AP Physics C: Mechanics Practice Test 4 – 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 body moves with velocity v(t) = 6t + 6 m/s, where t is in seconds. What is its displacement from t = 0 to t = 4 s?
- (A) 62 m
- (B) 90 m
- (C) 67 m
- (D) 72 m
Show Answer
Answer: D – Displacement = ∪v dt from 0 to 4 = (6/2)t2 + 6t evaluated from 0 to 4 = 48 + 24 = 72 m.
Q2. A body starts with velocity 4 m/s and has acceleration a(t) = 4t + 2 m/s2. What is its velocity at t = 3 s?
- (A) 7 m/s
- (B) 23 m/s
- (C) 14 m/s
- (D) 28 m/s
Show Answer
Answer: D – v = v0 + ∪a dt = 4 + (4/2)t2 + 2t from 0 to 3 = 4 + 18 + 6 = 28 m/s.
Q3. A projectile is launched from level ground with speed 36 m/s at 30θ above the horizontal. What is its horizontal range? Take g = 9.8 m/s2.
- (A) 114.53 m
- (B) 91.62 m
- (C) 124.53 m
- (D) 68.72 m
Show Answer
Answer: A – Range R = v02 sin(2θ)/g = 362 × sin(60θ) ÷ 9.8 = 114.53 m.
Q4. A stone is dropped from rest from a height of 50 m. How long does it take to reach the ground? Take g = 9.8 m/s2.
- (A) 3.19 s
- (B) 3.51 s
- (C) 2.4 s
- (D) 2.56 s
Show Answer
Answer: A – h = (1/2)gt2, so t = √(2h/g) = √(2×50÷9.8) = 3.19 s.
Q5. A particle moves with velocity v(t) = 3t2 + 2 m/s, where t is in seconds. What is its displacement from t = 0 to t = 4 s?
- (A) 82 m
- (B) 108 m
- (C) 54 m
- (D) 72 m
Show Answer
Answer: D – Displacement = ∪v dt from 0 to 4 = (3/3)t3 + 2t from 0 to 4 = 64 + 8 = 72 m.
Q6. A car moves with velocity v(t) = 20 – 4t m/s, where t is in seconds. At what time does the car momentarily come to rest?
- (A) 4 s
- (B) 7.5 s
- (C) 15 s
- (D) 5 s
Show Answer
Answer: D – Set v = 0: 20 – 4t = 0, so t = 20÷4 = 5 s.
Q7. A particle has velocity v(t) = 3t2 + 6t m/s, where t is in seconds. What is its acceleration at t = 4 s?
- (A) 25 m/s2
- (B) 30 m/s2
- (C) 18 m/s2
- (D) 35 m/s2
Show Answer
Answer: B – a = dv/dt = 6t + 6. At t = 4 s, a = 6×4 + 6 = 30 m/s2.
Unit 2: Force and Translational Dynamics
Q8. An object of mass 11 kg is on a planet where the acceleration due to gravity is 2 times that on Earth. What is its weight on that planet? Take g = 9.8 m/s2 on Earth.
- (A) 215.6 N
- (B) 323.4 N
- (C) 161.7 N
- (D) 237.16 N
Show Answer
Answer: A – Weight = m×gplanet = 11×(2×9.8) = 215.6 N.
Q9. A net force acts on a 12 kg mass and gives it an acceleration of 2 m/s2. What is the magnitude of the net force?
- (A) 19.2 N
- (B) 18 N
- (C) 24 N
- (D) 30 N
Show Answer
Answer: C – F = ma = 12×2 = 24 N.
Q10. Two horizontal forces, 37 N to the right and 7 N to the left, act on a 6 kg block on a frictionless surface. What is the acceleration of the block?
- (A) 7.5 m/s2
- (B) 2.5 m/s2
- (C) 15 m/s2
- (D) 5 m/s2
Show Answer
Answer: D – Net force = 37 – 7 = 30 N to the right. a = F/m = 30÷6 = 5 m/s2.
Q11. A 10 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) 29.4 N
- (B) 88.2 N
- (C) 44.1 N
- (D) 14.7 N
Show Answer
Answer: A – Maximum static friction = μN = μmg = 0.3×10×9.8 = 29.4 N.
Q12. A block slides down a plane inclined at 45θ to the horizontal. The coefficient of kinetic friction is 0.2. What is the acceleration of the block down the plane? Take g = 9.8 m/s2.
