A bat strikes a 0.145kg. baseball. Just before impact, theball is traveling hori

Question

A bat strikes a 0.145kg. baseball. Just before impact, theball is traveling horizontally to the right at 50.0 m/s, and itleaves the bat traveling to the left at an angle of 30.0 degreesaboe the horizontal witha a speed of 65.0 m/s.
a) What is the horizontal component of he impulse the batimparts to the ball?
b) What is the vertical component of the impulse the batimparts to the ball?
c) If the ball and bat are in contact for 1.75 ms, find thehorizontal component of the average force on the ball.
d) If the ball and bat are in contact for 1.75 ms, find thevertical component of the average force on the ball.

Explanation / Answer

The initial momentum is
p1x=mv1x=(o.145 kg)(50 m/s)=7.25 kg m/s to the right

The final momentum is

p2x=mv2x=mv2cos=(0.145 kg)(65 m/s cos30)=8.16 kg m/s to the left

p2y=mv2y=mv2sin=(0.145 kg)(65 m/s sin 30)=4.71 kg m/s upward

a) the horizontal component of the impulse is

Jx = p2x - p1x = 8.16 kg m/s - 7.25 kg m/s = 0.91 kg m/s to the left

b) the vertical component of the impulse is

Jy = p2y - p1y = 4.71 kg m/s - 0 kg m/s = 4.71 kg m/s upward

c) the horizontal component of the force is

Fx = Jx / t = (0.91 kg m/s) / (1.75 x 10-3 s) = 520 N to the left

d) the vertical component of the force is

Fy = Jy / t = (4.71 kg m/s) /(1.75 x 10-3 s) = 2693 N upward

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