AP Physics C: Mechanics Practice Test 10 – 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 has velocity v(t) = 8t – 1t2 m/s. What is its displacement from t = 0 until it first comes to rest again (apart from t = 0)?
- (A) 256 m
- (B) 85.3 m
- (C) 42.7 m
- (D) 171 m
Show Answer
Answer: B – Set v = 0: t(8 – 1t) = 0, so t = 8 s. Displacement = ∪v dt = 8t2/2 – 1t3/3 evaluated at that t, giving 85.3 m.
Q2. 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 2 s?
- (A) 24 m
- (B) 15.2 m
- (C) 27.1 m
- (D) 21.2 m
Show Answer
Answer: B – Solve dv/dt = -0.5v to get v(t) = 12 e-0.5t. Integrating gives distance = (12/0.5)(1 – e-0.5(2)) = 15.2 m.
Q3. Two runners start from opposite ends of a straight track. Runner A has position xA = 1t2 m and runner B has position xB = 40 – 10t m, measured from A’s start. At what time do they meet?
- (A) 4.59 s
- (B) 4 s
- (C) 6.32 s
- (D) 3.06 s
Show Answer
Answer: D – Set xA = xB: 1t2 + 10t – 40 = 0. The positive root is t = 3.06 s.
Q4. A sled starts from rest with acceleration a(t) = 4 + 2t m/s2. How far does it travel in the first 3 s?
- (A) 45 m
- (B) 13.5 m
- (C) 18 m
- (D) 27 m
Show Answer
Answer: D – Integrate twice: v(t) = 4t + 2t2/2 and x(t) = 4t2/2 + 2t3/6. At t = 3 s, x = 27 m.
Q5. A particle moves along the x-axis with velocity v(t) = 1 + 6t + 2t2 m/s, where t is in seconds. What is its displacement between t = 1 s and t = 5 s?
- (A) 79.3 m
- (B) 159 m
- (C) 238 m
- (D) 163 m
Show Answer
Answer: B – Integrate v(t): displacement = ∪v dt = [1t + 6t2/2 + 2t3/3] from 1 to 5 s, which gives 159 m.
Q6. A cart starts with velocity 4 m/s and has acceleration a(t) = 2 + 2t m/s2. What is its velocity at t = 3 s?
- (A) 28 m/s
- (B) 19 m/s
- (C) 28.5 m/s
- (D) 10 m/s
Show Answer
Answer: B – Integrate a(t): v(t) = v0 + ∪a dt = 4 + 2t + 2t2/2. At t = 3 s this gives 19 m/s.
Q7. The position of a block is x(t) = 3t3 – 1t2 + 5t metres. What is its velocity at t = 3 s?
- (A) 132 m/s
- (B) 40 m/s
- (C) 80 m/s
- (D) 86 m/s
Show Answer
Answer: C – Differentiate: v(t) = dx/dt = 9t2 – 2t + 5. At t = 3 s, v = 80 m/s.
Unit 2: Force and Translational Dynamics
Q8. Two blocks 3 kg and 2 kg connected by a light string rest on a frictionless table. A horizontal force of 20 N pulls the 3 kg block. What is the tension in the string?
- (A) 12 N
- (B) 20 N
- (C) 16 N
- (D) 8 N
Show Answer
Answer: D – System acceleration a = 20/(3 + 2) = 4 m/s2. Tension accelerates the 2 kg block: T = 2 × 4 = 8 N.
Q9. A 0.5 kg mass moves in a horizontal circle of radius 1 m at speed 8 m/s, held by a string. What is the tension in the string?
- (A) 4 N
- (B) 64 N
- (C) 16 N
- (D) 32 N
Show Answer
Answer: D – Tension provides the centripetal force: T = mv2/r = 0.5 × 82/1 = 32 N.
Q10. A 3 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) 3.12 m/s2
- (B) 6.93 m/s2
- (C) 6.24 m/s2
- (D) 7.62 m/s2
Show Answer
Answer: C – Along the slope, a = g(sin θ – μ cos θ) = 9.8(sin 45° – 0.1 cos 45°) = 6.24 m/s2.
