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Malcolm Gladwell made a claim in a recent talk that a sling with a stone going at 30m/s has the same stopping power as a .45 calibre handgun.

How would I verify whether or not this claim is true - even given some assumptions, like the stone has a weight of 1lb or 2 lbs or whatever.

Not quite sure how to work this out.

Assume that the bullet is a 185grain 0.45 calibre bullet.

Qmechanic
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marcamillion
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    You start by asking someone to define "stopping power", discover that the definitions vary but that they are all intimately wrapped up in the physiology of shock. And realize that Physics SE may be able to help with some parts of the problem but is the wrong place to just drop the whole question. – dmckee --- ex-moderator kitten Mar 06 '14 at 00:07
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    Stopping power is a complex issue and not something you can just calculate from ballistic first principles. It depends on the materials of both the projectile and the target, the biological structure of the target, and a host of other of things. The notion of "stopping" in this sense isn't even really precisely stated. It doesn't seem that far-fetched though. A 30m/s rock certainly sounds like it could do some serious damage. But I'd be wary of Malcolm Gladwell's claims; he's a story teller, not a scientist. He's basically admitted to mining data to suit his purposes. –  Mar 06 '14 at 00:08
  • I agree with both of you above, I guess I would ask you to define some assumptions - so we can know where to start - and just run some numbers about how one might approach this problem. Assume that stopping power is basically causing the person to stop running towards you. – marcamillion Mar 06 '14 at 00:10
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    You don't begin with physics. That's approaching the problem from the wrong end. The system to understand is a biological one and that is where all the complexity lives. You have to master the complexity of injury and shock in the human system to even get started. Why do you think people are always basing these decisions on stories about how badly the 38 revolver did in the Philippines? Or about how their buddy knew a cop who had a partner who was killed when his [insert round you want to disparage] didn't stop the [scary drug of the moment]-crazed maniac. – dmckee --- ex-moderator kitten Mar 06 '14 at 00:14
  • Why don't you stop it and calculate the power needed to do so? – evil999man Mar 06 '14 at 14:24
  • Concerning question based on videos, see this meta post. – Qmechanic Mar 06 '14 at 17:42

1 Answers1

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While I agree with the caveats made by dmckee in his comments, there is an obvious interpretation of stopping power as the change in momentum caused by the projectile.

Collision

The mass and velocity of the projectile are $m$ and $v$ respectively, and the mass of the target is $M$. Since the target is stationary the initial momentum is just $mv$.

Assuming the bullet transfers all its energy to the target (i.e. it stops within it and doesn't go through and out the other side) then after the impact the projectile and the target will be moving with some velocity $v'$, so the total momentum is $(m + M)v'$. Conservation of momentum tells us that the initial and final momenta must be the same so:

$$ (m + M) v' = mv $$

and rearranging this gives:

$$ v' = \frac{m}{m + M} v $$

Let's say the stone weighs 0.1kg. My mass is about 70kg, so if the stone hit me then final velocity would be:

$$ v' = \frac{0.1}{0.1 + 70} 30 \approx 0.043 m/s $$

Now for the bullet. I guess it depends on which pistol you choose, but the .45 ACP has a muzzle velocity of about 350 m/sec and a 185 grain bullet weighs about 0.012kg so:

$$ v' = \frac{0.012}{0.012 + 70} 350 \approx 0.06 m/s $$

which is actually pretty close to the stone.

Really the key bit of the calculation is the initial momentum of the projectile, $mv$. The stone weighs about ten times as much as the bullet but is moving at about one tenth of the speed, so the two factors of ten cancel out and the initial momentum is roughly the same. This is presumably the basis of the claim about stopping power. But attend to dmckee's comments. The velocity changes calculated above are around 0.04 m/s, and since humans can run at several metres per second you're not literally going to stop a running human with either the gun or the sling. Presumably a gun stops someone because it makes them fall, and friction with the ground does the stopping, in which case you should probably avoid trying to shoot axe wielding maniacs when standing on an ice rink.

John Rennie
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