AP Physics 1 Practice Test 3 – 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 commuter train starts with a velocity of 0 m/s and accelerates uniformly at 4 m/s2 for 3 s. What is its final velocity? (Ignore air resistance.)

  • (A) 12 m/s
  • (B) 4 m/s
  • (C) 24 m/s
  • (D) 18 m/s
Show Answer

Answer: A – v = v0 + at = 0 + (4 × 3) = 0 + 12 = 12 m/s. (Simplification: air resistance is ignored, as stated.)

Q2. A skateboarder moves with an initial velocity of 6 m/s and a constant acceleration of 2 m/s2. How far does the skateboarder travel in 5 s?

  • (A) 30 m
  • (B) 55 m
  • (C) 80 m
  • (D) 25 m
Show Answer

Answer: B – x = v0t + 0.5at2 = (6 × 5) + 0.5 × 2 × 52 = 30 + 25 = 55 m.

Q3. A car travelling at 18 m/s brakes with a constant deceleration of 3 m/s2. How far does it travel before it stops?

  • (A) 108 m
  • (B) 6 m
  • (C) 54 m
  • (D) 27 m
Show Answer

Answer: C – v2 = v02 + 2aΔx with v = 0 gives 0 = 182 – 2 × 3 × Δx, so Δx = 324 ÷ 6 = 54 m.

Q4. A small rock is dropped from rest from a tower. Ignoring air resistance, what is its velocity after 4 s?

  • (A) 19.6 m/s
  • (B) 156.8 m/s
  • (C) 9.8 m/s
  • (D) 39.2 m/s
Show Answer

Answer: D – v = v0 + gt = 0 + (9.8 × 4) = 39.2 m/s downward. (Simplification: air resistance is ignored, as stated.)

Q5. A rubber ball is dropped from rest from the roof of a apartment tower and hits the ground after 4 s. Ignoring air resistance, how tall is the apartment tower?

  • (A) 39.2 m
  • (B) 156.8 m
  • (C) 78.4 m
  • (D) 19.6 m
Show Answer

Answer: C – h = 0.5gt2 = 0.5 × 9.8 × 42 = 0.5 × 9.8 × 16 = 78.4 m. (Simplification: air resistance is ignored, as stated.)

Q6. A jogger runs 180 m east in 15 s, then stops. What is the jogger’s average velocity for the run?

  • (A) 12 m/s east
  • (B) 2700 m/s east
  • (C) 0.08 m/s east
  • (D) 6 m/s east
Show Answer

Answer: A – Average velocity = displacement ÷ time = 180 ÷ 15 = 12 m/s east.

Q7. A marble rolls horizontally off a table 19.6 m high with a horizontal velocity of 15 m/s. Ignoring air resistance, how far from the base of the table does it land?

  • (A) 15 m
  • (B) 60 m
  • (C) 2 m
  • (D) 30 m
Show Answer

Answer: D – Vertical fall: t = √(2h ÷ g) = √(39.2 ÷ 9.8) = 2 s. Horizontal range = vx × t = 15 × 2 = 30 m. (Simplification: air resistance is ignored, as stated.)

Q8. A truck moving at 36 m/s brakes uniformly at 6 m/s2. How long does it take to come to rest?

  • (A) 216 s
  • (B) 6 s
  • (C) 3 s
  • (D) 9 s
Show Answer

Answer: B – v = v0 + at with v = 0 gives 0 = 36 – 6t, so t = 36 ÷ 6 = 6 s.

Unit 2: Force and Translational Dynamics

Q9. A net force acts on a 2 kg crate and gives it an acceleration of 5 m/s2. What is the magnitude of the net force?

  • (A) 7 N
  • (B) 10 N
  • (C) 0.4 N
  • (D) 20 N
Show Answer

Answer: B – Fnet = ma = 2 × 5 = 10 N.

Q10. What is the weight of a 12 kg backpack on Earth?

  • (A) 12 N
  • (B) 120 N
  • (C) 1.22 N
  • (D) 117.6 N
Show Answer

Answer: D – Weight = mg = 12 × 9.8 = 117.6 N.

Q11. A 15 kg box slides on a horizontal floor. The coefficient of kinetic friction is 0.4. What is the magnitude of the kinetic friction force?

