Answer
$v=\frac{c}{\sqrt{1+(\frac{m_{\circ}c}{p})^2}}$
Work Step by Step
We know that the relativistic momentum is given as
$p=\frac{m_{\circ}v}{\sqrt{1-\frac{v^2}{c^2}}}$
Squaring both sides, we obtain:
$p^2=\frac{m_{\circ}^2v^2}{1-\frac{v^2}{c^2}}$
This simplifies to:
$p^2=v^2[m_{\circ}^2+\frac{p^2}{c^2}]$
This can be rearranged as
$v^2=\frac{p^2}{m_{\circ}^2+\frac{p^2}{c^2}}$
$\implies v=\frac{c}{\sqrt{1+(\frac{m_{\circ}c}{p})^2}}$
