AP Physics 1 Practice Test 5 – 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 electric scooter starts with a velocity of 5 m/s and accelerates uniformly at 3 m/s2 for 4 s. What is its final velocity? (Ignore air resistance.)

  • (A) 17 m/s
  • (B) 8 m/s
  • (C) 12 m/s
  • (D) 29 m/s
Show Answer

Answer: A – v = v0 + at = 5 + (3 × 4) = 5 + 12 = 17 m/s. (Simplification: air resistance is ignored, as stated.)

Q2. A skateboarder moves with an initial velocity of 8 m/s and a constant acceleration of 2 m/s2. How far does the skateboarder travel in 4 s?

  • (A) 32 m
  • (B) 48 m
  • (C) 64 m
  • (D) 16 m
Show Answer

Answer: B – x = v0t + 0.5at2 = (8 × 4) + 0.5 × 2 × 42 = 32 + 16 = 48 m.

Q3. A car travelling at 22 m/s brakes with a constant deceleration of 5 m/s2. How far does it travel before it stops?

  • (A) 96.8 m
  • (B) 4.4 m
  • (C) 48.4 m
  • (D) 24.2 m
Show Answer

Answer: C – v2 = v02 + 2aΔx with v = 0 gives 0 = 222 – 2 × 5 × Δx, so Δx = 484 ÷ 10 = 48.4 m.

Q4. A marble is dropped from rest from a rooftop edge. Ignoring air resistance, what is its velocity after 3 s?

  • (A) 14.7 m/s
  • (B) 88.2 m/s
  • (C) 9.8 m/s
  • (D) 29.4 m/s
Show Answer

Answer: D – v = v0 + gt = 0 + (9.8 × 3) = 29.4 m/s downward. (Simplification: air resistance is ignored, as stated.)

Q5. A golf ball is dropped from rest from the roof of a warehouse and hits the ground after 3 s. Ignoring air resistance, how tall is the warehouse?

  • (A) 29.4 m
  • (B) 88.2 m
  • (C) 44.1 m
  • (D) 14.7 m
Show Answer

Answer: C – h = 0.5gt2 = 0.5 × 9.8 × 32 = 0.5 × 9.8 × 9 = 44.1 m. (Simplification: air resistance is ignored, as stated.)

Q6. A jogger runs 200 m east in 20 s, then stops. What is the jogger’s average velocity for the run?

  • (A) 10 m/s east
  • (B) 4000 m/s east
  • (C) 0.1 m/s east
  • (D) 5 m/s east
Show Answer

Answer: A – Average velocity = displacement ÷ time = 200 ÷ 20 = 10 m/s east.

Q7. A marble rolls horizontally off a table 44.1 m high with a horizontal velocity of 14 m/s. Ignoring air resistance, how far from the base of the table does it land?

  • (A) 21 m
  • (B) 84 m
  • (C) 3 m
  • (D) 42 m
Show Answer

Answer: D – Vertical fall: t = √(2h ÷ g) = √(88.2 ÷ 9.8) = 3 s. Horizontal range = vx × t = 14 × 3 = 42 m. (Simplification: air resistance is ignored, as stated.)

Q8. A truck moving at 28 m/s brakes uniformly at 4 m/s2. How long does it take to come to rest?

  • (A) 112 s
  • (B) 7 s
  • (C) 3.5 s
  • (D) 4 s
Show Answer

Answer: B – v = v0 + at with v = 0 gives 0 = 28 – 4t, so t = 28 ÷ 4 = 7 s.

Unit 2: Force and Translational Dynamics

Q9. A net force acts on a 3 kg crate and gives it an acceleration of 6 m/s2. What is the magnitude of the net force?

  • (A) 9 N
  • (B) 18 N
  • (C) 0.5 N
  • (D) 36 N
Show Answer

Answer: B – Fnet = ma = 3 × 6 = 18 N.

Q10. What is the weight of a 20 kg backpack on Earth?

  • (A) 20 N
  • (B) 200 N
  • (C) 2.04 N
  • (D) 196 N
Show Answer

Answer: D – Weight = mg = 20 × 9.8 = 196 N.

Q11. A 8 kg box slides on a horizontal floor. The coefficient of kinetic friction is 0.5. What is the magnitude of the kinetic friction force?

