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Question
If the speed of a particle moving at a relativistic speed is doubled, its linear momentum will _____________ .
Options
become double
become more than double
remain equal
become less than double
Solution
become more than double
If a particle is moving at a relativistic speed v, its linear momentum (p) is given as,
\[p = \frac{m_o v}{\sqrt{1 - \frac{v^2}{c^2}}}\]
\[ \Rightarrow p = m_o v \left( 1 - \frac{v^2}{c^2} \right)^\frac{- 1}{2} \]
Expanding binomially and neglecting higher terms we have,
\[p \simeq m_o v\left( 1 + \frac{v^2}{2 c^2} \right)\]
\[ \Rightarrow p \simeq m_o v + \frac{m_o v^3}{2 c^2}\]
If the speed is doubled, such that it is travelling with speed 2v ,linear momentum will be given as
\[p' = \frac{m_o (2v)}{\sqrt{1 - \frac{4 v^2}{c^2}}}\]
\[ \Rightarrow p' = 2 m_o v \left( 1 - \frac{4 v^2}{c^2} \right)^\frac{- 1}{2} \]
Expanding binomially and neglecting higher terms we have,
\[p' \simeq 2 m_o v\left( 1 + \frac{4 v^2}{2 c^2} \right) \]
\[ \Rightarrow p' \simeq 2 m_o v + \frac{4 m_o v^3}{c^2}\]
\[ \therefore p' \simeq 2p + \frac{3 m_o v^3}{c^2}, \frac{3 m_o v^3}{c^2} > 0\]
Therefore, p' is more than double of p.
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