AP Physics 1 Practice Test 11 – 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 cyclist travels 120 m in 8 s at a steady pace. What is the average speed?

  • (A) 15 m/s
  • (B) 30 m/s
  • (C) 7.5 m/s
  • (D) 22.5 m/s
Show Answer

Answer: A – average speed = distance ÷ time = 120 ÷ 8 = 15 m/s.

Q2. A car starts at 5 m/s and accelerates uniformly at 2 m/s2 for 6 s. What is its final speed?

  • (A) 34 m/s
  • (B) 17 m/s
  • (C) 8.5 m/s
  • (D) 25.5 m/s
Show Answer

Answer: B – v = v0 + at = 5 + (2 × 6) = 5 + 12 = 17 m/s.

Q3. A trolley starts at 3 m/s and accelerates at 2 m/s2 for 5 s. How far does it travel?

  • (A) 80 m
  • (B) 20 m
  • (C) 40 m
  • (D) 60 m
Show Answer

Answer: C – displacement = v0t + 0.5at2 = (3 × 5) + 0.5 × 2 × 52 = 15 + 25.0 = 40 m.

Q4. A sprinter starts at 4 m/s and accelerates at 3 m/s2 over a distance of 20 m. What is the final speed?

  • (A) 23.32 m/s
  • (B) 5.83 m/s
  • (C) 17.49 m/s
  • (D) 11.66 m/s
Show Answer

Answer: D – v2 = v02 + 2aΔx = 42 + 2 × 3 × 20 = 136, so v = √136 = 11.66 m/s.

Q5. A stone is dropped from rest from a tall tower. Ignoring air resistance, what is its speed after 2 s?

  • (A) 19.6 m/s
  • (B) 39.2 m/s
  • (C) 9.8 m/s
  • (D) 29.4 m/s
Show Answer

Answer: A – v = gt = 9.8 × 2 = 19.6 m/s. (Simplification: air resistance is ignored, as stated.)

Q6. A metal ball is dropped from rest. Ignoring air resistance, how far does it fall in 2 s?

  • (A) 39.2 m
  • (B) 19.6 m
  • (C) 9.8 m
  • (D) 29.4 m
Show Answer

Answer: B – distance = 0.5gt2 = 0.5 × 9.8 × 22 = 19.6 m. (Simplification: air resistance is ignored, as stated.)

Q7. A ball rolls horizontally off a cliff 19.6 m high with a speed of 8 m/s. Ignoring air resistance, how far from the base of the cliff does it land?

  • (A) 32 m
  • (B) 8 m
  • (C) 16 m
  • (D) 24 m
Show Answer

Answer: C – Time to fall: t = √(2h ÷ g) = √(39.2 ÷ 9.8) = 2 s. Horizontal range = vxt = 8 × 2 = 16 m. (Simplification: air resistance is ignored, as stated.)

Q8. A car travelling at 12 m/s brakes with a constant deceleration of 3 m/s2 until it stops. What is its stopping distance?

  • (A) 48 m
  • (B) 12 m
  • (C) 36 m
  • (D) 24 m
Show Answer

Answer: D – 0 = v02 – 2aΔx, so Δx = v02 ÷ 2a = 122 ÷ 6 = 24 m.

Unit 2: Force and Translational Dynamics

Q9. A net force accelerates a 4 kg crate at 3 m/s2 across a smooth floor. What is the size of the net force?

  • (A) 12 N
  • (B) 24 N
  • (C) 6 N
  • (D) 18 N
Show Answer

Answer: A – F = ma = 4 × 3 = 12 N.

Q10. A 10 kg box is pushed across a rough floor at constant speed by a horizontal force of 40 N. What is the coefficient of kinetic friction between the box and the floor?

  • (A) 0.10
  • (B) 0.41
  • (C) 0.80
  • (D) 4.0
Show Answer

Answer: B – At constant speed the net force is zero, so kinetic friction equals the applied force: f = 40 N. The normal force is N = mg = 10 × 9.8 = 98 N. So μ = f / N = 40 / 98 = 0.41.

Q11. A 10 kg box slides on a horizontal floor where the coefficient of kinetic friction is 0.2. What is the kinetic friction force?

