AP Physics 1 Practice Test 8 – 60 Questions with Answers

Test yourself with 60 original AP Physics 1 practice questions covering all 8 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, oscillations, and fluids. Every answer includes the full working. All questions are algebra-based (no calculus). Take g = 9.8 m/s2 unless stated otherwise.

Unit 1: Kinematics

Q1. A go-kart starts at 8 m/s and accelerates uniformly at 2 m/s2 for 4 s. How far does it travel in that time?

  • (A) 32
  • (B) 64
  • (C) 58
  • (D) 48
Show Answer

Answer: D – Use x = v0t + (1/2)at2 = (8)(4) + (1/2)(2)(4)2 = 32 + 16 = 48 m. (Simplification: air resistance is ignored, as stated.)

Q2. A tennis ball is dropped from rest from a bridge 44.1 m above the water. Ignoring air resistance, how long does it take to reach the water?

  • (A) 4
  • (B) 3
  • (C) 2.5
  • (D) 6
Show Answer

Answer: B – For free fall from rest, h = (1/2)gt2, so t = √(2h/g) = √(2 x 44.1 / 9.8) = √(9) = 3 s. (Simplification: air resistance is ignored, as stated.)

Q3. A softball is thrown straight upward at 14 m/s. Ignoring air resistance, what is its maximum height above the release point?

  • (A) 10
  • (B) 20
  • (C) 5
  • (D) 14.7
Show Answer

Answer: A – At maximum height v = 0, so v2 = v02 – 2gh gives h = v02 / (2g) = (14)2 / (2 x 9.8) = 196 / 19.6 = 10 m. (Simplification: air resistance is ignored, as stated.)

Q4. A motorcycle travelling at 24 m/s brakes uniformly at 6 m/s2. What is its stopping distance?

  • (A) 96
  • (B) 24
  • (C) 48
  • (D) 144
Show Answer

Answer: C – Use v2 = v02 – 2ad with final v = 0, so d = v02 / (2a) = (24)2 / (2 x 6) = 576 / 12 = 48 m. (Simplification: the braking force is treated as constant.)

Q5. A bearing rolls off a horizontal table 19.6 m high with a horizontal speed of 12 m/s. Ignoring air resistance, how far from the base of the table does it land?

  • (A) 235.2
  • (B) 12
  • (C) 14
  • (D) 24
Show Answer

Answer: D – Vertical fall time: t = √(2h/g) = √(2 x 19.6 / 9.8) = 2 s. Horizontal distance = vxt = (12)(2) = 24 m. (Simplification: air resistance is ignored, as stated.)

Q6. A ferry moving at 10 m/s accelerates at 1 m/s2 for 4 s. What is its final speed?

  • (A) 40
  • (B) 14
  • (C) 13
  • (D) 20.18
Show Answer

Answer: B – Use v = v0 + at = 10 + (1)(4) = 10 + 4 = 14 m/s.

Q7. A jogger covers 100 m in 20 s and then 60 m in 15 s. What is the average speed for the whole trip?

  • (A) 4.5
  • (B) 4.57
  • (C) 5.57
  • (D) 5
Show Answer

Answer: B – Average speed = total distance / total time = (100 + 60) / (20 + 15) = 160 / 35 = 4.57 m/s.

Unit 2: Force and Translational Dynamics

Q8. A net force accelerates a 8 kg box at 2 m/s2 on a frictionless surface. What is the magnitude of the net force?

  • (A) 10
  • (B) 8
  • (C) 24
  • (D) 16
Show Answer

Answer: D – Newton’s second law: Fnet = ma = (8)(2) = 16 N. (Simplification: the surface is frictionless, as stated.)

Q9. A 20 kg storage box slides on a horizontal floor with coefficient of kinetic friction 0.3. What is the kinetic friction force?

  • (A) 58.8
  • (B) 6
  • (C) 196
  • (D) 29.4
Show Answer

Answer: A – Friction f = μN = μmg = (0.3)(20)(9.8) = 58.8 N.

Q10. A 5 kg block rests on a frictionless incline at 30° to the horizontal. What is the component of its weight parallel to the incline?

  • (A) 49
  • (B) 42.44
  • (C) 24.5
  • (D) 12.25
Show Answer

Answer: C – Parallel component = mg sinΘ = (5)(9.8)(sin 30°) = 49 x 0.5 = 24.5 N. (Simplification: the incline is frictionless, as stated.)

Q11. A person of mass 50 kg stands in a lift accelerating upward at 1 m/s2. What is the normal force on the person?

