A major league catcher catches a fastball moving at 80.0mi/h and his hand and gl

Question

A major league catcher catches a fastball moving at 80.0mi/h and his hand and glove recoil 10.0 cm in bringing the ball to rest.

Part A

If it took 0.00500s to bring the ball (with a mass of 300g ) to rest in the glove, what is the of the change in momentum of the ball?

Part B

What is the direction of the change in momentum of the ball?

Part C

Find the magnitude of the average force the ball exerts on the hand and glove?

Part D

Find the direction of the average force the ball exerts on the hand and glove?

Explanation / Answer

1) change in momentum = M*V1 - M*V2

V1 = 0 and V2 = 80 mi/hr

change in momentum = 300 * 80 = 2400 gm*mi/hr = 2400*1.6/3600 = 1.06 kg*m/sec

B)change in momentum is the same as the initial velocity of the ball

c)ball comes to rest travelling 10 cm in 0.005 sec

acceleration = 0.1/0.005^2 = 4000 m/sec

force = 300 * 10^-3 * 4000 = 1200 N

D)the direction of the average force the ball exerts on the hand and glove is the same as the initial velocity of the ball

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.