- (A) 16.63 m/s2
- (B) 15.54 m/s2
- (C) 6.93 m/s2
- (D) 5.54 m/s2
Show Answer
Answer: D – a = g(sinθ – μcosθ) = 9.8×(0.71 – 0.2×0.71) = 5.54 m/s2.
Q13. A 7 kg block rests on a plane inclined at 30θ to the horizontal. What is the normal force on the block? Take g = 9.8 m/s2.
- (A) 59.41 N
- (B) 47.53 N
- (C) 74.26 N
- (D) 14.85 N
Show Answer
Answer: A – Normal force N = mg cosθ = 7×9.8×0.87 = 59.41 N.
Q14. Two masses, 2 kg and 5 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) 4.2 m/s2
- (B) 5.25 m/s2
- (C) 4.62 m/s2
- (D) 6.3 m/s2
Show Answer
Answer: A – a = (m2 – m1)g/(m1 + m2) = (5 – 2)×9.8÷(2 + 5) = 4.2 m/s2.
Q15. A person of mass 69 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) 441.6 N
- (B) 883.2 N
- (C) 1766.4 N
- (D) 878.2 N
Show Answer
Answer: B – Normal force N = m(g + a) = 69×(9.8 + 3) = 883.2 N.
Q16. A person of mass 50 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) 400 N
- (B) 292.5 N
- (C) 390 N
- (D) 234 N
Show Answer
Answer: C – Normal force N = m(g – a) = 50×(9.8 – 2) = 390 N.
Q17. A 4 kg object moves in a horizontal circle of radius 4 m at a constant speed of 6 m/s. What is the magnitude of the centripetal force on it?
- (A) 9 N
- (B) 36 N
- (C) 54 N
- (D) 21.6 N
Show Answer
Answer: B – F = mv2/r = 4×62÷4 = 36 N.
Q18. A spring of spring constant 50 N/m is stretched by 0.4 m from its natural length. What is the magnitude of the spring force?
- (A) 16 N
- (B) 10 N
- (C) 20 N
- (D) 60 N
Show Answer
Answer: C – F = kx = 50×0.4 = 20 N.
Q19. A planet has surface gravity 20 m/s2. At a height of 2 planet radii above the surface, what is the acceleration due to gravity there?
- (A) 2.78 m/s2
- (B) 2.22 m/s2
- (C) 3.33 m/s2
- (D) 2.44 m/s2
Show Answer
Answer: B – g’ = g(R/(R + h))2 = 20×(1/3)2 = 2.22 m/s2.
Unit 3: Work, Energy, and Power
Q20. A force of 48 N acts at 60θ to the direction of motion as a block moves 4 m. How much work is done by the force?
- (A) 96 J
- (B) 72 J
- (C) 48 J
- (D) 57.6 J
Show Answer
Answer: A – W = Fd cosθ = 48×4×0.5 = 96 J.
Q21. A variable force F(x) = 40x N stretches a spring from x = 0 to x = 0.4 m. How much work is done? Here x is in metres.
- (A) 2.4 J
- (B) 3.2 J
- (C) 9.6 J
- (D) 3.52 J
Show Answer
Answer: B – W = ∪F dx from 0 to 0.4 = (1/2)kX2 = 0.5×40×0.42 = 3.2 J.
Q22. A force F(x) = 6x2 N acts on a particle as it moves from x = 0 to x = 4 m. How much work is done by the force?
- (A) 32 J
- (B) 138 J
- (C) 140.8 J
- (D) 128 J
Show Answer
Answer: D – W = ∪F dx from 0 to 4 = (6/3)x3 from 0 to 4 = 128 J.
Q23. What is the kinetic energy of a 9 kg object moving at 10 m/s?
- (A) 450 J
- (B) 460 J
- (C) 1350 J
- (D) 270 J
Show Answer
Answer: A – KE = (1/2)mv2 = 0.5×9×102 = 450 J.
Q24. A spring of spring constant 80 N/m is compressed by 0.15 m. What is the elastic potential energy stored in the spring?
- (A) 0.22 J
- (B) 0.99 J
- (C) 0.9 J
- (D) 2.7 J
Show Answer
Answer: C – PE = (1/2)kx2 = 0.5×80×0.152 = 0.9 J.
Q25. A 9 kg object is raised vertically by 8 m. By how much does its gravitational potential energy increase? Take g = 9.8 m/s2.
- (A) 2116.8 J
- (B) 176.4 J
- (C) 705.6 J
- (D) 695.6 J
Show Answer
Answer: C – Change in PE = mgh = 9×9.8×8 = 705.6 J.
Q26. An object is dropped from rest from a height of 30 m. Ignoring air resistance, what is its speed just before it hits the ground? Take g = 9.8 m/s2.
