When the speed of a particle travelling at a relativistic speed is...

The relativistic momentum of a particle is given by $p = \dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

When the speed is $v$, the initial momentum is $p_1 = \dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

When the speed is doubled to $2v$, the new momentum is $p_2 = \dfrac{m(2v)}{\sqrt{1 - \dfrac{4v^2}{c^2}}}$

Comparing $p_2$ with $2p_1$:

$2p_1 = \dfrac{2mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Since $1 - \dfrac{4v^2}{c^2} < 1 - \dfrac{v^2}{c^2}$, it follows that $\sqrt{1 - \dfrac{4v^2}{c^2}} < \sqrt{1 - \dfrac{v^2}{c^2}}$

Therefore, $\dfrac{1}{\sqrt{1 - \dfrac{4v^2}{c^2}}} > \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

This implies $p_2 > 2p_1$

Thus, the linear momentum becomes more than double.