6) Joe throws a 0.15-kg rubber ball directly downward to the floor. The ball’s speed just before impact is 6.5 m/s and just after the impact

Question

6) Joe throws a 0.15-kg rubber ball directly downward to the floor. The ball’s speed just before impact is 6.5 m/s and just after the impact is 3.5 m/s. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force that the floor exerts on the ball during the time of contact? A. 133 N B. 60 N C. 30 N D. 18 N E. 3.0 N

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Thiên Hương 4 years 2021-07-20T10:11:19+00:00 1 Answers 16 views 0

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  1. mocmien2
    0
    2021-07-20T10:12:22+00:00
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    Answer:

    B. 60 N

    Explanation:

    Change in momentum after ball bounces back = m ( v – u )

    v is velocity after the bounce and u is before the bounce

    = .15 ( 6.5 + 3.5 )

    1.5 kg m /s

    If F be the average force acting on the ball which changed its velocity

    impulse of force = force x time

    = F x .025

    Impulse of force = change in momentum

    F x .025  = 1.5

    F = 1.5 / .025

    = 60 N .

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