  • (A) 58.8 N
  • (B) 6 N
  • (C) 147 N
  • (D) 29.4 N
Show Answer

Answer: A – Normal force N = mg = 15 × 9.8 = 147 N. Friction f = μN = 0.4 × 147 = 58.8 N.

Q12. A 8 kg sled is pulled to the right with a 60 N force while a 20 N friction force acts to the left. What is the sled’s acceleration?

  • (A) 10 m/s2 to the right
  • (B) 7.5 m/s2 to the right
  • (C) 5 m/s2 to the right
  • (D) 5 m/s2 to the left
Show Answer

Answer: C – Fnet = 60 – 20 = 40 N to the right. a = Fnet ÷ m = 40 ÷ 8 = 5 m/s2 to the right.

Q13. A 4 kg block sits on a frictionless ramp inclined at 37° to the horizontal. Using sin 37° = 0.6, what is the component of the block’s weight down the ramp?

  • (A) 39.2 N
  • (B) 47.04 N
  • (C) 15.68 N
  • (D) 23.52 N
Show Answer

Answer: D – Weight = mg = 4 × 9.8 = 39.2 N. Component down the ramp = mg sin Θ = 39.2 × 0.6 = 23.52 N. (Simplification: the ramp is frictionless, as stated.)

Q14. A 50 kg person stands on a scale in an elevator that accelerates upward at 2 m/s2. What does the scale read, in newtons?

  • (A) 490 N
  • (B) 390 N
  • (C) 590 N
  • (D) 100 N
Show Answer

Answer: C – Apparent weight = m(g + a) = 50 × (9.8 + 2) = 50 × 11.8 = 590 N.

Q15. Two blocks of masses 2 kg and 3 kg are in contact on a frictionless table. A horizontal force of 20 N pushes the pair. What is the acceleration of the pair?

  • (A) 10 m/s2
  • (B) 4 m/s2
  • (C) 6.67 m/s2
  • (D) 8 m/s2
Show Answer

Answer: B – Treat the pair as one mass: total mass = 2 + 3 = 5 kg. a = F ÷ m = 20 ÷ 5 = 4 m/s2. (Simplification: the table is frictionless, as stated.)

Q16. A 5 kg crate is pulled horizontally with a 50 N force. The coefficient of kinetic friction is 0.4. What is the crate’s acceleration?

  • (A) 6.08 m/s2
  • (B) 10 m/s2
  • (C) 3.92 m/s2
  • (D) 13.92 m/s2
Show Answer

Answer: A – Friction = μmg = 0.4 × 5 × 9.8 = 19.6 N. Net force = 50 – 19.6 = 30.4 N. a = Fnet ÷ m = 30.4 ÷ 5 = 6.08 m/s2.

Unit 3: Work, Energy, and Power

Q17. A student pushes a box with a constant horizontal force of 40 N through a distance of 3 m. How much work is done by the push?

  • (A) 120 J
  • (B) 43 J
  • (C) 13.33 J
  • (D) 240 J
Show Answer

Answer: A – W = Fd = 40 × 3 = 120 J.

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

  • (A) 20 J
  • (B) 100 J
  • (C) 25 J
  • (D) 50 J
Show Answer

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

Q19. A 5 kg cart starts from rest. A net work of 100 J is done on it. What is its final speed?

  • (A) 20 m/s
  • (B) 4.47 m/s
  • (C) 6.32 m/s
  • (D) 40 m/s
Show Answer

Answer: C – Work-energy: W = ΔKE = 0.5mv2 – 0, so v = √(2W ÷ m) = √(200 ÷ 5) = √40 = 6.32 m/s.

Q20. A 2 kg textbook is lifted 10 m above the floor. By how much does its gravitational potential energy increase?

  • (A) 20 J
  • (B) 196 J
  • (C) 19.6 J
  • (D) 98 J
Show Answer

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

Q21. A motor lifts a 10 kg load 5 m in 2 s at constant speed. What is the average power output of the motor?

  • (A) 490 W
  • (B) 980 W
  • (C) 245 W
  • (D) 122.5 W
Show Answer

Answer: C – Work = mgh = 10 × 9.8 × 5 = 490 J. Power = W ÷ t = 490 ÷ 2 = 245 W.