  • (A) 39.2 N
  • (B) 4 N
  • (C) 78.4 N
  • (D) 19.6 N
Show Answer

Answer: A – Normal force N = mg = 8 × 9.8 = 78.4 N. Friction f = μN = 0.5 × 78.4 = 39.2 N.

Q12. A 8 kg sled is pulled to the right with a 70 N force while a 30 N friction force acts to the left. What is the sled’s acceleration?

  • (A) 12.5 m/s2 to the right
  • (B) 8.75 m/s2 to the right
  • (C) 5 m/s2 to the right
  • (D) 5 m/s2 to the left
Show Answer

Answer: C – Fnet = 70 – 30 = 40 N to the right. a = Fnet ÷ m = 40 ÷ 8 = 5 m/s2 to the right.

Q13. A 6 kg block sits on a frictionless ramp inclined at 53° to the horizontal. Using sin 53° = 0.8, what is the component of the block’s weight down the ramp?

  • (A) 58.8 N
  • (B) 94.08 N
  • (C) 11.76 N
  • (D) 47.04 N
Show Answer

Answer: D – Weight = mg = 6 × 9.8 = 58.8 N. Component down the ramp = mg sin Θ = 58.8 × 0.8 = 47.04 N. (Simplification: the ramp is frictionless, as stated.)

Q14. A 65 kg person stands on a scale in an elevator that accelerates upward at 2 m/s2. What does the scale read, in newtons?

  • (A) 637 N
  • (B) 507 N
  • (C) 767 N
  • (D) 130 N
Show Answer

Answer: C – Apparent weight = m(g + a) = 65 × (9.8 + 2) = 65 × 11.8 = 767 N.

Q15. Two blocks of masses 6 kg and 4 kg are in contact on a frictionless table. A horizontal force of 30 N pushes the pair. What is the acceleration of the pair?

  • (A) 5 m/s2
  • (B) 3 m/s2
  • (C) 7.5 m/s2
  • (D) 6 m/s2
Show Answer

Answer: B – Treat the pair as one mass: total mass = 6 + 4 = 10 kg. a = F ÷ m = 30 ÷ 10 = 3 m/s2. (Simplification: the table is frictionless, as stated.)

Q16. A 25 kg crate is pulled horizontally with a 120 N force. The coefficient of kinetic friction is 0.2. What is the crate’s acceleration?

  • (A) 2.84 m/s2
  • (B) 4.8 m/s2
  • (C) 1.96 m/s2
  • (D) 6.76 m/s2
Show Answer

Answer: A – Friction = μmg = 0.2 × 25 × 9.8 = 49 N. Net force = 120 – 49 = 71 N. a = Fnet ÷ m = 71 ÷ 25 = 2.84 m/s2.

Unit 3: Work, Energy, and Power

Q17. A student pushes a box with a constant horizontal force of 50 N through a distance of 4 m. How much work is done by the push?

  • (A) 200 J
  • (B) 54 J
  • (C) 12.5 J
  • (D) 400 J
Show Answer

Answer: A – W = Fd = 50 × 4 = 200 J.

Q18. What is the kinetic energy of a 1 kg ball moving at 10 m/s?

  • (A) 10 J
  • (B) 100 J
  • (C) 25 J
  • (D) 50 J
Show Answer

Answer: D – KE = 0.5mv2 = 0.5 × 1 × 102 = 0.5 × 1 × 100 = 50 J.

Q19. A 10 kg cart starts from rest. A net work of 80 J is done on it. What is its final speed?

  • (A) 8 m/s
  • (B) 2.83 m/s
  • (C) 4 m/s
  • (D) 16 m/s
Show Answer

Answer: C – Work-energy: W = ΔKE = 0.5mv2 – 0, so v = √(2W ÷ m) = √(160 ÷ 10) = √16 = 4 m/s.

Q20. A 4 kg textbook is lifted 5 m above the floor. By how much does its gravitational potential energy increase?

  • (A) 20 J
  • (B) 196 J
  • (C) 39.2 J
  • (D) 98 J
Show Answer

Answer: B – ΔPE = mgh = 4 × 9.8 × 5 = 196 J.

