Question

Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy (1/2mv2) when a particle has a speed of (a) 2.71 x 10-3c. and (b) 0.855c.

Answer 1

equation to determine the value of relativistic kinetic energy

`KE_r = m_oc^2({\gamma - 1})`

`KE_r = m_oc^2({\frac{1}{\sqrt{1-(v^2)/(c^2)}} - 1}})`

`KE = (m_ov^2)/(2)`

the required ratio

`(KE_r)/(KE) = \frac{m_oc^2({\frac{1}{\sqrt{1-(v^2)/(c^2)}} - 1})}{(m_ov^2)/(2)}`

`(KE_r)/(KE) = \frac{2c^2({\frac{1}{\sqrt{1-(v^2)/(c^2)}} - 1})}{v^2}`

a)when particle has speed of v = 2.71 x 10⁻³ c

`(KE_r)/(KE) = \frac{2c^2({\frac{1}{\sqrt{1-((2.71* 10^(-3))^2)/(c^2)}} - 1})}{(2.71* 10^(-3))^2}`

`(KE_r)/(KE) =1`

b) when particle has speed of v = 0.855 c

`(KE_r)/(KE) = \frac{2c^2({\frac{1}{\sqrt{1-((0.855)^2)/(c^2)}} - 1})}{(0.855)^2}`

`(KE_r)/(KE) = 2.54`