A 68.6 kg archer stands at rest on frictionless ice and fires a 0.46 kg arrow horizontally. He fires and slides back -0.37m/s. He fires another arrow while he is still sliding that is going 76m/s. What will be his velocity after he fires the second arrow?
m1v1f + m2v2f = -0.37
The Attempt at a Solution
I solved for v1f and plugged in the numbers to get
v1f = -.37/68.6 - .46(76)/68.6 = -.515