For instance, take the pliability of the log out of the equation. Assume they are both hitting a solid block of hardened steel. The answer to this equation implies that the rubber bullet would knock that over easier also since it would rebound as well. I cannot buy that. I know that studies have been done and the answer is correct (that the rubber bullet knocks it over easier) but I think it has to do with the lead bullet wasting it's energy spreading the log vs the addition of rebound energy.

But what do I know?

I think conservation of energy was violated here. Since the two bullets had the same mass and velocity, they had the same amount of energy when they struck the wood. That's a given. Now the rubber bullet somehow has more energy? The wood had no energy to begin with, remember.

Some of the energy that *would have been* imparted to the log was used to compress the rubber bullet. This actually would have resulted in *less* energy (force) going into the wood. Since you don't get something for nothing, heat, etc. would have been generated in the rubber bullet as it compressed and so the amount of stored energy available for it spring back into shape and essentially "bounce off" the wood would have been diminished somewhat. The net effect is less force imparted to the wood.

Still, on a very fine level these effects may have some impact, but I still maintain that essentially it's a wash.

I hate stuff like this. I can't stop thinking about it. :>)

Mike

The lead will not compress like rubber will, instead it will continue on with its intended force. The rubber upon hitting is compressed and although it is traveling at the same V it can't continue onward to its intended destination because of deformation has taken place. The energy has to go someplace.

An interesting think also would be to consider any energy that comes off as heat. I wonder what you would see if you looked at the process with a high intensity IR camera.

I've had to do impact calculations before. It really all comes down to how quick the "mass" is decelerated. Maybe that's the key here. The lead bullet penetrates and decelerates at a slower rate than the rubber bullet, which decelerates more rapidly.

I think I just proved myself wrong. Oh well. BTDT.

Mike

Your first paragraph doesn't help me, but the second certainly reinforces my point. I agree the lead will indeed create heat while deforming.

OK, maybe this will not help as well:

It's a hot afternoon; you've just finished mowing the lawn and are in need of a nice, cold beer. Prior to starting your yard work, you put the beer in the freezer so it would be very cold — and refreshing — when you were ready to drink it. You take the beer from the freezer, and it is indeed very cold (but not frozen). But before drinking it, you add a little lime juice. All of a sudden, ice crystals form and work their way down through the beer. What happened?

Not true. The wood has potential energy stored in it due to its position. It has been set on end so that it can then be knocked over by the bullet.

You introduced a nidus for crystal formation into the beer, thus allowing the ice crystals to form. Same thing happens when you have a super saturated solution of sugar like in candy making. It stays liquid until you introduce a single grain of sugar into it, then instant crystals form.

The other option is that the lime juice changed the freezing temperature of the beer and thus allowed the crystals to form. Same idea as adding salt to water to raise the boiling point.

So how was this potential energy used to help knock itself over?

Sounds like something for the Mythbusters to try. :D
-Chris

OK, I'm confused by this answer. Not to say it's wrong, but I don't get it.

First, "force" equals mass x acceleration; we've stipulated equal mass and equal velocity, so are the bullets "supplying... force"? I don't think I'd say it that way. But the last physics course I took was 18 years ago.

Whence the energy that allows the rubber bullet to rebound? it has to come from somewhere ... so it comes from the energy the rubber bullet was carrying to the wood. So the wood doesn't absorb the energy that causes the rubber to rebound. That energy remains with the rubber bullet.

All of this assumes that the energy involved is sufficient to knock over our piece of wood, and that the rubber used is more elastic than lead. There are some very non-"springy" rubber compounds.

I've read this thread at CR4, and I'd still say that, if the lead bullet enters the wood at all, but does not fully penetrate, the wood *has* to take all the energy. If the rubber bullet bounces, it is taking some of its energy back with it.

And.. I think any issue regarding the rubber bullet applying its force "over more time" is inapplicable if the lead bullet does not fully penetrate the wood.

JP

Yep.

I too would love an explanation that resolves this.

Too many variables - this is a silly thread.

How about this. A high speed bullet hits the wood with a small hole - losing speed and energy. As it expands it blows a chunk of the back of the log out equal to the force on the front. What is that about equal and opposite reaction?? Firewood stands.

"for every action, there is an equal and opposite reaction" so the rubber bullet bouncing off would generate log speeds in proportion to the the relative masses of the objects.

if the bullet penetrates the wood, the wood (and energy) is separated in a direction 90deg off from the intitial direction, thus disippating the energy in a perpendicular direction.

That's my thinking and i'm sticking to it. (hopefully)

Bounce a rubber ball on the pavement. The force you deliver to that ball is reciprocated back in the bounce back - but not 100%. Try dropping a rubber ball instead of throwing it down. It will bounce back up but not to the height you dropped it at - it can't (unless a chemical reaction ensued on the way down that changed its mass on the way back - different discussion though) There is lost energy? It had to go someplace. With a bouncing ball the energy that seems lost is so small it is un-noticed but that is not the same for a projectile traveling at 800 to 1500 ft/sec. The bullet absorbs so much but the wood (or any surface has too as well). The total energy has to be accounted for. It this case it is transfered to the surface where action has to occur (and some may even be given off in the form of heat).

I'm not saying it knocked itself over, I'm saying that there IS energy in the piece of wood in the form of potential energy. There are also forces acting on the wood to keep it upright.

Answer me this. How can an object rebound against something that offers no resistance? If the wood did not "push back" against the rubber bullet, the wood would simple instantly fall over and the bullet would continue on its path, having lost no energy at all.

"There is no Spoon."

can't be answered , there's not enough data

what caliber
what distance
how big a log

several variables, that when combined could go either way.

bigger calibers, big log Rubber
smaller calibers small log, Lead
longer distance, medium log, Lead
short distance, medium log, Rubber

footpounds...

The wood is offering resistance - not complete resistance but enough.

What about the expansion (mushroom effect) that takes place with a lead bullet upon impact? Isn't that a large factor in the "knockdown" power of a lead projectile? If not, then why are hollowpoint bullets so effective?

They simply prevent the bullet from passing through and not imparting it's full potential on the target.