Running at 2.0 m/s, Bruce, a 45.0- kg soccer player, collides with Biff, the 90.0-kg opponent, who is traveling at 7.0 m/s in the other direction. Upon collision, Biff continues to travel forward at 1.0 m/s. How fast is Bruce knocked backwards?
Answer (1)
The answer is 14.0 m/s.Bruce's momentum:[tex]$$p_{\text{Bruce}} = 45.0 \, \text{kg} \cdot 2.0 \, \text{m/s} = 90.0 \, \text{kg m/s}$$[/tex]Biff's momentum:[tex]$$p_{\text{Biff}} = 90.0 \, \text{kg} \cdot (-7.0 \, \text{m/s}) = -630.0 \, \text{kg m/s}$$[/tex] Total initial momentum:[tex]$$p_{\text{initial}} = 90.0 - 630.0 = -540.0 \, \text{kg m/s}$$[/tex] Biff's momentum after the collision:[tex]$$p'_{\text{Biff}} = 90.0 \, \text{kg} \cdot 1.0 \, \text{m/s} = 90.0 \, \text{kg m/s}$$[/tex] Total final momentum:[tex]$$p_{\text{final}} = (45.0 \cdot v'_{\text{Bruce}}) + 90.0$$[/tex]Set Initial Equal to Final Momentum[tex]$$-540.0 = 45.0 \cdot v'_{\text{Bruce}} + 90.0$$[/tex] Solve for Bruce's Final Velocity[tex]$$45.0 \cdot v'_{\text{Bruce}} = -540.0 - 90.0$$[/tex][tex]$$v'_{\text{Bruce}} = \frac{-630.0}{45.0} = -14.0 \, \text{m/s}$$[/tex]
