In the figure shown, the heavy ball of mass `2m`, rests on the horizontal surface and the lighter ball of mass `m` is dropped from a height `h gt 2l`. At the instant the string gets taut, the upward the velocity of the heavy ball will be

Options

A. `2/3sqrt(gl)`
B. `4/3sqrt(gl)`
C. `1/3sqrt(gl)`
D. `1/2sqrt(gl)`

Answer

A

Explanation

Velocity of light ball, whenit is descended by height `2l`
`v=sqrt(2g2l)=2sqrt(gl)`
From conservastion of linear momentum
`m2sqrt(gl)=2mv'+mv'`
`2msqrt(gl)=3mv'`
`2/3sqrt(gl)=v'`