Q22. A spring with spring constant 50 N/m is compressed 0.4 m from equilibrium. How much elastic potential energy is stored?

  • (A) 20 J
  • (B) 4 J
  • (C) 8 J
  • (D) 6 J
Show Answer

Answer: B – PE = 0.5kx2 = 0.5 × 50 × 0.42 = 0.5 × 50 × 0.16 = 4 J.

Q23. A ball is dropped from rest from a height of 10 m. Ignoring air resistance, what is its speed just before it hits the ground?

  • (A) 14 m/s
  • (B) 98 m/s
  • (C) 9.9 m/s
  • (D) 7 m/s
Show Answer

Answer: A – Energy: mgh = 0.5mv2, so v = √(2gh) = √(2 × 9.8 × 10) = √196 = 14 m/s. (Simplification: air resistance is ignored, as stated.)

Q24. A tractor pulls a trailer with a constant force of 50 N while moving at a constant speed of 6 m/s. What power is delivered by the pulling force?

  • (A) 8.33 W
  • (B) 56 W
  • (C) 600 W
  • (D) 300 W
Show Answer

Answer: D – P = Fv = 50 × 6 = 300 W.

Unit 4: Linear Momentum

Q25. What is the magnitude of the momentum of a 2 kg bowling ball rolling at 9 m/s?

  • (A) 18 kg·m/s
  • (B) 11 kg·m/s
  • (C) 162 kg·m/s
  • (D) 9 kg·m/s
Show Answer

Answer: A – p = mv = 2 × 9 = 18 kg·m/s.

Q26. A constant force of 60 N acts on a puck for 0.2 s. What is the impulse delivered to the puck?

  • (A) 300 N·s
  • (B) 12 N·s
  • (C) 60.2 N·s
  • (D) 24 N·s
Show Answer

Answer: B – J = FΔt = 60 × 0.2 = 12 N·s.

Q27. A 5 kg tennis ball changes velocity by 4 m/s during a 0.4 s contact with a racket. What is the average force exerted by the racket?

  • (A) 20 N
  • (B) 8 N
  • (C) 50 N
  • (D) 25 N
Show Answer

Answer: C – Impulse = change in momentum = mΔv = 5 × 4 = 20 kg·m/s. Average force = impulse ÷ time = 20 ÷ 0.4 = 50 N.

Q28. A 6 kg cart moving at 4 m/s collides with a stationary 4 kg cart and the two stick together. What is their common velocity after the collision? (Ignore friction.)

  • (A) 4 m/s
  • (B) 6 m/s
  • (C) 4.8 m/s
  • (D) 2.4 m/s
Show Answer

Answer: D – Momentum: m1v1 = (m1 + m2)vf, so vf = (6 × 4) ÷ (6 + 4) = 24 ÷ 10 = 2.4 m/s. (Simplification: friction is ignored, as stated.)

Q29. Two trolleys move in the same direction and stick together on collision. The first has mass 4 kg and velocity 5 m/s; the second has mass 1 kg and velocity 5 m/s. What is their velocity after the collision? (Ignore friction.)

  • (A) 10 m/s
  • (B) 7.5 m/s
  • (C) 5 m/s
  • (D) 3.75 m/s
Show Answer

Answer: C – Total momentum before = 4 × 5 + 1 × 5 = 20 + 5 = 25 kg·m/s. Total mass = 5 kg, so vf = 25 ÷ 5 = 5 m/s. (Simplification: friction is ignored, as stated.)

Q30. Two skaters stand at rest on frictionless ice and push each other apart. The first skater has mass 40 kg and moves away at 3 m/s. The second skater has mass 20 kg. How fast does the second skater move?

  • (A) 6 m/s
  • (B) 3 m/s
  • (C) 1.5 m/s
  • (D) 12 m/s
Show Answer

Answer: A – Total momentum stays zero, so mAvA = mBvB. vB = (40 × 3) ÷ 20 = 120 ÷ 20 = 6 m/s. (Simplification: the ice is frictionless, as stated.)


Q31. A 0.4 kg rubber ball hits a wall at 5 m/s and rebounds at 5 m/s in the opposite direction. What is the magnitude of the change in the ball’s momentum?

