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25 November, 22:24

A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at 1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her

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  1. 26 November, 01:42
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    - - As she lands on the air mattress, her momentum is (m v)

    Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

    - - As she leaves it after the bounce,

    Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

    - - The impulse (change in momentum) is

    Change = (60 kg-m/s up) - (300 kg-m/s down)

    Magnitude of the change = 360 km-m/s

    The direction of the change is up / /.
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