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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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1 Answer 1

It is. At low speeds, the energy of a particle can be suitably approximated as $E = mc^2 + \frac{1}{2}mv^2$.

The true formula, which is valid at any speed, is

$$E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

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.

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