Question

In relativistic motion, a body quadruples its momenta, after its speed is

doubled to the initial one. Calculate the initial speed in unit of c.​

Answer 1

The initial speed in unit of c is c/√5

Explanation:

Let the initial speed be $u$ and final speed be $v$

As, the motion is relativistic, the momentum of s particle of mass m and speed v is given by

$p=\gamma mv$

Where,

$\gamma=1/\sqrt{1-\frac{v^2}{c^2}}$

or, $\frac{1}{\gamma}=\sqrt{1-\frac{v^2}{c^2}}$

Given that

$v=2u$

And

$p_f=4p_i$

Thus,

$\gamma_f mv=4\gamma_i mu$

$\implies \frac{mv}{\gamma_i}=4\frac{mu}{\gamma_f}$

$\implies (\sqrt{1-\frac{u^2}{c^2}})mv=4(\sqrt{1-\frac{v^2}{c^2}})mu$

$\implies (\sqrt{1-\frac{u^2}{c^2}})(2u)=4u\sqrt{1-\frac{(2u)^2}{c^2}}$

$\implies \sqrt{1-\frac{u^2}{c^2}}=2\sqrt{1-\frac{4u^2}{c^2}}$

$\implies 1-\frac{u^2}{c^2}=4(1-\frac{4u^2}{c^2})$

$\implies 15\frac{u^2}{c^2}=3$

$\implies \frac{u^2}{c^2}=\frac{3}{15}$

$\implies \frac{u^2}{c^2}=\frac{1}{5}$

$\implies \frac{u}{c}=\frac{1}{\sqrt{5}}$

$\implies u=\frac{c}{\sqrt{5}}$

Hope this answer is helpful.