Answer
$\theta=tan^{-1} \sqrt{\frac{7}{5}e}$
Work Step by Step
We can determine the required angle as follows:
We know that
$e=\frac{v_{s_2}-v_{b_2}}{v_{b_1}-v_{s_1}}$
$\implies =\frac{0-v_2sin\theta}{-v_1cos\theta-0}$
This simplifies to:
$v_1=\frac{v_2tan\theta}{e}$ [eq(1)]
Now, we apply the conservation of angular momentum
$mv_{b_1}r_1=mv_{b_2}r_2+I\omega_2$ [eq(2)]
As $I=\frac{2}{5} mr^2$
$\omega_2=\frac{v_2cos\theta}{r}$
$r_1=rsin\theta$
and $r_2=rcos\theta$
Now, from eq(2) $mv_1rsin\theta=\frac{2}{5} mr^2(\frac{v_2cos\theta}{r})+mv_2rcos\theta$
This simplifies to:
$v_2=\frac{5}{7}v_1tan\theta$ [eq(3)]
We plug in the known values from eq(3) into eq(1) to obtain:
$v_1=\frac{\frac{5}{7}v_1tan\theta\times tan\theta}{e}$
This simplifies to:
$\theta=tan^{-1} \sqrt{\frac{7}{5}e}$
