Chapter 26, Problem 67GP

To determine

The speed of one particle relative to the other.

Answer to Problem 67GP

  $0.96c$

Explanation of Solution

Given:

Speed of each particle is $0.75c$

Formula used:

It is known that

  $u=\frac{\left(v+u\text{'}\right)}{1+\frac{vu\text{'}}{c^2}}$

Where,

  $m$ is relativistic mass

  $v$ is velocity (with the respect to the earth)

  $u\text{'}$ is a speed (Or speed respect to the first speed)

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

If $u\text{'}$ is in the opposite direction from $v$ then $u\text{'}$ must have a minus sign in the above mentioned expression.

Calculation:

From one particle’s point of view, the laboratory and the other particle are moving in the same direction.

  $u=\frac{\left(0.75c+0.75c\right)}{\left[1+\left(0.75c\right)\left(0.75c\right)\right]}=0.96c$

Conclusion: The required value is $0.96c$ .