I will go mostly with Chad's argumentation. Larger equals slower, which should excite relatively more low frequency sound. Also note that the larger the blast (hopefully) the further away the observer is. And air is not a perfectly elastic acoustic medium, some energy is lost, and the higher frequencies attenuate quicker than shorter, so distance will selectively filter out the higher frequencies.
Also, note, explosion usually means detonation. A detonation is an exothermic reaction which spreads by the compression (adiabatic) heating from the shockwave, and the chemical energy maintains the shockwave. A shockwave is essentially a highly nonlinear soundwave, and as the overpressure decays with distance from the source is will grade into a soundwave. Fireworks (pyrotechnics) are not explosives, but are the (relatively) slow reaction of chemicals (combustables, and an oxidizer) due to heat. Fireworks may generate shockwaves in air, if the package ruptures at sufficiently high pressure. Likewise volcanic blasts are not detonations, but shockwaves formed by the escape of high pressure gas.
If an explosion is fast compared to the sound frequencies the detector is sensitive to (probably human ears in your case), then we might be able to model the explosion as a delta-function in time. A delta function should equally excite all frequencies, so it should be a simple matter of distance attenuation of sound waves.
An explosion close to a solid surface, creates an amazing effect I've heard called a mach-stem (although wikipedia does not produce anything useful for this term). In any case, at fixed distance the near surface shock wave is much stronger than the shockwave at height above ground. In essence the ground effect part of the shockwave weakens roughly only as 1/R rather that the 1/R**2 one would expect for a free air spherical blastwave. I don't know what effect this has on the sound spectrum (pitch), but you need to be at a much larger standoff distance from a near groundblast than from a samesized high altitude blast because of it.
In the simplest approximation, an explosion is a shockwave moving out from some locus. The shockwave may be a compression front in a ambient medium, or may be a wave of gas propagating from the explosive into a vacuum.
So that's the first thing you need to tell us: in air, water, vacuum, or what?
When the shockwave arrives at some material thing, it is the pressure exerted by the shockwave that transfers momentum (i.e. applies a force) to the target. The target object then accelrates as per Newton's law:
$$ \vec{F} = m\vec{a} .$$
The vector part of the above is the trigonometry that you show. I'm simply going to assume that you have your coordinate system squared away. However, we still haven't said how much force. To a first approximation it goes by the shock pressure $P$ times the area $A$ the object presents to the shock wave. So that gets us to
$$ \vec{a} = \frac{P A}{m} \hat{n} $$
where the unit vector $\hat{n}$ is normal to the surface of the shockwave.
We're still not done because we don't know $P$. Again, we'll take the simplest approximation. Assume you know the shock pressure $P_0$ at the surface of the explosive (this is presumably something you can look up for chemical explosives). When the shockwave has total area $\mathcal{A}$, it's pressure is
$$ P = P_0 \frac{\mathcal{A}_0}{\mathcal{A}} $$
where $\mathcal{A_0}$ is the area over which $P_0$ applied. For small sources (like a bomb or stick of dynamite) this will give you a $1/r^2$ type dependence for the pressure. For long linear sources you get a $1/\rho$ dependence; and for large surface charges a pressure that is independent of distance.
That isn't a good approximation, because you can expect the shockwave to spread out of it's own accord over time, and I have not yet modeled that. The really, really simple thing to do here is to choose a maximum range $l = \tau v_s$ equal to some time interval times the speed of shock propagation, and apply a hard cutoff at that point.
Now we know how much force as a function of distance from the explosive and geometry of the explosive, but to get to change in velocity we need to know for how long the force is applied. Obviously that will be related to the depth of the shockwave and to it's speed of propagation: $ \Delta t \approx v_s * d $. Unfortunately I'm hazy on how to estimate $d$. For a simple explosive I might go with half the "size" of the charge (radius, thickness, whatever). The speed of propagation depends on the medium. Again, for game purposes you could simple chose some number. Once we have $\Delta t$ we get
$$ \Delta v = \Delta t * a . $$
Since you say that your engine works on a applied impulse, you get
$$ I = \Delta t * P * A = \Delta t * \frac{P_0 \mathcal{A}_0}{\mathcal{A}} * A, $$
which is actually pretty simple.
And that should get you on your way.
Best Answer
If you start with a finite amount of gas in the inner sphere and then deposit a massive amount of energy, the molecules of the gas begin moving rapidly outwards and piling up, creating the blast wave. However, the rate at which the gas is moving outwards may not be balanced by the amount of gas molecules being created by the explosive. If this is the case, then the pressure must decrease below ambient as the molecules are pushed outwards with the blast wave.
You can see this in videos of blast waves. The initial wave continues to move outwards, but the smoke/dirt/debris caused by the explosive will move outwards initially, then inwards as the lower pressure region sucks it back in towards the center. There is actually considerably banging that goes on where the low pressure behind the blast wave moves inwards and outwards until it relaxes back to atmospheric pressure.
Here is a great video that shows the blast and resulting banging as the pressure relaxes.