  • (A) 39.2 N
  • (B) 9.8 N
  • (C) 19.6 N
  • (D) 29.4 N
Show Answer

Answer: C – Normal force N = mg = 10 × 9.8 = 98 N. Friction = μN = 0.2 × 98 = 19.6 N.

Q12. A horizontal force of 50 N pushes a 10 kg sled on a rough surface (μ = 0.2). What is the acceleration of the sled?

  • (A) 6.08 m/s2
  • (B) 1.52 m/s2
  • (C) 4.56 m/s2
  • (D) 3.04 m/s2
Show Answer

Answer: D – Friction = μmg = 0.2 × 10 × 9.8 = 19.6 N. Net force = 50 – 19.6 = 30.4 N. Acceleration = Fnet ÷ m = 30.4 ÷ 10 = 3.04 m/s2.

Q13. A 10 kg block rests on a frictionless slope inclined at 30° to the horizontal. What is the component of its weight down the slope? (sin 30° = 0.5)

  • (A) 49 N
  • (B) 98 N
  • (C) 24.5 N
  • (D) 73.5 N
Show Answer

Answer: A – Component down the slope = mg sinΘ = 10 × 9.8 × 0.5 = 49 N. (Simplification: the slope is frictionless, as stated.)

Q14. Two masses of 3 kg and 5 kg are joined by a light string over a frictionless pulley. What is the magnitude of their acceleration? (The 5 kg mass is heavier.)

  • (A) 4.9 m/s2
  • (B) 2.45 m/s2
  • (C) 1.23 m/s2
  • (D) 3.68 m/s2
Show Answer

Answer: B – Net force = (5 – 3)g = 2 × 9.8 = 19.6 N. Total mass = 8 kg. Acceleration = 19.6 ÷ 8 = 2.45 m/s2. (Simplification: the pulley and string are massless and frictionless, as stated.)

Q15. A 2 kg ball on a string moves in a horizontal circle of radius 3 m at a speed of 6 m/s. What is the centripetal force on the ball?

  • (A) 48 N
  • (B) 12 N
  • (C) 24 N
  • (D) 36 N
Show Answer

Answer: C – F = mv2 ÷ r = 2 × 62 ÷ 3 = 2 × 36 ÷ 3 = 24 N.

Unit 3: Work, Energy, and Power

Q16. A worker pushes a crate with a constant horizontal force of 20 N over a distance of 15 m. How much work is done?

  • (A) 600 J
  • (B) 150 J
  • (C) 450 J
  • (D) 300 J
Show Answer

Answer: D – W = Fd = 20 × 15 = 300 J.

Q17. A rope pulls a box with a force of 50 N at 37° above the horizontal while the box moves 10 m. How much work is done by the rope? (cos 37° = 0.8)

  • (A) 400 J
  • (B) 800 J
  • (C) 200 J
  • (D) 600 J
Show Answer

Answer: A – W = Fd cosΘ = 50 × 10 × 0.8 = 400 J.

Q18. What is the kinetic energy of a 4 kg bowling ball moving at 5 m/s?

  • (A) 100 J
  • (B) 50 J
  • (C) 25 J
  • (D) 75 J
Show Answer

Answer: B – KE = 0.5mv2 = 0.5 × 4 × 52 = 0.5 × 4 × 25 = 50 J.

Q19. A 2 kg book is lifted to a shelf 10 m above the floor. By how much does its gravitational potential energy increase?

  • (A) 392 J
  • (B) 98 J
  • (C) 196 J
  • (D) 294 J
Show Answer

Answer: C – ΔPE = mgh = 2 × 9.8 × 10 = 196 J.

Q20. A 4 kg cart starts from rest. A net force does work on it until it reaches 6 m/s. How much work was done on the cart?

  • (A) 144 J
  • (B) 36 J
  • (C) 108 J
  • (D) 72 J
Show Answer

Answer: D – By the work-energy theorem, W = ΔKE = 0.5mv2 – 0 = 0.5 × 4 × 62 = 72 J.

Q21. A motor does 600 J of work in 10 s. What is its average power output?

  • (A) 60 W
  • (B) 120 W
  • (C) 30 W
  • (D) 90 W
Show Answer

Answer: A – P = W ÷ t = 600 ÷ 10 = 60 W.