Q11. Two masses 4 kg and 6 kg hang from a light pulley (Atwood machine). What is the magnitude of their acceleration?
- (A) 3.92 m/s2
- (B) 3.27 m/s2
- (C) 1.96 m/s2
- (D) 3.69 m/s2
Show Answer
Answer: C – Net force = (6 – 4)g and total mass = 10 kg, so a = (6 – 4)(9.8)/(10) = 1.96 m/s2.
Q12. A horizontal force of 40 N is applied at 30° above the horizontal to a 2 kg block on a frictionless floor. What is the block’s horizontal acceleration?
- (A) 10 m/s2
- (B) 26 m/s2
- (C) 17.3 m/s2
- (D) 20 m/s2
Show Answer
Answer: C – Horizontal component = 40 cos 30° = 34.6 N. Then a = Fx/m = 34.6/2 = 17.3 m/s2.
Q13. A 2 kg object experiences a time-varying force F(t) = 20 + 5t N. What is its acceleration at t = 2 s?
- (A) 7.5 m/s2
- (B) 20 m/s2
- (C) 15 m/s2
- (D) 10 m/s2
Show Answer
Answer: C – At t = 2 s, F = 20 + 5(2) = 30 N, so a = F/m = 30/2 = 15 m/s2.
Q14. A 3 kg mass hangs at rest from a vertical spring of constant 100 N/m. By how much is the spring stretched from its natural length?
- (A) 0.294 m
- (B) 0.588 m
- (C) 0.03 m
- (D) 0.147 m
Show Answer
Answer: A – At equilibrium kx = mg, so x = (3)(9.8)/100 = 0.294 m.
Q15. A car travels on a flat circular bend of radius 20 m. The coefficient of static friction is 0.3. What is the maximum speed without slipping?
- (A) 11.5 m/s
- (B) 5.42 m/s
- (C) 14 m/s
- (D) 7.67 m/s
Show Answer
Answer: D – Friction provides the centripetal force: μmg = mv2/r, so v = √(μgr) = √(0.3 × 9.8 × 20) = 7.67 m/s.
Q16. A road is banked so that a car at 20 m/s can round a curve of radius 50 m with no sideways friction. What banking angle is required?
- (A) 58.8°
- (B) 50.8°
- (C) 19.6°
- (D) 39.2°
Show Answer
Answer: D – tan θ = v2/(gr) = 202/(9.8 × 50) = 0.816, so θ = 39.2°.
Q17. A planet has mass 2 times Earth’s mass and radius 2 times Earth’s radius. What is the gravitational field strength at its surface? (Earth’s g = 9.8 m/s2.)
- (A) 4.9 m/s2
- (B) 5.54 m/s2
- (C) 39.2 m/s2
- (D) 19.6 m/s2
Show Answer
Answer: A – g = GM/R2, so gplanet = 9.8 × 2/22 = 4.9 m/s2.
Q18. A 50 kg person stands on a scale in a lift accelerating upwards at 3 m/s2. What is the scale reading (normal force)?
- (A) 640 N
- (B) 768 N
- (C) 490 N
- (D) 340 N
Show Answer
Answer: A – N – mg = ma, so N = m(g + a) = 50(9.8 + 3) = 640 N.
Q19. A 4 kg crate on a rough floor (μk = 0.2) is pushed by a horizontal force of 30 N. What is its acceleration?
- (A) 5.54 m/s2
- (B) 7.5 m/s2
- (C) 2.77 m/s2
- (D) 9.46 m/s2
Show Answer
Answer: A – Friction = μmg = 0.2 × 4 × 9.8 = 7.84 N. Then a = (30 – 7.84)/4 = 5.54 m/s2.
Unit 3: Work, Energy, and Power
Q20. A force F(x) = 10 + 6x N acts on a particle as it moves from x = 2 m to x = 5 m along the x-axis. How much work is done?
