Let $v$ be the desired speed. Then, from the relativistic velocity addition formula, the speed of one space ship as measured by the other is

$\begin{align*}\frac{2v}{1 + v^2/c^2}\end{align*}$
The factor by which the meterstick will be contracted is
$\begin{align*}\frac{1}{\gamma} = \sqrt{1 - \frac{1}{c^2}\cdot \frac{4 v^2}{1 + 2 \left(v/c \right)^2 + \left(v/c \right)^4}} ...$
From the problem statement we know that this factor is
$\begin{align*}\frac{60 \mbox{ cm}}{100 \mbox{ cm}} = \frac{3}{5}\end{align*}$
So,
$\begin{align*}&&\frac{1}{\gamma} &= \frac{3}{5} \\&\Longrightarrow& 1 - \frac{1}{c^2}\cdot \frac{4 v^2}{1...$
We can now apply the quadratic formula to find that
$\begin{align*}\left(v/c\right)^2 = \frac{17 \pm \sqrt{17^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4}&= \frac{17 \pm \sqrt{289 - 64...$
$\left(v/c\right)^2$ must be less than $1$ , so we take the minus sign. This gives $\left(v/c\right)^2 = 1/4$ and $\boxed{v/c = 1/2}$ . Therefore, answer (B) is correct.

Solution to 2008 Problem 23