Q22. A car engine exerts a constant driving force of 200 N while the car moves at a steady 3 m/s. What power does the engine deliver?

  • (A) 1200 W
  • (B) 600 W
  • (C) 300 W
  • (D) 900 W
Show Answer

Answer: B – P = Fv = 200 × 3 = 600 W.

Q23. A spring of spring constant 200 N/m is compressed by 0.2 m from its natural length. What is its elastic potential energy?

  • (A) 8 J
  • (B) 2 J
  • (C) 4 J
  • (D) 6 J
Show Answer

Answer: C – PE = 0.5kx2 = 0.5 × 200 × 0.22 = 0.5 × 200 × 0.04 = 4 J.

Unit 4: Linear Momentum

Q24. What is the momentum of a 3 kg cart moving at 6 m/s?

  • (A) 36 kg m/s
  • (B) 9 kg m/s
  • (C) 27 kg m/s
  • (D) 18 kg m/s
Show Answer

Answer: D – p = mv = 3 × 6 = 18 kg m/s.

Q25. A constant force of 40 N acts on a hockey puck for 0.5 s. What impulse does it deliver?

  • (A) 20 N s
  • (B) 40 N s
  • (C) 10 N s
  • (D) 30 N s
Show Answer

Answer: A – J = FΔt = 40 × 0.5 = 20 N s.

Q26. A 2 kg ball changes velocity by 6 m/s during a 0.3 s contact with a bat. What is the average force exerted by the bat?

  • (A) 80 N
  • (B) 40 N
  • (C) 20 N
  • (D) 60 N
Show Answer

Answer: B – Impulse = mΔv = 2 × 6 = 12 kg m/s. Average force = impulse ÷ time = 12 ÷ 0.3 = 40 N.

Q27. A 4 kg cart moving at 5 m/s hits a stationary 6 kg cart and the two stick together. What is their common speed after the collision? (Ignore friction.)

  • (A) 4 m/s
  • (B) 1 m/s
  • (C) 2 m/s
  • (D) 3 m/s
Show Answer

Answer: C – Momentum before = 4 × 5 = 20 kg m/s. Total mass = 10 kg, so v = 20 ÷ 10 = 2 m/s. (Simplification: friction is ignored, as stated.)

Q28. Two skaters at rest on frictionless ice push apart. One skater has mass 50 kg and moves at 2 m/s. The other has mass 25 kg. How fast does the other skater move?

  • (A) 8 m/s
  • (B) 2 m/s
  • (C) 6 m/s
  • (D) 4 m/s
Show Answer

Answer: D – Total momentum stays zero, so m1v1 = m2v2. v2 = (50 × 2) ÷ 25 = 100 ÷ 25 = 4 m/s. (Simplification: the ice is frictionless, as stated.)

Q29. A 0.5 kg rubber ball hits a wall at 6 m/s and rebounds at 6 m/s in the opposite direction. What is the magnitude of the change in its momentum?

  • (A) 6 kg m/s
  • (B) 12 kg m/s
  • (C) 3 kg m/s
  • (D) 9 kg m/s
Show Answer

Answer: A – Taking the rebound direction as positive, Δp = m(vf – vi) = 0.5 × (6 – (-6)) = 0.5 × 12 = 6 kg m/s.

Q30. A 3.0 kg cart moving at 4.0 m/s on a frictionless track collides with a stationary 2.0 kg cart, and the two carts stick together. What is their common speed after the collision?

  • (A) 1.2 m/s
  • (B) 2.4 m/s
  • (C) 4.0 m/s
  • (D) 6.0 m/s
Show Answer

Answer: B – Momentum is conserved: (3.0 × 4.0) + 0 = (3.0 + 2.0) × v, so 12 = 5.0 × v and v = 12 / 5.0 = 2.4 m/s.


Unit 5: Torque and Rotational Dynamics

Q31. A force of 20 N is applied perpendicular to a wrench at a distance of 0.5 m from the pivot. What is the torque about the pivot?

  • (A) 20 N m
  • (B) 5 N m
  • (C) 10 N m
  • (D) 15 N m
Show Answer

Answer: C – τ = rF = 0.5 × 20 = 10 N m (force perpendicular to the lever arm).

