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?

Question

Zachary Bean

Physics

7 July, 12:17

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?

+4

Answers (1)

Know the Answer?

Rodney Leach · Answer 1

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