CBSE-XI-Physics
47: The Special Theory of Relativity
- #5If 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.digAnsr: bAns : (b) becomes more than double
If a particle is moving at a relativistic speed v, its linear momentum `` \left(p\right)`` is given as,
`` 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.}``
`` ``
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\text{'}=\frac{{m}_{o}\left(2v\right)}{\sqrt{1-{\displaystyle \frac{4{v}^{2}}{{c}^{2}}}}}``
`` \Rightarrow p\text{'}=2{m}_{o}v{\left(1-\frac{4{v}^{2}}{{c}^{2}}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}``
`` ``
Expanding binomially and neglecting higher terms we have,
`` p\text{'}\simeq 2{m}_{o}v\left(1+\frac{4{v}^{2}}{2{c}^{2}}\right)``
`` \Rightarrow p\text{'}\simeq 2{m}_{o}v+\frac{4{m}_{o}{v}^{3}}{{c}^{2}}``
`` \therefore p\text{'}\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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