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Q.
For a body moving with relativistic speed if the velocity is doubled, then
ManipalManipal 2013
Solution:
From Einsteins theorem
Relativistic momentum $=\frac{m_{0} v}{\sqrt{1-v^{2} / C^{2}}}$
Putting $v=2 v$
we get $\rho=\frac{m_{0}(2 v)}{\sqrt{1-(2 v)^{2} / c^{2}}}$
$=2\left(\frac{m_{0} v}{\sqrt{1-(2 v)^{2} / c^{2}}}\right)$
So the momentum becomes more than the double.