A famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_{0}$ 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:
$m=\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}$
Guess where to put the missing $c$
A famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_{0}$ 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:
$m=\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}$
Guess where to put the missing $c$
Given the relation, $m=\frac{m_{0}}{\left(1-v^{2}\right)^{\frac{1}{2}}}$
Dimension of $m= M ^{1} L ^{0} T ^{0}$
Dimension of $m_{0}= M ^{1} L ^{0} T ^{0}$
Dimension of $v= M ^{0} L ^{1} T ^{-1}$
Dimension of $v^{2}= M ^{0} L ^{2} T ^{-2}$
Dimension of $c= M ^{0} L ^{1} T ^{-1}$
The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S.
This is only possible when the factor, $\left(1-v^{2}\right)^{1 / 2}$ is dimensionless i.e., $\left(1-v^{2}\right)$ is dimensionless. This is only possible if $v^{2}$ is divided by $c^{2} .$
Hence, the correct relation is
$m=\frac{m_{0}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}}$
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