A bit rusty, tried using $E=mc^2$ to figure out how much kinetic energy would 1kg of mass convert to, and then work backwards to figure out what would be the final velocity of a 1kg mass having kinetic energy equivalent to 1kg mass as kinetic energy.

Not sure how correct this is but here it goes: $E = 1kg\times c^2$ is the (kinetic) energy equivalent of 1kg mass. I am hazy about the next step, 1kg mass having kinetic energy of $1kg\times c^2$ w.r.t. to a stationary point would appear to be moving at what (relativistic) speed? (not sure even if the first step is correct)


If you take a mass $m=1\:\rm kg$ and turn it straight into energy, by whatever means, then indeed you will get an energy $E=mc^2$ (a.k.a. its rest energy) out of it.

If you then use all of that energy to accelerate a second $1\:\rm kg$ mass, that second mass will definitely be accelerated to relativistic domains. This means that you cannot use the old relationship $E_\mathrm{kin}=\frac12 mv^2$ between the particle's velocity $v$ relative to its initial rest frame and its kinetic energy $E_\mathrm{kin}$. Instead, you need to use the relativistic version for the total energy, which reads $$ E=\gamma mc^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}}, $$ where $\gamma=1/\sqrt{1-v^2/c^2}$ is known as the Lorentz factor and reduces to $1$ at zero velocity (thereby giving the rest energy $E=mc^2$).

In your case, the second mass already has its rest energy, and you're doubling this, so $$ E=\gamma mc^2 = 2mc^2, $$ which then gives you the equation $$\frac{1}{\sqrt{1-v^2/c^2}}=2$$ that you can solve for $v$ to give $v=\frac{\sqrt{3}}{2}c\approx 0.8660\,c$.


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