MythBusters: How do you characterize a collision?

I thought the same thing when I watched, but really though, in the bat situation how different are the cases? You are talking about a very short interval for the actual collision in both cases and pretty much inelastic collisions. So even though I agree with you in the sense they maybe should have categorized it as momentum and time of collision, and how elastic the collision was, this I think was close enough to get a feel.
In the gun shooting sideways, likewise the interval for momentum transfer when the gun went off is still pretty darn short so somewhat similar to an impact.

The interesting part for me was how the bullet shattered and would have caused so much shrapnel damage. Pretty nasty in any event!

How could we measure the force of bullet on gun? How about a ballistics pendulum?

Hang the gun from a string and shoot it. Measure the time of collision with high speed camera and calculate the speed after collision based on how high it swings.

Still, I think this would be too high since the bullet pushes the gun and hand. So, now what? Add the some mass to the gun to compensate for the hand? But bigger hands might have better grips-- aagh!

So we ballpark the new force and then go back to the double handle gun (because, let's face it, it was cool) but this time can we tune in the powder in the bullet so if fires with less force (is that even possible?- I don't know enough about ammo).

@Dale,

Even if you measured the time of the collision between the bullet and the block of the pendulum (which would be a very short time), that would tell you the force the bullet exerts on the block. It would not tell you the force the bullet would exert hitting a gun.

There is something missing from your analysis because the last time they categorized a collision using your method they got the answer wrong. I want to say its because in order to measure the force imparted by the object you have to measure its momentum after the collision.

A problem I had was that the pieces of bullet kept going in roughly the same direction as the original bullet. This must have carried a significant chunk of the total energy/momentum so that it was not actually delivered to the target gun.

No, not a block. Use the GUN as the pendulum. Shoot a bullet at a gun hanging from a string(s).

@Dale,

OH! I get it. However, that still wouldn't tell you about the forces involved in the collision. It would just tell you about the momentum after the collision. If you have a long collision bullet (squishy) and a hard one that were otherwise identical, the pendulum gun would behave the same way.

Then we divide the momentum by time of the collision which we find from the high speed video.

Of course, this would give you average force during the collision not the peak force.

I'm afraid they get it very right by looking at kinetic energy...

To shoot a gun out of someone's hand you basically have to exert a force which is higher than the hand's grasp, over a distance that is larger than the fingers' range of motion. Call them F_h and x_f. If you are going to get that done shooting an object at the gun, the single most important thing to consider is the kinetic energy of the object shot, which has to exceed F_h.x_f.

Think of the bullet as a mass M, with a spring with constant k in front. Make k infinite, and you have a perfectly rigid body, make k small, and you are shooting a marshmallow... If you wanted to take into account dissipation of energy during the collision, you could add a damper in parallel with the spring, but I'll forgo that.

With this simple model, very rigid objects will exert a very high force initially, but the corresponding reaction will drain their kinetic energy correspondingly fast. And unless they have enough energy, they will be stopped before freeing the gun from the hand.

On the other hand, very elastic bodies will initially not generate enough force to move the gun. But the correspondingly small reaction force will do little to stop the elastic bullet, which will continue to deform against the contact point, increasing the force exerted. Eventually, if the energy stored is sufficiently high, the force exerted will overcome F_h. And if there's enough energy in store, the force will remain above F_h while travelling x_f until the gun is released from the hand.

Comparable situations arise when ionizing an atom, and the standard measure is an energy of ionization. Or when molecules evaporate from a liquid into a vapor, and again you have an energy, the latent heat. Resilience, the ability of a material to withstand impacts, which is a measure of (lack of) brittleness is also measured in energy units, the area below the strain-stress curve for the material.

And more down to Earth, if you get hit with a padded bat, it may not break your skull or your ribs, because the padding will reduce the peak impact force. But it will send you flying just as if it was a wooden bat. Get shot in the chest with a rubber ball, and you'll fall on you back, just as if it were a real bullet, only you'll live to tell it.

It'd be interesting to model the thing more rigurously, taking into account loss of energy due to plastic deformation of the projectile. That is, adding the damper I commented earlier. But my guess is that it will not make things change too significantly.

Always assuming that "shooting the gun from his hand" is primarily a phenomenon in gross physics.

From all the analysis I read ages ago (hey, growing up in the 50s with westerns!) the main reason that people drop weapons which are struck by bullets [1] is that the bullet->gun impact is pretty elastic and very short. The result is high-frequency acoustic propagation in the gun (much higher than when firing) and the main lossy element is the person holding the gun.

In other words, it hurts like hell. Reflexive release of the hand and the gun falls.

[1] It has happened -- primarily to riflemen in combat.