Circular motion with freely moving balls 
Two identical small balls of mass $m$ each are connected to each
other with the help of an inextensible massless string of length
$L$. The whole arrangement is placed on a horizontal frictionless
table with the string just slack. Now, ball B is given a sudden
horizontal velocity $v_0$ at angle $60°$ to the string.

I did not understand what will happen in this condition, there are two balls connected with a string and one has a horizontal velocity and a vertical velocity. Now I understand that circular motion will take place, so my question is how can we calculate the tension in the string when it becomes taut?
 A: Initial Impulse
Assume ball A is at $x=y=0$ and B is at $(1,0)$. B is given an “impulse” (instantaneous transfer of momentum) in the polar direction $\theta = 60^o$ or $v_B= |v_B| (cos 60 \hat{i} + sin 60 \hat{j})$ where $v$ is a vector and $|v|$ its magnitude.
That is the instantaneous initial velocity at $t=0$ for B. What is it for A? Any string can only transfer force (or impulse) along its direction, so the only component of $v_B$ given to A is along the direction of the string, or the $x$ component: $v_A= v_{B,x} = |v_B|cos 60 \hat{i}$
Description of Motion
This can be considered its initial condition, and how it responds from here will depend on the masses of A and B. Initially, from a still frame this system has translation of its center of mass (com) and rotation about its com (the $\omega$ vector is pointing out of the page at us).
The initial translation momentum is $m_Av_A + m_Bv_B$ and by conservation of momentum that will stay at that value if frictionless. This sets the velocity vector of the system com. The angular momentum (around the system’s com) is also conserved. But from the perspective of each ball, the string will pull on it (and on the other ball the opposite direction and same magnitude) in the direction of the string. But we don’t need to worry about that we can model the whole system by the translation of its com and rotation about that com, and know where each ball is.
Calculations
Initial momentum:
$$m_Av_A + m_Bv_B =m_A |v_B|cos 60 \hat{i} +m_B |v_B| (cos 60 \hat{i} + sin 60 \hat{j})$$
$$=|v_B| [(m_A+m_B) cos 60 \hat{i} + m_B sin 60 \hat{j}]$$
To get the velocity from that momentum vector, just divide by the system’s mass $m_A+m_B$: $$v_{\text{com}}= |v_B|[cos 60 \hat{i} + \tfrac{m_B}{m_A+m_B}sin 60 \hat{j}]$$ The com will stay on that path indefinitely
But what exactly is following that path? Because we put A at origin and normalized string to 1: $$\text{com}= \tfrac{m_B}{m_A+m_B} \hat{i}$$
And always $\tfrac{m_B}{m_A+m_B}$ distance from A. That is the point that travels along the $v_{\text{com}}$ vector. A and B will rotate around it in a moving frame as if it were a fixed point, always with a relative velocity perpendicular to the string. For example at the beginning, $v_A$ in the moving frame is $$v_A-v_ {\text{com}} = -\tfrac{m_B}{m_A+m_B}sin 60 \hat{j}$$ The angular velocity is given by $\omega=|\tfrac{v}{r}|$
$$ \omega = \tfrac{ \tfrac{m_B}{m_A+m_B}sin 60 \hat{j}}{\tfrac{m_B}{m_A+m_B}} = sin 60 $$
The com and $v_ {\text{com}} $ and $\omega$ characterize the whole system.
