Question

Two rods of the same proper length $l_0$ move toward each other parallel to a common horizontal axis. In the reference frame fixed to one of the rods the time interval between the moments, when the right and left ends of the rods coincide, is equal to $\Delta t$. What is the velocity of one rod relative to the other?

Answer 1

The coordinates of the ends of the rods in the frame fixed to the left rod are shown below.
The points $B$ and $D$ coincides when
$l_0=c_1-v{t}_0$ or $t_0=\frac{c_1-l_0}{v}$
The points $A$ and $E$ coincides when
$0=c_1+l_0\sqrt{1-{\beta }^2}-v{t}_1$, $t_1=\frac{c_1+l_0\sqrt{1-{\beta }^2}}{v}$
Thus $\Delta t=t_1-t_0=\frac{l_0}{v}(1+\sqrt{1-{\beta }^2})$
or ${(\frac{v\Delta t}{l_0}-1)}^2=1-{\beta }^2=1-\frac{v^2}{c^2}$
From
this $v=\frac{2c^2\frac{\Delta t}{l_0}}{1+c^2\frac{\Delta t^2}{l_0^2}}=\frac{2\frac{l_0}{\Delta t}}{1+{\left(\frac{l_0}{c\Delta t}\right)}^2}$