AP Physics C: Mechanics Practice Test 7 – 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 projectile is launched at 20 m/s at 60° above the horizontal. What is its speed at t = 3 s? Take g = 9.8 m/s2.

  • (A) 15.7 m/s
  • (B) 18.8 m/s
  • (C) 22.6 m/s
  • (D) 20 m/s
Show Answer

Answer: A – vx = 20 cos 60° = 10 m/s and vy = 20 sin 60° – 9.8(3) = -12.1 m/s, so speed = √(vx2 + vy2) = 15.7 m/s.

Q2. A particle has velocity v(t) = 6t – 2t2 m/s. What is its displacement from t = 0 until it first comes to rest again (apart from t = 0)?

  • (A) 4.5 m
  • (B) 18 m
  • (C) 27 m
  • (D) 9 m
Show Answer

Answer: D – Set v = 0: t(6 – 2t) = 0, so t = 3 s. Displacement = ∪v dt = 6t2/2 – 2t3/3 evaluated at that t, giving 9 m.

Q3. A glider experiences acceleration a = -0.5v, where v is its velocity in m/s. It starts at 12 m/s. How far does it travel in the first 1 s?

  • (A) 12 m
  • (B) 13.2 m
  • (C) 24 m
  • (D) 9.44 m
Show Answer

Answer: D – Solve dv/dt = -0.5v to get v(t) = 12 e-0.5t. Integrating gives distance = (12/0.5)(1 – e-0.5(1)) = 9.44 m.

Q4. Two runners start from opposite ends of a straight track. Runner A has position xA = 1t2 m and runner B has position xB = 60 – 10t m, measured from A’s start. At what time do they meet?

  • (A) 7.75 s
  • (B) 6.33 s
  • (C) 6 s
  • (D) 4.22 s
Show Answer

Answer: D – Set xA = xB: 1t2 + 10t – 60 = 0. The positive root is t = 4.22 s.

Q5. A sled starts from rest with acceleration a(t) = 4 + 3t m/s2. How far does it travel in the first 3 s?

  • (A) 18 m
  • (B) 49.5 m
  • (C) 15.8 m
  • (D) 31.5 m
Show Answer

Answer: D – Integrate twice: v(t) = 4t + 3t2/2 and x(t) = 4t2/2 + 3t3/6. At t = 3 s, x = 31.5 m.

Q6. A particle moves along the x-axis with velocity v(t) = 1 + 2t + 3t2 m/s, where t is in seconds. What is its displacement between t = 2 s and t = 5 s?

  • (A) 70.5 m
  • (B) 141 m
  • (C) 212 m
  • (D) 155 m
Show Answer

Answer: B – Integrate v(t): displacement = ∪v dt = [1t + 2t2/2 + 3t3/3] from 2 to 5 s, which gives 141 m.

Q7. A cart starts with velocity 2 m/s and has acceleration a(t) = 2 + 3t m/s2. What is its velocity at t = 4 s?

  • (A) 51 m/s
  • (B) 10 m/s
  • (C) 34 m/s
  • (D) 58 m/s
Show Answer

Answer: C – Integrate a(t): v(t) = v0 + ∪a dt = 2 + 2t + 3t2/2. At t = 4 s this gives 34 m/s.

Unit 2: Force and Translational Dynamics

Q8. A 4 kg object experiences a time-varying force F(t) = 20 + 2t N. What is its acceleration at t = 2 s?

  • (A) 6 m/s2
  • (B) 3 m/s2
  • (C) 5 m/s2
  • (D) 7 m/s2
Show Answer

Answer: A – At t = 2 s, F = 20 + 2(2) = 24 N, so a = F/m = 24/4 = 6 m/s2.

Q9. A 2 kg mass hangs at rest from a vertical spring of constant 200 N/m. By how much is the spring stretched from its natural length?

  • (A) 0.196 m
  • (B) 0.098 m
  • (C) 0.01 m
  • (D) 0.049 m
Show Answer

Answer: B – At equilibrium kx = mg, so x = (2)(9.8)/200 = 0.098 m.

Q10. A car travels on a flat circular bend of radius 20 m. The coefficient of static friction is 0.6. What is the maximum speed without slipping?

