Answer
a) $v=0.43c$
b) $L_{\circ}=9.97m$
Work Step by Step
(a) We know that
$L=L_{\circ}\sqrt{1-\frac{v^2}{c^2}}$
$\implies 9.00m=L_{\circ}\sqrt{1-\frac{v^2}{c^2}}$........eq(1)
for velocity $2v$, we can write the above equation as
$5.00m=L_{\circ}\sqrt{1-\frac{(2v)^2}{c^2}}$
$5.00m=L_{\circ}\sqrt{1-\frac{4v^2}{c^2}}$.....eq(2)
Dividing eq(1) by eq(2), we obtain:
$\frac{9}{5}=\sqrt{\frac{1-\frac{v^2}{c^2}}{1-\frac{4v^2}{c^2}}}$
$\implies \frac{81}{25}=\frac{c^2-v^2}{c^2-4v^2}$
This simplifies to:
$v=0.43c$
(b) We know that
$L_{\circ}=\frac{9.00m}{\sqrt{1-\frac{v^2}{c^2}}}$
$L_{\circ}=\frac{9.00m}{\sqrt{1-(\frac{0.43c}{c})^2}}$
$\implies L_{\circ}=9.97m$
