Bullets & Firewood: Newsletter Challenge (09/05/06)

From a purely theoretical point of view, I would argue that the more elastic rubber bullet would have more propensity to knock over the wood, since it would bounce off the wood, in an elastic collision, causing the wood to begin travelling in the opposite direction at a speed proportional to the mass ratio, whilst the inelastic lead bullet is more likely to travel through the wood, and therefore more likely to leave it standing.

ok - real world, even the lead bullet will knock it over due to the "wind" effect and off centre impacts and so on.

And it's not impossible for the rubber bullet to penetrate the wood - we were shown film during our 'O' level physics lessons of a candle being fired from a rifle and passing through a wooden door

(the point of the film was to show the effects of rifling -

plain bore vs rifled bore =>

plain: candle tumbles end over end and leaves a long tear in the door

rifled: candle spins and leaves a circular hole in door)

I say the lead bullet. The two components that are the same for the bullets, mass and velocity, is not the determining factor. The collision of the bullets with the wood is what needs to be considered. When the two bullets leave the gun barrel they have the same amount of energy, the lead bullet, which will penetrate the wood, will transfer more of that energy to the wood because while it is penetrating, it deforms causing the energy to be dissipated over a larger surface area, all the while tearing its way through the fibers of the wood block. The rubber bullet would just deform slighly on impact and head off in another direction while retaining most of its energy and velocity. Or you could consider it this way. If the lead bullet does not pass completely through the wood block, then it has transferred all of its energy to the wood, if the rubber bullet bounces off the wood block, it has not transferred all of its energy to the wood block. Since the initial energy of the two bullets are the same, then the lead bullet wins the game of energy transfer. Exactly the reason why rubber bullets are not used for deer hunting.

Really? Then riddle me this. Would you rather be shot with a...cue stick, or pushed over with a bullet?

I would rather not be shot at all, nor pushed over. Would you rather be clubbed over the head with a rubber bat or a lead pipe? Both of equal mass and traveling at the same velocity.

The answer to your question--insulting though it was intended to be--is: rubber bat. But had you thought about what the answer should be?

I regret that my riddle was not taken with the spirit of humor in which it given but, rather, literally and defensively--when it was intended only to propose a farcical analogy, to evoke some practical reasoning, by which I might engage you in friendly (perhaps even collegial) discussion of the topic. I would like to believe you would not have seen a need for insults had you realized this when you posted. But, for not anticipating the possibility, I offer an apology.

That said, I leave it to you to decide whether you would prefer for me to explain my little "riddle," or another participant in the challenge discussion. Let me know....

Truly I did not intend to be insulting, and I did not take your question defensively, but I failed to provide you with an answer, which was rude, I wish to offer an apology. However, how can I answer such a question? Aren't "pushed" and "shot" relative terms? How hard is pushed, and how fast is shot? Why is your answer a rubber bat? The bats are the same mass, and traveling at the same velocity, and according to posts further down, the rubber bat will transfer almost twice the momentum as the lead bat. Does momentum hurt? Maybe the question should have been: Would you rather be beaten to DEATH by a lead bat or rubber bat, given both have the same mass and travel at the same velocity? Now that is humorous. BTW, yes I would like for you to explain your riddle.

As specified, the rubber bullet must have a much larger diameter than the lead one – thus the rubber will transfer much more momentum to the wood before (perhaps) penetrating it.

It was specified that the bullets were lead and rubber, with the same mass and velocity, not the same diameter. The rubber bullet could just be longer, and the same diameter. Commonly accepted practice in ballistics is to ignore potential engergy, thermal energy and concentrate on kinetic energy. KE=1/2 m v^2, which both bullets have the same amount when leaving the muzzle of the gun. The effiency of the transfer of this energy is the key to the solution. If we assume that the lead bullet penetrates and does not exit the wood, then it has transferred all of its KE to the wood. If the rubber bullet does not penetrate the wood, then it has only transferred a portion of its KE to the wood. The diameter of the bullets only dictates how large of an area their respective amounts of energy are transferred. On the other hand, if the bullets are fired at sufficient velocity that both bullets penetrate the wood, it is likely that the lead bullet will pass through entirely and the rubber bullet will remain lodged in the wood. In that case the rubber bullet transferred the most energy to the wood block, since it was stopped by the wood, it expended all of its KE in the wood, while the lead retained some of it's KE on exiting the wood. If the bullets were fired at a velocity that resulted in exiting the wood in both cases, then internal ballistics would be a factor and bullet diameter would have to be considered.

