Question

Two massess M and 2M, are arranged to be at the same initial height H, above the ground, by using a massless inextensible string and a smooth pulley. The masses are ‘let-go’, at t = 0, and the string is suddenly cut when the mass M has risen through a height (H/2). The velocity, with which the mass 2M would hit the ground, is given by

(1) \(\sqrt{\frac{4gH}{5}}\)

(2) \(\sqrt{\frac{3gH}{4}}\)

(3) \(\frac{4gH}{3}\)

(4) \(\frac{gH}{2}\)

Answer 1

 (3) \(\frac{4gH}{3}\)

The equations of motion of the two masses are T – Mg = Ma

and 2Mg – T = 2Ma

∴ a = Mg/3M = g/3

When the mass M rises through a height H/2 , the mass 2M falls down through a distance H/2 .

The downward velocity of mass 2M, when it has moved downward through a distance H/2 , is, given by

At this instant the string is cut. The mass, 2m, therefore, falls subsequently as a freely falling mass with an acceleration g. It is, at this instant, at a height H/2 above the ground. Hence the velocity, V, with which it would hit the ground, is given by