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Option C: The Physics of Sports

Applying the impulse equation in sport

  • 3 things to know
  • A2 Unit 4 Section B

Overview

Recall questions on Applying the impulse equation in sport.

What you need to know

Answer each one before you reveal it.

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Key term

Write Newton's second law in the impulse form and define each symbol with its unit.

Key term

What does the area under a force against time graph represent for a body being struck?

Key term

Why does a squash ball bounce much higher after a few minutes of play than when it is first taken out of the box?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Write Newton's second law in the impulse form and define each symbol with its unit.

    Model answer

    The law is Ft = mv − mu, where F is the average resultant force in N, t the time for which it acts in s, m the mass in kg, u the initial velocity in m s⁻¹ and v the final velocity in m s⁻¹.

  2. 2

    What does the area under a force against time graph represent for a body being struck?

    Model answer

    The area represents the impulse of the force. Impulse is equal to the change in momentum of the body, measured in kg m s⁻¹.

  3. 3

    A cricket ball of mass 0.16 kg travelling at 30 m s⁻¹ is caught and brought to rest in 0.12 s. Calculate the average force on the ball.

    Model answer

    The change in momentum is 0.16 × 30 = 4.8 kg m s⁻¹, so F = 4.8 ÷ 0.12 = 40 N. Substituting back, 40 × 0.12 = 4.8 kg m s⁻¹.

  4. 4

    A cricket ball of mass 0.16 kg travelling at 30 m s⁻¹ is stopped by rigid hands in only 0.030 s. Calculate the average force and compare it with a 0.12 s catch.

    Model answer

    The change in momentum is still 4.8 kg m s⁻¹, so F = 4.8 ÷ 0.030 = 160 N. That is four times the 40 N needed when the catch takes 0.12 s, because the time is four times shorter.

  5. 5

    Why does a fielder pull the hands back while taking a fast catch?

    Model answer

    The change in momentum of the ball is fixed by the speed at which it arrives, so lengthening the time over which that change happens is the only way to reduce the force. Pulling the hands back stretches the contact time and cuts the peak force on the fingers.

  6. 6

    A tennis ball of mass 0.058 kg arrives at 25 m s⁻¹ and leaves in the opposite direction at 35 m s⁻¹. The contact lasts 5.0 ms. Calculate the average force from the racket.

    Model answer

    Taking the outgoing direction as positive, the change in momentum is 0.058 × (35 + 25) = 3.48 kg m s⁻¹, so F = 3.48 ÷ 5.0 × 10⁻³ = 696 N in the outgoing direction. Substituting back, 696 × 0.0050 = 3.48 kg m s⁻¹.

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