Question

Calculate the relativistic momentum of a particle of mass $1.76\times {{10}^{-27}}kg$ if the relativistic energy is equal to three times the rest energy.
a)$2.86\times {{10}^{-18}}kgm/s$
b)$9.68\times {{10}^{-18}}kgm/s$
c)$2.05\times 10kgm/s$
d)$1.29\times 10kgm/s$

Answer 1

Relativistic case is used when particle velocity is compared to the speed of light. The relativistic momentum is defined as the $\gamma$ times rest mass times velocity of the particle.

Formula used:
1. Relativistic momentum $p$ is,
$p=\gamma {{m}_{0}}v$
Here,$\gamma =\dfrac{1}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$, $v$ is velocity of particle, $m$ rest mass of the particle,$c$ is speed of light i.e. $3\times {{10}^{8}}m/s$ .
2. Relativistic kinetic energy ${{E}_{k}}$,
${{E}_{k}}=\left( \gamma -1 \right){{m}_{0}}{{c}^{2}}$
Here, ${{m}_{0}}{{c}^{2}}$ is rest mass energy ${{E}_{0}}$ of the particle.

Complete step by step answer:
You have given,
Mass ${{m}_{0}}=1.76\times {{10}^{-27}}kg$
The Relativistic kinetic energy ${{E}_{k}}$is three times the rest mass energy ${{E}_{0}}$,
i.e. ${{E}_{k}}=3{{E}_{0}}......(1)$
you have to find relativistic momentum $p$.
$p=\gamma {{m}_{0}}v......(2)$
To find the relativistic momentum $p$first you have to calculate the velocity $v$and the relativistic factor $\gamma$,
The relativistic kinetic energy ${{E}_{k}}$ ,
${{E}_{k}}=\left( \gamma -1 \right){{m}_{0}}{{c}^{2}}......(3)$
The rest mass energy ${{E}_{0}}$ of the particle is
${{E}_{0}}={{m}_{0}}{{c}^{2}}......(4)$
Put the values of equation 2 and 3 in equation 1,
$\left( \gamma -1 \right){{m}_{0}}{{c}^{2}}=3{{m}_{0}}{{c}^{2}}$
Expand,
$\gamma {{m}_{0}}{{c}^{2}}-{{m}_{0}}{{c}^{2}}=3{{m}_{0}}{{c}^{2}}$
Solve for $\gamma$
$\begin{aligned} & \gamma {{m}_{0}}{{c}^{2}}=4{{m}_{0}}{{c}^{2}} \\\ & \Rightarrow \gamma =4......(5) \\\ \end{aligned}$
Now, calculate velocity $v$,
You know,
$\gamma =\dfrac{1}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$
Put value of $\gamma$
$4=\dfrac{1}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$
Take reciprocal of equation,
$\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}=\dfrac{1}{4}$
Square both sides,
$1-\dfrac{{{v}^{2}}}{{{c}^{2}}}=\dfrac{1}{16}$
Simplify,
$\dfrac{{{v}^{2}}}{{{c}^{2}}}=\dfrac{15}{16}$
Solve for $v$, multiply both sides by ${{c}^{2}}$ and take square root,
$v=\sqrt{\dfrac{15}{16}}c......(5)$
Put the values of equation 5 and 6 in equation 2
$p=4\times 1.76\times {{10}^{-27}}\times \sqrt{\dfrac{15}{16}}3\times {{10}^{8}}$
$\therefore p=2.86\times {{10}^{-18}}kgm/s$.

So, the correct answer is “Option A”.

Note:
Students generally get confused with rest mass energy and rest mass kinetic energy.
So, keep clear in mind that rest mass kinetic energy means the energy of particle at rest i.e. kinetic energy is zero but the rest mass energy term comes from relativistic physics that is given by the Einstein formula of energy mass conservation,
i.e. rest mass energy ${{E}_{0}}={{m}_{0}}{{c}^{2}}$