Answer
It is safer for the collision to be inelastic.
Work Step by Step
The impulse delivered to an object by an average force $\vec{\mathrm{F}}_{\mathrm{a}\mathrm{v}}$ acting for a time $\Delta t$ is
$\vec{\mathrm{I}}=\vec{\mathrm{F}}_{\mathrm{a}\mathrm{v}}\Delta t$
Impulse is a vector, proportional to the force vector. By Newton's second law, the impulse delivered to an object is equal to the change in its momenta:
$\vec{\mathrm{I}}=\vec{\mathrm{F}}_{\mathrm{a}\mathrm{v}}\Delta t=\Delta\vec{\mathrm{p}}$
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In an inelastic collision (objects hit and stick together), the pole delivers enough impulse to the car so that it loses the momentum it initially has.
In an elastic collision, the final kinetic energy is equal to the initial kinetic energy,
so the pole delivers more impulse, one part is to make the car stop, and another to force it back with the same speed it initially had when it collided.
A greater force is applied, so the driver and passengers (and the car) experience greater acceleration. Greater acceleration leads to greater damage (and pain).
It is thus safer for the collision to be inelastic.
