AP Physics C: Mechanics Practice Test 6 – 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 particle moves along the x-axis with velocity v(t) = 3 + 4t + 2t2 m/s, where t is in seconds. What is its displacement between t = 2 s and t = 6 s?

  • (A) 107 m
  • (B) 215 m
  • (C) 234 m
  • (D) 322 m
Show Answer

Answer: B – Integrate v(t): displacement = ∪v dt = [3t + 4t2/2 + 2t3/3] from 2 to 6 s, which gives 215 m.

Q2. A cart starts with velocity 0 m/s and has acceleration a(t) = 3 + 2t m/s2. What is its velocity at t = 3 s?

  • (A) 9 m/s
  • (B) 27 m/s
  • (C) 18 m/s
  • (D) 30.5 m/s
Show Answer

Answer: C – Integrate a(t): v(t) = v0 + ∪a dt = 0 + 3t + 2t2/2. At t = 3 s this gives 18 m/s.

Q3. The position of a block is x(t) = 4t3 – 1t2 + 4t metres. What is its velocity at t = 2 s?

  • (A) 94 m/s
  • (B) 24 m/s
  • (C) 52 m/s
  • (D) 48 m/s
Show Answer

Answer: D – Differentiate: v(t) = dx/dt = 12t2 – 2t + 4. At t = 2 s, v = 48 m/s.

Q4. A projectile is launched at 25 m/s at 30° above the horizontal. What is its speed at t = 1 s? Take g = 9.8 m/s2.

  • (A) 26.2 m/s
  • (B) 21.8 m/s
  • (C) 25 m/s
  • (D) 28.2 m/s
Show Answer

Answer: B – vx = 25 cos 30° = 21.7 m/s and vy = 25 sin 30° – 9.8(1) = 2.7 m/s, so speed = √(vx2 + vy2) = 21.8 m/s.

Q5. A particle has velocity v(t) = 10t – 1t2 m/s. What is its displacement from t = 0 until it first comes to rest again (apart from t = 0)?

  • (A) 167 m
  • (B) 83.3 m
  • (C) 500 m
  • (D) 333 m
Show Answer

Answer: A – Set v = 0: t(10 – 1t) = 0, so t = 10 s. Displacement = ∪v dt = 10t2/2 – 1t3/3 evaluated at that t, giving 167 m.

Q6. A glider experiences acceleration a = -2v, where v is its velocity in m/s. It starts at 15 m/s. How far does it travel in the first 2 s?

  • (A) 10.3 m
  • (B) 7.5 m
  • (C) 7.36 m
  • (D) 30 m
Show Answer

Answer: C – Solve dv/dt = -2v to get v(t) = 15 e-2t. Integrating gives distance = (15/2)(1 – e-2(2)) = 7.36 m.

Q7. Two runners start from opposite ends of a straight track. Runner A has position xA = 2t2 m and runner B has position xB = 60 – 10t m, measured from A’s start. At what time do they meet?

  • (A) 6 s
  • (B) 3.52 s
  • (C) 5.28 s
  • (D) 5.48 s
Show Answer

Answer: B – Set xA = xB: 2t2 + 10t – 60 = 0. The positive root is t = 3.52 s.

Unit 2: Force and Translational Dynamics

Q8. A 2 kg block slides down a rough incline at 45° to the horizontal. The coefficient of kinetic friction is 0.2. What is its acceleration down the slope?

  • (A) 5.54 m/s2
  • (B) 2.77 m/s2
  • (C) 8.32 m/s2
  • (D) 6.93 m/s2
Show Answer

Answer: A – Along the slope, a = g(sin θ – μ cos θ) = 9.8(sin 45° – 0.2 cos 45°) = 5.54 m/s2.

Q9. Two masses 3 kg and 6 kg hang from a light pulley (Atwood machine). What is the magnitude of their acceleration?

  • (A) 6.53 m/s2
  • (B) 3.27 m/s2
  • (C) 5.54 m/s2
  • (D) 4.9 m/s2
Show Answer

Answer: B – Net force = (6 – 3)g and total mass = 9 kg, so a = (6 – 3)(9.8)/(9) = 3.27 m/s2.