  • (A) 490
  • (B) 440
  • (C) 540
  • (D) 640
Show Answer

Answer: C – N – mg = ma, so N = m(g + a) = (50)(9.8 + 1) = (50)(10.8) = 540 N.

Q12. Two blocks of masses 4 kg and 6 kg are pushed together on a frictionless surface by a 48 N horizontal force. What is the acceleration of the pair?

  • (A) 4.8
  • (B) 12
  • (C) 8
  • (D) 9.6
Show Answer

Answer: A – Treat the pair as one mass: a = F / (m1 + m2) = 48 / (4 + 6) = 48 / 10 = 4.8 m/s2. (Simplification: friction is ignored, as stated.)

Q13. A 60 N horizontal force pulls a 10 kg box while kinetic friction of 10 N opposes the motion. What is the acceleration?

  • (A) 6
  • (B) 1
  • (C) 7
  • (D) 5
Show Answer

Answer: D – Fnet = 60 – 10 = 50 N, so a = Fnet/m = 50 / 10 = 5 m/s2.

Q14. A 2 kg cyclist moves in a horizontal circle of radius 5 m at 6 m/s. What is the centripetal force on it?

  • (A) 2.4
  • (B) 14.4
  • (C) 72
  • (D) 7.2
Show Answer

Answer: B – Fc = mv2/r = (2)(6)2 / 5 = 72 / 5 = 14.4 N.

Unit 3: Work, Energy, and Power

Q15. A force of 40 N at 0° to the horizontal pushes an object 10 m along the floor. How much work does the force do?

  • (A) 400
  • (B) 450
  • (C) 200
  • (D) 549
Show Answer

Answer: A – W = Fd cosΘ = (40)(10)(cos 0°) = 400 J.

Q16. What is the kinetic energy of a 2 kg puck moving at 8 m/s?

  • (A) 128
  • (B) 8
  • (C) 64
  • (D) 32
Show Answer

Answer: C – KE = (1/2)mv2 = (1/2)(2)(8)2 = 64 J.

Q17. What is the gravitational potential energy of a 10 kg book held 10 m above the ground, relative to the ground?

  • (A) 100
  • (B) 98
  • (C) 490
  • (D) 980
Show Answer

Answer: D – PE = mgh = (10)(9.8)(10) = 980 J.

Q18. A motor does 400 J of work in 20 s. What is its average power output?

  • (A) 8000
  • (B) 20
  • (C) 10
  • (D) 40
Show Answer

Answer: B – P = W/t = 400 / 20 = 20 W.

Q19. A 2 kg cart moving at 5 m/s receives 40 J of net work on a frictionless surface. What is its final speed?

  • (A) 25
  • (B) 8.06
  • (C) 6.32
  • (D) 9.06
Show Answer

Answer: B – Work-energy: (1/2)mv2 – (1/2)mv02 = W, so v = √(v02 + 2W/m) = √(52 + 2 x 40 / 2) = √(65) = 8.06 m/s. (Simplification: friction is ignored, as stated.)

Q20. A spring of constant 200 N/m is compressed by 0.3 m. How much elastic potential energy is stored?

  • (A) 60
  • (B) 30
  • (C) 18
  • (D) 9
Show Answer

Answer: D – PE = (1/2)kx2 = (1/2)(200)(0.3)2 = 9 J.

Q21. A worker lifts a 50 kg load through a height of 5 m in 5 s at constant speed. What is the average power delivered?

  • (A) 490
  • (B) 2450
  • (C) 245
  • (D) 50
Show Answer

Answer: A – Work = mgh = (50)(9.8)(5) = 2450 J, so P = W/t = 2450 / 5 = 490 W.

Unit 4: Linear Momentum

Q22. What is the magnitude of the momentum of a 8 kg tennis ball moving at 6 m/s?

  • (A) 14
  • (B) 24
  • (C) 48
  • (D) 96
Show Answer

Answer: C – p = mv = (8)(6) = 48 kg m/s.

Q23. A constant force of 50 N acts on a 5 kg object moving at 10 m/s in the opposite direction until it stops. For how long does the force act?

  • (A) 2500
  • (B) 2
  • (C) 1
  • (D) 0.5
Show Answer

Answer: C – Impulse: FΔt = mΔv = (5)(10) = 50 kg m/s, so Δt = 50 / 50 = 1 s.

Q24. A 3 kg cart moving at 6 m/s collides with and sticks to a stationary 4 kg cart on a frictionless track. What is their common speed after the collision?

  • (A) 2.57
  • (B) 6
  • (C) 4.5
  • (D) 5.14
Show Answer

Answer: A – Momentum is conserved: m1v1 = (m1 + m2)v, so v = (3)(6) / (3 + 4) = 18 / 7 = 2.57 m/s. (Simplification: friction is ignored, as stated.)

