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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?
      ^
     / \
      |
      |
-------------
  \   |   /

  -   *   -

  /   |   \

edit: updated diagram:

       ^
      / \
       |
 ______|______
|   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.

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    $\begingroup$ This is more complicated than it may seem at first blush. It depends on a lot of details. Is it a low explosive or a high explosive? Is it a thermobaric bomb? How heavy is the plane it is hitting? Etc. $\endgroup$ – Chris Mar 17 '18 at 21:29
  • $\begingroup$ the mass of the plane it is hitting shouldn't affect the % of energy transferred. Let's assume a uniform sphere of energy and an infinite plane. $\endgroup$ – xaxxon Mar 17 '18 at 23:02
  • $\begingroup$ With an infinite plane, 0% of the energy will go into the kinetic energy of the plane: it will all go into heat. This should be relatively obvious: some finite momentum is transferred to the plane, but the plane has infinite mass, so $T={p^2\over 2m}=0$. $\endgroup$ – Chris Mar 18 '18 at 0:04
  • $\begingroup$ @Chris: 0% of the energy will go into the kinetic energy of the plane. Wrong. There is a finite speed of sound in the wall. At any given moment there is a finite mass of (part of) the wall participating in the momentum exchange. Not to mention that energy carried away by elastic waves is not the same as going into heat regardless of its ultimate fate. $\endgroup$ – A.V.S. Mar 18 '18 at 5:42
  • $\begingroup$ @A.V.S. That's getting a little pedantic. You agree that the plane will not gain any translational kinetic energy, yes? That 0% of the energy will go into pushing the plane, in the long run? $\endgroup$ – Chris Mar 18 '18 at 5:50
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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).

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  • $\begingroup$ I don't understand why the elastic collision equation you come up with "J=4π2ρv" cares about the p or v. shouldn't the final equation just be J(interesting) = J(total)*(some-scaling-factor)? how does anything else about a spherical explosion matter? whether it's big slow chunks or smaller fast chunks... or am I missing something and the p*v aspect of it is actually the total joules of the explosion? $\endgroup$ – xaxxon Mar 18 '18 at 3:27
  • $\begingroup$ Let us assume vacuum, infinite plane. This is already wrong. Unless you place explosive directly at the wall, most momentum transferred to it would come from air carried by the shock wave of the explosion. If a dimensionless parameter $A=4\pi/3 \rho_\text{air} a^3/m_\text{bomb}$ (where $a$ is a distance from explosion to the wall) is larger than 1 your solution is completely off. For example for 100kg bomb at 20m from the wall we would have A>400 more than 2 orders of magnitude. $\endgroup$ – A.V.S. Mar 18 '18 at 5:17
  • $\begingroup$ Agreed, looking at my answer when fully awake rather than one in the morning, it does not work that well. Will do a radical overhaul after I checked my books on explosives. $\endgroup$ – Anders Sandberg Mar 18 '18 at 15:13

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