A 69.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 33.0

Question

A 69.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 33.0 m/s. A second person, with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 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.
thrower m/s
catcher m/s

I keep getting response of being withing 10% correct, however not good enough. Please show steps so I may compare as to what i am doing wrong. Thank you!

Explanation / Answer

momentum is always conserved , so when first person throws the ball (69 + 0.043 ) X 2.20 = 69 X v1 + 0.043 X 33 69v1 = 150.4756 v1 = 2.181 m/s ............... speed of 1st person after throwing ball after second person cataches ball, 0.043 X 33 + 60 X 0= (60 + 0.043) X v2 v2 = 0.0236 m/s ...............speed of person who catches ball

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