physics
posted by rohit on .
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 they catch the frisbee, what is the relative velocity between you and your friend in m/s?

m₁=m= 70 kg
m₃=0.175 kg, v₃=10 m/s
Let v₃ be positive and directed to the right
0=  m₁v₁+m₃v₃
v₁=m₃v₃/m₁=0.175•10/70=0.025 m/s (directed to the left)
m₃v₃ =(m₂+m₃)v
v= m₃v₃/(m₂+m₃)=0.175•10/(70+0.175)=0.0249 m/s (directed to the right)
V(relative) =
=v₁+v=0.025+0.0249 =0.0499 m/s 
thankyou