Question

Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction. The impulse on the ball caused by the bat will bo in the negative x direction. Part E Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of -8.4 N s to the baseball? Enter your answer numerically in meters per second using two significant figures.

Answer 1

Answer:25 m/s

Explanation:

Given

mass of ball $m=0.145 kg$

Initial speed of ball $u=32 m/s$

$Impulse=-8.4 N s$

Now we know impulse is equal to  change in momentum

$Impulse=\Delta P_f-\Delta P_i$

$-8.4=m(v_f-v_i)$

$v_f-v_i=-\frac{8.4}{0.145}$

$v_f-32=-57.93$

$v_f=32-57.93$

$v_f=25.93 m/s$

with two significant figures 25 m/s