Q25. A 0.5 kg projectile is fired at 10 m/s from a stationary launcher of mass 5 kg on a frictionless surface. What is the recoil speed of the launcher?

  • (A) 10
  • (B) 5
  • (C) 2
  • (D) 1
Show Answer

Answer: D – Total momentum stays zero: MV = mv, so V = mv/M = (0.5)(10) / 5 = 1 m/s. (Simplification: friction is ignored, as stated.)

Q26. A 0.3 kg ball hits a wall at 10 m/s and rebounds at 5 m/s in the opposite direction. What is the magnitude of the impulse on the ball?

  • (A) 3
  • (B) 4.5
  • (C) 1.5
  • (D) 2.25
Show Answer

Answer: B – Impulse = mΔv = (0.3)(5 – (-10)) = (0.3)(15) = 4.5 N s.

Q27. An object at rest on a frictionless surface splits into a 4 kg part moving at 10 m/s and a 3 kg part. What is the speed of the 3 kg part?

  • (A) 13.33
  • (B) 10
  • (C) 40
  • (D) 6.67
Show Answer

Answer: A – Momentum is conserved from zero: m1v1 = m2v2, so v2 = (4)(10) / 3 = 13.33 m/s. (Simplification: friction is ignored, as stated.)

Q28. A 2 kg object changes speed by 10 m/s in 5 s under a constant net force. What is the magnitude of that force?

  • (A) 20
  • (B) 100
  • (C) 4
  • (D) 2
Show Answer

Answer: C – F = Δp/Δt = mΔv/Δt = (2)(10) / 5 = 4 N.

Unit 5: Torque and Rotational Dynamics

Q29. A force of 30 N acts perpendicular to a wrench at a distance of 0.5 m from the pivot. What is the torque about the pivot?

  • (A) 60
  • (B) 30.5
  • (C) 7.5
  • (D) 15
Show Answer

Answer: D – τ = rF sinΘ = (0.5)(30)(sin 90°) = 15 N m.

Q30. A net torque of 8 N m acts on an object with moment of inertia 2 kg m2. What is its angular acceleration?

  • (A) 16
  • (B) 4
  • (C) 2
  • (D) 5
Show Answer

Answer: B – τ = Iα, so α = τ/I = 8 / 2 = 4 rad/s2.


Q31. A disk starts with angular speed 2 rad/s and accelerates uniformly at 2 rad/s2 for 3 s. What is its final angular speed?

  • (A) 5
  • (B) 8
  • (C) 12
  • (D) 6
Show Answer

Answer: B – ω = ω0 + αt = 2 + (2)(3) = 8 rad/s.

Q32. A flywheel with initial angular speed 0 rad/s undergoes angular acceleration 4 rad/s2 for 3 s. Through what angle does it rotate?

  • (A) 36
  • (B) 12
  • (C) 9
  • (D) 18
Show Answer

Answer: D – Θ = ω0t + (1/2)αt2 = (0)(3) + (1/2)(4)(3)2 = 18 rad.

Q33. What is the moment of inertia of a 2 kg point mass at a distance of 1 m from the rotation axis?

  • (A) 2
  • (B) 3
  • (C) 4
  • (D) 3.74
Show Answer

Answer: A – I = mr2 = (2)(1)2 = 2 kg m2. (Simplification: the mass is treated as a point mass, as stated.)

Q34. A seesaw is balanced when a 40 N force acts 0.5 m from the pivot on one side. What force must act 1.5 m from the pivot on the other side to keep it balanced?

  • (A) 40
  • (B) 120
  • (C) 13.33
  • (D) 26.67
Show Answer

Answer: C – Balanced torques: F1d1 = F2d2, so F2 = (40)(0.5) / 1.5 = 13.33 N.

Q35. A solid disk of mass 2 kg and radius 0.4 m experiences a net torque of 4 N m. What is its angular acceleration? (Disk: I = (1/2)MR2.)

  • (A) 2
  • (B) 5
  • (C) 25
  • (D) 12.5
Show Answer

Answer: C – I = (1/2)MR2 = (1/2)(2)(0.4)2 = 0.16 kg m2. Then α = τ/I = 4 / 0.16 = 25 rad/s2.

Q36. A wheel with initial angular speed 2 rad/s accelerates at 4 rad/s2 through an angle of 8 rad. What is its final angular speed?

