Answer
$(v_b)_2=3.36~ft/s$
Work Step by Step
The required velocity can be determined as follows:
We know that
$e=\frac{(v_b)_2-(v_B)_2}{(v_B)_1-(v_b)1}$
$\implies 0.8=\frac{(v_b)_2-(v_B)_2}{0-12}$
$\implies (v_b)_2=(v_B)_2-9.6$ [eq(1)]
We also know that the angular momentum is conserved about A
$\implies H_{A_1}=H_{A_2}$
$\implies m_b(v_b)_1 r_b=I_A(\omega_B)_2+m_b(v_b)_2r_b$ [eq(2)]
As $I_A=\frac{m_rl^2}{12}+m_r r^2$
$\implies I_A=\frac{4}{32.2}\times \frac{(3)^2}{12}+\frac{4}{32.2}(1.5)^2=0.37267 slug/ft^2$
We plug in the known values in eq (2) to obtain:
$\frac{2}{32.2}\times 12\times 3=0.37267(\omega_B)_2+(\frac{2}{32.2}\times (v_b)_2\times 3)$
This simplifies to:
$2.236=0.12422(v_B)_2+0.18633(v_b)_2$ [eq(3)]
We plug in the values from eq(1) into eq(3) to obtain:
$2.36=0.12422(v_B)_2+0.18633(v_B)_2-9.6$
This simplifies to:
$(v_B)_2=12.959ft/s$
Now, from eq(1), we obtain:
$(v_b)_2=12.959-9.6=3.36~ft/s$