Q10. A horizontal force of 20 N is applied at 45° above the horizontal to a 3 kg block on a frictionless floor. What is the block’s horizontal acceleration?

  • (A) 4.71 m/s2
  • (B) 5.33 m/s2
  • (C) 6.67 m/s2
  • (D) 7.07 m/s2
Show Answer

Answer: A – Horizontal component = 20 cos 45° = 14.1 N. Then a = Fx/m = 14.1/3 = 4.71 m/s2.

Q11. A 4 kg object experiences a time-varying force F(t) = 10 + 5t N. What is its acceleration at t = 3 s?

  • (A) 3.12 m/s2
  • (B) 6.25 m/s2
  • (C) 2.5 m/s2
  • (D) 13.8 m/s2
Show Answer

Answer: B – At t = 3 s, F = 10 + 5(3) = 25 N, so a = F/m = 25/4 = 6.25 m/s2.

Q12. A 2 kg mass hangs at rest from a vertical spring of constant 400 N/m. By how much is the spring stretched from its natural length?

  • (A) 0.005 m
  • (B) 0.0245 m
  • (C) 0.098 m
  • (D) 0.049 m
Show Answer

Answer: D – At equilibrium kx = mg, so x = (2)(9.8)/400 = 0.049 m.

Q13. A car travels on a flat circular bend of radius 25 m. The coefficient of static friction is 0.3. What is the maximum speed without slipping?

  • (A) 8.57 m/s
  • (B) 15.7 m/s
  • (C) 6.06 m/s
  • (D) 12.9 m/s
Show Answer

Answer: A – Friction provides the centripetal force: μmg = mv2/r, so v = √(μgr) = √(0.3 × 9.8 × 25) = 8.57 m/s.

Q14. A road is banked so that a car at 25 m/s can round a curve of radius 50 m with no sideways friction. What banking angle is required?

  • (A) 77.9°
  • (B) 38.1°
  • (C) 26°
  • (D) 51.9°
Show Answer

Answer: D – tan θ = v2/(gr) = 252/(9.8 × 50) = 1.28, so θ = 51.9°.

Q15. A planet has mass 3 times Earth’s mass and radius 1 times Earth’s radius. What is the gravitational field strength at its surface? (Earth’s g = 9.8 m/s2.)

  • (A) 37.5 m/s2
  • (B) 9.8 m/s2
  • (C) 29.4 m/s2
  • (D) 33.2 m/s2
Show Answer

Answer: C – g = GM/R2, so gplanet = 9.8 × 3/12 = 29.4 m/s2.

Q16. A 70 kg person stands on a scale in a lift accelerating upwards at 2 m/s2. What is the scale reading (normal force)?

  • (A) 826 N
  • (B) 991 N
  • (C) 686 N
  • (D) 546 N
Show Answer

Answer: A – N – mg = ma, so N = m(g + a) = 70(9.8 + 2) = 826 N.

Q17. A 4 kg crate on a rough floor (μk = 0.3) is pushed by a horizontal force of 50 N. What is its acceleration?

  • (A) 4.78 m/s2
  • (B) 9.56 m/s2
  • (C) 12.5 m/s2
  • (D) 15.4 m/s2
Show Answer

Answer: B – Friction = μmg = 0.3 × 4 × 9.8 = 11.8 N. Then a = (50 – 11.8)/4 = 9.56 m/s2.

Q18. A 1 kg bob on a string of length 1 m swings in a horizontal circle with the string at 60° to the vertical. What is the tension in the string?

  • (A) 29.4 N
  • (B) 19.6 N
  • (C) 9.8 N
  • (D) 4.9 N
Show Answer

Answer: B – Vertically, T cos θ = mg, so T = mg/cos θ = (1)(9.8)/cos 60° = 19.6 N.

Q19. A 3 kg object falls through air with drag force 1v N, where v is in m/s. What is its acceleration when its speed is 5 m/s?

  • (A) 8.13 m/s2
  • (B) 9.8 m/s2
  • (C) 12.2 m/s2
  • (D) 1.67 m/s2
Show Answer

Answer: A – Net force = mg – kv = 3(9.8) – 1(5) = 24.4 N, so a = F/m = 8.13 m/s2.

