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 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:

Options

A. `m/M V_(r)`
B. `m/(2M)v_(r)`
C. `M/mv_(r)`
D. `(2M)/m v_(r)`

Answer

B

Explanation

The masses of load , ladder and man are M, M - m and m respectively . Their velocities are v (upward) , `-v` and `cv_(r)-v` respectively
` :. V_(cm) = (Sigmam_(i)v_(i))/(Sigmam_(i))`
` (M(v) + (M_m)(-v ) +m (v - v) )/(2M) = m/(2M) v_(r)`