Chemistry, asked by abdulshukur973, 9 months ago

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:

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Answered by Anonymous
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\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}} }

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