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17 December, 04:30

Now assume that the pitcher in Part D throws a 0.145-kgkg baseball parallel to the ground with a speed of 32 m/sm/s in the + x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of - 8.4N⋅s-8.4N⋅s to the baseball?

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  1. 17 December, 06:23
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    V2 = - 25.93 m/s

    Explanation:

    First of all, we know that;

    Momentum = mass x change in velocity

    Thus;

    Momentum = m (v2 - v1)

    Also, we know that;

    Impulse = force x time

    And also that;

    Momentum = Impulse

    From the question, m = 0.145kg and V1 = 32m/s while impulse = - 8.4 N. s

    Thus;

    0.145 (v2 - 32) = - 8.4

    Now,

    0.145v2 - 4.64 = - 8.4

    0.145v2 = - 8.4 + 4.64

    0.145v2 = - 3.76

    v2 = - 3.76/0.145

    = - 25.931 m/s and to two significant figures gives - 25.93 m/s
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