What % of the energy from a spherical explosion pushes away a perpendicular plane? I'm trying to figure out how much energy from an explosion pushes away an object vs either completely missing it or having its horizontal component cancelled out by an opposing horizontal force from the explosion (presumably trying to stretch the plane and being turned into heat).
I know how to take joules to convert it into m/s given the object's mass, I just don't know how to get the relevant portion of the energy.
The below diagram is a 2D representation of my actual 3D question:
 What % energy
 pushes this way?
      ^
     / \
      |
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  \   |   /

  -   *   -

  /   |   \

edit: updated diagram:
       ^
      / \
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 ______|______
|   Finite    |
|  Ma ___ ss  |
|  .-' | '-.  |
| |  \ | /  | |
|_| -  *  - |_|
     / | \

If the object had a semi-circle cutout perfect centered (unlike my ascii art) around the explosion (or somehow infinitely close to achieve a similar result).  The object is strong enough to resist any splitting forces by the horizontal component of the energy absorbed converts it to heat.
 A: Let us assume vacuum, infinite plane, perfect elastic collisions, no gravity, etc. 
Suppose the explosion is a sphere of $N$ mass $m$ particles moving away at velocity $v$. Half of the particles will miss the plane. The other ones will sooner or later bounce off the plane. 
Each bounce will have vertical momentum $mv \cos(\theta)$ where $\theta$ is the angle between the particle velocity and the plane normal. Each bounce will cause an instantaneous impulse of magnitude $2mv\cos(\theta)$. 
So the total impulse provided to the plane will be $J=N \int_{0}^{\pi/2} 2mv\cos(\theta) (2\pi \sin(\theta)) d\theta = \pi m N v$. 
So if we say the explosion has an area density $\rho$ we subdivide into $N$ pieces of mass $4\pi\rho/N$ we get $J=4 \pi^2 \rho v$ (neatly independent of $N$).
Now, this just measures the total push. Note that it is not distributed evenly nor does it arrive at the same time (the spread is going to follow a $\sim 1/(d^2+r^2)$ Cauchy distribution, hit times are going to scale as $\sim \csc(t)$).
In terms of absorbed kinetic energy, that unfortunately depends. A fully elastic collision that does not move the wall means that there was no energy transfer. One where every particle inelastically lodges in the wall absorbs half of the kinetic energy. If the wall has mass $M$ momentum conservation gives a new speed $w$ so that $Mw =2N \int_{0}^{\pi/2} mv\cos(\theta) (2\pi \sin(\theta)) d\theta$, or $w = 2\pi N m v/M$. With the density that is $w=8\pi^2 \rho v/M$. That kinetic energy as a fraction of the initial total kinetic energy $(1/2)\rho v^2$ is $(1/2) M (8\pi^2 \rho v/M)^2/(1/2)\rho v^2 = 64 \pi^4 \rho / M$. So the bigger $M$ is, the smaller the energy transfer (and once it becomes comparable to the mass of the explosion our approximations are invalid). 