  • (A) 16.3 m/s
  • (B) 14 m/s
  • (C) 10.8 m/s
  • (D) 7.67 m/s
Show Answer

Answer: C – Friction provides the centripetal force: μmg = mv2/r, so v = √(μgr) = √(0.6 × 9.8 × 20) = 10.8 m/s.

Q11. A road is banked so that a car at 25 m/s can round a curve of radius 80 m with no sideways friction. What banking angle is required?

  • (A) 57.8°
  • (B) 51.4°
  • (C) 38.6°
  • (D) 19.3°
Show Answer

Answer: C – tan θ = v2/(gr) = 252/(9.8 × 80) = 0.797, so θ = 38.6°.

Q12. A planet has mass 4 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) 44.3 m/s2
  • (B) 39.2 m/s2
  • (C) 50.1 m/s2
  • (D) 9.8 m/s2
Show Answer

Answer: B – g = GM/R2, so gplanet = 9.8 × 4/12 = 39.2 m/s2.

Q13. A 60 kg person stands on a scale in a lift accelerating upwards at 2 m/s2. What is the scale reading (normal force)?

  • (A) 588 N
  • (B) 468 N
  • (C) 850 N
  • (D) 708 N
Show Answer

Answer: D – N – mg = ma, so N = m(g + a) = 60(9.8 + 2) = 708 N.

Q14. A 4 kg crate on a rough floor (μk = 0.2) is pushed by a horizontal force of 40 N. What is its acceleration?

  • (A) 8.04 m/s2
  • (B) 10 m/s2
  • (C) 12 m/s2
  • (D) 4.02 m/s2
Show Answer

Answer: A – Friction = μmg = 0.2 × 4 × 9.8 = 7.84 N. Then a = (40 – 7.84)/4 = 8.04 m/s2.

Q15. A 2 kg bob on a string of length 0.8 m swings in a horizontal circle with the string at 30° to the vertical. What is the tension in the string?

  • (A) 22.6 N
  • (B) 19.6 N
  • (C) 33.9 N
  • (D) 17 N
Show Answer

Answer: A – Vertically, T cos θ = mg, so T = mg/cos θ = (2)(9.8)/cos 30° = 22.6 N.

Q16. A 3 kg object falls through air with drag force 0.5v N, where v is in m/s. What is its acceleration when its speed is 5 m/s?

  • (A) 0.833 m/s2
  • (B) 13.4 m/s2
  • (C) 9.8 m/s2
  • (D) 8.97 m/s2
Show Answer

Answer: D – Net force = mg – kv = 3(9.8) – 0.5(5) = 26.9 N, so a = F/m = 8.97 m/s2.

Q17. Two blocks 4 kg and 3 kg connected by a light string rest on a frictionless table. A horizontal force of 40 N pulls the 4 kg block. What is the tension in the string?

  • (A) 40 N
  • (B) 22.9 N
  • (C) 17.1 N
  • (D) 34.3 N
Show Answer

Answer: C – System acceleration a = 40/(4 + 3) = 5.71 m/s2. Tension accelerates the 3 kg block: T = 3 × 5.71 = 17.1 N.

Q18. A 0.5 kg mass moves in a horizontal circle of radius 0.5 m at speed 4 m/s, held by a string. What is the tension in the string?

  • (A) 32 N
  • (B) 16 N
  • (C) 4 N
  • (D) 8 N
Show Answer

Answer: B – Tension provides the centripetal force: T = mv2/r = 0.5 × 42/0.5 = 16 N.

Q19. A 5 kg block slides down a rough incline at 45° to the horizontal. The coefficient of kinetic friction is 0.1. What is its acceleration down the slope?

  • (A) 6.24 m/s2
  • (B) 7.62 m/s2
  • (C) 3.12 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° – 0.1 cos 45°) = 6.24 m/s2.

Unit 3: Work, Energy, and Power

Q20. A net work of 100 J is done on a 4 kg object initially moving at 0 m/s. What is its final speed?

