A tennis player practices against a wall, hitting a 0.1 kg ball towards the wall with a velocity of +20 m/s. The ball bounces straight

Question

A tennis player practices against a wall, hitting a 0.1 kg ball towards the
wall with a velocity of +20 m/s. The ball bounces straight back from the
wall with a velocity of -20 m/s. What is the average force the wall exerts on
the ball if the ball is in contact with the wall for 0.05 s?*

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Hồng Cúc 4 years 2021-08-11T18:14:13+00:00 1 Answers 11 views 0

Answers ( )

  1. Jezebel
    0
    2021-08-11T18:15:29+00:00
    Reply

    Answer: 80 Newton

    Explanation:

    Initial velocity of ball = +20 m/s.

    Final velocity of ball = -20 m/s

    Mass of ball = 0.1kg

    Time taken = 0.05 seconds

    Average force = (Change in momentum of moving ball / Time taken)

    Since, change in momentum = Mass (final velocity – initial velocity)

    Change in momentum =0.1 x (-20 – (+20))

    = 0.1 x (-20-20)

    = 0.1 x (-40)

    = -4.0 kgm/s

    Then, put -4.0 kgm/s in the equation of force when Average Force = (Change in momentum / Time taken)

    = (-4.0kgm/s / 0.05 seconds)

    = 80Newton (note that the negative sign does not reflect on the magnitude of force)

    Thus, the average force exerted on the ball is 80N

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