Question

At what speed is a particle's momentum 2.5 times more than its newtonian value?

Answer 1

The relativistic momentum of a particle moving with speed v is equal to:
$p=\gamma m_0 v$
where
$m_0$ is the rest mass of the particle
$\gamma= \frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }$ is the relativistic factor, with c being the speed of light.

The Newtonian momentum is instead
$p=m_0 v$

For the particle in our problem, the relativistic momentum is 2.5 times the Newtonian momentum: this means $\gamma=2.5$. If we re-arrange the formula of $\gamma$, we get:
$v=c \sqrt{1- \frac{1}{\gamma^2} }$
and by using $\gamma=2.5$, we find the particle velocity:
$v=(3 \cdot 10^8 m/s) \sqrt{1- \frac{1}{(2.5)^2}} =0.917 c = 2.75 \cdot 10^8 m/s$