Two identical particles  of mass $m$ carry a charge $Q$ each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first particle from a large distance with a speed $v$. If d is the distance of closest approach then

$\frac{1}{2}m{v}^2=\frac{Q^2}{4\pi {\epsilon }_{0}d}$+$\frac{1}{2}\left(2m\right){\left(\frac{v}{2}\right)}^2$[As both of the particles are movable,by conservation of momentum,at the minimum separation 

                                                       their common velocity is v'=$\frac{mv}{2m}=\frac{v}{2}]$

$\Rightarrow d=\frac{Q^2}{\pi {\epsilon }_{0}m{v}^2}$

Hence, (1),(2) and (4) are correct. between x=0 to $x=\frac{2\alpha }{\beta }$ with mean position