Q21. A motor lifts a 15 kg load 6 m in 3 s at constant speed. What is the average power output of the motor?

  • (A) 882 W
  • (B) 2646 W
  • (C) 294 W
  • (D) 147 W
Show Answer

Answer: C – Work = mgh = 15 × 9.8 × 6 = 882 J. Power = W ÷ t = 882 ÷ 3 = 294 W.

Q22. A spring with spring constant 80 N/m is compressed 0.5 m from equilibrium. How much elastic potential energy is stored?

  • (A) 40 J
  • (B) 10 J
  • (C) 20 J
  • (D) 15 J
Show Answer

Answer: B – PE = 0.5kx2 = 0.5 × 80 × 0.52 = 0.5 × 80 × 0.25 = 10 J.

Q23. A ball is dropped from rest from a height of 15 m. Ignoring air resistance, what is its speed just before it hits the ground?

  • (A) 17.15 m/s
  • (B) 147 m/s
  • (C) 12.12 m/s
  • (D) 8.57 m/s
Show Answer

Answer: A – Energy: mgh = 0.5mv2, so v = √(2gh) = √(2 × 9.8 × 15) = √294 = 17.15 m/s. (Simplification: air resistance is ignored, as stated.)

Q24. A tractor pulls a trailer with a constant force of 40 N while moving at a constant speed of 10 m/s. What power is delivered by the pulling force?

  • (A) 4 W
  • (B) 50 W
  • (C) 800 W
  • (D) 400 W
Show Answer

Answer: D – P = Fv = 40 × 10 = 400 W.

Unit 4: Linear Momentum

Q25. What is the magnitude of the momentum of a 4 kg bowling ball rolling at 7 m/s?

  • (A) 28 kg·m/s
  • (B) 11 kg·m/s
  • (C) 196 kg·m/s
  • (D) 14 kg·m/s
Show Answer

Answer: A – p = mv = 4 × 7 = 28 kg·m/s.

Q26. A constant force of 100 N acts on a puck for 0.3 s. What is the impulse delivered to the puck?

  • (A) 333.33 N·s
  • (B) 30 N·s
  • (C) 100.3 N·s
  • (D) 60 N·s
Show Answer

Answer: B – J = FΔt = 100 × 0.3 = 30 N·s.

Q27. A 10 kg tennis ball changes velocity by 3 m/s during a 0.6 s contact with a racket. What is the average force exerted by the racket?

  • (A) 30 N
  • (B) 18 N
  • (C) 50 N
  • (D) 25 N
Show Answer

Answer: C – Impulse = change in momentum = mΔv = 10 × 3 = 30 kg·m/s. Average force = impulse ÷ time = 30 ÷ 0.6 = 50 N.

Q28. A 5 kg cart moving at 6 m/s collides with a stationary 5 kg cart and the two stick together. What is their common velocity after the collision? (Ignore friction.)

  • (A) 6 m/s
  • (B) 4.5 m/s
  • (C) 2.25 m/s
  • (D) 3 m/s
Show Answer

Answer: D – Momentum: m1v1 = (m1 + m2)vf, so vf = (5 × 6) ÷ (5 + 5) = 30 ÷ 10 = 3 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 3 kg and velocity 10 m/s; the second has mass 2 kg and velocity 5 m/s. What is their velocity after the collision? (Ignore friction.)

  • (A) 7.5 m/s
  • (B) 10 m/s
  • (C) 8 m/s
  • (D) 16 m/s
Show Answer

Answer: C – Total momentum before = 3 × 10 + 2 × 5 = 30 + 10 = 40 kg·m/s. Total mass = 5 kg, so vf = 40 ÷ 5 = 8 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 55 kg and moves away at 2 m/s. The second skater has mass 22 kg. How fast does the second skater move?

  • (A) 5 m/s
  • (B) 2 m/s
  • (C) 0.8 m/s
  • (D) 10 m/s
Show Answer

Answer: A – Total momentum stays zero, so mAvA = mBvB. vB = (55 × 2) ÷ 22 = 110 ÷ 22 = 5 m/s. (Simplification: the ice is frictionless, as stated.)


