Answer
$v_c=5.04m/s \leftarrow$
Work Step by Step
We can determine the required final velocity as follows:
According to the conservation of energy
$T_A+V_A=T_B+V_B$
$\implies \frac{1}{2}(m_c+m_b) v_A^2+(m_c+m_b) gh_A=\frac{1}{2}(m_c+m_b) v_B^2+(m_c+m_b) gh_A$
We plug in the known values to obtain:
$\frac{1}{2}(3+0.5)(0)^2+(3+0.5)(9.81)(1.25)=\frac{1}{2}(3+0.5)v_B^2+(3+0.5)(9.81)(0)$
This simplifies to:
$v_B=4.95m/s$
Now according to the conservation of linear momentum
$(m_c+m_b) v_B=m_cv_c+m_bv_b$
$\implies (3+0.5)(4.95)=3v_c-0.5v_b$
$\implies v_b=6v_c-34.66$eq(1)
We know that:
$v_{b/c}=v_b-v_c$
$\implies -0.6=-v_b-v_c$
$\implies v_b=0.6-v_c$
From eq(1), we plug in the value of $v_b$ into the above equation to obtain:
$6v_c-34.66=0.6-v_c$
This simplifies to:
$v_c=5.04m/s \leftarrow$
