AS Level Mathematics chapter 2: Mechanics MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

40 exam-style questions on Mechanics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

40 questions · 5 topics · 10 per test

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What chapter 2 covers

Ten Mechanics questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. Particles of mass 3 kg and 2 kg are connected by a light inextensible string over a smooth fixed pulley and released from rest. Find the tension in the string. (g = 10 m s−2)

  1. A
    20 N
  2. B
    30 N
  3. C
    24 N
  4. D
    25 N
Show answer

Answer: C. Whole system: 30 − 20 = 5a, so a = 2 m s−2. For the 2 kg mass: T − 20 = 2 × 2, so T = 24 N. The tension lies between the two weights.

2. A car of mass 1200 kg tows a trailer of mass 300 kg on a level road using a light rigid tow-bar. The driving force is 3600 N and the resistances are 300 N on the car and 300 N on the trailer. Find the tension in the tow-bar.

  1. A
    600 N
  2. B
    900 N
  3. C
    1200 N
  4. D
    3000 N
Show answer

Answer: B. Whole system: 3600 − 600 = 1500a, so a = 2 m s−2. Trailer alone: T − 300 = 300 × 2, so T = 900 N. Forgetting the trailer resistance gives 600 N.

3. A block of mass 5 kg rests in limiting equilibrium on a rough plane inclined at 30° to the horizontal. Find the coefficient of friction, to 3 significant figures. (g = 10 m s−2)

  1. A
    0.577
  2. B
    0.5
  3. C
    0.866
  4. D
    1.73
Show answer

Answer: A. Along the plane: F = 50 sin 30° = 25 N. Perpendicular: R = 50 cos 30° = 43.3 N. Limiting, so μ = F/R = 25/43.3 = 0.577 (= tan 30°). Using R = 50 N gives the wrong 0.5.

4. A car accelerates uniformly from 5 m s−1 to 20 m s−1 in 6 s. Find its acceleration.

  1. A
    3.33 m s−2
  2. B
    2.5 m s−2
  3. C
    15 m s−2
  4. D
    0.4 m s−2
Show answer

Answer: B. v = u + at: 20 = 5 + 6a, so a = 15/6 = 2.5 m s−2. Use the change in velocity, not the final velocity.

5. A box of mass 4 kg is pulled along a rough horizontal floor by a horizontal force of 20 N. The frictional force is 8 N. Find the acceleration of the box.

  1. A
    5 m s−2
  2. B
    3 m s−2
  3. C
    2 m s−2
  4. D
    7 m s−2
Show answer

Answer: B. Resultant force = 20 − 8 = 12 N. F = ma gives 12 = 4a, so a = 3 m s−2. Use the net force, not the pulling force alone.

6. A car of mass 1000 kg climbs a hill inclined at angle α to the horizontal, where sin α = 1/20, at a constant speed of 20 m s−1. The resistance is 300 N. Find the power output of the engine. (g = 10 m s−2)

  1. A
    6 kW
  2. B
    16 kW
  3. C
    10 kW
  4. D
    22 kW
Show answer

Answer: B. Constant speed: driving force = resistance + component of weight down the slope = 300 + 10 000 × (1/20) = 800 N. Power = 800 × 20 = 16 000 W = 16 kW. Leaving out the weight component gives 6 kW.

7. A particle of mass 0.5 kg is projected vertically upwards. Air resistance of constant magnitude 1 N acts on it. Find the magnitude of its deceleration while it is moving upwards. (g = 10 m s−2)

  1. A
    8 m s−2
  2. B
    10 m s−2
  3. C
    12 m s−2
  4. D
    2 m s−2
Show answer

Answer: C. While it rises, the weight (5 N) and the resistance (1 N) both act downwards: 5 + 1 = 0.5a, so a = 12 m s−2. On the way down the resistance acts upwards and the acceleration is (5 − 1)/0.5 = 8 m s−2.

8. A particle moves in a straight line with velocity v = 6t − 3t2 m s−1 at time t seconds. Find the total distance travelled from t = 0 to t = 3.

  1. A
    8 m
  2. B
    0 m
  3. C
    4 m
  4. D
    12 m
Show answer

Answer: A. v = 3t(2 − t) changes sign at t = 2. Displacement from 0 to 2: [3t2 − t3] = 4 m; from 2 to 3: (27 − 27) − 4 = −4 m. Total distance 4 + 4 = 8 m; the single integral from 0 to 3 gives the displacement 0.

9. A block of mass 3 kg rests in equilibrium on a rough horizontal surface while a horizontal force of 12 N acts on it. Find the magnitude of the total contact force exerted by the surface on the block, to 3 significant figures. (g = 10 m s−2)

  1. A
    32.3 N
  2. B
    30 N
  3. C
    42 N
  4. D
    27.5 N
Show answer

Answer: A. The contact force has a normal component R = 30 N and a frictional component F = 12 N (balancing the applied force). Magnitude = √(302 + 122) = √1044 = 32.3 N. The components are at right angles, so adding them to 42 N is wrong.

10. Find the kinetic energy of a car of mass 800 kg moving at 15 m s−1.

  1. A
    180 kJ
  2. B
    90 kJ
  3. C
    6 kJ
  4. D
    12 kJ
Show answer

Answer: B. KE = (1/2)mv2 = (1/2) × 800 × 225 = 90 000 J = 90 kJ. Leaving out the 1/2 gives 180 kJ.

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How many Mechanics MCQs are there?

40, each with the answer and a short explanation, across 5 topics: Forces and equilibrium, Kinematics of motion in a straight line, Momentum, Newton’s laws of motion, Energy, work and power. Every chapter test draws ten of them, so you can take it many times.

Is the AS Level Mathematics chapter 2 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

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