Yep, you're right about the diameters, but it would probably have to be an awkwardly long rubber bullet to have the same diameter than the lead bullet!

There's another scenario where the rubber bullet would transfer more energy than the 'stopped inside the wood' lead bullet. If the rubber bullet bounces back, it may impart more kinetic energy to the wood than it had before it hit. OK, we are ignoring energy lost as heat in both cases...

Let us agree that the velocity of both bullets is reasonable, and that the diameter of the bullets are reasonable, such that the lead bullet is the type that is available today, and the rubber bullet is of a larger diameter scaled up from the shape of the lead bullet so as to have the same mass of the lead bullet. I am picturing a rather large .50 cal lead ball which is 1/2 inch in diameter and a larger rubber ball like a golf ball or tennis ball size. I can't say that the masses are equivalent at these scales but I can say that the rubber ball will be larger. If these are fired at the same velocity, a velocity that will allow the lead bullet to penetrate and be stopped and the rubber ball to bounce off, then they start with the same Kinetic Energy from the muzzle, with the rubber ball being larger, it will lose energy faster because of air resistance and a larger surface area, however for argument sake let us agree that the range at which the wood is hit is sufficiently small that we can neglect this KE loss for both bullets. Now we are at the point of impact, the lead bullet expends all of its KE to the wood block. If the rubber ball bounces, it does not transfer all of its energy to the wood block. The rubber will deform, and then bounce back, which is a conversion of KE to potential energy stored as in a spring, which is immediately released for the bounce effect, so that is just a give and take, no increase. When the rubber ball moves after this it is from it original KE it recieved at the muzzle. In this scenario, the only way I can see the rubber ball delivering all of it's KE to the wood, is if it stuck itself to the wood, no bouncing but a good hard splat and stick.

Maybe KE was a bad choice of words and should be replaced by momentum. The total momentum of the bullet + wood system remains constant in the reference frame, ignoring other losses (e.g. heat).

Say the bullet has an initial momentum = +MV and after bouncing back with perfect elasticity, a final momentum = -MV, with M and V having their usual meanings. The total change in momentum of the bullet is then = -2MV, which it imparts onto the wood with opposite sign in order to satisfy conservation of momentum. This is twice what a bullet that got stuck into the wood would impart.

In a real system, it would be less than twice, but could easily exceed unity. It is similar to catching a ball and recoiling a bit. Then throw the ball back to where it came from and recoil another bit...

That would be correct in a one dimensional perfectly elastic collision. If that happens you better duck when you pull the trigger.

Can't remember login. Rouellette13 Anyway, you are right about the elastic collision, however the rubber bullet would bounce of the wood in the opposite direction at a slightly lower velocity. and the log would absorb minimal momentum. In turn this would transfer very little momentum to the log. With a lead bullet some momentum will be spent in piercing the wood but the majority would then be conserved in turn knocking the wood over. equation for inelastic collision: P=momentum M=mass V=velocity P(in)=M x V then add mass of bullet to mass of log, M1+M2=M(total), M(total X V=P(out) manipulated to, P(out)/M(total)=V of log

Well the real answer is without knowing the projectiles mass and kinetic energy applied to such mass. It would be impossible to predict which would be the most efficent in knocking over the said wood also the shape of the wood would no doubt play a part in such an experiment, also might I add if the said wood was destroyed by the lead bullet how would you make a statical review of the experiment .. the true answer is there is not enough information to determine the results as we could use balsa wood and I sure a ping pong ball at the right speed would do the trick ...

forgot login, but dependent on velocity, a lead bullet in the real world would most likely turn the piece of wood into a spray of splinters

I believe the point of the riddle is comparing a non-elastic impact with an elastic impact. It is assumed that the bullet with momentum MV strikes the wood coming to a rest within it. The bullet's momentum is absorbed by the wood. The rubber ball with identical mass and velocity strikes the wood and bounces off. The ball now has momentum in the opposite direction with magnitude b*MV where "b" is some value between 0 and 1 representing the elasticity of the impact with 1 being perfect elasticity. Because momentum is a vector and is conserved, the total momentum imparted to the wood is (1 + b)*MV. Which is more than the lead bullet would impart.

