In the shown figure, pulleys and strings are ideal and horizontal surface is smooth. The block C (mass 2m) is given a horizontal velocity of ${\mathrm{V}}_0=3\mathrm{m}/\mathrm{s}$  Just after the strings regain tension, velocities of A , B and C are $V_A, V_B and V_c respectively.$ Then    (   $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$ )
 

 

Before the strings get taut, block C moves with constant velocity of 3 m/s and the blocks A and B (along with the pulley) fall with an acceleration g. The strings recover tension, when distance travelled by C = distance travelled by the falling pulley
$$\begin{gathered} \therefore \frac{1}{2}\times g{t}^2=3t \\ \Rightarrow t=\frac{6}{10}=0.6s \end{gathered}$$
Speed of A and B at this instant is 6 m/s. During the short interval in which tension is regained, let the impulse of string tension on 
$\mathrm{C}\text{ be }\int \mathrm{Tdt}=\mathrm{I}(\to )$
Impulse on A and B both will be $\frac{\mathrm{I}}{2}(\uparrow )$
For (C)
$$\begin{gathered} \mathrm{I}=2{\mathrm{mV}}_{\mathrm{C}}-2\mathrm{m}.3 \\ \frac{\mathrm{I}}{\mathrm{m}}=2{\mathrm{V}}_{\mathrm{C}}-6 ............\left(\mathrm{i}\right) \end{gathered}$$     [VC =velocity of C after string regains tension]
For A , $-\frac{\mathrm{I}}{2}=2{\mathrm{mV}}_{\mathrm{A}}+2\mathrm{m}.6$
     $\therefore \frac{\mathrm{I}}{\mathrm{m}}=24-4{\mathrm{V}}_{\mathrm{A}} .....\left(\mathrm{ii}\right)$
for B $-\frac{\mathrm{I}}{2}={\mathrm{mV}}_{\mathrm{B}}-\mathrm{m}.6$
     $\frac{\mathrm{I}}{\mathrm{m}}=12-2{\mathrm{V}}_{\mathrm{B}} .......\left(\mathrm{iii}\right)$
And after the strings are taut, velocity of A and B relative to the falling pulley must be equal and opposite.

$\therefore {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{C}}=-\left({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{C}}\right)$
$\therefore {\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}=2{\mathrm{V}}_{\mathrm{C}}\cdots \cdots \cdots \cdots \cdots \cdots \cdots \text{ (iv) }$
      From (i) and (ii) $2{\mathrm{V}}_{\mathrm{C}}-6=24-4{\mathrm{V}}_{\mathrm{A}}$
$\therefore {\mathrm{V}}_{\mathrm{C}}+2{\mathrm{V}}_{\mathrm{A}}=15$     …..(v)

 From (iv) ,(iii) $2V_C-6=12-2V_B$

$\therefore {\mathrm{V}}_{\mathrm{C}}+{\mathrm{V}}_{\mathrm{B}}=9\dots \dots \dots \dots \dots \dots \dots \left(\mathrm{vi}\right)$
Solving (iv), (v) and (vi)
${\mathrm{V}}_{\mathrm{C}}=\frac{33}{7}\mathrm{m}/\mathrm{s},{\mathrm{V}}_{\mathrm{A}}=\frac{36}{7}\mathrm{m}/\mathrm{s},{\mathrm{V}}_{\mathrm{B}}=\frac{30}{7}\mathrm{m}/\mathrm{s}$