Question

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

(a) become double
(b) become more than double
(c) remain equal
(d) become less than double.

Answer 1

(b) becomes more than double

If a particle is moving at a relativistic speed v, its linear momentum $\left(p\right)$ is given as,
$$\begin{gathered} p=\frac{m_{o}v}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \Rightarrow p=m_{o}v{\left(1-\frac{v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$
Expanding binomially and neglecting higher terms we have,
$$\begin{gathered} p\simeq m_{o}v\left(1+\frac{v^2}{2c^2}\right) \\ \Rightarrow p\simeq m_{o}v+\frac{m_o{v}^3}{2c^2} \end{gathered}$$
If the speed is doubled, such that it is travelling with speed 2v ,linear momentum will be given as
$$\begin{gathered} p\text{'}=\frac{m_o\left(2v\right)}{\sqrt{1-{\displaystyle \frac{4v^2}{c^2}}}} \\ \Rightarrow p\text{'}=2m_{o}v{\left(1-\frac{4v^2}{c^2}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$
Expanding binomially and neglecting higher terms we have,
$$\begin{gathered} p\text{'}\simeq 2m_{o}v\left(1+\frac{4v^2}{2c^2}\right) \\ \Rightarrow p\text{'}\simeq 2m_{o}v+\frac{4m_o{v}^3}{c^2} \\ \therefore p\text{'}\simeq 2p+\frac{3m_o{v}^3}{c^2},\frac{3m_o{v}^3}{c^2}>0 \end{gathered}$$
Therefore, p' is more than double of p.