A 69.5-kg person throws a 0.0440-kg snowball forward with a ground speed of 33.0

Question

A 69.5-kg person throws a 0.0440-kg snowball forward with a ground speed of 33.0 m/s. A second person, with a mass of 56.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.95 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice. (Take the direction the snowball is thrown to be the positive direction. Indicate the direction with the sign of your answer.)

Explanation / Answer

Applying momentum (mv ) consrvation for first man and snowball,

initial momentum = final momentum

69.5 x 2.95 + 0.0440 x 0 = 69.5 x v1 + 0.0440 x 33

v1 = 2.93 m/s forward. .........final velocity of first person,

now using for second person and ball,

56 x 0 + 0.0440 x 33 = (56 + 0.0440) x v2

v2 = 0.026 m/s forward ..............velocity of 2nd person

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