Why does a capacitor act like a short-circuit during a current impulse?
The basic equation for a capacitor is charge, Q = voltage, V x capacitance, C. If this is differentiated we get: -

\$\dfrac{dQ}{dt} = C\cdot\dfrac{dV}{dt}\$

And rate of change of charge is current, therefore: -

\$I = C\cdot\dfrac{dV}{dt}\$

A voltage impulse has very high dV/dt therefore I is also very high.

This to me, represents a short when dV/dt is infinite.

I'll leave you to use the formulas for an inductor to satisfy your curiosity and my laziness.