I am nitpicking, but even lead has a modulus of elasticity of 14 GPa compared to 0.0015 GPa for natural ruber.

i agree with that, but wouldnt that make it more likely that a wooden bullet could very well penitrate the wood or split the wood without knocking it over where the rubber bullet would have a better chance of knocking it over without too much damage to the log itself?

The lead bullet wins. Keeping mass constant, the rubber bullet would have to be larger then the lead one. Assuming that both have the same muzzle velocity, at that point they have equal momentum HOWEVER in reality, the rubber will deform in flight and its larger surface area creates more drag and unstable flight when compared to the lead bullet, decreasing its velocity, meaning less momentum at time of impact with the firewood. One could argue that a rifled gun bore would improve the "spin" on the rubber bullet, but as anyone who has watched any crime show in the last decade knows, rifled bores score even steel jacketed bullets, a rubber bullet would probably attain a greater degree of surface roughness due to the lands and grooves. Even if we ignore drag and unstable flight, we consider the force of impact = dp/dt. Because the rubber bullet will bounce off the wood, it still has energy and momentum, meaning it could not have imparted as much force to the block of wood to tip it over. Of course, much of this assumes resonable distance between the gun and the wood, for short distances, you are more likely to penetrate the wood with the lead bullet thus not imparting enough momentum to the wood, and decreasing the rate at which the rubber bullet will lose velocity, improving its chances of knocking over the wood.

A rubber bullet, when impacting a target, will recoil (bounce) from it. This secondary action that is not usually present in lead shot, adds more than the initial kinetic energy to the target by a factor of 1 + sqrt(rebound efficiency).
This extra energy is not conjured from the great beyond, but was part of the initial acceleration imparted to the gun from the bullet.
Dependant factors are of course energy loss on travel to target (kinetic energy at impact), the rebound efficiency of the elastic bullet, and the exit velocity of the lead bullet if it does indeed exit the wood.
If we assume the lead bullet does not exit the wood and the rubber bullet doesn't penetrate the wood, as well as having similar kinetic energies on impact, then the rubber bullet most definately will impart more energy to the wood in effort to tip it over.

I can see a scenario that would have the lead bulled pass clean through the wood and not transfer as much energy as the rubber bullet. Sort of like pulling the table cloth out from underneath the dishes...

Lead bullets are specifically designed to pass through objects, not disperse energy. Rubber bullets are designed specifically to disperse the energy onto the target. Hence, a rubber bullet would more likely knock the wood over... and a lead bullet would most likely destroy it, but not knock it. IMO.

agreed - if the lead bullet is able to penetrate the wood then some of its energy will be absorbed by the wood - therefore the rubber bullet should expend all of its energy on the surface of the wood - knocking it over

Momentum transfer is maximum when mass of the two objects are alike, however, the pressure is greater when area is small. Hence, lead bullet simply with dril a hole without making it to fall, while rubber bullet will make it to fall due to lower pressure per unit area even though force is identical.

Needle can easily go though skin but rubber can not make it even with same force if area is large.

Shyam

Surely the momentum remains the same regardless (assuming the lead bullet doesn't pass straight through) and is identical to the momentum absorbed by the shooter? If the rubber bullet bounces back it does so by storing some of the impact in it's compression which it then releases as it bounces off. Nett transfer of momentum remains the same. Of course you might have to look at the speed of transfer V what does it take to knock a block of wood over. Gut feeling is that the slower tranfer of momentum from the rubber would have more "knocking over" effect.

