Two identical small balls are interconnected with a light and inextensible thread having length L. The system is on a smooth horizontal table with the thread just taut. Each ball is imparted a velocity v, one towards the other ball and the other in a direction that is perpendicular to the velocity given to the first ball.

 After time T the thread becomes taut again? and  the kinetic energy of the system after the string gets taut is K. Then

(a) We will study the motion of second particle in the reference frame attached to the first particle. The velocity of second particle makes an angle of 45° with the initial line joining the two particles (see fig 2). The thread is loose before the distance between particles again becomes L. Fig.3 shows the situation just before the string gets taut.
Required time is $\mathrm{t}=\frac{\mathrm{L}\sqrt{2}}{\mathrm{v}\sqrt{2}}=\frac{\mathrm{L}}{\mathrm{v}}$
(b) In the reference frame of ground, velocities just before the string gets taut, has been shown in fig.4. The velocity component for the two particles along the string will be same for both particles after the string is taut. Fig.5 shows the situation immediately after the string gets taut. 
 

Now, total K.E. = $\frac{1}{2}\frac{{\mathrm{mv}}^2}{4}+\frac{1}{2}{\mathrm{mv}}^2\left(1^2+\frac{1}{2^2}\right)=\frac{3}{4}{\mathrm{mv}}^2$