Answer
$v_A=0.16m/s\leftarrow$
$v_B=8.24m/s\rightarrow$
Work Step by Step
We can determine the required velocity as follows:
According to the conservation of linear momentum
$m_Av_{A_1}+m_Bv_{B_1}=m_Av_{A_2}+m_Bv_{B_2}$
We plug in the known values to obtain:
$3(8)+(2)(-4)=3v_{A_2}+2v_{B_2}$
This simplifies to:
$3v_A+2v_B=16~~~$[eq(1)]
We know that:
$e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$
$\implies 0.7=\frac{v_{B_2}-v_{A_2}}{8-(-4)}$
$\implies 8.4=v_{B_2}-v_{A_2}~~~$[eq(2)]
After solving eq(1) and eq(2), we obtain:
$v_A=0.16m/s\leftarrow$
and $v_B=8.24m/s\rightarrow$