- (A) 156 J
- (B) 93 J
- (C) 46.5 J
- (D) 30 J
Show Answer
Answer: B – W = ∪F dx = [10x + 6x2/2] from 2 to 5 = 93 J.
Q21. A variable force F(x) = 10x2 N moves an object from x = 2 m to x = 3 m. What work is done?
- (A) 190 J
- (B) 63.3 J
- (C) 25 J
- (D) 31.7 J
Show Answer
Answer: B – W = ∪10x2 dx = 10x3/3 from 2 to 3 = 63.3 J.
Q22. A spring of constant 200 N/m is compressed by 0.4 m from equilibrium. How much elastic potential energy is stored?
- (A) 16 J
- (B) 40 J
- (C) 36.2 J
- (D) 32 J
Show Answer
Answer: A – U = (1/2)kx2 = 0.5 × 200 × 0.42 = 16 J.
Q23. A net work of 150 J is done on a 3 kg object initially moving at 2 m/s. What is its final speed?
- (A) 10 m/s
- (B) 15.3 m/s
- (C) 52 m/s
- (D) 10.2 m/s
Show Answer
Answer: D – W = ΔK, so (1/2)m v2 – (1/2)m(2)2 = 150. Solving gives v = 10.2 m/s.
Q24. A constant force of 30 N pushes an object in the direction of its motion at 3 m/s. What power is delivered?
- (A) 180 W
- (B) 90 W
- (C) 33 W
- (D) 45 W
Show Answer
Answer: B – P = Fv = 30 × 3 = 90 W.
Q25. An object is dropped from rest from a height of 20 m. Ignoring air resistance, what is its speed just before hitting the ground?
- (A) 14 m/s
- (B) 196 m/s
- (C) 19.8 m/s
- (D) 9.9 m/s
Show Answer
Answer: C – Energy conservation: mgh = (1/2)mv2, so v = √(2gh) = √(2 × 9.8 × 20) = 19.8 m/s.
Q26. A small bead slides on a frictionless vertical loop of radius 1 m. What is the minimum speed it must have at the top of the loop to stay in contact?
- (A) 4.43 m/s
- (B) 4.95 m/s
- (C) 6.26 m/s
- (D) 3.13 m/s
Show Answer
Answer: D – At the top, mg = mv2/r for the limiting case, so v = √(gr) = √(9.8 × 1) = 3.13 m/s.
Q27. A 3 kg block attached to a spring (k = 100 N/m) is released from rest when the spring is compressed 0.3 m. What is its maximum speed?
- (A) 10 m/s
- (B) 0.866 m/s
- (C) 5.77 m/s
- (D) 1.73 m/s
Show Answer
Answer: D – Energy: (1/2)kx2 = (1/2)mv2, so v = x√(k/m) = 0.3 × √(100/3) = 1.73 m/s.
Q28. A 4 kg block slides 10 m across a rough horizontal surface with coefficient of kinetic friction 0.1. What is the magnitude of the work done by friction?
- (A) 19.6 J
- (B) 4 J
- (C) 392 J
- (D) 39.2 J
Show Answer
Answer: D – Magnitude = μmgd = 0.1 × 4 × 9.8 × 10 = 39.2 J.
Q29. A motor lifts a 50 kg load through a height of 10 m in 5 s at constant speed. What is the average power output?
- (A) 490 W
- (B) 980 W
- (C) 4.9e+03 W
- (D) 100 W
Show Answer
Answer: B – Work = mgh = 50 × 9.8 × 10 = 4.9e+03 J, so average power = Work/t = 4.9e+03/5 = 980 W.
Q30. A force F(x) = 25 cos(πx/(2 × 2)) N, where x is in metres, acts from x = 0 to x = 2 m. How much work is done? Use π = 3.14.
