(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 γL0Vc2 seconds before lamp B in frame F, we compute t2−t1=γ[(t′2−t′1)−(x′2−x′1)−Vc2]=γL0Vc2>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 Δt between the two lamps, first consider an observer OA positioned at lamp A on the moving rod. OA sees lamp B come on L0c seconds after lamp A. We treat it as a pulse with period T′=L0c enamating from end A of the rod. Applying Doppler effect, observer at O observes that Δt=√c+Vc−VT′=√c+Vc−VL0c.
(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′=V−u1−uc2V and V′=u, this will make both rods have the same length. Solving for u, we have u=c2v(1−2√1−v2c2) reject the other quadratic solution since that will make u≥c.
(c)
Total energy twice as rest energy, this means γmc2=2mc2⟹γ=2.
Solving 12=√1−β2 we have u=√32c.