Slippery catch
You and your friend, both of mass 70 kg, are playing catch with a frisbee on frictionless ice. Initially, you are both at rest. You throw a 0.175 kg frisbee at 10 m/s horizontally at your friend, who catches it. After he catches the frisbee, what is the relative velocity between you and your friend in m/s ?
The answer is 0.05.
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1 solution
When you throw the frisbee, the total momentum is conserved. We therefore have p I = 0 = m f v f + m y o u v y o u . You therefore have a velocity with respect to the ground of 0.025 m/s. When your friend catches the frisbee, the total momentum is conserved again, − m f v f = ( m f + m f r i e n d ) v f r i e n d which gives them a velocity of -0.249 m/s. The relative velocity is therefore 0.0499 or 0.05 m/s.