In Fig. 37-26a, particle P is to move parallel to the x and $\mathit{x}\mathbf{\text{'}}$ axes of reference frames S and $\mathit{S}\mathbf{\text{'}}$ , at a certain velocity relative to frame S. Frame $\mathit{S}\mathbf{\text{'}}$$$" width="9">x axis of frame S at velocity v. Figure 37-26b gives the velocity localid="1664359069513" $\mathit{u}\mathbf{\text{'}}$of the particle relative to frame localid="1664359072841" $\mathit{S}\mathbf{\text{'}}$ for a range of values for v. The vertical axis scale is set by ${\mathit{u}}_{\mathbf{a}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0.800}\mathit{c}$. What value will$\mathit{u}\mathbf{\text{'}}$ have if (a)$\mathit{v}\mathbf{=}\mathbf{0.90}\mathit{c}$ and (b) $\mathit{v}\mathbf{\to }\mathit{c}$?

The relativistic velocity of the particle is given by,

$\mathbf{u}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{u}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}\frac{\mathbf{uv}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{.}\mathbf{.....}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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$-C$ .