When you say bullet, you mean very high momentum and there is no time to bounce back as it reaches a plastic reason and no more elastic effect can be seen. Only effect is due to area, which rubber will have more than lead. If entire lead is trapped inside the wood then both rubber and lead will show same impact and momentum will go to wood. If lead crosses the wood then it also carries part of the energy with it and hence less is the energy transfer to the wood. Difference thus is only if any of the bullet crosses the thickness of the wood or not and what is the exit velocity. Rest of the momentum goes to the wood. Nothing else seriously matters.

Shyam

I presume you mean high velocity as the momentum in bullets is not overly high, depending on how big the cannon is of course! Do you have any thoughts on the way the momentum is transfered to the wood as, whilst the time is brief, the rubber surely will transfer its momentum morely "gently " than the lead?

In fact when density of two materials is closer, the momentum transfer is maximum. Material in contact if has same mass then 100% momentum transfer takes place. Remember the billiard ball collision.

When you say wood, then we have no idea about its size, mass and depth etc. All these factors play a role. Good example is to shoot the bullet in sand and you already know the result now.

Shyam

Momentum is a vector product of mass and velocity and hence has a direction due to velocity vector. It is not dependent on size. When you fire a bullet from a gun then both bullet and gun have same momentum but different velocities and directions due to different mass and action-reaction. Gun does not hurt you but bullet does. It is as simple as that.

If Gun is made to move at the same velocity as bullet then it will be a cannon shot and will sure hurt as its momentum will be perhaps 100000 times to the bullet.

Momentum does not hurt if you have same velocity and same direction. Hence momentum of earth does not hurt us. Imagine bullet of the size of earth travelling across the sun hitting us

Shyam

It's the rubber bullet. The lead bullet would just go through. The rubber bullet wouldn't go through, it would just hit the wood and tip it over.

Rubber bullets used to disperse people hurt less because then have less momentum (mV) due to less mass and large area. They just break the skin surface and don't go deep. They do not bounce back at all as they have enough energy to behave like hard solid material even though for less force they have elastic effect.

Heavy metal like lead drills deeper due to smaller size and greater momentum.

If we assume same momentum or same mass and velocity, then small size like to cross the body and hence, rubber bullet will be more damaging. If momentum is too small then body can take impact of rubber but not that of lead bullet. Bouncing back is only at very very low velocities. Are you talking about Tennis ball impact? Well, that is different.

Shyam

What is the mass of the wood?????

Perhaps a wodden plank or tree trunk. I don't think this was for some tiny wood block experiment. Please ask the originator fo the query. I am in the tail of this que.

Shyam

Simple really...the mass that does not shatter or go through the wood; the one that expends it's energy and forward momentum on the surface of the wood, not on the surface, within, and behind the wood (one of these of which impart no "knock-over" impetus to the wood...you guessed it: it's the rubber bullet.

....Not the lead bullet rebounding of the log, but the rubber bullet rebounding of the log; and the log off of it... Here's an analogy. You have an axe with a blade edge and a blunt edge. Both have the same mass and momentum when you swing the axe. Think of what is most likely to happen--which is more likely to topple a log--when you strike with the blunt edge as opposed to when you strike with the blade. In the former instance the axe head bounces back and the log bounces forward; in the latter there's no recoil because the blade penetrates...dispersing its energy between propelling the log and cleaving the log. Since both swings had the same energy at the point of impact, the one whose energy was expended only to causing rebound--it was not divided between two efforts, rebounding and cutting...that's the one that carries more "toppling" force.

What no one has mentioned is how the momentum is tranfered from the lead or rubber to the wood. As the lead bullet penetrates, the momentum that is transfered is not straight ahead to the wood. The lead bullet, because of its shape and its ability to hold that shape (at least better than the rubber), displaces the wood radially as it penetrates dispersing a considerable portion of its energy at right angles to the plane of the wood resulting in less energy being transfered to the forward vector leaving less energy available to "knock over" the wood. The rubber probably does not penetrate therefore less energy is lost making a hole leaving more available to knock over the wood.