- (A) 15.9 J
- (B) 31.8 J
- (C) 63.7 J
- (D) 50 J
Show Answer
Answer: B – W = ∪F dx = 25 × (2 × 2/π) = 31.8 J.
Unit 4: Linear Momentum
Q31. A 2 kg ball moving at 8 m/s collides elastically head-on with a stationary 2 kg ball. What is the second ball’s velocity after the collision?
- (A) 9.04 m/s
- (B) 4 m/s
- (C) 8 m/s
- (D) 0 m/s
Show Answer
Answer: C – For an elastic collision with the second mass at rest, v2‘ = 2m1v1/(m1 + m2) = 2 × 2 × 8/(2 + 2) = 8 m/s.
Q32. A 3 kg cart at 6 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) 7.2 m/s
- (B) 2.4 m/s
- (C) 4.8 m/s
- (D) 3.6 m/s
Show Answer
Answer: C – Momentum conservation: (3 × 6 + 2 × 3)/(3 + 2) = 4.8 m/s.
Q33. A 0.02 kg bullet at 200 m/s embeds in a 2 kg wooden block hanging as a pendulum. How high does the block rise?
- (A) 0.2 m
- (B) 0.4 m
- (C) 0.452 m
- (D) 20.2 m
Show Answer
Answer: A – First, momentum gives V = 0.02 × 200/(0.02 + 2) = 1.98 m/s. Then energy gives h = V2/(2g) = 0.2 m.
Q34. An object at rest explodes into a 2 kg fragment moving at 4 m/s and a 2 kg fragment. What is the speed of the second fragment?
- (A) 4 m/s
- (B) 5.11 m/s
- (C) 8 m/s
- (D) 4.52 m/s
Show Answer
Answer: A – Momentum conservation: 2 × 4 = 2 × v2, so v2 = 4 m/s.
Q35. Two masses 3 kg at x = 0 m and 4 kg at x = 6 m lie on the x-axis. Where is their centre of mass?
- (A) 3.43 m
- (B) 3 m
- (C) 5.14 m
- (D) 12 m
Show Answer
Answer: A – xcm = (3 × 0 + 4 × 6)/(3 + 4) = 3.43 m.
Q36. A 1 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) 20 N·s
- (B) 10 N·s
- (C) 5 N·s
- (D) 11.3 N·s
Show Answer
Answer: A – Δp = m(vf – vi) = 1(10 – (-10)) = 20 N·s in magnitude.
Q37. A force F(t) = 20 sin(πt/4) N acts from t = 0 to t = 4 s. What impulse does it deliver? Use π = 3.14.
- (A) 28.8 N·s
- (B) 25.5 N·s
- (C) 80 N·s
- (D) 50.9 N·s
Show Answer
Answer: D – J = ∪20 sin(πt/4) dt from 0 to 4 = 2 × 20 × 4/π = 50.9 N·s.
Q38. A constant force of 20 N acts for 3 s on a 2 kg object already moving at 5 m/s in the force direction. What is its final speed?
- (A) 17.5 m/s
- (B) 35 m/s
- (C) 65 m/s
- (D) 20 m/s
Show Answer
Answer: B – Impulse J = 20 × 3 = 60 N·s, so Δv = J/m = 60/2 = 30 m/s, giving vf = 5 + 30 = 35 m/s.
Unit 5: Torque and Rotational Dynamics
Q39. Point masses 2 kg at 1 m and 2 kg at 1 m from an axis have what total moment of inertia about that axis?
- (A) 4.52 kg·m2
- (B) 2 kg·m2
- (C) 4 kg·m2
- (D) 8 kg·m2
Show Answer
Answer: C – I = Σmr2 = 2 × 12 + 2 × 12 = 4 kg·m2.
Q40. A uniform rod of mass 4 kg and length 2 m has moment of inertia 1.33 kg·m2 about its centre. What is its moment of inertia about a parallel axis 0.5 m from the centre?
