All subjects · Mathematics · WME01 (M1)

When the forces change part-way through

The specification includes motion where a force changes from one value to another, such as a particle hitting soft ground, a string breaking, or an engine being switched off. Split the motion into stages.

Stages again

Each stage has its own forces, so its own acceleration from F = ma. The velocity at the end of one stage is the starting velocity of the next.

Hitting the ground

A particle falls under gravity, then sinks into soft ground against a constant resistance R. In the ground: resultant upward force = R - mg, which causes the deceleration. So R = m(a + g), where a is the size of the deceleration.

A string breaks or an engine stops

Once the string breaks, its tension disappears: a particle moving up keeps rising under gravity alone until v = 0. When an engine is switched off, only the resistances act and the vehicle decelerates.

Worked example

A 2 kg ball is dropped from 5 m onto soft ground and sinks 0.14 m. Find the resistance of the ground.

  1. Speed on hitting: v2 = 2(9.8)(5) = 98.
  2. In the ground: 0 = 98 - 2a(0.14), so a = 350 m/s2 (deceleration).
  3. R - 2g = 2a, so R = 720 N.

Answer: R = 720 N

Key idea: New forces mean a new acceleration. Link stages by velocity. In the ground, R - mg = ma (deceleration a).

Check yourself

  1. A pile driver of mass 500 kg falls 2.5 m from rest onto a post. Find its speed as it hits the post.
    Show answer

    7 m/s. v2 = 2(9.8)(2.5) = 49, so v = 7 m/s.

  2. A 0.5 kg stone hits soft mud at 14 m/s and comes to rest after sinking 0.35 m. Find the constant resistance of the mud, to 3 significant figures.
    Show answer

    145 N. Deceleration a = 196/0.7 = 280 m/s2. R - 0.5g = 0.5(280), so R = 145 N.

  3. A 3 kg particle is pulled vertically upwards from rest by a string of tension 39.4 N for 2 s, then the string breaks. Find its speed when the string breaks.
    Show answer

    6.67 m/s. a = (39.4 - 29.4)/3 = 10/3 m/s2, so v = 2(10/3) = 6.67 m/s (3 s.f.).

  4. After the string breaks, how much further does the 3 kg particle rise, to 3 significant figures?
    Show answer

    2.27 m. 0 = (20/3)2 - 2(9.8)s, so s = 2.27 m.

  5. A car of mass 900 kg is moving at 20 m/s when the engine is switched off. A constant resistance of 450 N acts. Find the distance it travels before stopping.
    Show answer

    400 m. Deceleration = 450/900 = 0.5 m/s2. 0 = 400 - 2(0.5)s, so s = 400 m.

  6. A ball falls and sinks into sand. While it is sinking and slowing down, how does the resistance R compare with its weight?
    • R equals the weight
    • R is greater than the weight
    • R is less than the weight
    • R is zero
    Show answer

    R is greater than the weight. To slow the ball, the resultant force must be upwards, so R > mg.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.