Ah, but don't overlook: that there is more to the rubber bullet that just momentum. With its momentum, the lead bullet carries kinetic energy, which dissipates on and within the log as it strikes, penetrates, and distorts; and releases heat. The rubber bullet, on the other hand, also carries the same kinetic energy (its momentum) but, unlike the lead bullet, also carries potential energy--energy bound up in the chemical bonds that hold the rubber bullet together, and also which permit it to recover its original shape (instead of deforming or fragmenting like the lead bullet)...this energy which, moreover, in addition to its kinetic energy of motion, the rubber bullet converts to kinetic energy which is imparted to the log at the instant of impact. So, as I see it, the rebound effect cannot be discounted. For example, if we imagined the rubber bullet to be, say, a plastic or wooden bullet (one whose chemical bonds would break instead of recover) then my "vote" would more likely be with the lead bullet.

Get a clue!!!!!

I agree that the rubber bullet is more likely to topple the wood, but I am troubled by some of your reasoning.

Conservation of momentum requires that the momentum of the bullet is tranfered to the wood in the same direction as the bullet travels, unless the bullet is deflected sideways.

I am also concerned that you say "resulting in less energy being transfered to the forward vector". Energy is not a vector. It's just not.

Force is a force is a force. There is no toppling force. Toppling is an unbalanced state of the material on which force is applied.

What is more force? there is no more force then applied force and in fact only part of the force is used and part is wasted.

Shyam

A toppling force is a force that brings about toppling...as of a log erected on its end. But, thank you for helping make my point (???) Or what exactly was your point?

LOL, what a subject! And my. what a group of answers. Statistics would show the rubber bullet pushing over the standing wood verses the lead bullet. The bullet is subject to a complete penetration and thus there is a 50/50 chance of not completing the task. It would also be concidered in the caliber of the projectile. The smaller the caliber, the more greater the chance of not toppling the standing wood. Only if the standing wood is of a larger mass area per XX:XX size ratio in round fired to diameter of wood. Rubber bullet would have a greater chance of completing the task only because it does not have the penetrating coeffiency in its material. The rubber bullet would actually expand on impact and thus cover a greater surface area, thus unloading nearly 100% of its enertial velocity upon impact. Which the bullet is designed to deep penetration and then dispering once it hits a solid point. Plus the rotation of the round develops heat and actually would burn through the piece of wood, depending on the type of wood it is, such as density, composition of moisture in acounts to how long it has been sitting to drying out. There are literally more veriables using a lead bullet verses a rubber bullt. Rubber bullet has a greater effect over the lead bullet.

You stated thy are the same mass at the same speed though which presents some difficulties. physically when they hit the target they will have the same result. (same mass same speed) The advantage of same speed is the lead bullet speed is limited by the ability of the rubber's projectile to travel given heat and friction in the air. By that I mean if the rubber is going slow enough not to melt then the lead is going slow enough not to penetrate. But equal mass? That will be one big rubber projectile and will certainly not fit in the same barrel so it will nescessitate a different launching method to the target. When launched however the same physics apply and the result is the same they will both have the same result. Even though they are equally likely to produce the same result the lead projectile with its greater mass per cubic inch is the more efficient choice since the mass can be smaller for the same speed.E = MC^2 "A piece of firewood is standing on end. Which is more likely to knock the piece of wood over – a lead bullet fired from a gun or a rubber bullet of the same mass traveling at the same speed (and hitting the same spot)?"

I would say the rubber bullet. Why? Well, all the forces have to balance. We have no information on the type of wood or the velocity of the projectiles. A bullets velocity varies from muzzle velocity right down to nothing at the end of it's travel so it's not unreasonable to say that both bullets are travelling at a speed low enough that the lead bullet will not go straight through in a cloud of splinters. In this instance there are two forces, one exerted on the rifle and it's holder and the other on the block of wood. These have to match (disregarding all the friction losses which I really should be deducting)This is the case for both the lead bullet and the rubber bullet. However in the case of the lead bullet all of it's energy is converted into heat etc whereas the rubber bullet has the ability to store some of the energy. If the rubber bullet stuck fast they'd be the same but if the rubber releases that stored energy by bouncing off there's another force requiring balancing so the firewood has now the extra force of the bouncing bullet to counter as well as the intiial force of impact.

