# (a)

Let event 1 be lamp \(A\) turning on and event 2 for lamp \(B\). Let frame \(F'\) be the moving along the rod, at velocity \(V\) away from frame \(F\) that is stationary with respect to \(O\).

To see that lamp \(A\) turns on \(\frac{\gamma L_0V}{c^2}\) seconds before lamp \(B\) in frame \(F\), we compute \[ t_2 - t_1 = \gamma\left[(t_2'-t_1') - (x_2'-x_1')\frac{-V}{c^2}\right] = \frac{\gamma L_0V}{c^2} > 0. \]

Light from lamp \(A\) will reach the observer \(O\) first as it comes on first and the light travels a shorter distance from lamp \(A\).

To find time delay \(\Delta t\) between the two lamps, first consider an observer \(O_A\) positioned at lamp \(A\) on the moving rod. \(O_A\) sees lamp \(B\) come on \(\frac{L_0}c\) seconds after lamp \(A\). We treat it as a pulse with period \(T' = \frac{L_0}{c}\) enamating from end \(A\) of the rod. Applying Doppler effect, observer at \(O\) observes that \[ \Delta t = \sqrt{\frac{c+V}{c-V}}T' = \sqrt{\frac{c+V}{c-V}}\frac{L_0}{c}. \]

# (b)

Start from the frame \(F\) where rod 1 is at rest and the other is moving at speed \(V\). We want to find a frame \(F'\) moving at velocity \(u\) away from rod 1 such that \[ V' = \frac{V - u}{1 - \frac{u}{c^2}V} \] and \(V' = u\), this will make both rods have the same length. Solving for \(u\), we have \[ u = \frac{c^2}{v}\left(1 - 2\sqrt{1 - \frac{v^2}{c^2}}\right) \] reject the other quadratic solution since that will make \(u\geq c\).

# (c)

Total energy twice as rest energy, this means \[ \gamma mc^2 = 2 mc^2 \implies \gamma = 2. \]

Solving \[ \frac12 = \sqrt{1 - \beta^2} \] we have \[ u = \frac{\sqrt{3}}2 c. \]