Question Description
A 67.0-kg person throws a 0.0470-kg snowball forward with a ground speed of 32.0 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 2.00 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 |
Final Answer
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we can apply force momentum equation as fallows
to snow ball and 2nd person
0.047*32=(56+0.047)*v
v=0.047*32/56.047
v=0.026 m/s
so the catcher speed 0.026 m/s
applying force momentum for ball and first person
67*2-0.047*32=67*u
u=67*2-0.047*32/67
u=1.98 m/s
so the thrower speed=1.98 m/s
Europe(Continental)