Unit 3: Work, Energy, and Power

Q20. A force F(x) = 20 + 4x N acts on a particle as it moves from x = 2 m to x = 4 m along the x-axis. How much work is done?

  • (A) 88 J
  • (B) 64 J
  • (C) 40 J
  • (D) 32 J
Show Answer

Answer: B – W = ∪F dx = [20x + 4x2/2] from 2 to 4 = 64 J.

Q21. A variable force F(x) = 10x2 N moves an object from x = 2 m to x = 4 m. What work is done?

  • (A) 60 J
  • (B) 560 J
  • (C) 93.3 J
  • (D) 187 J
Show Answer

Answer: D – W = ∪10x2 dx = 10x3/3 from 2 to 4 = 187 J.

Q22. A spring of constant 100 N/m is compressed by 0.2 m from equilibrium. How much elastic potential energy is stored?

  • (A) 10 J
  • (B) 4.52 J
  • (C) 2 J
  • (D) 4 J
Show Answer

Answer: C – U = (1/2)kx2 = 0.5 × 100 × 0.22 = 2 J.

Q23. A net work of 150 J is done on a 4 kg object initially moving at 0 m/s. What is its final speed?

  • (A) 13 m/s
  • (B) 37.5 m/s
  • (C) 8.66 m/s
  • (D) 9.79 m/s
Show Answer

Answer: C – W = ΔK, so (1/2)m v2 – (1/2)m(0)2 = 150. Solving gives v = 8.66 m/s.

Q24. A constant force of 40 N pushes an object in the direction of its motion at 3 m/s. What power is delivered?

  • (A) 60 W
  • (B) 120 W
  • (C) 240 W
  • (D) 43 W
Show Answer

Answer: B – P = Fv = 40 × 3 = 120 W.

Q25. An object is dropped from rest from a height of 8 m. Ignoring air resistance, what is its speed just before hitting the ground?

  • (A) 12.5 m/s
  • (B) 6.26 m/s
  • (C) 78.4 m/s
  • (D) 8.85 m/s
Show Answer

Answer: A – Energy conservation: mgh = (1/2)mv2, so v = √(2gh) = √(2 × 9.8 × 8) = 12.5 m/s.

Q26. A small bead slides on a frictionless vertical loop of radius 2 m. What is the minimum speed it must have at the top of the loop to stay in contact?

  • (A) 6.26 m/s
  • (B) 8.85 m/s
  • (C) 4.43 m/s
  • (D) 7 m/s
Show Answer

Answer: C – At the top, mg = mv2/r for the limiting case, so v = √(gr) = √(9.8 × 2) = 4.43 m/s.

Q27. A 3 kg block attached to a spring (k = 50 N/m) is released from rest when the spring is compressed 0.2 m. What is its maximum speed?

  • (A) 4.08 m/s
  • (B) 0.408 m/s
  • (C) 3.33 m/s
  • (D) 0.816 m/s
Show Answer

Answer: D – Energy: (1/2)kx2 = (1/2)mv2, so v = x√(k/m) = 0.2 × √(50/3) = 0.816 m/s.

Q28. A 2 kg block slides 5 m across a rough horizontal surface with coefficient of kinetic friction 0.2. What is the magnitude of the work done by friction?

  • (A) 2 J
  • (B) 9.8 J
  • (C) 98 J
  • (D) 19.6 J
Show Answer

Answer: D – Magnitude = μmgd = 0.2 × 2 × 9.8 × 5 = 19.6 J.

Q29. A motor lifts a 80 kg load through a height of 10 m in 5 s at constant speed. What is the average power output?

  • (A) 160 W
  • (B) 7.84e+03 W
  • (C) 784 W
  • (D) 1.57e+03 W
Show Answer

Answer: D – Work = mgh = 80 × 9.8 × 10 = 7.84e+03 J, so average power = Work/t = 7.84e+03/5 = 1.57e+03 W.

Q30. A force F(x) = 20 cos(πx/(2 × 3)) N, where x is in metres, acts from x = 0 to x = 3 m. How much work is done? Use π = 3.14.

  • (A) 38.2 J
  • (B) 19.1 J
  • (C) 60 J
  • (D) 76.4 J
Show Answer

Answer: A – W = ∪F dx = 20 × (2 × 3/π) = 38.2 J.


