If the speed of a particle moving at a relativistic speed is doubled, its linear momentum will _____________ .

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become more than double If a particle is moving at a relativistic speed v, its linear momentum (p) is given as, p = m_o v/$\sqrt$(1 - v^2/c^2) \Rightarrow p = m_o v \left( 1 - v^2/c^2 \right)^- $\frac{1}{2}$ Expanding binomially and neglecting higher terms we have, p \simeq m_o v\left( 1 + v^$\frac{2}{2}$ c^2 \right) \Rightarrow p \simeq m_o v + m_o v^$\frac{3}{2}$ 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 - 4 v^2/c^2)} \Rightarrow p' = 2 m_o v \left( 1 - 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 + 4 v^$\frac{2}{2}$ c^2 \right) \Rightarrow p' \simeq 2 m_o v + 4 m_o v^3/c^2 \therefore p' \simeq 2p + 3 m_o v^3/c^2, 3 m_o v^3/c^2 > 0 Therefore, p' is more than double of p.

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