Answer
$\frac{1}{4}mv^2$
Work Step by Step
We can find the required final kinetic energy of the system as follows:
After collision the total mass $m^{\prime}=m+m=2m$
Initial momentum is given as
$P_i=mv+(m\times 0m/s)=mv$
and final momentum is given as $P_f=2mv^{\prime}$
According to law of conservation of momentum
$P_i=P_f$
$\implies mv=2mv^{\prime}$
$\implies v^{\prime}=\frac{v}{2}$
Now the total kinetic energy of the system is
$K.E=\frac{1}{2}m^{\prime}v^{\prime 2}$
$\implies K.E=\frac{1}{2}(2m)(\frac{v}{2})^2$
$K.E=\frac{1}{4}mv^2$
