Question

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of 0.890c. Both particles travel at the same speed as measured in the lab- oratory. What is the speed of each particle, as measured in the laboratory?

Short answer

The speed of each particle is 0.6108c.

Step-by-step solution

Formulas used to solve the question

Particle’s velocity in frame S:

${\mathit{v}}_{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{x}}\mathbf{-}\mathbf{u}}{\mathbf{1}\mathbf{-}\mathbf{n}{\mathbf{v}}_{\mathbf{x}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}$

Particle’s velocity in frame $\mathit{S}\mathbf{\text{'}}$:

${\mathbf{v}}_{\mathbf{x}}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{u}}{\mathbf{1}\mathbf{-}{\mathbf{uv}}_{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}$

Determine the speed of each particle

Let S be the frame of Lab and $\mathit{S}\mathbf{\text{'}}$ be the frame of particle 1.

Speed of particle 2 in frame$\mathit{S}\mathbf{\text{'}}$ is${\mathit{v}}_{\mathbf{x}\mathbf{-}\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{890}\mathit{c}$

The negative sign shows that speed of particle 2 in opposite direction to particle .

The speed of particle in frame of lab S is$v_{x\_2}=-u$

Now,

${\mathit{v}}_{\mathbf{x}\mathbf{\_}\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{890}\mathit{c}$

And

${\mathit{v}}_{\mathbf{x}\mathbf{\_}\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{6108}\mathit{c}$

Thus, the speed of each particle is $\left|v\right|\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6108}\mathit{c}$.