  • (A) 7.99 m/s
  • (B) 7.07 m/s
  • (C) 10.6 m/s
  • (D) 25 m/s
Show Answer

Answer: B – W = ΔK, so (1/2)m v2 – (1/2)m(0)2 = 100. Solving gives v = 7.07 m/s.

Q21. A constant force of 20 N pushes an object in the direction of its motion at 4 m/s. What power is delivered?

  • (A) 160 W
  • (B) 24 W
  • (C) 80 W
  • (D) 40 W
Show Answer

Answer: C – P = Fv = 20 × 4 = 80 W.

Q22. An object is dropped from rest from a height of 30 m. Ignoring air resistance, what is its speed just before hitting the ground?

  • (A) 12.1 m/s
  • (B) 294 m/s
  • (C) 17.1 m/s
  • (D) 24.2 m/s
Show Answer

Answer: D – Energy conservation: mgh = (1/2)mv2, so v = √(2gh) = √(2 × 9.8 × 30) = 24.2 m/s.

Q23. A small bead slides on a frictionless vertical loop of radius 2.5 m. What is the minimum speed it must have at the top of the loop to stay in contact?

  • (A) 7.83 m/s
  • (B) 7 m/s
  • (C) 9.9 m/s
  • (D) 4.95 m/s
Show Answer

Answer: D – At the top, mg = mv2/r for the limiting case, so v = √(gr) = √(9.8 × 2.5) = 4.95 m/s.

Q24. A 3 kg block attached to a spring (k = 100 N/m) is released from rest when the spring is compressed 0.2 m. What is its maximum speed?

  • (A) 0.577 m/s
  • (B) 1.15 m/s
  • (C) 5.77 m/s
  • (D) 6.67 m/s
Show Answer

Answer: B – Energy: (1/2)kx2 = (1/2)mv2, so v = x√(k/m) = 0.2 × √(100/3) = 1.15 m/s.

Q25. A 4 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) 39.2 J
  • (B) 4 J
  • (C) 19.6 J
  • (D) 196 J
Show Answer

Answer: A – Magnitude = μmgd = 0.2 × 4 × 9.8 × 5 = 39.2 J.

Q26. A motor lifts a 80 kg load through a height of 10 m in 10 s at constant speed. What is the average power output?

  • (A) 392 W
  • (B) 784 W
  • (C) 7.84e+03 W
  • (D) 80 W
Show Answer

Answer: B – Work = mgh = 80 × 9.8 × 10 = 7.84e+03 J, so average power = Work/t = 7.84e+03/10 = 784 W.

Q27. A force F(x) = 25 cos(πx/(2 × 4)) N, where x is in metres, acts from x = 0 to x = 4 m. How much work is done? Use π = 3.14.

  • (A) 100 J
  • (B) 127 J
  • (C) 31.8 J
  • (D) 63.7 J
Show Answer

Answer: D – W = ∪F dx = 25 × (2 × 4/π) = 63.7 J.

Q28. A 2 kg block starts down a frictionless incline at 6 m/s and descends a vertical height of 5 m. What is its speed at the bottom?

  • (A) 5.79 m/s
  • (B) 9.9 m/s
  • (C) 11.6 m/s
  • (D) 15.9 m/s
Show Answer

Answer: C – Energy: (1/2)mv2 + mgh = (1/2)mvf2, so vf = √(62 + 2 × 9.8 × 5) = 11.6 m/s.

Q29. A force F(x) = 20 + 4x N acts on a particle as it moves from x = 1 m to x = 3 m along the x-axis. How much work is done?

  • (A) 40 J
  • (B) 56 J
  • (C) 28 J
  • (D) 72 J
Show Answer

Answer: B – W = ∪F dx = [20x + 4x2/2] from 1 to 3 = 56 J.

Q30. A variable force F(x) = 6x2 N moves an object from x = 2 m to x = 4 m. What work is done?

  • (A) 36 J
  • (B) 112 J
  • (C) 56 J
  • (D) 336 J
Show Answer

Answer: B – W = ∪6x2 dx = 6x3/3 from 2 to 4 = 112 J.


Unit 4: Linear Momentum

Q31. A 3 kg cart at 4 m/s catches up and sticks to a 4 kg cart moving at 3 m/s in the same direction. What is their common velocity?

