Answer
18.6 percent
Work Step by Step
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 $
$1-.814=\fbox{18.6 percent}$
