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 x-component of velocity just after leaving the bat if the bat applies an impulse of - 8.4N⋅s to the baseball?

Question

Benny

Physics

24 September, 05:27

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 x-component of velocity just after leaving the bat if the bat applies an impulse of - 8.4N⋅s to the baseball?

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Alexa Griffith · Answer 1

25.9 m/s

Explanation:

mass of ball, m = 0.145 kg

initial velocity, u = + 32 m/s

It bounce back with the velocity but in opposite direction so final velocity,

v = - v

Impulse, I = - 8.4 Ns

Impulse is defined as the change in momentum

I = m v - mu = m (v - u)

- 8.4 = 0.145 x ( - v - 3 2)

- 57.9 = - v - 32

v = 57.9 - 32 = 25.9 m/s

Thus, the final speed of the ball is 25.9 m/s