Q32. A force of 40 N acts at 30° to a lever arm of length 0.5 m. What is the torque about the pivot? (sin 30° = 0.5)

  • (A) 20 N m
  • (B) 5 N m
  • (C) 15 N m
  • (D) 10 N m
Show Answer

Answer: D – τ = rF sinΘ = 0.5 × 40 × 0.5 = 10 N m.

Q33. Two forces act on opposite sides of a pivot: 20 N at 0.6 m on one side and 10 N at 0.4 m on the other side, tending to rotate it oppositely. What is the magnitude of the net torque?

  • (A) 8 N m
  • (B) 16 N m
  • (C) 4 N m
  • (D) 12 N m
Show Answer

Answer: A – Torques are 20 × 0.6 = 12 N m and 10 × 0.4 = 4 N m in opposite directions. Net torque = 12 – 4 = 8 N m.

Q34. A net torque of 12 N m acts on a wheel with moment of inertia 4 kg m2. What is its angular acceleration?

  • (A) 6 rad/s2
  • (B) 3 rad/s2
  • (C) 1.5 rad/s2
  • (D) 4.5 rad/s2
Show Answer

Answer: B – α = τ ÷ I = 12 ÷ 4 = 3 rad/s2.

Q35. A flywheel starts with angular velocity 2 rad/s and angular acceleration 3 rad/s2. What is its angular velocity after 4 s?

  • (A) 28 rad/s
  • (B) 7 rad/s
  • (C) 14 rad/s
  • (D) 21 rad/s
Show Answer

Answer: C – ω = ω0 + αt = 2 + (3 × 4) = 2 + 12 = 14 rad/s.

Q36. A disc starts with angular velocity 2 rad/s and angular acceleration 2 rad/s2. Through what angle does it rotate in 5 s?

  • (A) 70 rad
  • (B) 17.5 rad
  • (C) 52.5 rad
  • (D) 35 rad
Show Answer

Answer: D – Θ = ω0t + 0.5αt2 = (2 × 5) + 0.5 × 2 × 52 = 10 + 25.0 = 35 rad.

Q37. A 2 kg point mass is attached at a distance of 0.5 m from an axis. What is its moment of inertia about that axis? (Treat it as a point mass.)

  • (A) 0.5 kg m2
  • (B) 1 kg m2
  • (C) 0.25 kg m2
  • (D) 0.75 kg m2
Show Answer

Answer: A – I = mr2 = 2 × 0.52 = 2 × 0.25 = 0.5 kg m2. (Simplification: the object is treated as a point mass, as stated.)

Q38. What net torque is needed to give a rotor of moment of inertia 4 kg m2 an angular acceleration of 3 rad/s2?

  • (A) 24 N m
  • (B) 12 N m
  • (C) 6 N m
  • (D) 18 N m
Show Answer

Answer: B – τ = Iα = 4 × 3 = 12 N m.

Unit 6: Energy and Momentum of Rotating Systems

Q39. A spinning disc has moment of inertia 4 kg m2 and angular velocity 5 rad/s. What is its rotational kinetic energy?

  • (A) 100 J
  • (B) 25 J
  • (C) 50 J
  • (D) 75 J
Show Answer

Answer: C – KErot = 0.5Iω2 = 0.5 × 4 × 52 = 0.5 × 4 × 25 = 50 J.

Q40. A wheel has moment of inertia 3 kg m2 and spins at 6 rad/s. What is its angular momentum?

  • (A) 36 kg m2/s
  • (B) 9 kg m2/s
  • (C) 27 kg m2/s
  • (D) 18 kg m2/s
Show Answer

Answer: D – L = Iω = 3 × 6 = 18 kg m2/s.

Q41. A figure skater spins with moment of inertia 6 kg m2 at 4 rad/s, then pulls her arms in so her moment of inertia becomes 3 kg m2. What is her new angular velocity? (Ignore external torques.)

  • (A) 8 rad/s
  • (B) 16 rad/s
  • (C) 4 rad/s
  • (D) 12 rad/s
Show Answer

Answer: A – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = (6 × 4) ÷ 3 = 24 ÷ 3 = 8 rad/s. (Simplification: external torques are ignored, as stated.)