  • (A) 5.14 m/s
  • (B) 3.43 m/s
  • (C) 0 m/s
  • (D) 1.71 m/s
Show Answer

Answer: B – Momentum conservation: (3 × 4 + 4 × 3)/(3 + 4) = 3.43 m/s.

Q32. A 0.01 kg bullet at 200 m/s embeds in a 2 kg wooden block hanging as a pendulum. How high does the block rise?

  • (A) 0.0505 m
  • (B) 0.114 m
  • (C) 0.101 m
  • (D) 10.2 m
Show Answer

Answer: A – First, momentum gives V = 0.01 × 200/(0.01 + 2) = 0.995 m/s. Then energy gives h = V2/(2g) = 0.0505 m.

Q33. An object at rest explodes into a 1 kg fragment moving at 4 m/s and a 3 kg fragment. What is the speed of the second fragment?

  • (A) 1.33 m/s
  • (B) 4 m/s
  • (C) 2.67 m/s
  • (D) 12 m/s
Show Answer

Answer: A – Momentum conservation: 1 × 4 = 3 × v2, so v2 = 1.33 m/s.

Q34. Two masses 2 kg at x = 1 m and 6 kg at x = 4 m lie on the x-axis. Where is their centre of mass?

  • (A) 2 m
  • (B) 13 m
  • (C) 3.25 m
  • (D) 4.88 m
Show Answer

Answer: C – xcm = (2 × 1 + 6 × 4)/(2 + 6) = 3.25 m.

Q35. A 1 kg ball hits a wall at 10 m/s and rebounds at 20 m/s in the opposite direction. What is the magnitude of the impulse from the wall?

  • (A) 30 N·s
  • (B) 10 N·s
  • (C) 11.3 N·s
  • (D) 15 N·s
Show Answer

Answer: A – Δp = m(vf – vi) = 1(20 – (-10)) = 30 N·s in magnitude.

Q36. A force F(t) = 30 sin(πt/2) N acts from t = 0 to t = 2 s. What impulse does it deliver? Use π = 3.14.

  • (A) 21.6 N·s
  • (B) 38.2 N·s
  • (C) 60 N·s
  • (D) 19.1 N·s
Show Answer

Answer: B – J = ∪30 sin(πt/2) dt from 0 to 2 = 2 × 30 × 2/π = 38.2 N·s.

Q37. A constant force of 20 N acts for 2 s on a 3 kg object already moving at 5 m/s in the force direction. What is its final speed?

  • (A) 11.7 m/s
  • (B) 9.17 m/s
  • (C) 18.3 m/s
  • (D) 45 m/s
Show Answer

Answer: C – Impulse J = 20 × 2 = 40 N·s, so Δv = J/m = 40/3 = 13.3 m/s, giving vf = 5 + 13.3 = 18.3 m/s.

Q38. A force F(t) = 20 + 5t N acts on an object from t = 0 to t = 4 s. What impulse does it deliver?

  • (A) 80 N·s
  • (B) 60 N·s
  • (C) 120 N·s
  • (D) 160 N·s
Show Answer

Answer: C – J = ∪F dt = [20t + 5t2/2] from 0 to 4 = 120 N·s.

Unit 5: Torque and Rotational Dynamics

Q39. 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 0.5 m from the centre?

  • (A) 0.753 kg·m2
  • (B) 0.667 kg·m2
  • (C) 1.33 kg·m2
  • (D) 0.167 kg·m2
Show Answer

Answer: B – Parallel-axis theorem: I = Icm + md2 = 0.167 + 2 × 0.52 = 0.667 kg·m2.

Q40. A net torque of 20 N·m acts on a body with moment of inertia 2 kg·m2. What is its angular acceleration?

  • (A) 20 rad/s2
  • (B) 40 rad/s2
  • (C) 10 rad/s2
  • (D) 5 rad/s2
Show Answer

Answer: C – τ = Iα, so α = 20/2 = 10 rad/s2.

Q41. A 1 kg particle moves at 3 m/s at a perpendicular distance 0.5 m from a point. What is its angular momentum about that point?

