Question

According to the Theory of Relativity, the mass mm of a particle depends on its velocity vv. That is

m=mo1v2c2m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}

where mom_o is the mass when the particle is at rest and cc is the speed of light. Find the limit of the mass, mm, as vv, approaches cc^{-}.

Solution

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Answered 2 years ago
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If we have the information that vv approaches to cc^-, this means that vv approaches to cc from the left side, and vv is slightly less than cc, but vcv\neq c.

Based on this, we can say that the value vc\dfrac{v}{c} will be slightly less than 11, and vc1\dfrac{v}{c}\neq1. Also (vc)2\left({\dfrac{v}{c}}\right)^2 will be slightly less than 11, but it won't be equal to 11.

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