Question

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 15 m/s. The ball is hit straight back at the pitcher with a final speed of 29 m/s. (Assume the direction of the initial motion of the baseball to be positive.) (a) What is the impulse delivered to the ball? kg·m/s (b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0 ✕ 10-3 s. N

Answer 1

J = impulse =I
p = momentum
Fnet = net force or average force
t = time in seconds =2x10^-3 s
given m = 0.15 kg
v = 15 m/s

impulse I = change in momentum
p = mxv

a) first momentum is (0.15 kg)(15 m/s) m1= 2.25 (kg)(m/s)
second momentum is (0.15 kg)(29 m/s) m2= 4.35 (kg)(m/s)
change in momentum=m2-m1

                                  = 2.1 (kg)(m/s)

therefore impulse =2 .1 (kg) (m/s)

b) I = Fnet x t
Fnet=I/t

       =2.1/(2x10^-3 )

=1050 N

Therefore Favg=1050 N