# If a nuclear bomb is dropped, do we calculate $+p^2c^2$ too?

Imagine a nuclear bomb is dropped from $10 \: \mathrm{km}$ with the speed of $v$, since $E=\sqrt{m^2c^4+p^2c^2}$ , we have speed which means $p$ is greater than zero, but isn't the speed already calculated in $m^2c^4$ part, since $m$ is relativistic? Don't we only calculate $p^2c^2$ when it's photon?

m is the (velocity independent) mass of the system. You would most likely call it rest mass (but there is no need to call it that way, as there is no other mass - just call it mass). So the kinetic energie of the falling bomb is only in the $p^2\,c^2$ part of the equation.
What you call relativistic mass is an old concept which sadly still made it to text books and into student's minds. With it, the used equation would look like $E=m_{rel}\,c^2$. The equation you used is already split in the rest energy and kinetic energy part. The momentum in this equation can behave relativistically if you go to velocities close to the speed of light. So you will see, that the second term is almost negligible if you are at normal drop speeds of a bomb and that the rest energy term is the dominant one.
• relativistic mass as an "old concept"? Authoritative source requested. Working with collisions between a mass at the speed of $0.8c$ and another at $0.01c$, you'd want to work in COM ref frame, but then you need to calculate the values of the masses in a relativistic reference frame. Feb 14, 2018 at 13:39
• See for example the answer to this question and the references given there. Clearly, if you use the equation $E = \sqrt(m^2c^4+p^2c^2)$ the m is the invariant mass, rest mass, or just mass, whatever you wanna call it. Feb 14, 2018 at 13:44