A 0.174-kg softball is pitched horizontally at 26.0 m/s. The ball moves in the opposite direction at 38.0 m/s after it is hit by the bat. A.

Question

A 0.174-kg softball is pitched horizontally at 26.0 m/s. The ball moves in the opposite direction at 38.0 m/s after it is hit by the bat. A. What is the change in momentum of the ball? B. What is the impulse delivered by the bat? C. If the bat and ball are in contact for 0.80 ms, what is the average force the bat exerts on the ball?

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Thanh Hà 5 years 2021-07-29T17:51:03+00:00 1 Answers 1406 views 0

Answers ( )

  1. taiduc
    2
    2021-07-29T17:52:47+00:00
    Reply

    Answer:

    A) -11.136

    B) -11.136

    C) 139.2 N

    Explanation:

    A )

    Change in momentum is equal to Final momentum – Initial momentum

    Final momentum

    = 0.174 * 26\\= 4.524\\

    Initial Momentum

    = 0.174 * -38\\= - 6.612

    Thus, Change in momentum

    = (-6.612- 4.525 )\\= - 11. 136\\

    B) Impulse delivered by bat is equal to the change in momentum i.e - 11. 136

    C) Force is equal to change in momentum divided by time

    Force = \frac{-11.136}{80 * 10^{-3}} \\= 139.2

    Force exerted by bat on ball is139.2 N

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