  • (A) 2 kg·m/s
  • (B) 0 kg·m/s
  • (C) 6 kg·m/s
  • (D) 4 kg·m/s
Show Answer

Answer: D – Take the rebound direction as positive: Δp = m(vf – vi) = 0.4 × (5 – (-5)) = 0.4 × 10 = 4 kg·m/s in magnitude.

Q32. Two objects move in the same direction. One has mass 4 kg and velocity 2 m/s; the other has mass 1 kg and velocity 5 m/s. What is the total momentum of the system?

  • (A) 8 kg·m/s
  • (B) 13 kg·m/s
  • (C) 5 kg·m/s
  • (D) 18 kg·m/s
Show Answer

Answer: B – Total p = m1v1 + m2v2 = (4 × 2) + (1 × 5) = 8 + 5 = 13 kg·m/s.

Unit 5: Torque and Rotational Dynamics

Q33. A wrench handle is 0.3 m long. A force of 60 N is applied perpendicular to the handle at its end. What is the torque about the bolt?

  • (A) 200 N·m
  • (B) 18 N·m
  • (C) 60.3 N·m
  • (D) 36 N·m
Show Answer

Answer: B – τ = rF sin Θ with Θ = 90°, so τ = 0.3 × 60 = 18 N·m.

Q34. A force of 50 N is applied at a distance of 0.6 m from a pivot, at an angle of 53° to the lever arm. Using sin 53° = 0.8, what is the torque about the pivot?

  • (A) 30 N·m
  • (B) 12 N·m
  • (C) 48 N·m
  • (D) 24 N·m
Show Answer

Answer: D – τ = rF sin Θ = 0.6 × 50 × 0.8 = 24 N·m.

Q35. A wheel starts with an angular velocity of 1 rad/s and undergoes a constant angular acceleration of 4 rad/s2. What is its angular velocity after 3 s?

  • (A) 13 rad/s
  • (B) 12 rad/s
  • (C) 5 rad/s
  • (D) 26 rad/s
Show Answer

Answer: A – ω = ω0 + αt = 1 + (4 × 3) = 1 + 12 = 13 rad/s.

Q36. A fan blade rotates with an initial angular velocity of 6 rad/s and a constant angular acceleration of 2 rad/s2. Through what angle does it turn in 2 s?

  • (A) 12 rad
  • (B) 20 rad
  • (C) 16 rad
  • (D) 4 rad
Show Answer

Answer: C – Θ = ω0t + 0.5αt2 = (6 × 2) + 0.5 × 2 × 22 = 12 + 4 = 16 rad.

Q37. A 4 kg point mass is attached to a light rod 0.3 m from the rotation axis. What is its moment of inertia about that axis? (Treat the mass as a point mass.)

  • (A) 1.2 kg·m2
  • (B) 0.72 kg·m2
  • (C) 4 kg·m2
  • (D) 0.36 kg·m2
Show Answer

Answer: D – I = mr2 = 4 × 0.32 = 4 × 0.09 = 0.36 kg·m2. (Simplification: the mass is treated as a point mass, as stated.)

Q38. A net torque of 20 N·m acts on a wheel with a moment of inertia of 4 kg·m2. What is the wheel’s angular acceleration?

  • (A) 80 rad/s2
  • (B) 0.2 rad/s2
  • (C) 5 rad/s2
  • (D) 10 rad/s2
Show Answer

Answer: C – α = τnet ÷ I = 20 ÷ 4 = 5 rad/s2.

Q39. A child of mass 20 kg sits 3 m from the pivot of a seesaw. A second child of mass 30 kg sits on the other side. Where must the second child sit for the seesaw to balance?

  • (A) 3 m from the pivot
  • (B) 2 m from the pivot
  • (C) 4.5 m from the pivot
  • (D) 4 m from the pivot
Show Answer

Answer: B – Balance needs m1d1 = m2d2, so d2 = (20 × 3) ÷ 30 = 60 ÷ 30 = 2 m from the pivot.

Unit 6: Energy and Momentum of Rotating Systems

Q40. A flywheel has a moment of inertia of 3 kg·m2 and rotates at 4 rad/s. What is its rotational kinetic energy?

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

Answer: A – KErot = 0.5Iω2 = 0.5 × 3 × 42 = 0.5 × 3 × 16 = 24 J.

