A 66.0-kg person throws a 0.0415-kg snowball forward with a ground speed of 34.0

Question

A 66.0-kg person throws a 0.0415-kg snowball forward with a ground speed of 34.0 m/s. A second person, with a mass of 57.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.30 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.)

thrower _____ m/s (Give your answer to at least three decimal places.)

catcher _____ m/s

Explanation / Answer

This is conservation of momentum.

Initially thrower has a momentum of mv which is:

mtotalv =

(66+0.0415)(3.3) = 217.936 kg.s

When he throws it at 34.0m/s, that constitutes a momentum of

(0.0415)(34) = 1.411 kg.s

So now his momentum is 217.936 – 1.411 = 216.525 kg.s

So now if we want to solve for his velocity:

216.525 = mv

216.525 = (66)(v) {he lost the weight of the snowball}

216.525/66 = 3.280 m/s after throwing snow ball {thrower}

Now the reciever will gain the 1.411 kg.s momentum. He initially has NO momentum because he was not moving, but now he caught the 1.411 kg.s momentum in the snowball.

1.411 = mv

1.411 = (57.5 + 0.0415 )v

v = 0.024 m/s {catcher}

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