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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 ' $\mathrm{c}$ '. (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.
$$
m=\frac{m_0}{\left(1-v^2\right)^{1 / 2}}
$$
Guess where to put the missing c.
Solution:
2061 Upvotes
Verified Answer
From the given equation, $\frac{m_0}{m}=\sqrt{1-v^2}$
Left hand side is dimensionless.
Therefore, right hand side should also be dimensionless.
It is possible only when $\sqrt{1-v^2}$ should be $\sqrt{1-\frac{v^2}{c^2}}$ Thus, the correct formula is
$$
m=m_0\left(1-\frac{v^2}{c^2}\right)^{-1 / 2}
$$
Left hand side is dimensionless.
Therefore, right hand side should also be dimensionless.
It is possible only when $\sqrt{1-v^2}$ should be $\sqrt{1-\frac{v^2}{c^2}}$ Thus, the correct formula is
$$
m=m_0\left(1-\frac{v^2}{c^2}\right)^{-1 / 2}
$$
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