1. we can know the ultimate speed. As I said before the speed of both projectiles is limited by the rubber projectile and its ability to travel through the air without burning away. So we know the lead bullet will be forced to travel slower than it normally does. which means penetration of the wood is no longer a factor. 2.X grains of rubber is quite a bit larger in size than X grains of lead so let's forget about the rifle and let's just say they were launched and they are going the same speed. 3. Let us also say they strike the target at the highest point giving them equal and maximum fulcrum. 4. now we get to the meat of the challenge they are both about to deliver the same amount of energy to the target and both will need to overcome the same amount of inertia. The lead will have follow through and the rubber will have snap. The rubber will compress as it strikes and while doing so its striking point will increase in area which will help give it more overall fulcrum than the lead but it will be fighting against its own energy delivery during the compression phase. That will slow the energy delivery to the target which should impede the ability of the rubber in it's job. Also the stored energy is delivered last and it is partly delivered back to the rubber projectile which means it was never delivered to the target and that means the lead projectile delivered a greater amount of energy to the target. FOR ME IT IS ALL ABOUT THE ENERGY DELIVERY.

These guys should go back to PHYSICS 101. In elastic collisions most momentum is conserved in the object that strikes and bounces off. Leaving the object it hits basically undisturbed. In a non-elastic the momentum is conserved by transfering into the object that the original source hits. (by the way in my REAL WORLD EXPERIENCE a bullet tends to imbed itself into a log unless of coarse we are speaking of a small log or a high powered rifle, in which either of the 2 bullets would tip the log over) One person was right we do not have enough info, however I remain unconvinced of the rubber bullet. My mony is on the lead. Rouellette13 (1312) my user name no longer exists and yet I cannot create a new account because my name and email are already in use. GO FIGURE!!

I think your a bit confused about conservation of momentum.

When the rubber bullet contacts the wood it is compressed. For the rubber bullet to then 'bounce off' with similar speed requires that the wood is just going to stand there and provide a reaction force. It can't!

Wood mass is m1, bullet mass is m2, u=initial velocity, v=final velocity:

m1u1+m2u1=m1v1+m2v2

m1(u1-v1)=m2(v2-u2)

or m1/m2=(v2-u2)/(u1-v1)

Since m1 & m2 are set values the change in velocity of the wood will be proportional to and opposite in sign to the change in velocity of the bullet.

It is true that an object can virtually "bounce back" if the ratio of m1/m2 is low enough, such as a ball bouncing off a concrete floor.

I would estimate that the wood weighs 3kg, bullet 5g. (Pick whatever values you like...). Then m2/m1=3000/5=600. Say the bullet arrives at 300m/s and bounces back at -300m/s, change in velocity is 600m/s. The change in velocity of wood would 1m/s.

Someone should go back to high school physics and learn the principles, not just watch the fun experiments.

Since no one bothered to mention the size of the firewood (height and diameter), no mention of material (hard or soft wood), was it new wood or aged. No information was given about the durometer of the rubber (eraser or bowling ball). How about the distance the bullet has to travel before reaching the proposed piece of firewood? I hope everyone doesn't reach conclusions before having all of the required information. I don't believe anyone can make a accurate conjecture.

I saw an interesting demonstration related to this a long time ago: A piston was attached to a crackshaft and was free to move up and down within a cylinder. The piston-crankshaft assembly moved freely with very little effort. The demonstrator had two hammers, one made of steel and one made of rubber. If he struck the top of the piston with each hammer while the piston was at rest just past TDC, which hammer would have the greatest effect? First he struck the piston with the steel hammer. The piston moved down a little. Next he struck the piston with the rubber hammer. The piston and crank assembly spun through a couple of revolutions. Why? The rubber hammer distributed the force of the blow over a longer period and was overcame the inertia of the system. He was actually talking about knocking in gasolene engines, but this seems applicable here.

Has anyone considered elevation in their thoughts? If area of the impact is low, say 5cm from the base, wouldn't the lead bullet have a greater chance of penetrating the wood?

you guys are making it way to complicated...

F*t = m*change in v (impulse=change in momentum)

lead buller burrows in, which takes more time. same mass of wood, meens greater t=greater change in v

so the lead bullet makes the wood move more, better chance of falling over