A mass 2 m rests on a horizontal table . It is attached to a light inextensible string which passes over a smooth pulley and carries a mass m is raised vertically through a distance h and is then dropped , then the speed with which the mass 2 m begins to rise is

Explanation

Velocity of the mass `m` just before string becomes tight.

`v=sqrt(2gh)` ……….i
Using impulse ` =` change in momentum
For mass `2m`
`J=2mv_(1)`……….ii
For mass `m`
`mv-J=mv_(1)` ………..iii
Solving eqn i , ii and iii we get `v=sqrt((2gh)/3`