Question

A 1.0 g particle has momentum 400,000 kgm/s. What is the particle’s speed in m/s?

Short answer

The speed of particle is$2.4\times {10}^{8}m/s$.

Step-by-step solution

Given information 

We have given that, a 1.0 g particle has momentum 400,000 kgm/s.

Step 1.

We will use relativistic momentum formula to get the value of photon's velocity.

Formula of relativistic momentum is :

$\rho =\frac{mv}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

Here, m = mass of photon

v = velocity of photon

c = speed of light

$\rho$= relativistic density

Step 2.

Substituting the values in the formula :

$400,000kgm/s=\frac{(1.0\times {10}^{-3}kg)\times v}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$\sqrt{1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}=\frac{1.0\times {10}^{-3}kg}{400,000kgm/s}c\left(\frac{v}{c}\right)$

$\sqrt{1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}=\frac{3}{4}\left(\frac{v}{c}\right)$

$1-\frac{v^2}{c^2}=\frac{9}{16}\frac{v^2}{c^2}$

$v=2.4\times {10}^{8}m/s$