The kinetic energy possessed by a body of mas | vedclass

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(B) The classical expression for kinetic energy,$K = \frac{1}{2}mv^2$,is derived from Newtonian mechanics.According to Einstein's theory of special re

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The body moves with velocities negligible compared to the speed of light

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(B) The classical expression for kinetic energy,$K = \frac{1}{2}mv^2$,is derived from Newtonian mechanics.According to Einstein's theory of special relativity,the total energy of a body is given by $E = \gamma mc^2$,where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$.The kinetic energy is the difference between the total energy and the rest energy: $K = E - mc^2 = mc^2(\gamma - 1)$.Using the binomial expansion for $\gamma$ when $v \ll c$,we get $\gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2}$.Substituting this back,$K \approx mc^2(1 + \frac{1}{2}\frac{v^2}{c^2} - 1) = \frac{1}{2}mv^2$.Thus,the classical formula is valid only when the velocity of the body is negligible compared to the speed of light.

The kinetic energy possessed by a body of mass $m$ moving with a velocity $v$ is equal to $\frac{1}{2}mv^2$,provided

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