A rope thrown over a pulley has a ladder with a man of mass m  on one of its ends and a counter balancing mass M  on its other end. The man climbs with a velocity $V_r$  relative to ladder. Ignoring the masses of pulley and the rope as well as the friction on the pulley axis, the velocity of the centre of mass of this system is 

(Ladder + Man) system is counterbalanced by block of mass M, Hence Mass of ladder $M=(M-m)$ 
There is no external force on system and rope tension is same both on the left and right hand side at every instant
Hence momentum of both sides are also equal 
 $Mv=(M-m)(-v)+m(v_r-v)$
 $\therefore \{v=\frac{m}{2M}{v}_r\}$
$\therefore$  Momentum of centre of mass
$2M{V}_{com}=Mv+Mv$

$V_{com}=\frac{m}{2M}{v}_r$