A Level Further Maths chapter 3: Further Mechanics MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

155 exam-style questions on Further 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.

155 questions · 6 topics · 10 per test

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What chapter 3 covers

Ten Further Mechanics questions with answers

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

1. A ball is projected from level ground with speed 24 m s−1 at 35° above the horizontal. Ignoring air resistance, the velocity of the ball as it returns to ground level is

  1. A
    19.7 m s−1 at 35° below the horizontal
  2. B
    24 m s−1 at 35° below the horizontal
  3. C
    24 m s−1 at 55° below the horizontal
  4. D
    19.7 m s−1 horizontally
Show answer

Answer: B. Energy is conserved, so the speed at the same level equals the launch speed, 24 m s−1. The horizontal component is unchanged and the vertical component is reversed, so the direction is 35° below the horizontal.

2. A particle moves in a circle of radius 0.40 m with a constant speed of 12 m s−1. Find its angular speed.

  1. A
    4.8 rad s−1
  2. B
    0.033 rad s−1
  3. C
    75.4 rad s−1
  4. D
    30 rad s−1
Show answer

Answer: D. v = rω gives ω = v/r = 12/0.40 = 30 rad s−1. The value 4.8 comes from multiplying instead of dividing.

3. A light elastic string of natural length 0.6 m stores 1.5 J of elastic potential energy when its extension is 0.3 m. Find its modulus of elasticity.

  1. A
    10 N
  2. B
    40 N
  3. C
    20 N
  4. D
    5 N
Show answer

Answer: C. E = λx2/(2l) gives λ = 2El/x2 = 2 × 1.5 × 0.6/0.09 = 20 N. Missing the 2 gives 10 N.

4. A ball is released from rest in a liquid and falls with acceleration (10 − 0.5v) m s−2 when its speed is v m s−1. Find the time taken to reach a speed of 10 m s−1.

  1. A
    0.693 s
  2. B
    1 s
  3. C
    1.39 s
  4. D
    2.77 s
Show answer

Answer: C. dv/dt = 10 − 0.5v, so ∫dv/(10 − 0.5v) = ∫dt, giving −2 ln(10 − 0.5v) = t + C. With v = 0 at t = 0: t = 2 ln(10/(10 − 0.5v)) = 2 ln 2 = 1.39 s. Using constant acceleration 10 gives 1.0 s.

5. A smooth sphere A of mass 2 kg moving at 5 m s−1 collides directly with a stationary smooth sphere B of mass 3 kg, with coefficient of restitution 0.4. The speeds after impact are 0.8 m s−1 and 2.8 m s−1 respectively. Find the kinetic energy lost in the impact.

  1. A
    12.4 J
  2. B
    25 J
  3. C
    12.6 J
  4. D
    13.2 J
Show answer

Answer: C. Before: ½ × 2 × 25 = 25 J. After: ½ × 2 × 0.64 + ½ × 3 × 7.84 = 0.64 + 11.76 = 12.4 J. Loss = 25 − 12.4 = 12.6 J. The value 12.4 J is the energy remaining.

6. A light elastic string of natural length 1.2 m is stretched by 0.3 m and the tension is 15 N. Find the modulus of elasticity of the string.

  1. A
    60 N
  2. B
    12 N
  3. C
    3.75 N
  4. D
    5.4 N
Show answer

Answer: A. From T = λx/l: λ = Tl/x = 15 × 1.2/0.3 = 60 N. Using the stretched length 1.5 m in place of the extension gives 12 N.

7. A particle moves along the x-axis so that its speed is v = x(4 − x) m s−1 when its displacement from the origin is x metres, for 0 < x < 4. Find its acceleration when x = 1.

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

Answer: B. a = v dv/dx with v = 4x − x2 and dv/dx = 4 − 2x. At x = 1: v = 3 and dv/dx = 2, so a = 3 × 2 = 6 m s−2. The value 3 is v and 2 is dv/dx: the product is needed.

8. ABCD is a square lamina of side 0.4 m. A force of 30 N acts along the diagonal AC. Find the moment of this force about the vertex B.

  1. A
    12 N m
  2. B
    6 N m
  3. C
    8.49 N m
  4. D
    17.0 N m
Show answer

Answer: C. B is at perpendicular distance BD/2 = 0.4/√2 = 0.283 m from line AC (half the diagonal 0.4√2). Moment = 30 × 0.283 = 8.49 N m. Using the side, half the side or the whole diagonal as the distance are wrong.

9. A particle of mass 2 kg starts from rest at the origin and moves along the x-axis under a force of 6t newtons at time t seconds. Find its displacement from the origin when t = 4.

  1. A
    32 m
  2. B
    24 m
  3. C
    64 m
  4. D
    96 m
Show answer

Answer: A. a = 6t/2 = 3t, so v = 1.5t2 (v = 0 at t = 0), and x = 0.5t3 (x = 0 at t = 0). At t = 4, x = 0.5 × 64 = 32 m. The value 24 m is the speed at that time; 64 m omits the division by the mass.

10. A light elastic string, natural length l, is extended by x and the tension is then T. Which expression gives the elastic potential energy stored?

  1. A
    ½Tx
  2. B
    Tx
  3. C
    T(l + x)/2
  4. D
    Tx2/(2l)
Show answer

Answer: A. Energy stored = work done = area under the tension–extension graph = ½ × T × x. The energy is not Tx: the tension increases from zero, so the average force is T/2.

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Further Mechanics: questions students ask

How many Further Mechanics MCQs are there?

155, each with the answer and a short explanation, across 6 topics: Motion of a projectile, Equilibrium of a rigid body, Circular motion, Hooke's law, Linear motion under a variable force, Momentum. Every chapter test draws ten of them, so you can take it many times.

Is the A Level Further Maths chapter 3 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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