Question

Find the speed at which the kinetic energy of a particle will differ by 1% from its nonrelativistic value \[\frac{1}{2} m_o v^2 .\]

Answer 1

If m₀ is the rest mass of a particle and c is the speed of light, then relativistic kinetic energy of particle = mc² − m₀c².   ........(1)

If \[m = \frac{m_0 c^2}{\sqrt{1 - v^2 / c^2}},\] then nonrelativistic kinetic energy of particle = \[\frac{1}{2} m_o v^2.\]

According to the question,

\[\frac{\left( \frac{m_0 c^2}{\sqrt{1 - v^2 / c^2}} - m_0 c^2 \right) - \frac{1}{2} m_0 v^2}{1/2 m_0 v^2} = 0 . 01\]
\[ \Rightarrow \left[ \frac{c^2 \left( 1 + \frac{v^2}{2 c^2} + \frac{1}{2} \times \frac{3}{4}\frac{v^2}{c^2} + \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6}\frac{v^6}{c^6} - 1 \right)}{\frac{1}{2} v^2} \right] = 0 . 01\]
\[ \Rightarrow \frac{\frac{1}{2} m_0 v^2 + \frac{3}{8} m_0 v^4 / c^2 + \frac{15}{96} m_0 v^6 / c^4 - \frac{1}{2} m_0 v^2}{\frac{1}{2} m_0 v^2} = 0 . 01\]
\[ \Rightarrow \frac{\frac{3}{8} m_0 v^2 / c^2 + \frac{15}{96} m_0 V^6 / c^4}{\frac{1}{2} m_0 v^2} = 0 . 01\]
\[ \Rightarrow \frac{3}{4} \frac{v^2}{c^2} + \frac{15 \times 2 v^4}{96}\frac{v^2}{c^2} = 0 . 01\]

Neglecting the v⁴ term as it is very small, we get

\[\frac{3}{4} \frac{v^2}{c^2} = 0 . 01\]
\[ \Rightarrow \frac{v^2}{c^2} = \frac{0 . 04}{3}\]
\[ \Rightarrow \frac{v}{c} = \frac{0 . 2}{\sqrt{3}}\]
\[ \Rightarrow v = \frac{2}{2\sqrt{3}}\]
\[c = \frac{0 . 02}{1 . 732} \times 3 \times {10}^8 = \frac{0 . 6}{1 . 722} \times {10}^8 \]
\[ = 0 . 345 \times 10 m/s = 3 . 46 \times {10}^7 m/s\]