A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling hor

Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0m/s , and it leaves the bat traveling to the left at an angle of 30? above horizontal with a speed of 65.0m/s . The ball and bat are in contact for 1.65ms .

Part A

Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right

Express your answer using two significant figures.

Part B

Find the vertical component of the average force on the ball.

Express your answer using two significant figures.

Explanation / Answer

Since we're only concerned with the x-axis, we need to break the velocities down into their respective x & y components.

Vox = initial velocity along the x-axis = 60 m/s
Vx = final velocity along the x-axis = 65m/s(cos 40) = -43.4 m/s

To find the force along the x-axis, we need to use the equation:

Fx = m(ax)

Where:
Fx = force acting along the x-axis = unknown
m = mass = 0.145 kg
ax = acceleration along the x-axis = unknown

To get ax, we need the kinematic equation of:

Vx = Vox + ax(t)

Where:
Vx = -43.4 m/s
Vox = 60.0 m/s
ax = acceleration along the x-axis = unknown
t = time = 1.65 s

So:
-43.4m/s = 60.0m/s + ax(1.65s)

ax = 62.7 m/s^2

Now, plug that back into the equation of:

Fx = m(ax)

Fx = (0.145kg)(62.7m/s^2)

******** Fx = (-9.09) N ********

or

******** Fx = (9.09) N to the left ********

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