This is a very complex question, unfortunately. There is blade deflection at the pivot upon impact, the fact it is rotating instead of moving in a straight line, the mass of the blade is not distributed evenly across the blade, motor on or off at time of impact, the whole heli is moving at some speed at the time of impact which will change the instaneous velocity of the blade at impact relative to the object being hit and where in the rotational period the impact occurs, etc. etc. etc.
Let's simplify things: assume no blade deflection at impact (totally rigid blade mounting), a 2 blade head, assume the 1/2 the mass of the blade is in the tip and part of the impact (attempt to compensate for the distribution issue), no heli velocity (strike in hover or on the ground).
As a starting point, it's not impact force, but a starting point is kinetic energy of the blade at the tip.
E = 1/2 m v^2 = .5 * (2 * 1/2 * .5kg) * (1.5m * 2200rev/min / 60sec/min) = .5 * .5kg * 55m/s = 13.75J
I used 2 * blade mass because the other blade's rotational energy will be transmitted into the strike as additional effective mass due to the rigid mounting and equal blade length on each side of the rotor.
Next, we need to know how far into the struck object the blade will wedge itself before coming to a stop. We will assume constant deceleration.
W = F s, where W is Work/Energy, F is deceleration force (aka force over the impact length), and s is the distance from impact to stop.
W = E, so:
F = E / s = 13.75J / s
Assume the blade stops in 2cm or .02m:
F = 13.75J / .02m = 687.5N
This stopping distance can be displacement into the struck object, it can be travel due to damage to the blade into an immovable object, etc. And we see, the impact force is inversely proportional to the distance after impact. This is why car makers started putting crumple zones in, the longer you can space that impact out the less effective force the occupants will experience and thus less injury on average.