- (A) 1.33 kg·m2
- (B) 4.67 kg·m2
- (C) 2.33 kg·m2
- (D) 5.33 kg·m2
Show Answer
Answer: C – Parallel-axis theorem: I = Icm + md2 = 1.33 + 4 × 0.52 = 2.33 kg·m2.
Q41. A net torque of 10 N·m acts on a body with moment of inertia 2 kg·m2. What is its angular acceleration?
- (A) 10 rad/s2
- (B) 2.5 rad/s2
- (C) 20 rad/s2
- (D) 5 rad/s2
Show Answer
Answer: D – τ = Iα, so α = 10/2 = 5 rad/s2.
Q42. A 2 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) 1.5 kg·m2/s
- (B) 3 kg·m2/s
- (C) 6.78 kg·m2/s
- (D) 6 kg·m2/s
Show Answer
Answer: B – L = mvr = 2 × 3 × 0.5 = 3 kg·m2/s.
Q43. A flywheel with I = 4 kg·m2 spins at 10 rad/s. A constant opposing torque of 4 N·m is applied. How long until it stops?
- (A) 5 s
- (B) 2.5 s
- (C) 20 s
- (D) 10 s
Show Answer
Answer: D – Angular impulse: τΔt = Iω, so Δt = 4 × 10/4 = 10 s.
Q44. A uniform rod of mass 2 kg and length 1 m is pivoted at one end and held at 60° to the vertical. What torque does gravity exert about the pivot?
- (A) 4.24 N·m
- (B) 9.8 N·m
- (C) 17 N·m
- (D) 8.49 N·m
Show Answer
Answer: D – Gravity acts at the centre, 0.5 m from the pivot, so τ = (mg)(L/2) sin θ = 2 × 9.8 × 0.5 × sin 60° = 8.49 N·m.
Q45. A tangential force of 20 N is applied at the rim of a wheel of radius 0.5 m and moment of inertia 2 kg·m2. What is the wheel’s angular acceleration?
- (A) 10 rad/s2
- (B) 5 rad/s2
- (C) 11.3 rad/s2
- (D) 2.5 rad/s2
Show Answer
Answer: B – τ = Fr = 20 × 0.5 = 10 N·m, so α = τ/I = 10/2 = 5 rad/s2.
Q46. A disk starts with angular velocity 0 rad/s and constant angular acceleration 4 rad/s2. Through what angle does it rotate in 5 s?
- (A) 100 rad
- (B) 50 rad
- (C) 113 rad
- (D) 25 rad
Show Answer
Answer: B – θ = ω0t + (1/2)αt2 = 0 × 5 + 0.5 × 4 × 52 = 50 rad.
Unit 6: Energy and Momentum of Rotating Systems
Q47. What is the orbital period, in hours, of a satellite at radius 20000 km from Earth’s centre? (GMEarth = 3.99 × 1014 m3/s2.)
- (A) 11.7 hours
- (B) 3.91 hours
- (C) 7.82 hours
- (D) 15.6 hours
Show Answer
Answer: C – T = 2π√(r3/GM) = 2π√((20000000)3/(3.99 × 1014)) = 7.82 hours.
Q48. A planet has mass 8 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) 5.6 km/s
- (B) 31.7 km/s
- (C) 22.4 km/s
- (D) 11.2 km/s
Show Answer
Answer: C – vesc = √(2GM/R), so it scales as √(M/R): 11.2 × √(8/2) = 22.4 km/s.
Q49. A 500 kg satellite orbits at radius 9000 km from Earth’s centre. What is the magnitude of its total mechanical energy, in GJ? (GMEarth = 3.99 × 1014 m3/s2.)
- (A) 11.1 GJ
- (B) 16.6 GJ
- (C) 22.1 GJ
- (D) 5.54 GJ
Show Answer
Answer: A – E = -GMm/(2r) = -(3.99 × 1014 × 500)/(2 × 9000000) = -11.1 GJ, so the magnitude is 11.1 GJ.
