Two mass-less strings of length $5m$ hang from the ceiling very near to each other as shown in the figure. Two balls $A$ and $B$ of masses $0.25kg$ and $0.5kg$ are attached to the string. The ball $A$ is released from rest at a height $0.45m$, as shown in the figure. The collision between two balls is completely elastic. Immediately after the collision, the kinetic energy of ball $B$ is $1J$. The velocity of the ball $A$, just after the collision is

The correct option is D $1m{s}^{-1}$ to the left
Given
$m_B=0.5Kg$
$m_A=0.25Kg$
$h=0.45m$
Velocity of ball A just before it collides with B is $v_{Ai}$

$m_{A}gh=\frac{1}{2}{m}_A(v_{Ai}{)}^2$

Hence $v_{Ai}=3$
$E_B=1J$
Now, after collision velocity of ball B can be obtained from

$E_B=\frac{1}{2}{m}_B(v_B{)}^2$

$\Rightarrow (v_B{)}^2=\frac{2}{m_B}$

$\therefore (v_B{)}^2=\frac{2}{0.5}$

$\therefore (v_B{)}^2=4$
$\therefore v_B=2$
Since, in collision elastic momentum is conserved hence,
$m_A{v}_{Ai}=m_A{v}_A+m_B{v}_B$
$.25\times 3=.25\times v_A+.5\times 2$
$.75=.25v_A+1$
$v_A=-1m{s}^{-1}$
Hence, the velocity of ball A just after the collision is $1m{s}^{-1}$, to the left.