A positively charged particle moves in mutually perpendicular electric field  $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ such that both are uniform and constant. At time, $t=0$  its velocity is ${\overrightarrow{v}}_o$  (${\overrightarrow{v}}_o\perp \overrightarrow{B}$ and  ${\overrightarrow{v}}_o\perp \overrightarrow{E};$  figure). It is given  $E=v_{o}B.$ Let the velocity vector of the particle at the moments of time when its velocity vector makes an angle ${180}^0$ with initial velocity vector ${\overrightarrow{v}}_o$  be $\overrightarrow{v}$ . Then find the value of  $\frac{\left|\overrightarrow{v}\right|}{\left|{\overrightarrow{v}}_o\right|}$

Let's write the equation of particle motion along the X axis (Fig.): 
 $m\frac{d{v}_x}{dt}=-q{v}_{y}B$
Where   $m$ is mass, $q$  -is  particle charge. The solution to this equation has the form   $v_x\left(t\right)=v_0-\frac{qB}{m}y$
At those moments in time $t=t_n,$  when the particle velocity $\overrightarrow{v}$  makes angle be180 with  the initial velocity vector,
  $v_x\left(t_n\right)=-vandy\left(t_n\right)=\frac{v_0+v}{\frac{qB}{m}}$
 According to the law of conservation of energy, $\frac{m{v}_0^2}{2}+qEy\left(t_n\right)=\frac{m{v}^2}{2}.$  Provided that $E=v_{o}B$ we get a quadratic equation $v^2-2v_{0}v-3v_0^2=0$ , from where we find the required speed:  $v=3v_0$