A charge +q is ejected from infinity with velocity v towards +Q charge ,it stops at a distance of R from the +Q charge.At what distance it will stop if it is ejected with 2v velocity

Dear Student,

In the Ist case, applying the conservation of the energy
Potential energy of change +q when it stop at distance R from change +Q = Initial kinetic energy of the charge +q
$\frac{Qq}{4\pi {\epsilon }_{0}R}=\frac{1}{2}{m}_q{v}^2 ....\left(i\right)$

Consider that, when change +q is projected with velocity 2v is stop at distance R' from charge +Q.
 applying the conservation of the energy, we get
$\frac{Qq}{4\pi {\epsilon }_{0}R\text{'}}=\frac{1}{2}{m}_q(2v{)}^2 ....(ii)$

Dividing equation (i) by (ii), we get
$R\text{'}=\frac{R}{4}$

Regards,