Answer to Question #88925 in Mechanics | Relativity for Sridhar

Question

A nuclear particle was observed to break into two fragments which moved in opposite direction the velocity of each fragment was found to be 2.5×10^8 m/s. relative to the laboratory the velocity of the one fragment relative to other is

Answer 1

Relative to one of the fragments, the laboratory moves with velocity $v = 2.5 \times 10^8\, \text{m/s}$, and, relative to the laboratory, another fragment moves with the same velocity in the same direction. Then the velocity $u$ of the second fragment relative to the first one is found by the formula of relativistic addition of velocities:

$$u = \frac{v + v}{1 + v \cdot v / c^2} = \frac{2 v}{1 + v^2/c^2} \, ,$$

where $c = 3 \times 10^8\, \text{m/s}$ is the speed of light. Substituting the numbers, we obtain $u = 2.95 \times 10^8\, \text{m/s}$.

Answer: $2.95 \times 10^8\, \text{m/s}$.