Q50. A 300 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) 1.71e+04 MJ
- (B) 1.02e+04 MJ
- (C) 4.27e+03 MJ
- (D) 8.54e+03 MJ
Show Answer
Answer: D – ΔU = -GMm(1/r2 – 1/r1) = 8.54e+03 MJ (positive, since it moves farther out).
Q51. A 4 kg mass falls 10 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) 14 m/s
- (B) 4.04 m/s
- (C) 8.08 m/s
- (D) 12.1 m/s
Show Answer
Answer: C – Energy: mgh = (1/2)mv2 + (1/2)I(v/r)2, so v = √(2mgh/(m + I/r2)) = 8.08 m/s.
Q52. Planet A orbits a star with period 3 years. Planet B orbits the same star at 8 times A’s orbital radius. What is B’s period?
- (A) 33.9 years
- (B) 192 years
- (C) 67.9 years
- (D) 24 years
Show Answer
Answer: C – Kepler’s third law: T2 proportional to r3, so TB = 3 × 83/2 = 67.9 years.
Q53. A body with moment of inertia 4 kg·m2 rotates at 10 rad/s. What is its rotational kinetic energy?
- (A) 452 J
- (B) 20 J
- (C) 200 J
- (D) 400 J
Show Answer
Answer: C – K = (1/2)Iω2 = 0.5 × 4 × 102 = 200 J.
Unit 7: Oscillations
Q54. A 2 kg mass on a spring (k = 100 N/m) oscillates with amplitude 0.2 m. What is its speed when its displacement is 0.1 m?
- (A) 1.41 m/s
- (B) 1.22 m/s
- (C) 0.612 m/s
- (D) 0.707 m/s
Show Answer
Answer: B – Energy: (1/2)kA2 = (1/2)kx2 + (1/2)mv2, so v = √((k/m)(A2 – x2)) = 1.22 m/s.
Q55. A particle in SHM obeys d2x/dt2 = -22x. What is the magnitude of its acceleration when its displacement is 0.1 m?
- (A) 0.4 m/s2
- (B) 0.8 m/s2
- (C) 0.226 m/s2
- (D) 0.2 m/s2
Show Answer
Answer: A – From d2x/dt2 = -ω2x, |a| = ω2|x| = 22 × 0.1 = 0.4 m/s2.
Q56. A 1 kg mass on a spring oscillates with period 1 s. What is the spring constant?
- (A) 19.7 N/m
- (B) 39.5 N/m
- (C) 44.6 N/m
- (D) 9.87 N/m
Show Answer
Answer: B – T = 2π√(m/k), so k = 4π2m/T2 = 4π2 × 1/12 = 39.5 N/m.
Q57. A particle in SHM has x(t) = 0.2 cos(2t) m. What is its position at t = 2 s?
- (A) -0.131 m
- (B) -0.151 m
- (C) -0.0291 m
- (D) 0.131 m
Show Answer
Answer: A – x(2) = 0.2 cos(2 × 2) = -0.131 m.
Q58. A pendulum of length 1 m has period 2.01 s. If its length is multiplied by 4, what is its new period?
- (A) 4.01 s
- (B) 2.01 s
- (C) 2.27 s
- (D) 8.03 s
Show Answer
Answer: A – T is proportional to √L, so Tnew = 2.01 × √4 = 4.01 s.
Q59. Two springs of constants 200 N/m and 200 N/m act in parallel on a 2 kg mass. What is the period of oscillation?
- (A) 0.444 s
- (B) 0.889 s
- (C) 1 s
- (D) 0.628 s
Show Answer
Answer: A – Parallel springs add: keff = 200 + 200 = 400 N/m. Then T = 2π√(m/keff) = 0.444 s.
Q60. A mass oscillates with x(t) = 0.2 cos(2t) m. What is its maximum speed?
- (A) 0.4 m/s
- (B) 0.8 m/s
- (C) 0.1 m/s
- (D) 0.904 m/s
Show Answer
Answer: A – vmax = Aω = 0.2 × 2 = 0.4 m/s.
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