Unit 4: Linear Momentum

Q31. A force F(t) = 20 + 2t N acts on an object from t = 0 to t = 2 s. What impulse does it deliver?

  • (A) 44 N·s
  • (B) 22 N·s
  • (C) 48 N·s
  • (D) 40 N·s
Show Answer

Answer: A – J = ∪F dt = [20t + 2t2/2] from 0 to 2 = 44 N·s.

Q32. A force-time graph is a triangle with peak force 200 N and base 0.3 s. What impulse does it deliver?

  • (A) 30 N·s
  • (B) 15 N·s
  • (C) 67.8 N·s
  • (D) 60 N·s
Show Answer

Answer: A – Impulse is the area under the F-t graph: J = (1/2) × 200 × 0.3 = 30 N·s.

Q33. A 2 kg ball moving at 6 m/s collides elastically head-on with a stationary 3 kg ball. What is the second ball’s velocity after the collision?

  • (A) -1.2 m/s
  • (B) 2.4 m/s
  • (C) 6 m/s
  • (D) 4.8 m/s
Show Answer

Answer: D – For an elastic collision with the second mass at rest, v2‘ = 2m1v1/(m1 + m2) = 2 × 2 × 6/(2 + 3) = 4.8 m/s.

Q34. A 3 kg cart at 4 m/s catches up and sticks to a 2 kg cart moving at 3 m/s in the same direction. What is their common velocity?

  • (A) 5.4 m/s
  • (B) 1.2 m/s
  • (C) 2.4 m/s
  • (D) 3.6 m/s
Show Answer

Answer: D – Momentum conservation: (3 × 4 + 2 × 3)/(3 + 2) = 3.6 m/s.

Q35. A 0.01 kg bullet at 200 m/s embeds in a 3 kg wooden block hanging as a pendulum. How high does the block rise?

  • (A) 0.0225 m
  • (B) 6.78 m
  • (C) 0.0509 m
  • (D) 0.0451 m
Show Answer

Answer: A – First, momentum gives V = 0.01 × 200/(0.01 + 3) = 0.664 m/s. Then energy gives h = V2/(2g) = 0.0225 m.

Q36. An object at rest explodes into a 1 kg fragment moving at 4 m/s and a 2 kg fragment. What is the speed of the second fragment?

  • (A) 4.52 m/s
  • (B) 4 m/s
  • (C) 8 m/s
  • (D) 2 m/s
Show Answer

Answer: D – Momentum conservation: 1 × 4 = 2 × v2, so v2 = 2 m/s.

Q37. Two masses 3 kg at x = 1 m and 6 kg at x = 4 m lie on the x-axis. Where is their centre of mass?

  • (A) 3 m
  • (B) 4.5 m
  • (C) 13.5 m
  • (D) 2 m
Show Answer

Answer: A – xcm = (3 × 1 + 6 × 4)/(3 + 6) = 3 m.

Q38. A 0.5 kg ball hits a wall at 10 m/s and rebounds at 10 m/s in the opposite direction. What is the magnitude of the impulse from the wall?

  • (A) 5 N·s
  • (B) 5.65 N·s
  • (C) 10 N·s
  • (D) 2.5 N·s
Show Answer

Answer: C – Δp = m(vf – vi) = 0.5(10 – (-10)) = 10 N·s in magnitude.

Unit 5: Torque and Rotational Dynamics

Q39. A force of 40 N is applied at a distance 0.5 m from a pivot, at 60° to the lever arm. What is the torque about the pivot?

  • (A) 10 N·m
  • (B) 17.3 N·m
  • (C) 20 N·m
  • (D) 34.6 N·m
Show Answer

Answer: B – τ = rF sin θ = 0.5 × 40 × sin 60° = 17.3 N·m.

Q40. A wheel starts from rest with angular acceleration α(t) = 2 + 2t rad/s2. What is its angular velocity at t = 2 s?

  • (A) 4.52 rad/s
  • (B) 8 rad/s
  • (C) 4 rad/s
  • (D) 12 rad/s
Show Answer

Answer: B – ω(t) = ∪α dt = 2t + 2t2/2. At t = 2 s, ω = 8 rad/s.

