A simple question about special relativity [closed]

Question

Assume that the table moves with the velocity $\vec{v} = v\hat{i}$.

For the observer at $x=0$, the event at $(ct',x')=(0,k)$ is observed at $(t,x)=(k\gamma\beta, k\gamma)$ using Lorentz's transformation $L(v) = \gamma\bigl(\begin{smallmatrix} 1&\beta\\ \beta&1 \end{smallmatrix} \bigr)$.

For the observer at $x=4$, I'm not confident about my answer. For him, the event at $(ct',x')=(0,k) (k=1,2,3)$ is probably observed at $(ct,x)=(-(4-k)\gamma\beta, (4-k)\gamma)$ using the Lorentz's transformation $L(-v)=\gamma\bigl(\begin{smallmatrix} 1&-\beta\\ -\beta&1 \end{smallmatrix} \bigr)$.

Is my answer for (a) correct?

Answer 1

You do not need to use Lorenz factor to answer (a). You only need to notice that light propagates at c and that firecrackers go off simultaneously (relative to table and to stationary observer at either end of the table). Then observer at x = 0 will notice light from the first firecracker after 1/c. (t = s/v that is after 1/c. The second after 2/c.)

The second question is a bit more tricky. What is simultaneous to one observer (call him stationary ) is not simultaneous to another observer moving relative to the first one. Here you will need to employ Special Relativity.