A batter hits a 0.140-kg baseball that was approaching him at 30 m/s and, as a result, the ball leaves the bat at 40 m/s in the reverse of its original direction. The ball remains in contact with the bat for 2.0 ms.

Required:
What is the magnitude of the average force exerted by the bat on the ball?

We know that the newton second law The force * time is directly proportional to the change in the momentum .It means force *time is equal to change in ,momentum

[tex]F* t = m * ( v2 -v1 )[/tex]

it can be written as

[tex]F = \frac{m*(v2-v1)}{t}[/tex]

Putting the value of m ,v2 ,v1 and t in the previous equation we get

[tex]= \frac{0.14*(40 - (-30)}{2*10^{-3} }[/tex]

On solving

[tex]= 4900\ N[/tex]

Eduard22sly Eduard22sly · Answer 2 · 2021-12-31T11:30:59+01:00

The magnitude of the average force exerted by the bat on the ball is 4900 N.

From the question given above, the following data were obtained:

Mass (m) = 0.140 Kg

Initial velocity (u) = 30 m/s

Final velocity (v) = 40 m/s

Time (t) = 2 ms = 2×10¯³ s

Force (F) =?

The magnitude of the force can be obtained as follow:

F = m(v + u) / t

F = 0.140 (40 + 30) / 2×10¯³

F = (0.140 × 70) / 2×10¯³

F = 9.8 / 2×10¯³

F = 4900 N

Thus, the magnitude of the force exerted is 4900 N

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