A) -2.5 rad/s2
  • (B) 15 rad/s2
  • (C) 8.25 rad/s2
  • (D) 7.5 rad/s2
Show Answer

Answer: D – α = τnet/I = 15÷2 = 7.5 rad/s2.

Unit 6: Energy and Momentum of Rotating Systems

Q47. A spinning skater with moment of inertia 3 kg·m2 rotates at 4 rad/s, then pulls in her arms so her moment of inertia becomes 2 kg·m2. What is her new angular velocity, assuming no external torque?

  • (A) 1.5 rad/s
  • (B) 6.6 rad/s
  • (C) 4.8 rad/s
  • (D) 6 rad/s
Show Answer

Answer: D – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = 3×4÷2 = 6 rad/s.

Q48. A satellite orbits a planet with period 8 days at orbital radius r. A second satellite orbits the same planet at radius 2r. What is the second satellite’s orbital period in days?

  • (A) 11.31 days
  • (B) 22.63 days
  • (C) 12.63 days
  • (D) 27.63 days
Show Answer

Answer: B – By Kepler’s third law, T2 is proportional to r3, so T2 = T1×(2)3/2 = 8×2.83 = 22.63 days.

Q49. A satellite in a circular orbit of radius r has orbital speed 4 km/s. What is the orbital speed of a satellite in a circular orbit of radius 3r around the same planet, in km/s?

  • (A) 6.93 km/s
  • (B) 1.15 km/s
  • (C) 1.85 km/s
  • (D) 2.31 km/s
Show Answer

Answer: D – Orbital speed v = √(GM/r), so v is proportional to 1/√r. v2 = 4÷√(3) = 2.31 km/s.

Q50. A solid disc of mass 3 kg rolls without slipping at 5 m/s. What is its total kinetic energy?

  • (A) 84.38 J
  • (B) 70.31 J
  • (C) 56.25 J
  • (D) 61.25 J
Show Answer

Answer: C – Total KE = translational + rotational = (1/2)mv2 + (1/2)Iω2 = (3/4)mv2 for a disc = 0.75×3×52 = 56.25 J.

Q51. A thin hoop of mass 7 kg rolls without slipping at 2 m/s. What is its total kinetic energy?

  • (A) 35 J
  • (B) 22.4 J
  • (C) 33 J
  • (D) 28 J
Show Answer

Answer: D – For a hoop, I = MR2, so total KE = (1/2)mv2 + (1/2)mv2 = mv2 = 7×22 = 28 J.

Q52. A torque τ(t) = 6t N·m acts on a body from t = 0 to t = 5 s. What is the change in its angular momentum?

  • (A) 82.5 kg·m2/s
  • (B) 75 kg·m2/s
  • (C) 85 kg·m2/s
  • (D) 56.25 kg·m2/s
Show Answer

Answer: B – ΔL = ∪τ dt from 0 to 5 = (6/2)t2 from 0 to 5 = 75 kg·m2/s.

Q53. How much work is needed to spin a solid disc of mass 5 kg and radius 0.2 m from rest to an angular velocity of 3 rad/s?

  • (A) 0.45 J
  • (B) 0.56 J
  • (C) 0.5 J
  • (D) 0.36 J
Show Answer

Answer: A – Work = change in rotational KE = (1/2)Iω2. I = (1/2)MR2 = 0.1 kg·m2, so W = 0.5×0.1×32 = 0.45 J.

Unit 7: Oscillations

Q54. A 1 kg mass oscillates on a spring of spring constant 100 N/m with amplitude 0.1 m. What is its maximum speed?

  • (A) 1.1 m/s
  • (B) 1.5 m/s
  • (C) 1 m/s
  • (D) 0.25 m/s
Show Answer

Answer: C – ω = √(k/m) = 10 rad/s. Maximum speed vmax = ωA = 10×0.1 = 1 m/s.

Q55. A 1 kg mass oscillates on a spring of spring constant 25 N/m with amplitude 0.2 m. What is the magnitude of its maximum acceleration?

  • (A) 5 m/s2
  • (B) 4 m/s2
  • (C) 3.75 m/s2
  • (D) 5.5 m/s2
Show Answer

Answer: A – ω = √(k/m) = 5 rad/s. Maximum acceleration amax = ω2A = 52×0.2 = 5 m/s2.

Q56. A particle in simple harmonic motion satisfies d2x/dt2 = -64x, where x is in metres and t is in seconds. What is the period of the motion?

  • (A) 0.79 s
  • (B) 1.57 s
  • (C) 0.47 s
  • (D) 0.2 s
Show Answer

Answer: A – Comparing with d2x/dt2 = -ω2x gives ω = √(64) = 8 rad/s. Period T = 2π/ω = 0.79 s.

Q57. A spring of spring constant 100 N/m oscillates with amplitude 0.2 m. What is the total mechanical energy of the oscillator?

  • (A) 1.2 J
  • (B) 2 J
  • (C) 1.5 J
  • (D) 2.2 J
Show Answer

Answer: B – E = (1/2)kA2 = 0.5×100×0.22 = 2 J.

Q58. An oscillator completes one full oscillation in 4 s. What is its frequency?

  • (A) 0.12 Hz
  • (B) 0.25 Hz
  • (C) 0.19 Hz
  • (D) 0.38 Hz
Show Answer

Answer: B – Frequency f = 1/T = 1÷4 = 0.25 Hz.

Q59. A simple pendulum has a period of 2 s for small oscillations. What is its length? Take g = 9.8 m/s2.

  • (A) 1.24 m
  • (B) 0.79 m
  • (C) 0.74 m
  • (D) 0.99 m
Show Answer

Answer: D – T = 2π√(L/g), so L = g(T/2π)2 = 9.8×(2÷2π)2 = 0.99 m.

Q60. A 1 kg mass on a spring of spring constant 25 N/m oscillates on a frictionless surface. What is the period of oscillation?

  • (A) 1.57 s
  • (B) 2.51 s
  • (C) 1.26 s
  • (D) 0.75 s
Show Answer

Answer: C – T = 2π√(m/k) = 2π√(1÷25) = 1.26 s.

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