- (A) 24.25 m/s
- (B) 19.4 m/s
- (C) 26.67 m/s
- (D) 14.55 m/s
Show Answer
Answer: A – Energy conservation: (1/2)mv2 = mgh, so v = √(2gh) = √(2×9.8×30) = 24.25 m/s.
Q27. A motor does 453 J of work in 10 s. What is its average power output?
- (A) 67.95 W
- (B) 27.18 W
- (C) 45.3 W
- (D) 49.83 W
Show Answer
Answer: C – P = W/t = 453÷10 = 45.3 W.
Q28. A constant force of 30 N moves an object at a steady speed of 2 m/s in the direction of the force. What power is delivered by the force?
- (A) 55 W
- (B) 120 W
- (C) 70 W
- (D) 60 W
Show Answer
Answer: D – P = Fv = 30×2 = 60 W.
Q29. A 5 kg block slides 8 m across a horizontal surface with coefficient of kinetic friction 0.2. What is the magnitude of the work done by friction? Take g = 9.8 m/s2.
- (A) 39.2 J
- (B) 78.4 J
- (C) 83.4 J
- (D) 62.72 J
Show Answer
Answer: B – Friction force = μmg = 0.2×5×9.8 = 9.8 N. Work magnitude = Fd = 9.8×8 = 78.4 J.
Q30. A 1 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) 2.12 m/s
- (B) 4.67 m/s
- (C) 3.39 m/s
- (D) 4.24 m/s
Show Answer
Answer: D – (1/2)kx2 = (1/2)mv2, so v = √(k/m)×x = √(200÷1)×0.3 = 4.24 m/s.
Unit 4: Linear Momentum
Q31. A rifle of mass 3 kg fires a bullet of mass 0.02 kg at 385 m/s. What is the recoil speed of the rifle?
- (A) 0.64 m/s
- (B) 7.7 m/s
- (C) 3.85 m/s
- (D) 2.57 m/s
Show Answer
Answer: D – Momentum conservation: Mvrecoil = mv = 0.023385 = 7.7. So vrecoil = 7.7÷3 = 2.57 m/s.
Q32. A 2 kg object moving at 4 m/s collides head-on with a 3 kg object moving at 5 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) -2.1 m/s
- (B) -1.05 m/s
- (C) -1.54 m/s
- (D) -1.4 m/s
Show Answer
Answer: D – Total momentum = 2×4 – 3×5 = -7 kg·m/s. Dividing by total mass 5 kg gives -1.4 m/s.
Q33. A 4 kg object changes its velocity by 4 m/s in 2 s. What is the average force acting on it?
- (A) 3 N
- (B) 18 N
- (C) 8 N
- (D) 8.8 N
Show Answer
Answer: C – Average force = Δp/Δt = mΔv/Δt = 4×4÷2 = 8 N.
Q34. A force acting on a body rises linearly from zero to 80 N over 2 s and then stops. What is the impulse delivered?
- (A) 60 N·s
- (B) 80 N·s
- (C) 20 N·s
- (D) 48 N·s
Show Answer
Answer: B – Impulse is the area under the force-time graph: a triangle of area (1/2)×80×2 = 80 N·s.
Q35. A 2 kg ball hits a wall at 4 m/s and rebounds at 3 m/s in the opposite direction. What is the magnitude of the change in its momentum?
- (A) 9 kg·m/s
- (B) 14 kg·m/s
- (C) 10.5 kg·m/s
- (D) 8.4 kg·m/s
Show Answer
Answer: B – Change in momentum = m(vf – vi) in magnitude = 2×(3 + 4) = 14 kg·m/s.
Q36. What is the momentum of a 5 kg object moving at 6 m/s?
- (A) 15 kg·m/s
- (B) 30 kg·m/s
- (C) 33 kg·m/s
- (D) 45 kg·m/s
Show Answer
Answer: B – p = mv = 5×6 = 30 kg·m/s.
Q37. A constant force of 25 N acts on an object for 3 s. What is the magnitude of the impulse delivered?
- (A) 75 N·s
- (B) 225 N·s
- (C) 85 N·s
- (D) 18.75 N·s
Show Answer
Answer: A – J = FΔt = 25×3 = 75 N·s.
Q38. A force F(t) = 2t + 9 N acts on a body from t = 0 to t = 3 s. What is the impulse delivered to the body?
- (A) 36 N·s
- (B) 39.6 N·s
- (C) 26 N·s
- (D) 9 N·s
Show Answer
Answer: A – J = ∪F dt from 0 to 3 = (2/2)t2 + 9t from 0 to 3 = 9 + 27 = 36 N·s.
