A baseball player hits a baseball (m = 0.142 kg) as shown in the figure above. T

Question

A baseball player hits a baseball (m = 0.142 kg) as shown in the figure above. The ball is initially traveling horizontally with speed of 35 m/s. The batter hits a fly ball as shown, with a speed vf = 54 m/s. (a) What are the magnitude and direction of the impulse imparted to the ball? magnitude _____ kg ? m/s direction _____? counterclockwise from the +x-axis (b) If the ball and the bat are in contact for a time of 8.0 ms, what is the magnitude of the average force of the bat on the ball? _____ N Compare this answer to the weight of the ball. average force on the ball / weight of the ball = ____ (c) What is the impulse imparted to the bat? magnitude _____ kg ? m/s direction _____ ? counterclockwise from the +x-axis

Explanation / Answer

(a)   Impulse = Change in Momentum, so this is the basic defn that we will use to find Impulse.

Since Momentum is a vector, we have to use x and y-components of Momentum.

x-component of Initial momentum =   - m v to left    =   - (0.142 kg) (35 m/s) = - 4.97 kg m/s.

y-component of inital momentum = zero   (since ball is traveling perfectly horizontally)

x-component of final momentum = m v cos 45 to right   = (0.142 kg)( 54 m/s) cos 45 = 5.42 kg m/s.

y-component of final momentum = m v sin 45 upward = 5.42 kg m/s.

Change in momentum for the x-components = final - initial = 5.42 - ( - 4.97) = 10.39 kg m/s to right

Change in momentum for the y-components = final - initial = 5.42 - 0 = 5.42 kg m/s upward.

The magnitude of the resultant change in mom = (10.39)^2 + (5.42)^2   = 11.72 kg m/s .

Thus the Magnitude of the Impulse on the ball is 11.72 kg m/s.

The angle is   inv tan (5.42 / 10.39) = 27.55 degrees counterclockwise from the + x-direction.

(b) Impulse = F (t ) = 11.72

                     F ( 8 x 10^-3 sec) = 11.72,   solving for F , we get F = 1465 N .

Weight of ball = m g = (0.142 kg) (9.8 m/s^2) = 1.39 N

So, the desired ratio is   1465N / 1.39N = 1053

(c) The impulse given to the bat is equal in magnitude but opposite in direction to the impulse on the ball.

So the magnitude of the bat's impulse = 11.72 kg m/s

The direction is 27.55 deg + 180 deg = 207.55 deg counterclockwise from the + x-axis.

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