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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Question:

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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Answer:

Answer:

?f=v%3D 25

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 32 m/s

Impulse applied by the bat to the baseball, J = -8.4 N-s

To find,

The ball’s x-component of velocity just after leaving the bat.

Solution,

The impulse imparted from one object to other is equal to the change in momentum. Mathematically, it is given by :

?f=J%3Dm(v u)

?f= 8.4%3D0

On solving the above equation, we can find the value of v as :

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Therefore, the velocity of the ball just after leaving the bat is -25.931 m/s. Hence, this is the required solution.

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