  • (A) 8.25
  • (B) 34
  • (C) 8
  • (D) 9.25
Show Answer

Answer: A – ω2 = ω02 + 2αΘ = 22 + 2(4)(8) = 68, so ω = √(68) = 8.25 rad/s.

Unit 6: Energy and Momentum of Rotating Systems

Q37. What is the rotational kinetic energy of an object with moment of inertia 6 kg m2 rotating at 5 rad/s?

  • (A) 150
  • (B) 15
  • (C) 103.75
  • (D) 75
Show Answer

Answer: D – KErot = (1/2)Iω2 = (1/2)(6)(5)2 = 75 J.

Q38. What is the angular momentum of a rotating body with moment of inertia 6 kg m2 and angular speed 2 rad/s?

  • (A) 3
  • (B) 12
  • (C) 24
  • (D) 6
Show Answer

Answer: B – L = Iω = (6)(2) = 12 kg m2/s.

Q39. A skater spins with moment of inertia 2 kg m2 at 8 rad/s, then changes shape so the moment of inertia becomes 4 kg m2. With no external torque, what is the new angular speed?

  • (A) 4
  • (B) 8
  • (C) 16
  • (D) 6.48
Show Answer

Answer: A – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = (2)(8) / 4 = 4 rad/s. (Simplification: external torque is ignored, as stated.)

Q40. A solid sphere of mass 3 kg rolls without slipping at 4 m/s. What is its total kinetic energy? (Sphere: I = (2/5)MR2.)

  • (A) 24
  • (B) 48
  • (C) 33.6
  • (D) 16.8
Show Answer

Answer: C – Total KE = (1/2)mv2 + (1/2)Iω2 with ω = v/R gives (7/10)mv2 = 0.7 x (3) x (4)2 = 33.6 J.

Q41. A solid disk of mass 4 kg and radius 0.3 m spins at 4 rad/s. What is its rotational kinetic energy? (Disk: I = (1/2)MR2.)

  • (A) 4.8
  • (B) 32
  • (C) 2.88
  • (D) 1.44
Show Answer

Answer: D – I = (1/2)(4)(0.3)2 = 0.18 kg m2, so KE = (1/2)Iω2 = (1/2)(0.18)(4)2 = 1.44 J.

Q42. A 1 kg ball moves at 5 m/s at a perpendicular distance of 2 m from a point. What is its angular momentum about that point?

  • (A) 5
  • (B) 10
  • (C) 2
  • (D) 14.7
Show Answer

Answer: B – L = mvr = (1)(5)(2) = 10 kg m2/s.

Q43. A turntable with moment of inertia 3 kg m2 rotates at 4 rad/s. A stationary object is dropped onto it, raising the total moment of inertia to 6 kg m2. What is the new angular speed if no external torque acts?

  • (A) 4
  • (B) 2
  • (C) 8
  • (D) 3
Show Answer

Answer: B – I1ω1 = I2ω2, so ω2 = (3)(4) / 6 = 2 rad/s. (Simplification: external torque is ignored, as stated.)

Q44. A solid cylinder of mass 4 kg rolls without slipping at 3 m/s. What is its total kinetic energy? (Cylinder: I = (1/2)MR2.)

  • (A) 18
  • (B) 36
  • (C) 54
  • (D) 27
Show Answer

Answer: D – Total KE = (1/2)mv2 + (1/2)Iω2 = (3/4)mv2 = 0.75 x (4) x (3)2 = 27 J.

Unit 7: Oscillations

Q45. A 8 kg mass on a spring of constant 200 N/m oscillates on a frictionless surface. What is the period of oscillation? (Use π = 3.14.)

  • (A) 1.26
  • (B) 31.4
  • (C) 0.63
  • (D) 2.51
Show Answer

Answer: A – T = 2π√(m/k) = 2(3.14)√(8/200) = 6.28 x 0.2 = 1.26 s. (Simplification: friction is ignored, as stated.)

Q46. What is the period of a simple pendulum with a wooden bob and length 1.44 m on Earth? (Use π = 3.14, g = 9.8 m/s2.)

  • (A) 16.38
  • (B) 1.2
  • (C) 2.41
  • (D) 4.3
Show Answer

Answer: C – T = 2π√(L/g) = 2(3.14)√(1.44/9.8) = 2.41 s. (Simplification: small-angle approximation is used.)

Q47. An tuning fork completes one oscillation in 0.5 s. What is its frequency?

  • (A) 0.5
  • (B) 12.56
  • (C) 2
  • (D) 4
Show Answer

Answer: C – f = 1/T = 1 / 0.5 = 2 Hz.

Q48. A mass of 2 kg on a spring (k = 200 N/m) oscillates with amplitude 0.2 m. What is its maximum speed?

