Question

A pitcher throws a 0.14-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of 48 m/s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat. a. What is the magnitude of the impulse delivered by the bat to the baseball? b. If the ball is in contact with the bat for 0.0050 s, what is the magnitude of the average force exerted by the bat on the ball?

Answer 1

Answer

given,

mass of base ball = 0.14 kg

speed before it made the contact with the ball (V i) = 42 m/s

speed after batter hit the ball(V f) = - 48 m/s

a)                                            

impulse = change in momentum

             = $m\times (V_f-V_i)$      

             =$0.14\times (-48-42)$

             = -12.6 Kg m/s

Magnitude of impulse = 12.6 Kg m/s

b)                                                        

Force = $\dfrac{impulse}{time}$

          =  $\dfrac{12.6}{0.005}$

Force = 2520 N