Question

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?

Answer 1

Answer:

4900 N

Explanation:

Given :

$m\ = \ 0.140\ kg\\V2= 40\ m/s\\v1\ = 30\ m/s\\t=\ 2.0 ms.$

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

$F* t = m * ( v2 -v1 )$

it can be written as

$F = \frac{m*(v2-v1)}{t}$

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

$= \frac{0.14*(40 - (-30)}{2*10^{-3} }$

On solving

$= 4900\ N$