  • (A) 2
  • (B) 0.4
  • (C) 20
  • (D) 4
Show Answer

Answer: A – ω = √(k/m) = √(200/2) = 10 rad/s, so vmax = ωA = (10)(0.2) = 2 m/s.

Q49. A mass of 4 kg on a spring (k = 100 N/m) oscillates with amplitude 0.1 m. What is its maximum acceleration?

  • (A) 0.5
  • (B) 10
  • (C) 1.25
  • (D) 2.5
Show Answer

Answer: D – ω = √(k/m) = 5 rad/s, so amax = ω2A = (5)2(0.1) = 2.5 m/s2.

Q50. What is the total mechanical energy of a spring oscillator with k = 200 N/m and amplitude 0.1 m?

  • (A) 20
  • (B) 1
  • (C) 10
  • (D) 2
Show Answer

Answer: B – E = (1/2)kA2 = (1/2)(200)(0.1)2 = 1 J.

Q51. A simple pendulum has a period of 2.5 s on Earth. What is its length? (Use π = 3.14, g = 9.8 m/s2.)

  • (A) 1.55
  • (B) 4.88
  • (C) 61.25
  • (D) 3.11
Show Answer

Answer: A – From T = 2π√(L/g), L = g(T/2π)2 = 9.8 x (2.5/6.28)2 = 1.55 m. (Simplification: small-angle approximation is used.)

Q52. A 1 kg mass oscillates on a spring with period 0.5 s. What is the spring constant? (Use π = 3.14.)

  • (A) 4
  • (B) 25.12
  • (C) 157.75
  • (D) 78.88
Show Answer

Answer: C – From T = 2π√(m/k), k = 4π2m/T2 = 4(3.14)2(1) / (0.5)2 = 157.75 N/m.

Unit 8: Fluids

Q53. A force of 150 N acts uniformly over an area of 0.25 m2. What is the pressure?

  • (A) 37.5
  • (B) 150.25
  • (C) 1200
  • (D) 600
Show Answer

Answer: D – P = F/A = 150 / 0.25 = 600 Pa.

Q54. What is the gauge pressure at a depth of 5 m in oil of density 1200 kg/m3? (g = 9.8 m/s2.)

  • (A) 6000
  • (B) 58800
  • (C) 11760
  • (D) 29400
Show Answer

Answer: B – P = ρgh = (1200)(9.8)(5) = 58800 Pa.

Q55. A metal block of volume 0.02 m3 is fully submerged in water (density 1000 kg/m3). What is the buoyant force on it?

  • (A) 20
  • (B) 196
  • (C) 0.2
  • (D) 98
Show Answer

Answer: B – Fb = ρVg = (1000)(0.02)(9.8) = 196 N.

Q56. Water flows at 3 m/s through a pipe of cross-sectional area 0.08 m2, then enters a narrower section of area 0.02 m2. What is the speed in the narrower section?

  • (A) 3
  • (B) 0.75
  • (C) 6
  • (D) 12
Show Answer

Answer: D – Continuity: A1v1 = A2v2, so v2 = (0.08)(3) / 0.02 = 12 m/s.

Q57. Water flows through a pipe of area 0.02 m2 at 4 m/s. What is the volume flow rate?

  • (A) 0.08
  • (B) 4.02
  • (C) 0.01
  • (D) 0.16
Show Answer

Answer: A – Q = Av = (0.02)(4) = 0.08 m3/s.

Q58. In a horizontal pipe, water speeds up from 2 m/s to 4 m/s. By how much does the pressure decrease? (Density of water = 1000 kg/m3.)

  • (A) 2000
  • (B) 1000
  • (C) 6000
  • (D) 3000
Show Answer

Answer: C – Bernoulli (horizontal): ΔP = (1/2)ρ(v22 – v12) = (1/2)(1000)(42 – 22) = 6000 Pa decrease.

Q59. What is the absolute pressure at a depth of 25 m in water? (Atmospheric pressure = 101000 Pa, density of water = 1000 kg/m3, g = 9.8 m/s2.)

  • (A) 245000
  • (B) 126000
  • (C) 346000
  • (D) 345000
Show Answer

Answer: C – P = Patm + ρgh = 101000 + (1000)(9.8)(25) = 101000 + 245000 = 346000 Pa.

Q60. A alloy block has mass 80 kg and volume 0.1 m3. What is its density?

  • (A) 800
  • (B) 8
  • (C) 0
  • (D) 400
Show Answer

Answer: A – ρ = m/V = 80 / 0.1 = 800 kg/m3.

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