Use the exact values you enter in previous answer(s) to make later calculation(s

Question

Use the exact values you enter in previous answer(s) to make later calculation(s).

A baseball player hits a baseball (m = 0.144 kg) as shown in the figure below. The ball is initially traveling horizontally with speed of 36 m/s. The batter hits a fly ball as shown, with a speed vf = 49 m/s.

(a) What are the magnitude and direction of the impulse imparted to the ball?

(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.

=

(c) What is the impulse imparted to the bat?

Explanation / Answer

impulse = chnage in momentum = Favg*dt

chnage in momentum along X-axis dpx = m*(vi+vfcos(45)) = 0.144*(36 + 49*0.707) = 10.2 kg*m/s

along Y-axis is dpy = m*(vf*sin(45)) = 0.144*49*0.707 = 4.98 kg*m/s

magnituder of the impulse is j = sqrt(4.98^2 + 10.2^2) = 11.35 N-s

direction is theta = atan(4.98/10.2) = 26 degrees

the required answer in direction is 180-26 = 154 degrees counter clockwise from the +x-axis

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B) Favg*dt = 11.35

Favg = 11.35/*0.008 = 1418.75 N

weight of the ball =m*g = 0.144*9.81 = 1.41264 N

Favg/weight of the ball = 148.75/1.41264 = 105.3

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C) impulse imparted to bat

magnitude is 11.35 N-s

direction is 154+180 = 334 degrees counterclockwise from the +X-axis

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