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

Respuesta :

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 = [tex]=0.145\times (40-(-50))=13.05kgm/sec[/tex]

Time is given as [tex]t=3ms=3\times 10^{-3}sec[/tex]

We know that change in momentum is given by impulse

So [tex]F\times t=13.05[/tex]

[tex]F\times 3\times 10^{-3}=13.05[/tex]

[tex]F=4350N[/tex]