Find the density function of the kinetic energy, $E = \frac 12 mV^2$

Question

A particle of mass m has a random velocity, $V$ , which is normally distributed with parameters $μ = 0$ and $σ$ . Find the density function of the kinetic energy, $E = \frac 12 mV^2$. (Rice, 2.64).

I already have the solution to this exercise, however i do not understand some of the steps:

\begin{align} F_E(x) & = P(\frac12 mV^2 \leq x) \\ & = P(V^2 \leq \frac {2x}{m})\\ & = P(-\sqrt{\frac {2x}{m}} \leq V \leq \sqrt{\frac {2x}{m}} )\\ & = 2 P(0\leq V \leq \sqrt{\frac {2x}{m}} )\\ & = 2 \Big(F_{V}(\sqrt{\frac {2x}{m}}) - \frac12\Big)\\ \end{align}

The zero in the fourth line is because the velocity cannot be smaller than 0. But where does the 2 come from ? also i do not understand the $\frac12$ in the last line? I know that & \begin{align} & = 2 P(0\leq V \leq \sqrt{\frac {2x}{m}} )\\ & = 2 \Big(F_{V}(\sqrt{\frac {2x}{m}}) - F_{V}(0)\Big)\\ \end{align}

and V is Normally distributed. But why does it yield $\frac12$?

Then to get the density function $f_E(x)$ from $F_E(x)$. I also dont uinderstand this part. \begin{align} f_E(x) & = 2 \Big(F_{V}'(\sqrt{\frac {2x}{m}}) \times \sqrt{\frac {2}{m} }\frac12 \frac {1}{\sqrt{x}} \Big) \end{align}

Answer 1

You get from the second line to the fourth line via: $$P(V^2\leq\frac{2x}{m})=P(|V|\leq\sqrt{\frac{2x}{m}})=P(-\sqrt{\frac{2x}{m}}\leq-|V|\leq0)+P(0\leq|V|\leq\sqrt{\frac{2x}{m}})=2P(0\leq|V|\leq\sqrt{\frac{2x}{m}})$$

You shouldn't ignore the possibility of movement in the negative direction, resulting in negative velocity