Q42. A rolling object of mass 4 kg moves at 6 m/s. Its moment of inertia is 2 kg m2 and its radius is 0.5 m. What is its total kinetic energy?

  • (A) 432 J
  • (B) 216 J
  • (C) 108 J
  • (D) 324 J
Show Answer

Answer: B – ω = v ÷ r = 6 ÷ 0.5 = 12 rad/s. Total KE = 0.5mv2 + 0.5Iω2 = 0.5 × 4 × 36 + 0.5 × 2 × 144 = 72 + 144 = 216 J.

Q43. A mass of 1 kg falls 5 m, turning a drum of moment of inertia 2 kg m2 with no slipping and no other energy loss. What is the angular velocity of the drum at the end of the fall?

  • (A) 14 rad/s
  • (B) 3.5 rad/s
  • (C) 7 rad/s
  • (D) 10.5 rad/s
Show Answer

Answer: C – mgh = 0.5Iω2, so ω = √(2mgh ÷ I) = √(98 ÷ 2) = √49 = 7 rad/s. (Simplification: other energy losses are ignored, as stated.)

Q44. A 2 kg particle moves at 6 m/s at a perpendicular distance of 3 m from a point. What is its angular momentum about that point?

  • (A) 72 kg m2/s
  • (B) 18 kg m2/s
  • (C) 54 kg m2/s
  • (D) 36 kg m2/s
Show Answer

Answer: D – L = mvr = 2 × 6 × 3 = 36 kg m2/s.

Q45. A rotor with moment of inertia 4 kg m2 speeds up from 2 rad/s to 6 rad/s. How much rotational work is done on it?

  • (A) 64 J
  • (B) 128 J
  • (C) 32 J
  • (D) 96 J
Show Answer

Answer: A – W = ΔKErot = 0.5I(ωf2 – ωi2) = 0.5 × 4 × (36 – 4) = 0.5 × 4 × 32 = 64 J.

Unit 7: Oscillations

Q46. A 1 kg mass on a spring of spring constant 100 N/m oscillates on a frictionless surface. What is the period of oscillation?

  • (A) 1.26 s
  • (B) 0.63 s
  • (C) 0.31 s
  • (D) 0.94 s
Show Answer

Answer: B – T = 2π√(m ÷ k) = 2π × √(1 ÷ 100) = 2π × 0.1 = 0.63 s. (Simplification: the surface is frictionless, as stated.)

Q47. A pendulum completes one oscillation in 2 s. What is its frequency?

  • (A) 1 Hz
  • (B) 0.25 Hz
  • (C) 0.5 Hz
  • (D) 0.75 Hz
Show Answer

Answer: C – f = 1 ÷ T = 1 ÷ 2 = 0.5 Hz.

Q48. A simple pendulum has length 0.98 m. What is its period for small oscillations? (Ignore air resistance.)

  • (A) 3.97 s
  • (B) 0.99 s
  • (C) 2.98 s
  • (D) 1.99 s
Show Answer

Answer: D – T = 2π√(L ÷ g) = 2π × √(0.98 ÷ 9.8) = 2π × 0.32 = 1.99 s. (Simplification: air resistance is ignored, as stated.)

Q49. A simple pendulum of length 0.64 m swings with a small amplitude on Earth. What is the period of the pendulum?

  • (A) 1.6 s
  • (B) 0.80 s
  • (C) 2.5 s
  • (D) 3.2 s
Show Answer

Answer: A – For small amplitudes, the period of a simple pendulum is T = 2π√(L/g) = 2π × 0.256 = 1.6 s. (Simplification: the small-angle approximation is used, as stated.)

Q50. A spring of spring constant 200 N/m oscillates with amplitude 0.1 m. What is the total mechanical energy of the oscillation?

  • (A) 2 J
  • (B) 1 J
  • (C) 0.5 J
  • (D) 1.5 J
Show Answer

Answer: B – E = 0.5kA2 = 0.5 × 200 × 0.12 = 0.5 × 200 × 0.01 = 1 J.

Q51. A pendulum of length 1 m has a period of 2 s. If its length is increased to 4 m, what is its new period?