Q41. Point masses 1 kg at 2 m and 3 kg at 1 m from an axis have what total moment of inertia about that axis?

  • (A) 3.5 kg·m2
  • (B) 12 kg·m2
  • (C) 7 kg·m2
  • (D) 5 kg·m2
Show Answer

Answer: C – I = Σmr2 = 1 × 22 + 3 × 12 = 7 kg·m2.

Q42. A uniform rod of mass 2 kg and length 1 m has moment of inertia 0.167 kg·m2 about its centre. What is its moment of inertia about a parallel axis 1 m from the centre?

  • (A) 0.667 kg·m2
  • (B) 4.33 kg·m2
  • (C) 2.17 kg·m2
  • (D) 0.167 kg·m2
Show Answer

Answer: C – Parallel-axis theorem: I = Icm + md2 = 0.167 + 2 × 12 = 2.17 kg·m2.

Q43. A net torque of 10 N·m acts on a body with moment of inertia 5 kg·m2. What is its angular acceleration?

  • (A) 1 rad/s2
  • (B) 50 rad/s2
  • (C) 2 rad/s2
  • (D) 4 rad/s2
Show Answer

Answer: C – τ = Iα, so α = 10/5 = 2 rad/s2.

Q44. A 1 kg particle moves at 4 m/s at a perpendicular distance 0.5 m from a point. What is its angular momentum about that point?

  • (A) 4.52 kg·m2/s
  • (B) 2 kg·m2/s
  • (C) 1 kg·m2/s
  • (D) 4 kg·m2/s
Show Answer

Answer: B – L = mvr = 1 × 4 × 0.5 = 2 kg·m2/s.

Q45. A flywheel with I = 6 kg·m2 spins at 12 rad/s. A constant opposing torque of 2 N·m is applied. How long until it stops?

  • (A) 6 s
  • (B) 18 s
  • (C) 72 s
  • (D) 36 s
Show Answer

Answer: D – Angular impulse: τΔt = Iω, so Δt = 6 × 12/2 = 36 s.

Q46. A uniform rod of mass 2 kg and length 2 m is pivoted at one end and held at 60° to the vertical. What torque does gravity exert about the pivot?

  • (A) 19.6 N·m
  • (B) 33.9 N·m
  • (C) 17 N·m
  • (D) 8.49 N·m
Show Answer

Answer: C – Gravity acts at the centre, 1 m from the pivot, so τ = (mg)(L/2) sin θ = 2 × 9.8 × 1 × sin 60° = 17 N·m.

Unit 6: Energy and Momentum of Rotating Systems

Q47. A body with moment of inertia 6 kg·m2 rotates at 5 rad/s. What is its rotational kinetic energy?

  • (A) 150 J
  • (B) 15 J
  • (C) 169 J
  • (D) 75 J
Show Answer

Answer: D – K = (1/2)Iω2 = 0.5 × 6 × 52 = 75 J.

Q48. A disk with I = 4 kg·m2 spinning at 6 rad/s couples to a stationary disk with I = 4 kg·m2 on the same axle, with no external torque. What is their common angular velocity?

  • (A) 6.78 rad/s
  • (B) 3 rad/s
  • (C) 6 rad/s
  • (D) 7.66 rad/s
Show Answer

Answer: B – Angular momentum conservation: 4 × 6 = (4 + 4)ω, so ω = 3 rad/s.

Q49. A satellite orbits Earth at a radius of 8000 km from Earth’s centre. What is its orbital speed? (GMEarth = 3.99 × 1014 m3/s2.)

  • (A) 7.06 km/s
  • (B) 3.53 km/s
  • (C) 10.6 km/s
  • (D) 9.98 km/s
Show Answer

Answer: A – v = √(GM/r) = √(3.99 × 1014 / 8000000) = 7.06 km/s.

Q50. What is the orbital period, in hours, of a satellite at radius 10000 km from Earth’s centre? (GMEarth = 3.99 × 1014 m3/s2.)

  • (A) 2.76 hours
  • (B) 1.38 hours
  • (C) 5.53 hours
  • (D) 4.15 hours
Show Answer

Answer: A – T = 2π√(r3/GM) = 2π√((10000000)3/(3.99 × 1014)) = 2.76 hours.

