Answer: \(\boxed{50\ \mathrm{m\ s^{-1}}}\).

Let the speed of the bullet as it leaves the barrier be \(v\ \mathrm{m\ s^{-1}}\).

The resisting force acts opposite to the motion, so its impulse is

\[Ft=1000(0.02)=20\ \mathrm{N\,s}.\]

This equals the loss of momentum of the bullet:

\[0.08(300-v)=20.\]

Divide by \(0.08\):

\[300-v=250.\]

Hence

\[v=50.\]

Therefore the bullet emerges with speed \(\boxed{50\ \mathrm{m\ s^{-1}}}\).