There is a well-known algebraic relationship between the Mach number before and after a shock.
[tex]M_2^2 = \dfrac{\left( \gamma - 1 \right)M_1^2 + 2}{2\gamma M_1^2 - \left( \gamma -1 \right)}[/tex]
where [itex]M_1[/itex] is the Mach number relative to an equivalent stationary shock wave of the flow into which the shock is advancing, [itex]M_2[/itex] is the Mach number relative to an equivalent stationary shock wave of the flow after the shock has passed, and [itex]\gamma[/itex] is the ratio of specific heats, which is 1.4 for diatomic gases like air. So, if you have simply the speed of your shock wave already and the temperature of the air into which it is advancing, then you already know [itex]M_1 = u_1/a_1 = V_{\text{shock}}/\sqrt{\gamma R T}[/itex] where [itex]R[/itex] is the specific gas constant (≈287.1 J kg-1 K-1 for air) and [itex]T[/itex] is the tempetature in Kelvin.
From there you just use the above equation To get [itex]M_2[/itex]. There exists a relationship for the temperature change across a shock as well
[tex]\dfrac{T_2}{T_1} = \left[ 1 + \dfrac{2\gamma}{\gamma + 1}\left( M_1^2 -1 \right) \right]\left[ \dfrac{2 + \left( \gamma-1 \right)M_1^2}{\left( \gamma + 1\right)M_1^2} \right].[/tex]
This will give you [itex]T_2[/itex], which, when combined with [itex]M_2[/itex] will give you the velocity after the shock, [itex]u_2[/itex]. Then it's just a matter of converting that back into a stationary reference frame.
Essentially what the shock does when it passes through a stationary medium is drag the air behind it along with it at some velocity, and it is that velocity that you are solving for here.
This works if the shock is of constant strength (constant Mach number) and that velocity of the air behind the shock absolutely will be constant. However, the shock strength won't likely be constant if we are talking about a detonation, as it will weaken as it expands. For that you would have to use knowledge of the pressures to track the changing strength of the shock as it expands, and that would be difficult if not impossible to do analytically, but you are welcome to try. As you may be able to guess though, there is a relationship between the pressures across the shock and Mach number, which is
[tex]\dfrac{p_2}{p_1} = 1 + \dfrac{2\gamma}{\gamma + 1}\left( M_1^2 -1\right)[/tex]
and a density relationship, too,
[tex]\dfrac{\rho_2}{\rho_1} = \dfrac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2 + 2}.[/tex]
You can find a lot of extra related equations in a report called NACA 1135 that you should be able to find for free pretty easily online, but it may be difficult to use if you don't know what you are doing.
At any rate, if your detonation that you are specifically talking about is in something like a tube, then this will work wonderfully because the velocity after the shock passes will be constant and you just need to estimate the drag on your particle in your favorite way. If it is a detonation out in the open, It is a more transient problem but you can likely get a decent estimate by assuming constant shock strength, especially if the particle is small and therefore likely to accelerate almost impulsively.