Q31. A 0.3 kg rubber ball hits a wall at 10 m/s and rebounds at 10 m/s in the opposite direction. What is the magnitude of the change in the ball’s momentum?

  • (A) 3 kg·m/s
  • (B) 0 kg·m/s
  • (C) 9 kg·m/s
  • (D) 6 kg·m/s
Show Answer

Answer: D – Take the rebound direction as positive: Δp = m(vf – vi) = 0.3 × (10 – (-10)) = 0.3 × 20 = 6 kg·m/s in magnitude.

Q32. Two objects move in the same direction. One has mass 5 kg and velocity 3 m/s; the other has mass 5 kg and velocity 2 m/s. What is the total momentum of the system?

  • (A) 15 kg·m/s
  • (B) 25 kg·m/s
  • (C) 10 kg·m/s
  • (D) 30 kg·m/s
Show Answer

Answer: B – Total p = m1v1 + m2v2 = (5 × 3) + (5 × 2) = 15 + 10 = 25 kg·m/s.

Unit 5: Torque and Rotational Dynamics

Q33. A wrench handle is 0.2 m long. A force of 100 N is applied perpendicular to the handle at its end. What is the torque about the bolt?

  • (A) 500 N·m
  • (B) 20 N·m
  • (C) 100.2 N·m
  • (D) 40 N·m
Show Answer

Answer: B – τ = rF sin Θ with Θ = 90°, so τ = 0.2 × 100 = 20 N·m.

Q34. A force of 25 N is applied at a distance of 1.2 m from a pivot, at an angle of 30° to the lever arm. Using sin 30° = 0.5, what is the torque about the pivot?

  • (A) 30 N·m
  • (B) 7.5 N·m
  • (C) 22.5 N·m
  • (D) 15 N·m
Show Answer

Answer: D – τ = rF sin Θ = 1.2 × 25 × 0.5 = 15 N·m.

Q35. A wheel starts with an angular velocity of 0 rad/s and undergoes a constant angular acceleration of 5 rad/s2. What is its angular velocity after 3 s?

  • (A) 15 rad/s
  • (B) 5 rad/s
  • (C) 30 rad/s
  • (D) 22.5 rad/s
Show Answer

Answer: A – ω = ω0 + αt = 0 + (5 × 3) = 0 + 15 = 15 rad/s.

Q36. A fan blade rotates with an initial angular velocity of 5 rad/s and a constant angular acceleration of 2 rad/s2. Through what angle does it turn in 4 s?

  • (A) 20 rad
  • (B) 52 rad
  • (C) 36 rad
  • (D) 16 rad
Show Answer

Answer: C – Θ = ω0t + 0.5αt2 = (5 × 4) + 0.5 × 2 × 42 = 20 + 16 = 36 rad.

Q37. A 1 kg point mass is attached to a light rod 1 m from the rotation axis. What is its moment of inertia about that axis? (Treat the mass as a point mass.)

  • (A) 2 kg·m2
  • (B) 1.5 kg·m2
  • (C) 0.75 kg·m2
  • (D) 1 kg·m2
Show Answer

Answer: D – I = mr2 = 1 × 12 = 1 × 1 = 1 kg·m2. (Simplification: the mass is treated as a point mass, as stated.)

Q38. A net torque of 12 N·m acts on a wheel with a moment of inertia of 3 kg·m2. What is the wheel’s angular acceleration?

  • (A) 36 rad/s2
  • (B) 0.25 rad/s2
  • (C) 4 rad/s2
  • (D) 8 rad/s2
Show Answer

Answer: C – α = τnet ÷ I = 12 ÷ 3 = 4 rad/s2.

Q39. A child of mass 25 kg sits 2.4 m from the pivot of a seesaw. A second child of mass 15 kg sits on the other side. Where must the second child sit for the seesaw to balance?

  • (A) 2.4 m from the pivot
  • (B) 4 m from the pivot
  • (C) 1.44 m from the pivot
  • (D) 8 m from the pivot
Show Answer

Answer: B – Balance needs m1d1 = m2d2, so d2 = (25 × 2.4) ÷ 15 = 60 ÷ 15 = 4 m from the pivot.

