A 0.2 kg baseball moving with a velocity of 20 m/s is struck by a bat. The baseball reverses its direction and moves with a velocity of 40 m

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A 0.2 kg baseball moving with a velocity of 20 m/s is struck by a bat. The baseball reverses its direction and moves with a velocity of 40 m/s . What is the average force exerted on the ball if the bat is in contact with the ball for 0.006 ?

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  1. Average force is 666.67 N

    Explanation:

    Given that :

    m = Mass of the baseball = 0.2 kg

    u = Initial velocity = 20 m/s  

    v = Final velocity = 40 m/s

    t = Time taken for change in velocity = 0.006 s


    We know:

    Force exerted = mass × acceleration = m ×a

    Acceleration can be found by means of dividing the change in velocity measured in m/s by the time taken in seconds.

    a = [tex]$\frac{v-u}{t}[/tex] = [tex]$a = \frac{40-20}{0.006} = 3,333.3[/tex] m/s²

    Now we have to find the force using the formula, F = mass × acceleration as,

    F = 0.2 kg ×3333.3 ms⁻² = 666.67 N

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