Two particles A and B each of mass m are attached by a light inextensible string of length 2l . The whole system lies on a smooth horizontal table with B initially at a distance l from A . The particle at end B is projected across the table with speed u perpendicular to AB . Velocity of ball A just after the string is taur , is

Explanation

When the string becomes taut, it can be depicted as in figure.
`sintheta-l/(2l)=1/2impliestheta=30^(@)`

Linear momentum will be conserved in any direction. Let us conserve the linear momentum in the direction of thread. then
`"mucos"theta=2mv`
`implies v=1/2ucostheta=1/2u sqrt(3)/2=(sqrt(3)/4)u`.......i
From the impulse diagram of `A` figure

`J=mvimpliesJ=1/2mucostheta`
`=1/2musqrt(3)/2=(sqrt(3)mu)/3`
velocity of ball `B` in the direction perpendicular to the string will remain constant. Hence
`v'=usintheta=u/2`
Hence net velocity of `B` is
`v_B=sqrt((u/2)^(2)+(v)^(2))=(sqrt(7)u)/4` and `v_(A)=(sqrt(3)u)/4`