  • (A) 8 s
  • (B) 2 s
  • (C) 4 s
  • (D) 6 s
Show Answer

Answer: C – Period is proportional to √L. Quadrupling L multiplies the period by √4 = 2, so T = 2 × 2 = 4 s.

Q52. A 1 kg mass on a spring (k = 100 N/m) oscillates with amplitude 0.2 m. What is its maximum acceleration?

  • (A) 40 m/s2
  • (B) 10 m/s2
  • (C) 30 m/s2
  • (D) 20 m/s2
Show Answer

Answer: D – ω = √(k ÷ m) = √(100 ÷ 1) = 10 rad/s. amax = ω2A = 100 × 0.2 = 20 m/s2.

Q53. A 1 kg block on a spring of spring constant 25 N/m oscillates on a frictionless surface. What is the frequency of oscillation?

  • (A) 0.8 Hz
  • (B) 1.59 Hz
  • (C) 0.4 Hz
  • (D) 1.19 Hz
Show Answer

Answer: A – f = (1 ÷ 2π) × √(k ÷ m) = (1 ÷ 2π) × √(25 ÷ 1) = (1 ÷ 2π) × 5 = 0.8 Hz. (Simplification: the surface is frictionless, as stated.)

Unit 8: Fluids

Q54. A block of mass 200 kg has a volume of 0.2 m3. What is its density?

  • (A) 2000 kg/m3
  • (B) 1000 kg/m3
  • (C) 500 kg/m3
  • (D) 1500 kg/m3
Show Answer

Answer: B – density = mass ÷ volume = 200 ÷ 0.2 = 1000 kg/m3.

Q55. A force of 200 N acts on an area of 0.5 m2. What pressure does it exert?

  • (A) 800 Pa
  • (B) 200 Pa
  • (C) 400 Pa
  • (D) 600 Pa
Show Answer

Answer: C – P = F ÷ A = 200 ÷ 0.5 = 400 Pa.

Q56. What is the gauge pressure due to water at a depth of 5 m below the surface? (Density of water = 1000 kg/m3.)

  • (A) 98000 Pa
  • (B) 24500 Pa
  • (C) 73500 Pa
  • (D) 49000 Pa
Show Answer

Answer: D – P = ρgh = 1000 × 9.8 × 5 = 49000 Pa.

Q57. A solid of volume 0.02 m3 is fully submerged in water (density 1000 kg/m3). What is the buoyant force on it?

  • (A) 196 N
  • (B) 392 N
  • (C) 98 N
  • (D) 294 N
Show Answer

Answer: A – Fb = ρVg = 1000 × 0.02 × 9.8 = 196 N.

Q58. Water flows through a pipe of cross-sectional area 0.04 m2 at 3 m/s, then enters a narrower section of area 0.02 m2. What is the speed in the narrower section?

  • (A) 12 m/s
  • (B) 6 m/s
  • (C) 3 m/s
  • (D) 9 m/s
Show Answer

Answer: B – By continuity, A1v1 = A2v2, so v2 = (0.04 × 3) ÷ 0.02 = 0.12 ÷ 0.02 = 6 m/s.

Q59. Water flows through a pipe of cross-sectional area 0.02 m2 at 3 m/s. What is the volume flow rate?

  • (A) 0.12 m3/s
  • (B) 0.03 m3/s
  • (C) 0.06 m3/s
  • (D) 0.09 m3/s
Show Answer

Answer: C – Q = Av = 0.02 × 3 = 0.06 m3/s.

Q60. Water speeds up from 2 m/s to 4 m/s in a horizontal pipe. By how much does the pressure drop? (Density of water = 1000 kg/m3.)

  • (A) 12000 Pa
  • (B) 3000 Pa
  • (C) 9000 Pa
  • (D) 6000 Pa
Show Answer

Answer: D – For horizontal flow, ΔP = 0.5ρ(v22 – v12) = 0.5 × 1000 × (16 – 4) = 500 × 12 = 6000 Pa.

Review every question you missed, work through the explanation step by step, and try the question again without looking at the answer before moving on to the next test.

📥 Prefer pen and paper? Take this test offline.

Download Printable PDF – Test 11

Free printable version: all 60 questions + answer key. Explanations stay here on the website.