Unit 5: Torque and Rotational Dynamics
Q39. A net torque of 18 N·m acts on a body with moment of inertia 3 kg·m2. What is its angular acceleration?
- (A) 7.5 rad/s2
- (B) 6 rad/s2
- (C) 9 rad/s2
- (D) 18 rad/s2
Show Answer
Answer: B – α = τnet/I = 18÷3 = 6 rad/s2.
Q40. A wheel starts with angular velocity 1 rad/s and has a constant angular acceleration of 2 rad/s2. What is its angular velocity after 4 s?
- (A) 19 rad/s
- (B) 9 rad/s
- (C) 9.9 rad/s
- (D) 2.25 rad/s
Show Answer
Answer: B – ω = ω0 + αt = 1 + 2×4 = 9 rad/s.
Q41. A disc starts with angular velocity 1 rad/s and angular acceleration 3 rad/s2. Through what angle does it turn in 2 s?
- (A) 12 rad
- (B) 8 rad
- (C) 16 rad
- (D) 6.4 rad
Show Answer
Answer: B – θ = ω0t + (1/2)αt2 = 1×2 + 0.5×3×22 = 8 rad.
Q42. The angular momentum of a body varies as L(t) = 3t2 kg·m2/s. What is the net torque on the body at t = 3 s?
- (A) 19.8 N·m
- (B) 10.8 N·m
- (C) 8 N·m
- (D) 18 N·m
Show Answer
Answer: D – τnet = dL/dt = 6t. At t = 3 s, τ = 6×3 = 18 N·m.
Q43. A force of 40 N is applied at a distance of 0.3 m from a pivot, at 90θ to the lever arm. What is the magnitude of the torque about the pivot?
- (A) 7 N·m
- (B) 2 N·m
- (C) 12 N·m
- (D) 9 N·m
Show Answer
Answer: C – τ = rF sinθ = 0.3×40×1 = 12 N·m.
Q44. Two forces act on a beam pivoted at its centre: 41 N at 0.4 m on one side and 16 N at 0.2 m on the other side, in opposite rotational senses. What is the magnitude of the net torque?
- (A) 39.6 N·m
- (B) 13.2 N·m
- (C) 18.2 N·m
- (D) 23.2 N·m
Show Answer
Answer: B – Net torque = |r1F1 – r2F2| = |0.4×41 – 0.2×16| = 13.2 N·m.
Q45. Two point masses, 2 kg at 1 m and 3 kg at 1 m from an axis, rotate about that axis. What is their total moment of inertia about the axis?
- (A) 1.25 kg·m2
- (B) 15 kg·m2
- (C) 3 kg·m2
- (D) 5 kg·m2
Show Answer
Answer: D – I = Σmr2 = 2×12 + 3×12 = 5 kg·m2.
Q46. A uniform rod of mass 6 kg and length 1 m rotates about an axis through one end, perpendicular to the rod. What is its moment of inertia?
- (A) 1.2 kg·m2
- (B) 2 kg·m2
- (C) 2.2 kg·m2
- (D) 6 kg·m2
Show Answer
Answer: B – For a rod about one end, I = (1/3)ML2 = (1/3)×6×12 = 2 kg·m2.
Unit 6: Energy and Momentum of Rotating Systems
Q47. How much work is needed to spin a solid disc of mass 7 kg and radius 0.2 m from rest to an angular velocity of 7 rad/s?
- (A) 2.06 J
- (B) 6.86 J
- (C) 3.43 J
- (D) 1.72 J
Show Answer
Answer: C – Work = change in rotational KE = (1/2)Iω2. I = (1/2)MR2 = 0.14 kg·m2, so W = 0.5×0.14×72 = 3.43 J.
Q48. A solid disc of mass 5 kg and radius 0.4 m rotates at 4 rad/s about its central axis. What is its rotational kinetic energy?
- (A) 4.8 J
- (B) 4 J
- (C) 3.2 J
- (D) 2.4 J
Show Answer
Answer: C – I = (1/2)MR2 = 0.4 kg·m2. KE = (1/2)Iω2 = 0.5×0.4×42 = 3.2 J.
Q49. A body with moment of inertia 2 kg·m2 rotates at 8 rad/s. What is its angular momentum?
- (A) 6 kg·m2/s
- (B) 12 kg·m2/s
- (C) 16 kg·m2/s
- (D) 21 kg·m2/s
Show Answer
Answer: C – L = Iω = 2×8 = 16 kg·m2/s.