Q41. What is the angular momentum of a spinning top with a moment of inertia of 5 kg·m2 and an angular velocity of 2 rad/s?

  • (A) 10 kg·m2/s
  • (B) 7 kg·m2/s
  • (C) 2.5 kg·m2/s
  • (D) 20 kg·m2/s
Show Answer

Answer: A – L = Iω = 5 × 2 = 10 kg·m2/s.

Q42. An ice skater spins with a moment of inertia of 4 kg·m2 at 6 rad/s. She pulls her arms in, reducing her moment of inertia to 8 kg·m2. With no external torque, what is her new angular velocity?

  • (A) 6 rad/s
  • (B) 12 rad/s
  • (C) 4.5 rad/s
  • (D) 3 rad/s
Show Answer

Answer: D – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = (4 × 6) ÷ 8 = 24 ÷ 8 = 3 rad/s.

Q43. A 1 kg ball moves in a circle of radius 0.5 m at a speed of 6 m/s. What is its angular momentum about the centre of the circle?

  • (A) 6 kg·m2/s
  • (B) 0.5 kg·m2/s
  • (C) 3 kg·m2/s
  • (D) 4.5 kg·m2/s
Show Answer

Answer: C – L = mvr = 1 × 6 × 0.5 = 3 kg·m2/s.

Q44. A wheel with a moment of inertia of 6 kg·m2 is accelerated from rest to 2 rad/s. How much rotational work was done on it?

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

Answer: B – Work = change in rotational kinetic energy = 0.5Iω2 – 0 = 0.5 × 6 × 22 = 12 J.

Q45. A cylinder of mass 2 kg rolls without slipping at 4 m/s. Its radius is 0.4 m and its moment of inertia is 1 kg·m2. What is its total kinetic energy?

  • (A) 16 J
  • (B) 50 J
  • (C) 66 J
  • (D) 132 J
Show Answer

Answer: C – ω = v ÷ r = 4 ÷ 0.4 = 10 rad/s. Translational KE = 0.5mv2 = 16 J. Rotational KE = 0.5Iω2 = 50 J. Total = 16 + 50 = 66 J. (Simplification: the cylinder rolls without slipping, as stated.)

Q46. A disk with a moment of inertia of 2 kg·m2 slows from 8 rad/s to 4 rad/s. By how much does its angular momentum change in magnitude?

  • (A) 16 kg·m2/s
  • (B) 8 kg·m2/s
  • (C) 4 kg·m2/s
  • (D) 12 kg·m2/s
Show Answer

Answer: B – ΔL = IΔω = 2 × (8 – 4) = 2 × 4 = 8 kg·m2/s in magnitude.

Unit 7: Oscillations

Q47. A mass of 2 kg hangs on a spring with spring constant 200 N/m. What is the period of its vertical oscillation?

  • (A) 0.63 s
  • (B) 62.83 s
  • (C) 0.1 s
  • (D) 1.26 s
Show Answer

Answer: A – T = 2π√(m ÷ k) = 2π × √(2 ÷ 200) = 2π × 0.1 = 0.63 s.

Q48. A simple pendulum has a length of 4 m. What is its period for small oscillations on Earth? (Treat it as a simple pendulum.)

  • (A) 9.83 s
  • (B) 0.64 s
  • (C) 8.03 s
  • (D) 4.01 s
Show Answer

Answer: D – T = 2π√(L ÷ g) = 2π × √(4 ÷ 9.8) = 2π × 0.64 = 4.01 s. (Simplification: the pendulum is treated as simple and oscillations are small, as stated.)

Q49. An oscillator completes one full cycle in 0.5 s. What is its frequency?

  • (A) 2 Hz
  • (B) 0.5 Hz
  • (C) 12.57 Hz
  • (D) 4 Hz
Show Answer

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

Q50. A block in simple harmonic motion oscillates with frequency 3 Hz and amplitude 0.05 m. What is its maximum speed?

  • (A) 18.85 m/s
  • (B) 0.94 m/s
  • (C) 0.15 m/s
  • (D) 1.88 m/s
Show Answer

Answer: B – ω = 2πf = 2π × 3 = 18.85 rad/s. vmax = ωA = 18.85 × 0.05 = 0.94 m/s.