Unit 6: Energy and Momentum of Rotating Systems

Q40. A flywheel has a moment of inertia of 6 kg·m2 and rotates at 3 rad/s. What is its rotational kinetic energy?

  • (A) 27 J
  • (B) 18 J
  • (C) 54 J
  • (D) 13.5 J
Show Answer

Answer: A – KErot = 0.5Iω2 = 0.5 × 6 × 32 = 0.5 × 6 × 9 = 27 J.

Q41. What is the angular momentum of a spinning top with a moment of inertia of 6 kg·m2 and an angular velocity of 5 rad/s?

  • (A) 30 kg·m2/s
  • (B) 11 kg·m2/s
  • (C) 1.2 kg·m2/s
  • (D) 60 kg·m2/s
Show Answer

Answer: A – L = Iω = 6 × 5 = 30 kg·m2/s.

Q42. An ice skater spins with a moment of inertia of 5 kg·m2 at 4 rad/s. She pulls her arms in, reducing her moment of inertia to 10 kg·m2. With no external torque, what is her new angular velocity?

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

Answer: D – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = (5 × 4) ÷ 10 = 20 ÷ 10 = 2 rad/s.

Q43. A 5 kg ball moves in a circle of radius 0.4 m at a speed of 2 m/s. What is its angular momentum about the centre of the circle?

  • (A) 10 kg·m2/s
  • (B) 2 kg·m2/s
  • (C) 4 kg·m2/s
  • (D) 8 kg·m2/s
Show Answer

Answer: C – L = mvr = 5 × 2 × 0.4 = 4 kg·m2/s.

Q44. A wheel with a moment of inertia of 5 kg·m2 is accelerated from rest to 6 rad/s. How much rotational work was done on it?

  • (A) 30 J
  • (B) 90 J
  • (C) 180 J
  • (D) 45 J
Show Answer

Answer: B – Work = change in rotational kinetic energy = 0.5Iω2 – 0 = 0.5 × 5 × 62 = 90 J.

Q45. A cylinder of mass 3 kg rolls without slipping at 5 m/s. Its radius is 0.5 m and its moment of inertia is 2 kg·m2. What is its total kinetic energy?

  • (A) 37.5 J
  • (B) 100 J
  • (C) 137.5 J
  • (D) 275 J
Show Answer

Answer: C – ω = v ÷ r = 5 ÷ 0.5 = 10 rad/s. Translational KE = 0.5mv2 = 37.5 J. Rotational KE = 0.5Iω2 = 100 J. Total = 37.5 + 100 = 137.5 J. (Simplification: the cylinder rolls without slipping, as stated.)

Q46. A disk with a moment of inertia of 3 kg·m2 slows from 10 rad/s to 4 rad/s. By how much does its angular momentum change in magnitude?

  • (A) 12 kg·m2/s
  • (B) 18 kg·m2/s
  • (C) 30 kg·m2/s
  • (D) 9 kg·m2/s
Show Answer

Answer: B – ΔL = IΔω = 3 × (10 – 4) = 3 × 6 = 18 kg·m2/s in magnitude.

Unit 7: Oscillations

Q47. A mass of 3 kg hangs on a spring with spring constant 300 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 × √(3 ÷ 300) = 2π × 0.1 = 0.63 s.

Q48. A simple pendulum has a length of 2 m. What is its period for small oscillations on Earth? (Treat it as a simple pendulum.)

  • (A) 13.91 s
  • (B) 0.45 s
  • (C) 5.68 s
  • (D) 2.84 s
Show Answer

Answer: D – T = 2π√(L ÷ g) = 2π × √(2 ÷ 9.8) = 2π × 0.45 = 2.84 s. (Simplification: the pendulum is treated as simple and oscillations are small, as stated.)

Q49. An oscillator completes one full cycle in 5 s. What is its frequency?

  • (A) 0.2 Hz
  • (B) 5 Hz
  • (C) 1.26 Hz
  • (D) 0.4 Hz
Show Answer

Answer: A – f = 1 ÷ T = 1 ÷ 5 = 0.2 Hz.

Q50. A block in simple harmonic motion oscillates with frequency 4 Hz and amplitude 0.1 m. What is its maximum speed?

