Answer
a) $v_{cm}=+1.1m/s$
b) $v_{cm}=+1.1m/s$
c) The answer in b) should be equal with $v_f$
Work Step by Step
We have $m_1=65\times10^3kg$, $m_2=92\times10^3kg$, $v_{01}=+0.8m/s$, $v_{02}=+1.3m/s$ and $v_f=+1.1m/s$
a) The velocity of the center of mass before the collision is $$v_{cm}=\frac{m_1v_{01}+m_2v_{02}}{m_1+m_2}=+1.1m/s$$
b) The velocity of the center of mass after the collision is $$v_{cm}=\frac{(m_1+m_2)v_f}{m_1+m_2}=v_f=+1.1m/s$$
c) The answer in b) should be equal with $v_f$, because as shown above $$v_{cm}=\frac{(m_1+m_2)v_f}{m_1+m_2}=v_f$$
