Find the speed at which the kinetic energy of a particle will differ by 1% from non-relativistic value` 1/ 2 m_0 V^(2)`.

Explanation

`((m_0c^(2) / (sqrt 1 -v^(2)/c^(2) ) - m_0c^(2) )-1 /2m_0v^(2) / 1 / 2 m_0v^(2) = 0.01))`
`rArr (m_0c^(2) (1 +v^(2)// 2 xx c^(2) + 1//2 xx 3 // 4v^(2) / c^(4) + 1//2 xx 3//4xx 5 // 6v^(2) // c^(2) + m_0c^(2)] - 1// 2 m_0v^(2)) /(( 1 // 2m_0v^(2)) = 0.1`
`rArr 1 /2 m_0v^(2) + 3 /8m_0v^(4) / c^(2) + 15 / 96m_0v^(6)/c^(4)- 1 /2m_0v^(2) / 1 / 2m_0v^(2) = 0.1 `
`rArr 1 /2 m_0v^(2) / c^(2)+ 15 / 96 m_0v^(6) / c^(4)- 1 / 2m_0v^(2) / 1 / 2 m_0v(2) = 0.01 `
` rArr (3 / 8 m_0v^(2)/ c^(2) + 15/96m_0V^(2)/ c^(4)) /( 1/ 2m_0v^(2))) = 0.01`
` rArr 3 / 4 v^(2) / c^(2) + 15 xx 2v^(2) / 96- v^(2) / c^(2) = 0.01`
Neglecting the `v^(4)` term as it is very small
`rArr (3 / 4) v^(2) / c^(2) = 0.01 `
` rArr v^(2) / c^(2) = 0.04 / 3`
`rArr v/ c = 0.2 / (sqrt 3)`
` rArr v = 2 / 2(sqrt 3)`
` c = 0.02 / 1.732 xx 3 xx 10^(8) = 0.6 / 10722 xx 10^(8) `
` = 0.345 xx 10m/s = 3.46 xx 10^(7)m/s` .