= Solution
In the shed frame, the leading end reaches the back when $0.8T+1.2=1.5$, giving $T_B=3/8\,\mathrm m$. Transforming both events yields
$$
\boxed{A:(T,x)=(0,0),\ (T',x')=(0,0);\qquad B:(T,x)=(3/8,3/2),\ (T',x')=(-11/8,2),}
$$
with every coordinate in metres. The invariant interval is
$$
\Delta s^2=(\Delta T)^2-(\Delta x)^2=\frac9{64}-\frac94=-\frac{135}{64}\,\mathrm{m^2}<0.
$$
Thus \b[the events are spacelike separated]. Event $A$ occurs first in the shed frame, but $B$ occurs first in the javelin frame; this is allowed for <spacelike separation>.
In the shed frame there is a pre-impact interval $0\le T<3/8$ during which both ends lie inside the shed, since its simultaneous moving length is only $1.2\,\mathrm m$. Those simultaneous endpoint events do not remain simultaneous after a boost. In the javelin frame the front hits the back before the rear enters, and there is no simultaneous pre-impact containment of the full $2\,\mathrm m$ rod in a $0.9\,\mathrm m$ shed. This is <rod containment and simultaneity>, not inconsistent length measurements.
After impact, the front's motion changes while the rear cannot respond instantaneously: stress signals propagate no faster than light. A claim that the entire javelin stops rigidly at event $B$ is inadmissible. Actual post-impact compression or fracture needs material data and is not represented by the straight-line extrapolations in the diagrams.
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