# Kinetic energy transfer in matter annihilation?

What happens to the kinetic energy of matter when it is annihilated? Is it released in the resultant explosion? In that case shouldn't it be $E=(mc^2 + \frac{1}{2} mv^2)$ ?

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It is. At low speeds, the energy of a particle can be suitably approximated as $E = mc^2 + \frac{1}{2}mv^2$.
$$E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Note that both of these answers are the same: the first is the first two terms in a Taylor approximation of the second. The case in which $E = mc^2 + \frac{1}{2}mv^2$ is not appropriate is the case when the third term in the series, $\frac{3mv^4}{8c^2}$ is large enough to be important.