A 0.15-kg baseball is pitched with a speed of 35 m/s (78 mph). Whenthe ball hits

Question

A 0.15-kg baseball is pitched with a speed of 35 m/s (78 mph). Whenthe ball hits the catcher's glove, the glove moves back by 5.0 cm(2 in) as it stops the ball.
a) What was the change in momentum of the ball? (Assume theball moves in the positive direction)
b) What impulse was applied to the baseball?
c) Assuming constant acceleration, what was the average forceapplied by the catcher's glove?
a) What was the change in momentum of the ball? (Assume theball moves in the positive direction)
b) What impulse was applied to the baseball?
c) Assuming constant acceleration, what was the average forceapplied by the catcher's glove?

Explanation / Answer

a) What was the change in momentump of the ball? (Assume the ball moves in the positivedirection) p= m1u1 -m1V1= p= m1u1 = 0.15x 35= p=5.25 kg m/s
b) Whatimpulse I was applied to the baseball? I=p
c) Assuming constant acceleration, whatwas the average force applied by the catcher's glove? S=0.5at2 while V=at then since t= V/a S=0.5a(V/a)2 S=0.5V2 /a and a = 0.5V2 /S Now since F=ma we have F= m 0.5V2 / S F=  0.5mV2 / S F= 0.5 x 0.15 (35)2 / 0.05 F=19, 300 N

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