Answer
$v_1=v/6$
$v_2=5v/6$
Work Step by Step
We know from conservation of momentum that:
$mv = mv_{1f}+mv_{2f}$
$v = v_{1f}+v_{2f}$
$v -v_{1f}= v_{2f}$
5/18 of the initial kinetic energy is lost, so:
$(1-5/18)(\frac{1}{2}mv^2)=\frac{1}{2}mv_{1f}^2+\frac{1}{2}mv_{2f}^2$
$(1-5/18)(\frac{1}{2}mv^2)=\frac{1}{2}mv_{1f}^2+\frac{1}{2}m(v -v_{1f})^2$
$(1-5/18)(v^2)=v_{1f}^2+(v -v_{1f})^2$
$-\frac{5v^2}{18}=2v_{1f}^2-2vv_{1f}$
To find the ratio, we set v equal to one:
$-\frac{5}{18}=2v_{1f}^2-2v_{1f}$
$v_{1f}=v/6$
This means:
$v_2=5v/6$
