91Ó°ÊÓ

The relativistic velocity of the particle is given by,

$\mathbf{u}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{u}\mathbf{−}\mathbf{v}}{\mathbf{1}\mathbf{−}\frac{\mathbf{uv}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}\mathbf{.}\mathbf{.....}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, u is the velocity of the particle as measured in S,$\mathbf{u}\mathbf{\text{'}}$is the velocity of the particle as measured in$\mathbf{S}\mathbf{\text{'}}$,is the velocity of the$\mathbf{S}\mathbf{\text{'}}$relative to S, and c is the velocity of the light.

(a)

From the figure, the velocity of particle relative to frame $S\text{'}$,$u\text{'}=0.8c$ when $v=0$. The velocity of the particle in frame$S$ is$u=0.8c$ . The velocity of the frame$S\text{'}$

With respect to frame$S$ is $v=0.9c$.

Substitute$0.8c$ for$u$ and $0.9c$for$v$ in equation (1).

$\begin{array}{c}u\text{'}=\frac{{\displaystyle 0.8c−0.9c}}{{\displaystyle 1−\frac{{\displaystyle \left(0.8c\right)\left(0.9c\right)}}{{\displaystyle c^2}}}}\\ =\frac{{\displaystyle (−0.1c)}}{{\displaystyle 1−0.72}}\\ =−\frac{{\displaystyle 0.1c}}{{\displaystyle 0.28}}\\ =−0.36c\end{array}$

Therefore, the value of$u\text{'}$ is$−0.36c$ .

(b)

Substitute$0.8c$ for$u$ and$c$ for$v$ in equation (1).

$\begin{array}{c}\mathrm{u}\text{'}=\frac{0.8\mathrm{c}−\mathrm{c}}{1−\frac{\left(0.8\mathrm{c}\right)\left(\mathrm{c}\right)}{{\mathrm{c}}^2}}\\ =\frac{(−0.2\mathrm{c})}{1−0.8}\\ =−\frac{0.2\mathrm{c}}{0.2}\\ =−\mathrm{c}\end{array}$

Therefore, the value of$u\text{'}$ is$-C$ .