CBSE-XI-Physics
47: The Special Theory of Relativity
Qstn# iv-28 Prvs-Qstn
- #28Find the speed at which the kinetic energy of a particle will differ by 1% from its nonrelativistic value
12 mov2.Ans : If m0 is the rest mass of a particle and c is the speed of light, then relativistic kinetic energy of particle = mc2 - m0c2. ...(1)
If `` m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}``, then nonrelativistic kinetic energy of particle = `` \frac{1}{2}{m}_{o}{v}^{2}``.
According to the question,
`` \frac{\left({\displaystyle \frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}}-{m}_{0}{c}^{2}\right)-{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}{1/2{m}_{0}{v}^{2}}=0.01``
`` \Rightarrow \left[\frac{{c}^{2}\left(1+{\displaystyle \frac{{v}^{2}}{2{c}^{2}}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}{\displaystyle \frac{{v}^{2}}{{c}^{2}}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}\times {\displaystyle \frac{5}{6}}{\displaystyle \frac{{v}^{6}}{{c}^{6}}}-1\right)}{{\displaystyle \frac{1}{2}}{v}^{2}}\right]=0.01``
`` \Rightarrow \frac{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}+{\displaystyle \frac{3}{8}}{m}_{0}{v}^{4}/{c}^{2}+{\displaystyle \frac{15}{96}}{m}_{0}{v}^{6}/{c}^{4}-{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}=0.01``
`` \Rightarrow \frac{{\displaystyle \frac{3}{8}}{m}_{0}{v}^{2}/{c}^{2}+{\displaystyle \frac{15}{96}}{m}_{0}{\,\mathrm{\,V\,}}^{6}/{c}^{4}}{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}=0.01``
`` \Rightarrow \frac{3}{4}\frac{{v}^{2}}{{c}^{2}}+\frac{15\times 2{v}^{4}}{96}\frac{{v}^{2}}{{c}^{2}}=0.01``
Neglecting the v4 term as it is very small, we get
`` \frac{3}{4}\frac{{v}^{2}}{{c}^{2}}=0.01``
`` \Rightarrow \frac{{v}^{2}}{{c}^{2}}=\frac{0.04}{3}``
`` \Rightarrow \frac{v}{c}=\frac{0.2}{\sqrt{3}}``
`` \Rightarrow v=\frac{2}{2\sqrt{3}}``
`` c=\frac{0.02}{1.732}\times 3\times {10}^{8}=\frac{0.6}{1.722}\times {10}^{8}``
`` =0.345\times 10\,\mathrm{\,m\,}/\,\mathrm{\,s\,}=3.46\times {10}^{7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``