Answer :
Answer
equation to determine the value of relativistic kinetic energy
[tex]KE_r = m_oc^2({\gamma - 1})[/tex]
[tex]KE_r = m_oc^2({\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} - 1}})[/tex]
[tex]KE = \dfrac{m_ov^2}{2}[/tex]
the required ratio
[tex]\dfrac{KE_r}{KE} = \dfrac{m_oc^2({\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} - 1})}{\dfrac{m_ov^2}{2}}[/tex]
[tex]\dfrac{KE_r}{KE} = \dfrac{2c^2({\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} - 1})}{v^2}[/tex]
a)when particle has speed of v = 2.71 x 10⁻³ c
[tex]\dfrac{KE_r}{KE} = \dfrac{2c^2({\dfrac{1}{\sqrt{1-\dfrac{(2.71\times 10^{-3})^2}{c^2}}} - 1})}{(2.71\times 10^{-3})^2}[/tex]
[tex]\dfrac{KE_r}{KE} =1[/tex]
b) when particle has speed of v = 0.855 c
[tex]\dfrac{KE_r}{KE} = \dfrac{2c^2({\dfrac{1}{\sqrt{1-\dfrac{(0.855)^2}{c^2}}} - 1})}{(0.855)^2}[/tex]
[tex]\dfrac{KE_r}{KE} = 2.54[/tex]