A major league catcher catches a fastball moving at 85.0 mi/h and his hand and g

ID: 1368486 • Letter: A

Question

A major league catcher catches a fastball moving at 85.0 mi/h and his hand and glove recoil 10.0 cm in bringing the ball to rest.

A) If it took 0.00420 s to bring the ball (with a mass of 200 g ) to rest in the glove, what is the of the change in momentum of the ball?

B) What is the direction of the change in momentum of the ball? (oppositite or same)

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

D) Find the direction of the average force the ball exerts on the hand and glove. (opposite or same)

A) 85 mi/h = 38 m/s

then chnage in momentum is m*(v-u)

here v = 0 m/s

u = 38 m/s

m = 0.2 kg

then change in momentum is dp = 0.2*(0-38) = -7.6 kg*m/s

B) opposite to initial direction of the ball

C) Favg*dt = chnage in momentum = -7.6

Favg = -7.6/0.0042 = -1809.52 N

D) ball applies the force in the same direction of its initial direction

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