(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 $$u = \frac{\sqrt{3}}2 c.$$