Q51. A planet has mass 8 times Earth’s mass and radius 1 times Earth’s radius. What is its escape velocity if Earth’s is 11.2 km/s?

  • (A) 35.8 km/s
  • (B) 11.2 km/s
  • (C) 15.8 km/s
  • (D) 31.7 km/s
Show Answer

Answer: D – vesc = √(2GM/R), so it scales as √(M/R): 11.2 × √(8/1) = 31.7 km/s.

Q52. A 1000 kg satellite orbits at radius 12000 km from Earth’s centre. What is the magnitude of its total mechanical energy, in GJ? (GMEarth = 3.99 × 1014 m3/s2.)

  • (A) 33.2 GJ
  • (B) 16.6 GJ
  • (C) 8.3 GJ
  • (D) 24.9 GJ
Show Answer

Answer: B – E = -GMm/(2r) = -(3.99 × 1014 × 1000)/(2 × 12000000) = -16.6 GJ, so the magnitude is 16.6 GJ.

Q53. A 100 kg satellite moves from radius 7000 km to radius 14000 km from Earth’s centre. By how much does its gravitational potential energy change, in MJ?

  • (A) 5.69e+03 MJ
  • (B) 3.42e+03 MJ
  • (C) 2.85e+03 MJ
  • (D) 1.42e+03 MJ
Show Answer

Answer: C – ΔU = -GMm(1/r2 – 1/r1) = 2.85e+03 MJ (positive, since it moves farther out).

Unit 7: Oscillations

Q54. A mass oscillates with x(t) = 0.2 cos(10t) m. What is its maximum speed?

  • (A) 20 m/s
  • (B) 4 m/s
  • (C) 0.02 m/s
  • (D) 2 m/s
Show Answer

Answer: D – vmax = Aω = 0.2 × 10 = 2 m/s.

Q55. A 2 kg mass on a spring (k = 400 N/m) oscillates horizontally. What is its period?

  • (A) 88.9 s
  • (B) 0.0707 s
  • (C) 0.444 s
  • (D) 0.222 s
Show Answer

Answer: C – T = 2π√(m/k) = 2π√(2/400) = 0.444 s.

Q56. A simple pendulum of length 0.5 m oscillates with small amplitude. What is its period?

  • (A) 27.8 s
  • (B) 2.84 s
  • (C) 1.42 s
  • (D) 0.226 s
Show Answer

Answer: C – T = 2π√(L/g) = 2π√(0.5/9.8) = 1.42 s.

Q57. A 1 kg mass on a spring (k = 100 N/m) oscillates with amplitude 0.1 m. What is its speed when its displacement is 0.05 m?

  • (A) 0.433 m/s
  • (B) 1 m/s
  • (C) 0.866 m/s
  • (D) 0.5 m/s
Show Answer

Answer: C – Energy: (1/2)kA2 = (1/2)kx2 + (1/2)mv2, so v = √((k/m)(A2 – x2)) = 0.866 m/s.

Q58. A particle in SHM obeys d2x/dt2 = -42x. What is the magnitude of its acceleration when its displacement is 0.1 m?

  • (A) 1.6 m/s2
  • (B) 3.2 m/s2
  • (C) 0.4 m/s2
  • (D) 0.8 m/s2
Show Answer

Answer: A – From d2x/dt2 = -ω2x, |a| = ω2|x| = 42 × 0.1 = 1.6 m/s2.

Q59. A 4 kg mass on a spring oscillates with period 1 s. What is the spring constant?

  • (A) 79 N/m
  • (B) 158 N/m
  • (C) 39.5 N/m
  • (D) 178 N/m
Show Answer

Answer: B – T = 2π√(m/k), so k = 4π2m/T2 = 4π2 × 4/12 = 158 N/m.

Q60. A particle in SHM has x(t) = 0.3 cos(2t) m. What is its position at t = 2 s?

  • (A) 0.196 m
  • (B) -0.227 m
  • (C) -0.0437 m
  • (D) -0.196 m
Show Answer

Answer: D – x(2) = 0.3 cos(2 × 2) = -0.196 m.

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