Answer
$v_{A_2}=1.53m/s \leftarrow$
$v_{B_2}=1.27m/s \rightarrow$
Work Step by Step
The required velocities can be determined 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:
$(2)(5)+4(-2)=2v_{A_2}+4v_{B_2}$
This simplifies to:
$v_{A_2}+2v_{B_2}=1~~~$[eq(1)]
We know that:
$e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$
$\implies 0.4=\frac{v_{B_2}-v_{A_2}}{5-(-2)}$
This simplifies to:
$0.8=v_{B_2}-v_{A_2}~~~$[eq(2)]
After solving eq(1) and eq(2), we obtain:
$v_{A_2}=1.53m/s \leftarrow$
and $v_{B_2}=1.27m/s \rightarrow$
