Answer
$E=[p^2c^2+m_{\circ}^2c^4]^{\frac{1}{2}}$
Work Step by Step
We know that
$E=\frac{m_{\circ}c^2}{\sqrt{1-\frac{v^2}{c^2}}}$
$\implies E^2=\frac{m_{\circ}^2c^4}{1-\frac{v^2}{c^2}}$
$\implies E^2=\frac{m_{\circ}^2c^4+m_{\circ}^2v^2c^2-m_{\circ}^2v^2c^2}{1-\frac{v^2}{c^2}}$
This simplifies to:
$E^2=[\frac{m_{\circ}v}{\sqrt{1-\frac{v^2}{c^2}}}]^2c^2+\frac{m_{\circ}^2c^4(1-\frac{v^2}{c^2})}{1-\frac{v^2}{c^2}}$......eq(1)
The relativistic momentum is given as $p=\frac{m_{\circ}v}{\sqrt{1-\frac{v^2}{c^2}}}$
Putting this value in eq(1), we obtain:
$E^2=p^2c^2+m_{\circ}^2c^4$
$\implies E=[p^2c^2+m_{\circ}^2c^4]^{\frac{1}{2}}$