  • (A) 25.13 m/s
  • (B) 2.51 m/s
  • (C) 0.4 m/s
  • (D) 5.03 m/s
Show Answer

Answer: B – ω = 2πf = 2π × 4 = 25.13 rad/s. vmax = ωA = 25.13 × 0.1 = 2.51 m/s.

Q51. A spring-mass system has spring constant 150 N/m and oscillates with amplitude 0.2 m. What is its total mechanical energy?

  • (A) 30 J
  • (B) 6 J
  • (C) 3 J
  • (D) 4.5 J
Show Answer

Answer: C – E = 0.5kA2 = 0.5 × 150 × 0.22 = 0.5 × 150 × 0.04 = 3 J.

Q52. A simple pendulum in a physics workshop on Earth has a period of 3 s. What is its length? (Treat it as a simple pendulum.)

  • (A) 88.2 m
  • (B) 0.92 m
  • (C) 4.47 m
  • (D) 2.23 m
Show Answer

Answer: D – T = 2π√(L ÷ g) gives L = g(T ÷ 2π)2 = 9.8 × (3 ÷ 6.28)2 = 9.8 × 0.23 = 2.23 m. (Simplification: the pendulum is treated as simple, as stated.)

Q53. A 0.5 kg mass oscillates on a spring with a period of 0.5 s. What is the spring constant?

  • (A) 1 N/m
  • (B) 0.25 N/m
  • (C) 78.96 N/m
  • (D) 39.48 N/m
Show Answer

Answer: C – T = 2π√(m ÷ k) gives k = m(2π ÷ T)2 = 0.5 × (12.57)2 = 78.96 N/m.

Unit 8: Fluids

Q54. A metal block has a mass of 250 kg and a volume of 0.25 m3. What is its density?

  • (A) 1000 kg/m3
  • (B) 62.5 kg/m3
  • (C) 0 kg/m3
  • (D) 500 kg/m3
Show Answer

Answer: A – ρ = m ÷ V = 250 ÷ 0.25 = 1000 kg/m3.

Q55. A force of 300 N is spread evenly over an area of 1.5 m2. What is the pressure exerted?

  • (A) 450 Pa
  • (B) 0.01 Pa
  • (C) 400 Pa
  • (D) 200 Pa
Show Answer

Answer: D – P = F ÷ A = 300 ÷ 1.5 = 200 Pa.

Q56. What is the gauge pressure at a depth of 4 m in a liquid of density 1200 kg/m3?

  • (A) 4800 Pa
  • (B) 47040 Pa
  • (C) 11760 Pa
  • (D) 23520 Pa
Show Answer

Answer: B – Gauge pressure = ρgh = 1200 × 9.8 × 4 = 47040 Pa.

Q57. A diver is 2 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) 19600 Pa
  • (B) 120600 Pa
  • (C) 101000 Pa
  • (D) 241200 Pa
Show Answer

Answer: B – Gauge pressure = ρgh = 1000 × 9.8 × 2 = 19600 Pa. Absolute pressure = 101,000 + 19600 = 120600 Pa.

Q58. A block displaces 0.04 m3 of liquid of density 900 kg/m3 when fully submerged. What is the buoyant force on the block?

  • (A) 36 N
  • (B) 0.39 N
  • (C) 176.4 N
  • (D) 352.8 N
Show Answer

Answer: D – Buoyant force = weight of displaced liquid = ρVg = 900 × 0.04 × 9.8 = 352.8 N.

Q59. Water flows through a pipe of cross-sectional area 0.03 m2 at 4 m/s. The pipe narrows to an area of 0.06 m2. What is the water speed in the narrow section? (Assume incompressible flow.)

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

Answer: A – Continuity: A1v1 = A2v2, so v2 = (0.03 × 4) ÷ 0.06 = 0.12 ÷ 0.06 = 2 m/s. (Simplification: the flow is incompressible, as stated.)

Q60. Water flows through a hose of cross-sectional area 0.02 m2 at 5 m/s. What is the volume flow rate?

  • (A) 0 m3/s
  • (B) 5.02 m3/s
  • (C) 0.1 m3/s
  • (D) 0.2 m3/s
Show Answer

Answer: C – Volume flow rate = Av = 0.02 × 5 = 0.1 m3/s.

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 5

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