Q50. A spinning skater with moment of inertia 2 kg·m2 rotates at 5 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) 15 rad/s
- (B) 10 rad/s
- (C) 8 rad/s
- (D) 12.5 rad/s
Show Answer
Answer: B – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = 2×5÷1 = 10 rad/s.
Q51. A satellite orbits a planet with period 1 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) 2.83 days
- (B) 3.54 days
- (C) 2.26 days
- (D) 2.12 days
Show Answer
Answer: A – By Kepler’s third law, T2 is proportional to r3, so T2 = T1×(2)3/2 = 1×2.83 = 2.83 days.
Q52. A satellite in a circular orbit of radius r has orbital speed 10 km/s. What is the orbital speed of a satellite in a circular orbit of radius 2r around the same planet, in km/s?
- (A) 7.07 km/s
- (B) -2.93 km/s
- (C) 10.61 km/s
- (D) 2.07 km/s
Show Answer
Answer: A – Orbital speed v = √(GM/r), so v is proportional to 1/√r. v2 = 10÷√(2) = 7.07 km/s.
Q53. A solid disc of mass 5 kg rolls without slipping at 4 m/s. What is its total kinetic energy?
- (A) 180 J
- (B) 75 J
- (C) 60 J
- (D) 90 J
Show Answer
Answer: C – Total KE = translational + rotational = (1/2)mv2 + (1/2)Iω2 = (3/4)mv2 for a disc = 0.75×5×42 = 60 J.
Unit 7: Oscillations
Q54. A 1 kg mass on a spring of spring constant 50 N/m oscillates on a frictionless surface. What is the period of oscillation?
- (A) 0.22 s
- (B) 0.71 s
- (C) 0.89 s
- (D) 1.11 s
Show Answer
Answer: C – T = 2π√(m/k) = 2π√(1÷50) = 0.89 s.
Q55. A simple pendulum of length 1.5 m oscillates with small amplitude. What is its period? Take g = 9.8 m/s2.
- (A) 3.07 s
- (B) 1.84 s
- (C) 0.61 s
- (D) 2.46 s
Show Answer
Answer: D – T = 2π√(L/g) = 2π√(1.5÷9.8) = 2.46 s.
Q56. A mass of 1 kg oscillates on a spring of spring constant 64 N/m. What is the angular frequency of the oscillation?
- (A) 18 rad/s
- (B) 13 rad/s
- (C) 8 rad/s
- (D) 4.8 rad/s
Show Answer
Answer: C – ω = √(k/m) = √(64÷1) = 8 rad/s.
Q57. A 1 kg mass oscillates on a spring of spring constant 49 N/m with amplitude 0.2 m. What is its maximum speed?
- (A) 2.1 m/s
- (B) 0.84 m/s
- (C) 1.4 m/s
- (D) 1.75 m/s
Show Answer
Answer: C – ω = √(k/m) = 7 rad/s. Maximum speed vmax = ωA = 7×0.2 = 1.4 m/s.
Q58. A 2 kg mass oscillates on a spring of spring constant 49 N/m with amplitude 0.3 m. What is the magnitude of its maximum acceleration?
- (A) 7.35 m/s2
- (B) 1.84 m/s2
- (C) 5.88 m/s2
- (D) 9.19 m/s2
Show Answer
Answer: A – ω = √(k/m) = 4.95 rad/s. Maximum acceleration amax = ω2A = 4.952×0.3 = 7.35 m/s2.
Q59. A particle in simple harmonic motion satisfies d2x/dt2 = -16x, where x is in metres and t is in seconds. What is the period of the motion?
- (A) 0.79 s
- (B) 3.14 s
- (C) 2.36 s
- (D) 1.57 s
Show Answer
Answer: D – Comparing with d2x/dt2 = -ω2x gives ω = √(16) = 4 rad/s. Period T = 2π/ω = 1.57 s.
Q60. A spring of spring constant 100 N/m oscillates with amplitude 0.25 m. What is the total mechanical energy of the oscillator?
- (A) 3.12 J
- (B) 4.69 J
- (C) 6.25 J
- (D) 2.34 J
Show Answer
Answer: A – E = (1/2)kA2 = 0.5×100×0.252 = 3.12 J.
Well done for finishing the test. Review every question you missed, work through its explanation again, and revisit the unit where you lost the most marks before you try the next practice test.
📥 Prefer pen and paper? Take this test offline.
Download Printable PDF – Test 4
Free printable version: all 60 questions + answer key. Explanations stay here on the website.
