Problem 9.19

9.19.
At v = .955c, we have
$\gamma = 3.3715$
First, we want to know how long the moving rod (in A's frame) is at rest. That would be the length of the rod measured by observer B in his frame. We use length contraction.
$L' = \gamma L$
L' = 3.3715 L₀
where L₀ is the length that A measured for the two rods. Now, to find out how long the rod that was stationary in A's frame is to observer B, we again use a length contraction, this time however we use the inverse transformation.
$L = \frac{1}{\gamma} L'$
L = .2966 L₀
So, the ratio of lengths measured in the rest frame of B is given by
$r = \frac{.2966 L_{0}}{3.3715 L_{0}}$
r = .0880