Q1 - A 0.30-kg softball has a velocity of 12 m/s at an angle of 28º below the ho

Question

Q1 - A 0.30-kg softball has a velocity of 12 m/s at an angle of 28º below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of 15 m/s horizontally back toward the pitcher?

A : 8.6 kg.m/s

B : 3.3 kg.m/s

C : 4.2 kg.m/s

D : 5.7 kg.m/s

E : 7.9 kg.m/s

Q2- A 1.2 kg rock is thrown through a glass window. It hits the window with a speed of +5.0 m/s, and after the window shatters, it has a speed of +3.0 m/s. If the glass breaks in 0.002 s, what force did the glass exert on the rock?

A : -1200 N

B : +1200 N

C : +4800 N

D : -4800 N

E : None of the choices are correct.

Explanation / Answer

change in momentum = final momentum - initial momentum

change in horizontal momentum = final horizontal momentum - initial horizontal mometum

change in horizontal momentum = 0.3 * 15 - (-0.3 * 12 * cos(28))

change in horizontal momentum = 7.6786 kg.m/s

change in vertical momentum = 0 - 0.3 * 12 * sin(28)

change in vertical momentum = -1.69 kg.m/s

change in momentum = sqrt(7.6786^2 + 1.69^2)

change in momentum = 7.9 kg.m/s

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