A light inelastic thread passes over a small frictionless pulley....

Let us assume that just after the collision, the speed of the particle plus $1$ kg block is $u$

$2u = 1 \times 10 ⇒ u = 5ms^{- 1}$

After the collision and until the string is taut again, the particle block system is in free fall. So just before the string is taut, the particle plus $1$ kg block will be moving downward with a velocity of $5ms^{- 1}$ and just after the string is taut, let us assume that the speed of both the blocks and the particle is $v$, then

$2 \times 5 = 5v$

$⇒ v = 2ms^{- 1}$