= Solution
For $\beta=0.99$, the <Lorentz factor> is $\gamma=1/\sqrt{1-\beta^2}=7.0888$. In the tracks' rest frame the train undergoes <length contraction> to $100/\gamma=14.1067\,\mathrm m$, which fits in the $90\,\mathrm m$ tunnel. There is a time interval during which it is entirely inside. In the train's rest frame the tunnel instead has length $90/\gamma=12.6961\,\mathrm m$, so it never simultaneously contains the whole $100\,\mathrm m$ train:
$$
\boxed{\text{Entirely inside at one track-frame time: yes. At one train-frame time: no.}}
$$
These refer to different sets of simultaneous events, as required by <relativity of simultaneity>. Quantitatively let $A$ be the rear entering the near portal, with $(x_A,t_A)=(0,0)$, and $B$ the front reaching the far portal. In the track frame,
$$
x_B=90\,\mathrm m,\qquad ct_B=\frac{90-100/\gamma}{\beta}=76.6599\,\mathrm m.
$$
The train is wholly inside between these events. Transforming their time separation gives
$$
c(t'_B-t'_A)=\gamma(ct_B-\beta x_B)=-88.1858\,\mathrm m<0.
$$
Thus the front leaves before the rear enters in the train frame. The two events are <spacelike separation>[spacelike separated], so their time order may reverse without a causal contradiction. No doors or forced acceleration need be added to the problem.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014-train-and-tunnel-events.png]
{title=Portal and train <worldlines> in the two inertial frames}
{description=Original <spacetime diagrams>, with vertical coordinates ct and ct-prime in metres. Events A and B exchange time order; the horizontal shaded time slice shows the different simultaneous configurations.}
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