Explanation:
It is given that,
The rest mass of the particle, [tex]m_o=10^{-30}\ kg[/tex]
Speed of the particle, v = 0.5c
Firstly calculating the relativistic factor. The formula is as follows :
[tex]\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]
[tex]\gamma=\dfrac{1}{\sqrt{1-\dfrac{(0.5c)^2}{c^2}}}[/tex]
[tex]\gamma=1.032[/tex]
1. Inertial mass is given by :
[tex]m=\gamma \times m_o[/tex]
[tex]m=1.032 \times 10^{-30}[/tex]
2. The momentum is given by :
[tex]p=\gamma\times mv[/tex]
[tex]p=1.032\times 1.032 \times 10^{-30}\times 0.5c[/tex]
[tex]p=1.59\times 10^{-22}\ kg-m/s[/tex]
3. Kinetic energy of the particle is given by :
[tex]E_k=(\gamma-1)mc^2[/tex]
[tex]E_k=(1.032-1)\times 1.032 \times 10^{-30}\times (3\times 10^8)^2[/tex]
[tex]E_k=2.97\times 10^{-15}\ J[/tex]
4. The total energy of the particle is :
[tex]E=\gamma mc^2[/tex]
[tex]E=1.032\times 1.032 \times 10^{-30}\times (3\times 10^8)^2[/tex]
[tex]E=9.58\times 10^{-14}\ J[/tex]
Hence, this is the required solution.