A. A 0.112 kg baseball traveling with a horizontal speed of 6.42 m/s is hit by a

Question

A. A 0.112 kg baseball traveling with a horizontal speed of 6.42 m/s is hit by a bat and then moves with a speed of 30.5 m/s in the opposite direction. What is the magnitude of the change in the ball's momentum?

B. A 63.8 kg man and his 37.4 kg daughter on skates stand together on a frozen lake. If they push apart and the father has a velocity of 0.557 m/s eastward, what is the magnitude of the velocity of the daughter? (Neglect friction.)

C. To get off a frozen, frictionless lake, a 70.3-kg person takes off a 0.253-kg shoe and thows it horizontally directly away from the shore with a speed of 1.80 m/s. If the person is 5.49m from the shore, how long does it take for him to reach it?

Explanation / Answer

A) change in momentum = m ( v - u)

v = final velocity= 30.5 m/s

u = initial velocity = - 6.42 m/s

change in momentum = 0.112 ( 30.5 - ( - 6.42)) = 4.13 kg.m/s

B) using momentum conservation,

initial momentum = final momentum

0 = 63.8x0.557 + ( 37.4v)

v = - 0.95 m/s

magnitude = 0.95 m/s

C) using momentum conservation,

initial momentum = final momentum

0 = 0.253x1.80 + ( 70.3v)

v = -0.0065 m/s

time = d / v   = 5.49 / 0.0065 = 847.49 sec

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