Newton’s laws of motion

AS Level Mathematics · Mechanics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Newton’s second law links force and motion: the resultant force on a particle equals its mass multiplied by its acceleration. If the resultant force is zero, the particle is at rest or moves with constant velocity.

Draw a clear force diagram. Typical forces are weight (mg, downwards), normal reaction, friction (opposing motion), tension and thrust. Resolve along the direction of motion and write resultant force = ma.

For connected particles joined by a light inextensible string or a rigid tow-bar, both have the same size of acceleration and the tension is the same at both ends. Write one equation for each particle, or treat the system as a whole.

Rules to remember

Common mistake

Writing F = ma with only one force instead of the resultant. Subtract every force that opposes the motion (friction, resistance, weight component) before setting the result equal to ma.

Worked example

Particle A of mass 3 kg rests on a smooth horizontal table. A light inextensible string joins A to particle B of mass 2 kg, which hangs freely over a smooth pulley at the edge of the table. The system is released. Find the tension in the string. (g = 10 m s−2)

  1. For B (downwards): 2 × 10 − T = 2a, so 20 − T = 2a.
  2. For A (along the table): T = 3a.
  3. Add the equations: 20 = 5a, so a = 4 m s−2.
  4. T = 3a = 3 × 4 = 12.

Answer: 12 N

Practice questions

Try each one, then open the answer.

1. 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.

2. 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.

3. 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.

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