# Q 2.15: A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_{o}$ of a particle in terms of its speed $v$ and the speed of light, $c$. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : $m = \frac{m_{o}}{(1-v^2)^{1/2}}$Guess where to put the missing c.

The relation given is $m = \frac{m_{o}}{(1-v^2)^{1/2}}$

Divide both sides by $m_{o}$;  $\therefore$ L.H.S becomes $m / m_{o}$ which is dimensionless.

Hence, R.H.S must be dimensionless too. (After Dividing by $m_{o}$ !)

$\frac{1}{(1-v^2)^{1/2}}$ can be dimensionless only when $v \rightarrow (v/c)$

Therefore, the dimensional equation is $m = \frac{m_{o}}{(1-(\frac{v}{c})^2)^{1/2}}$

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