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Question

According to the Theory of Relativity, the mass $m$ of a particle depends on its velocity $v$. That is

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

where $m_o$ is the mass when the particle is at rest and $c$ is the speed of light. Find the limit of the mass, $m$, as $v$, approaches $c^{-}$.

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 5If we have the information that $v$ approaches to $c^-$, this means that $v$ approaches to $c$ from the left side, and $v$ is slightly less than $c$, but $v\neq c$.

Based on this, we can say that the value $\dfrac{v}{c}$ will be slightly less than $1$, and $\dfrac{v}{c}\neq1$. Also $\left({\dfrac{v}{c}}\right)^2$ will be slightly less than $1$, but it won't be equal to $1$.

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