Q51. A spring-mass system has spring constant 400 N/m and oscillates with amplitude 0.1 m. What is its total mechanical energy?

  • (A) 40 J
  • (B) 4 J
  • (C) 2 J
  • (D) 3 J
Show Answer

Answer: C – E = 0.5kA2 = 0.5 × 400 × 0.12 = 0.5 × 400 × 0.01 = 2 J.

Q52. A simple pendulum in a university lab on Earth has a period of 1 s. What is its length? (Treat it as a simple pendulum.)

  • (A) 9.8 m
  • (B) 0.1 m
  • (C) 0.5 m
  • (D) 0.25 m
Show Answer

Answer: D – T = 2π√(L ÷ g) gives L = g(T ÷ 2π)2 = 9.8 × (1 ÷ 6.28)2 = 9.8 × 0.03 = 0.25 m. (Simplification: the pendulum is treated as simple, as stated.)

Q53. A 4 kg mass oscillates on a spring with a period of 1 s. What is the spring constant?

  • (A) 4 N/m
  • (B) 78.96 N/m
  • (C) 157.91 N/m
  • (D) 236.87 N/m
Show Answer

Answer: C – T = 2π√(m ÷ k) gives k = m(2π ÷ T)2 = 4 × (6.28)2 = 157.91 N/m.

Unit 8: Fluids

Q54. A metal block has a mass of 100 kg and a volume of 0.1 m3. What is its density?

  • (A) 1000 kg/m3
  • (B) 10 kg/m3
  • (C) 0 kg/m3
  • (D) 500 kg/m3
Show Answer

Answer: A – ρ = m ÷ V = 100 ÷ 0.1 = 1000 kg/m3.

Q55. A force of 100 N is spread evenly over an area of 0.25 m2. What is the pressure exerted?

  • (A) 25 Pa
  • (B) 0 Pa
  • (C) 800 Pa
  • (D) 400 Pa
Show Answer

Answer: D – P = F ÷ A = 100 ÷ 0.25 = 400 Pa.

Q56. What is the gauge pressure at a depth of 5 m in a liquid of density 800 kg/m3?

  • (A) 4000 Pa
  • (B) 39200 Pa
  • (C) 7840 Pa
  • (D) 19600 Pa
Show Answer

Answer: B – Gauge pressure = ρgh = 800 × 9.8 × 5 = 39200 Pa.

Q57. A diver is 20 m below the surface of water (density 1000 kg/m3). Taking atmospheric pressure as 101,000 Pa, what is the absolute pressure on the diver?

  • (A) 196000 Pa
  • (B) 297000 Pa
  • (C) 101000 Pa
  • (D) 594000 Pa
Show Answer

Answer: B – Gauge pressure = ρgh = 1000 × 9.8 × 20 = 196000 Pa. Absolute pressure = 101,000 + 196000 = 297000 Pa.

Q58. A block displaces 0.1 m3 of liquid of density 800 kg/m3 when fully submerged. What is the buoyant force on the block?

  • (A) 80 N
  • (B) 0.98 N
  • (C) 392 N
  • (D) 784 N
Show Answer

Answer: D – Buoyant force = weight of displaced liquid = ρVg = 800 × 0.1 × 9.8 = 784 N.

Q59. Water flows through a pipe of cross-sectional area 0.05 m2 at 1 m/s. The pipe narrows to an area of 0.01 m2. What is the water speed in the narrow section? (Assume incompressible flow.)

  • (A) 5 m/s
  • (B) 0.2 m/s
  • (C) 1 m/s
  • (D) 10 m/s
Show Answer

Answer: A – Continuity: A1v1 = A2v2, so v2 = (0.05 × 1) ÷ 0.01 = 0.05 ÷ 0.01 = 5 m/s. (Simplification: the flow is incompressible, as stated.)

Q60. Water flows through a hose of cross-sectional area 0.01 m2 at 4 m/s. What is the volume flow rate?

  • (A) 0 m3/s
  • (B) 4 m3/s
  • (C) 0.02 m3/s
  • (D) 0.04 m3/s
Show Answer

Answer: C – Volume flow rate = Av = 0.01 × 4 = 0.02 m3/s.

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