• # question_answer A rope thrown over a pulley has a ladder with a man of mass m on one of its ends and a counterbalancing mass Mon its other end. The man climbs with a velocity v, relative to ladder. Ignoring the masses of the pulley and the rope as well as the friction on the pulley axis, the velocity of the centre of mass of this system is A) $\frac{m}{(m+M)}v$                B) $\frac{m}{2M}v$ C) $\frac{2m}{M}v$          D)        $\frac{(m+M)}{2m}v$

[b] The rope tension is the same both on the left and right hand side at every instant, and consequently momentum of both sides are equal. $\therefore \,\,\,Mv=(M+m)(-v)+m({{v}_{r}}-v)$ Or         $v=\frac{m}{2M}{{v}_{F}}$ Momentum of the centre of mass is $P={{P}_{1}}+{{P}_{2}}$ Or     $2M{{v}_{com}}=Mv+Mv$ $\therefore \,\,\,\,{{v}_{com}}=v=\frac{m}{2M}{{v}_{r}}$