Two particles `A` and `B` of equal mass `m` are attached by a string of length `2l` and initially placed over a smooth horizontal table in the positoin shown in fig. particle `B` is projected across the table with speed `u` perpendicular to `AB` as shown in the figure. find the velocities of each particle after the string becomes taut and the magnitude of the impulse tension.

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`