Question

Now assume that the pitcher in Part D throws a 0.145-kgkg baseball parallel to the ground with a speed of 32 m/sm/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 x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s−8.4N⋅s to the baseball?

Answer 1

Answer:

V2 = -25.93 m/s

Explanation:

First of all, we know that;

Momentum = mass x change in velocity

Thus;

Momentum = m(v2 - v1)

Also, we know that;

Impulse = force x time

And also that;

Momentum = Impulse

From the question, m = 0.145kg and V1 = 32m/s while impulse = -8.4 N.s

Thus;

0.145(v2 - 32) = -8.4

Now,

0.145v2 - 4.64 = -8.4

0.145v2 = -8.4 + 4.64

0.145v2 = -3.76

v2 = -3.76/0.145

= -25.931 m/s and to two significant figures gives -25.93 m/s