Question

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 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:

Answer 1

$\bigstar$ Correct Question:

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo 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 :

$\sf m=\dfrac{m_0}{1-v^{2}}$

Guess where to put the missing c.

$\bigstar$ Given:

• The relation is $\sf m=\dfrac{m_0}{1-v^{2}}$

$\bigstar$ Solution:

We can get,

$\sf \dfrac{m_0}{m}= \sqrt{1-v^{2}} \: \: \dfrac{m_0}{m}$  is dimensionless. Therefore, the right hand side should also be dimensionless.

So, in this case

$\sf \sqrt{1-v^{2}}$ should become $\sf \sqrt{1-\dfrac{v^{2}}{c^{2}} }$

Therefore, $\sf m=m_0\sqrt{1-\dfrac{v^{2}}{c^{2}} }$