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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.