Past exam of the physics 1 paper of the University of Oxford / 2014 / 6 / 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: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,The train is wholly inside between these events. Transforming their time separation givesThus the front leaves before the rear enters in the train frame. The two events are spacelike separated, so their time order may reverse without a causal contradiction. No doors or forced acceleration need be added to the problem.