  • (A) 3 kg·m2/s
  • (B) 0.75 kg·m2/s
  • (C) 1.5 kg·m2/s
  • (D) 3.39 kg·m2/s
Show Answer

Answer: C – L = mvr = 1 × 3 × 0.5 = 1.5 kg·m2/s.

Q42. A flywheel with I = 4 kg·m2 spins at 12 rad/s. A constant opposing torque of 2 N·m is applied. How long until it stops?

  • (A) 24 s
  • (B) 12 s
  • (C) 48 s
  • (D) 6 s
Show Answer

Answer: A – Angular impulse: τΔt = Iω, so Δt = 4 × 12/2 = 24 s.

Q43. A uniform rod of mass 4 kg and length 1 m is pivoted at one end and held at 90° to the vertical. What torque does gravity exert about the pivot?

  • (A) 39.2 N·m
  • (B) 19.6 N·m
  • (C) 22.1 N·m
  • (D) 9.8 N·m
Show Answer

Answer: B – Gravity acts at the centre, 0.5 m from the pivot, so τ = (mg)(L/2) sin θ = 4 × 9.8 × 0.5 × sin 90° = 19.6 N·m.

Q44. A tangential force of 30 N is applied at the rim of a wheel of radius 0.4 m and moment of inertia 4 kg·m2. What is the wheel’s angular acceleration?

  • (A) 7.5 rad/s2
  • (B) 12 rad/s2
  • (C) 1.5 rad/s2
  • (D) 3 rad/s2
Show Answer

Answer: D – τ = Fr = 30 × 0.4 = 12 N·m, so α = τ/I = 12/4 = 3 rad/s2.

Q45. A disk starts with angular velocity 0 rad/s and constant angular acceleration 2 rad/s2. Through what angle does it rotate in 5 s?

  • (A) 25 rad
  • (B) 50 rad
  • (C) 12.5 rad
  • (D) 56.5 rad
Show Answer

Answer: A – θ = ω0t + (1/2)αt2 = 0 × 5 + 0.5 × 2 × 52 = 25 rad.

Q46. A force of 20 N is applied at a distance 2 m from a pivot, at 60° to the lever arm. What is the torque about the pivot?

  • (A) 69.3 N·m
  • (B) 40 N·m
  • (C) 20 N·m
  • (D) 34.6 N·m
Show Answer

Answer: D – τ = rF sin θ = 2 × 20 × sin 60° = 34.6 N·m.

Unit 6: Energy and Momentum of Rotating Systems

Q47. What is the orbital period, in hours, of a satellite at radius 42000 km from Earth’s centre? (GMEarth = 3.99 × 1014 m3/s2.)

  • (A) 47.6 hours
  • (B) 11.9 hours
  • (C) 23.8 hours
  • (D) 35.7 hours
Show Answer

Answer: C – T = 2π√(r3/GM) = 2π√((42000000)3/(3.99 × 1014)) = 23.8 hours.

Q48. A planet has mass 3 times Earth’s mass and radius 2 times Earth’s radius. What is its escape velocity if Earth’s is 11.2 km/s?

  • (A) 19.4 km/s
  • (B) 5.6 km/s
  • (C) 6.86 km/s
  • (D) 13.7 km/s
Show Answer

Answer: D – vesc = √(2GM/R), so it scales as √(M/R): 11.2 × √(3/2) = 13.7 km/s.

Q49. A 500 kg satellite orbits at radius 8000 km from Earth’s centre. What is the magnitude of its total mechanical energy, in GJ? (GMEarth = 3.99 × 1014 m3/s2.)

  • (A) 24.9 GJ
  • (B) 12.5 GJ
  • (C) 18.7 GJ
  • (D) 6.23 GJ
Show Answer

Answer: B – E = -GMm/(2r) = -(3.99 × 1014 × 500)/(2 × 8000000) = -12.5 GJ, so the magnitude is 12.5 GJ.

Q50. A 150 kg satellite moves from radius 7000 km to radius 21000 km from Earth’s centre. By how much does its gravitational potential energy change, in MJ?

