91Ó°ÊÓ

The velocity of the particle 1 is 0.99c.

The velocity of the particle 2 is 0.99c .

The expression for the Lorentz velocity transformation is given as follows.

$\mathbf{u}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{u}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{uv}}{{\mathbf{c}}^{\mathbf{2}}}}}$ …… (i)

Here, u' is the velocity of the object in frame s' , u is the velocity of object in frame S , v is the velocity of frame S' relative to S and c is the speed of the light.

Calculate the velocity of the particle which to observer moving with particle 2.

Substitute 0.99c for u and 0.99c for v into equation (i).

$$\begin{gathered} u\text{'}=\frac{0.99c-0.99c}{1-{\displaystyle \frac{\left(0.99c\right)\left(0.99c\right)}{c^2}}} \\ u\text{'}=\frac{0}{1-0.99×0.99} \\ u\text{'}=0 \end{gathered}$$

Therefore, the velocity of the particle 2 to the observer is 0 .

Calculate the velocity of the particle 1 to the observer.

Substitute -0.99c for u and 0.99c for v into equation (i).

$$\begin{gathered} u\text{'}=\frac{-0.99c-0.99c}{1-{\displaystyle \frac{\left(-0.99c\right)\left(0.99c\right)}{c^2}}} \\ u\text{'}=\frac{-1.98c}{1+0.9801} \\ u\text{'}=\frac{-1.98c}{1.9801} \\ u\text{'}=0.9999c \end{gathered}$$

Therefore, the velocity of the particle 1 to the observer is 0.9999c.

Hence the velocity of the particle 1 and 2 to the observer is 0 and 0.9999c respectively.