Answer
$v_A=3.29m/s$ ,$v_B=2.19m/s$
Work Step by Step
We can determine the required speed as follows:
We know that:
$\Sigma mv_1=\Sigma mv_2$
$\implies 0=m_Av_A+m_Bv_B$
$\implies 40v_A=60v_B$
$\implies v_A=1.5v_B~~~$eq(1)
The conservation of total energy before the blocks begin to rise is given as
$T_1+V_1=T_2+V_2$
$\implies \frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_Bv_{B_{\circ}}^2+\frac{1}{2}ks^2=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2+0$
$\implies ks^2=m_Av_A^2+m_Bv_B^2$
We plug in the known values to obtain:
$(180)(2)^2=40v_A^2+60v_B^2$
From eq(1), $v_A=1.5v_B$
$\implies (180)(2)^2=40(1.5v_B)^2+60v_B^2$
This simplifies to:
$v_B=2.19m/s$
We plug in this value in eq(1) to obtain:
$v_A=1.5(2.19)=3.29m/s$