  • (A) 1.14e+04 MJ
  • (B) 2.85e+03 MJ
  • (C) 6.83e+03 MJ
  • (D) 5.69e+03 MJ
Show Answer

Answer: D – ΔU = -GMm(1/r2 – 1/r1) = 5.69e+03 MJ (positive, since it moves farther out).

Q51. A 2 kg mass falls 5 m while unwinding a string from a pulley (I = 2 kg·m2, radius 0.5 m). What is its speed at the end, ignoring friction?

  • (A) 9.9 m/s
  • (B) 6.64 m/s
  • (C) 4.43 m/s
  • (D) 2.21 m/s
Show Answer

Answer: C – Energy: mgh = (1/2)mv2 + (1/2)I(v/r)2, so v = √(2mgh/(m + I/r2)) = 4.43 m/s.

Q52. Planet A orbits a star with period 2 years. Planet B orbits the same star at 8 times A’s orbital radius. What is B’s period?

  • (A) 45.3 years
  • (B) 128 years
  • (C) 22.6 years
  • (D) 16 years
Show Answer

Answer: A – Kepler’s third law: T2 proportional to r3, so TB = 2 × 83/2 = 45.3 years.

Q53. A body with moment of inertia 2 kg·m2 rotates at 8 rad/s. What is its rotational kinetic energy?

  • (A) 64 J
  • (B) 128 J
  • (C) 145 J
  • (D) 8 J
Show Answer

Answer: A – K = (1/2)Iω2 = 0.5 × 2 × 82 = 64 J.

Unit 7: Oscillations

Q54. A 1 kg mass on a spring (k = 200 N/m) oscillates with amplitude 0.2 m. What is its speed when its displacement is 0.1 m?

  • (A) 2.83 m/s
  • (B) 2.45 m/s
  • (C) 1.41 m/s
  • (D) 1.22 m/s
Show Answer

Answer: B – Energy: (1/2)kA2 = (1/2)kx2 + (1/2)mv2, so v = √((k/m)(A2 – x2)) = 2.45 m/s.

Q55. A particle in SHM obeys d2x/dt2 = -52x. What is the magnitude of its acceleration when its displacement is 0.2 m?

  • (A) 2.5 m/s2
  • (B) 1 m/s2
  • (C) 10 m/s2
  • (D) 5 m/s2
Show Answer

Answer: D – From d2x/dt2 = -ω2x, |a| = ω2|x| = 52 × 0.2 = 5 m/s2.

Q56. A 1 kg mass on a spring oscillates with period 2 s. What is the spring constant?

  • (A) 9.87 N/m
  • (B) 158 N/m
  • (C) 4.93 N/m
  • (D) 2.47 N/m
Show Answer

Answer: A – T = 2π√(m/k), so k = 4π2m/T2 = 4π2 × 1/22 = 9.87 N/m.

Q57. A particle in SHM has x(t) = 0.3 cos(3t) m. What is its position at t = 1 s?

  • (A) 0.288 m
  • (B) 0.297 m
  • (C) 0.0423 m
  • (D) -0.297 m
Show Answer

Answer: D – x(1) = 0.3 cos(3 × 1) = -0.297 m.

Q58. A pendulum of length 3 m has period 3.48 s. If its length is multiplied by 4, what is its new period?

  • (A) 3.93 s
  • (B) 13.9 s
  • (C) 6.95 s
  • (D) 3.48 s
Show Answer

Answer: C – T is proportional to √L, so Tnew = 3.48 × √4 = 6.95 s.

Q59. Two springs of constants 200 N/m and 200 N/m act in parallel on a 1 kg mass. What is the period of oscillation?

  • (A) 0.71 s
  • (B) 0.444 s
  • (C) 0.314 s
  • (D) 0.628 s
Show Answer

Answer: C – Parallel springs add: keff = 200 + 200 = 400 N/m. Then T = 2π√(m/keff) = 0.314 s.

Q60. A mass oscillates with x(t) = 0.2 cos(5t) m. What is its maximum speed?

  • (A) 1 m/s
  • (B) 0.04 m/s
  • (C) 2 m/s
  • (D) 5 m/s
Show Answer

Answer: A – vmax = Aω = 0.2 × 5 = 1 m/s.

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