Question

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

विकल्प

• become double

• become more than double

• remain equal

• become less than double

Answer 1

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.