Question

A pitcher throws a baseball with a velocity of 40 m/s. After being struck by a bat the ball travels in the opposite direction with a velocity of 50 m/s. If the ball has a mass of 0.145 kg and is in contact with the bat for 3.0 ms, the magnitude of the average force exerted by the bat on the ball is

Answer 1

Answer:

Force on the ball will be 4350 N

Explanation:

We have given that mass of the baseball m = 0.145 kg

Initial velocity of the baseball u = 40 m/sec

And final velocity v = - 50 m/sec ( in opposite direction )

So change in momentum = $=0.145\times (40-(-50))=13.05kgm/sec$

Time is given as $t=3ms=3\times 10^{-3}sec$

We know that change in momentum is given by impulse

So $F\times t=13.05$

$F\times 3\times 10^{-3}=13.05$

$F=4350N$