Question

Two particles moving in a laboratory frame of reference along the same straight line with the same velocity v=$(\dfrac{3}{4})c$ strike against a stationary target with the time interval $\Delta t$=50ns. Find the proper distance between the particles prior to their hitting the target.

Answer 1

Hint: Both the particles move with constant velocity and the second particle took 50ns more than the first particle to hit the target. Hence we took two frames of reference and derived the formula for the proper distance between the particles by using relativity.

Complete answer:
The position of the coordinate A and B in frame F is assumed by us to be
A: (0, 0, 0), B= (l, 0, 0)
In the frame K’, the coordinates are
A = (vt’, 0, 0), B= ($l\sqrt{1-{{\beta }^{2}}}+vt$, 0, 0), $\beta =\dfrac{v}{c}$
Suppose
B hits a stationary target in K’ after time $t{{'}_{B}}$ while A hits after time$t{{'}_{B}}+\Delta t$. Then,
$l\sqrt{1-{{\beta }^{2}}}+vt{{'}_{B}}=v(t{{'}_{B}}+\Delta t)$
So, $l=\dfrac{v\Delta t}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$
Therefore $l=\dfrac{\dfrac{3}{4}c\times 50ns}{\sqrt{1-\dfrac{9{{c}^{2}}}{16{{c}^{2}}}}}$= 1.7m
Hence the first particle is 1.7 m ahead of the second particle.

Additional Information:
The theory of relativity is the explanation of space and time and its movement of objects at a constant speed on a straight line path. Hence the speed of the object motion is comparable to the speed of light, the mass of the same object becomes infinite, and therefore, it is restricted to go at any faster speed than that of light.

Note:
The proper length between the particles must be larger than the observed length of the distance in the reference plane. This is because of the principle of relativity. It states that the relation between original size of the object and to the size which is observed in the reference plane.