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

Force on the ball will be 4350 N

Step-by-step 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* (40-(-50))=13.05kgm/sec`

Time is given as
`t=3ms=3* 10^(-3)sec`

We know that change in momentum is given by impulse

So
`F* t=13.05`

`F* 3* 10^(-3)=13.05`

`F=4350N`