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Assume we have an object of 1 kg, at rest and we invest 100 Joules of energy to accelerate it. The resultant velocity can be calculated by

$$ v = \sqrt{\frac{2K_{e}}{m}} $$

so, $$ \sqrt{\frac{2 (100)}{1}} \simeq 14.14 m/s $$

But because of relativity if we invest more and more energy, we won't get the same rise in the velocity as the relativistic mass goes on increasing and resultant velocity rises more and more slowly and it will approach at best the speed of light on spending infinite amount of energy. What is the formula or method to calculate the final relativistic velocity of an object of mass $m$ if I invest $K_{e}$ amount of energy to it?

Srinivas N Rao
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Your first formula is incorrect because it's using the newtonian version of the kinetic energy. In special relativity it becomes:

$$K_e=(\gamma-1)mc^2=\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)mc^2$$

After a bit of elbow grease, you get:

$$v=c\sqrt{1-\frac{1}{\left(1+\frac{K_e}{mc^2}\right)^2}}$$

Notice the approximation $K_e/mc^2\ll 1$ restores the formula of Newtonian mechanics.

Srinivas N Rao
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Miyase
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  • Your first formula is incorrect because it's using the newtonian version . Yes I know it, that is why I'm asking the question. Thank you for the answer. – Srinivas N Rao May 19 '22 at 11:11
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    Hey I worked out the formula given by you seem to be a missing a square. I have submitted an edit please accept it – Srinivas N Rao May 19 '22 at 11:21
  • You're right, I missed it when I typed the formula. Your edit was accepted. Thanks. – Miyase May 19 '22 at 11:30
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    Hey I have upvoted your answer, but my reputation is very less so it's not showing. Thanks for the quick and precise answer, really appreciate it – Srinivas N Rao May 19 '22 at 11:31