## Essential University Physics: Volume 1 (3rd Edition)

Using conservation of momentum, we find: $v=v_{1f}cos(37^{\circ})+2v_{2f}cos\theta$ $v=.799v_{1f}+2v_{2f}cos\theta$ We also know: $v_{1f}sin(37^{\circ})=v_{2f}sin\theta$ $v_{1f}=-1.66v_{2f}sin\theta$ Combining these equations gives: $v=-1.32v_{2f}sin\theta+2v_{2f}cos\theta$ We know from conservation of kinetic energy that: $\frac{1}{2}v^2 =\frac{1}{2}v_{1f}^2+v_{2f}^2$ Combining these equations, we find the value of the ratio of the kinetic energies: $\frac{K_{2f}}{K_{i}}=(\frac{v_{1i}}{1.11v_{2f}})^2=.813$ $1-.814=\fbox{18.6 percent}$