If you want this to be physically correct, then the force of the shockwave on an asteroid is inversely proportional to the distance of that asteroid from the center of the shockwave squared. (because at any given time the shockwave is a sphere, and the surface area of a sphere is proportional to the square of the radius)
In the simplest case, the shockwave is very thin, so you can just apply a one off impulse on each asteroid as the shockwave reaches it. The impulse will be proportional the surface area of the asteroid facing the center (which you can easily approximate by assuming the asteroid is roughly spherical) and inversely proportional to its distance squared. Of course that impulse is then divided by the asteroids mass to calculate the change in velocity.
If the shockwave is not thin then it will have an inner and an outer radius. The behaviour will be the same as before but instead of applying a single impulse, you apply a force for the duration that an asteroid is between the inner and outer radii of the shockwave. This will give the effect of asteroid "riding" the shockwave for a period of time before getting left behind.
So:
r = approximate radius of asteroid
R = distance of asteroid from centre of shockwave
s = approximate density of asteroid
M = strength of shockwave
Impulse (applied once):
change in velocity = M r²/(R² s r³) = M/(R² s r)
Force (applied while asteroid is in the shockwave):
acceleration = M/(R² s r)
As you can see from these formulae, denser and larger asteroids, and asteroids further from the center of the shockwave will be less affected.
The rest is just standard newtonian physics, ie. integrate acceleration to get velocity, integrate velocity to get displacement.
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