Is momentum conserved in a liquid? It seems to me that if you shoot a bullet into a tub of water less momentum will be transferred than if you shot into an empty tub. Most of the kinetic energy will decay to heat and with it a part of the momentum. When the bullet goes in it creates some kind of shock wave in the liquid. This shock wave is composed of particles moving back and forth. If you treat the particles like tiny little marbles it isn't impossible to see how some of the linear momentum would get scattered. The experiment - put a container of water on one of those air-tracks common to high school physics labs. Shoot at it with a bb gun empty and full and record momentum. Seems like momentum may get conserved but scattered; hence everything but the vector component of the momentum is conserved.

A bullet only pierces the liquid so far - it is slowed by friction. This turns kinetic energy into heat energy. The momentum of the bullet (which is considerably smaller than the kinetic energy) will also be transferred to the water molecules. Picture two marbles. One is stationary and the second is flung at it, in the x direction. The mometum in the x direction will not change. However, if the two marbles are flung at each other perfectly, they will both stop after coming into contact; their momentums cancel. In both cases the momentum of the whole system is conserved, however in the second case the momentum in the x direction is cancelled by the momentum coming from the -x direction. This is the same in a liquid - all liquids above 0 k are in thermal motion. If the water molecules are treated as marbles then shooting a bullet into a liquid will in some sense nullify the macroscopic momentum of the bullet because the liquid has a large amount of microscopic momentum according to boltzman distribution. The net microscopic momentum will indeed increase and momentum is not really destroyed.

Linear momentum is always conserved in a system that has spatial translation invariance. In other words, if you do the experiment in different places you should get the same result. Since that is quite a reasonable assumption, linear momentum is typically a conserved quantity. Considering a particle moving in a liquid is pretty much like not neglecting air resistance on a projectile. Momentum is still conserved, although some of the momentum of the particle is lost to the surroundings. Firstly, the plural of momentum is momenta. Secondly, your second scenario is not true - I can only assume you were too busy being a theoretical experimentalist to play with marbles as a child. If 2 particles with equal and opposite momenta collide they will have equal and opposite momenta after the collision. What you're assuming is that the only solution to \(p-p=0\) is \(p=0\) when quite obviously it's not.

If the bullet ends up stopped and the only place it's momentum can possibly be transferred to is the tub (i.e. water doesn't splash out or anything) then the momentum transfer is exactly the same. The time it takes is of course different so the forces involved are different, but if the bullet starts and ends with the same momentum and the tub's momentum is the only other momentum that changes, the amount of momentum transferred must be the same. Of course in practice a bullet will probably go richocheting off all over the place if you shoot it into an bathtub, wheras it will probably stay in the tub if the tub is full of water, so in that case you'd have more momentum transferred to the tub in the former case (because the bullet bounces out taking with it some "negative" momentum).

Hmm. True true. I suppose I have to upgrade my scenario a bit- Take these two marbles and coat them with velcro and anti-velcro so that they stick during the collision. Now, the outcome is \(p=0\) However, I think liquid particles would behave more like this - Imagine you have two pool balls. Both are shot at each other one going in the x direction the other in the -x with momentum z. They collide perfectly so they then move with momentum z in the y and -y directions. And I preferred legos to marbles. Yes, this is perfect for a physics class demonstration. The teacher does the shooting, of course. Please Register or Log in to view the hidden image! I don't think so, like I said above. I think the total momentum of the water molecules will increase to conserve overall momentum, but the tub will gain more momentum without water. I suppose the experiment should be modified by giving the bullet a firm surface to become lodged inside of in order to maximize momentum transfer, among other things. Its like the pool ball example - the bullet adds momentum to water molecules, which then collide with thermally agitated water molecules.

Okay, I admit defeat. The net momentum of water molecules is zero, so while some momentum will be cancelled for molecules the net effect is still a transfer of momentum. I thought the billiard balls could prove it.. but with the billiard balls the system starts out with 0 net momentum while in the case of a bullet there is net momentum. Please Register or Log in to view the hidden image! It may be worth testing though, its easy enough. Point is - in the liquid at equilibrium each molecule (on average) cancels the momentum of another molecule and so the net momentum of this system is zero. If you mess with this balance by adding net momentum (the bullet) it may be that the liquid molecules no longer cancel out just right and the system gains just as much momentum. I think this is the formal procedure - http://en.wikipedia.org/wiki/Momentum#Linear_momentum_of_a_system_of_particles

What about this scenario- Fire a cannon directly upwards. The earth will feel an initial recoil, and then another recoil when the cannonball returns to the ground. Net momentum will be in one direction. Hm. Does the the cannonball exert the same gravitational force on the earth as the earth does on the cannonball?

Your first statement is incorrect - because the answer to your final question is "yes." All the forces cancel out and return to baseline - exactly as they were initially.

Can you show some concrete calculations as to what will be Earth's acceleration and maximum velocity during this scenario?

It shouldn't be too difficult. First, you must decide on muzzle velocity and weight of the cannonball. Once you have the two you can determine momentum which is p=mv The momentum of the recoiling earth will be equal to this. You then divide it by mass to get velocity like p/m=v What is interesting is that the cannonball will lose some kinetic energy and momentum to air molecules due to friction, and this goes back to the first part of the thread. The air molecules will also rise and fall.

It all comes down to Newton's "equal and opposite forces". Assume both the earth and the cannonball are initially stationary: When you fire the cannonball, the cannonball is pushed up and the earth is pushed down. Total momentum is conserved. As the cannonball flies up, gravity pulls it towards the earth while pulling the earth towards the cannonball. Total momentum is conserved. At its zenith, the cannonball and the earth both have 0 momentum relative to their starting points. As the cannonball flies down, gravity pulls it towards the earth while pulling the earth up towards the cannonball. Total momentum is conserved. When the cannonball lands, its excess momentum is transferred to the earth, cancelling out the earth's upward velocity. Total momentum is conserved. Force is just the derivative of momentum with respect to time. So Newton's law of "equal and opposite forces" guarantees that every change in an object's momentum is balanced by an opposite change in another object's momentum. Total momentum is conserved.

Yes, it seems like this is the case. Sure, take this for example - http://www.economicexpert.com/a/Colt:Mk:12:cannon.htm p = mv .110 kg x 1,010 m/s = 111.1 kg · m/s for the bullet and -111.1 kg · m/s for the earth the acceleration on the shell due to gravity will be 9.81 m/s^2 and the force on both bullet and earth will be f=ma where mass is .110 kg and acceleration is 9.81 m/s^2 = 1.0791 N from which you can calculate the acceleration on the earth by plugging in the numbers a=f/m a = 1.0791 N / 5.9736 × 10^24 kg the momentum of the earth will also equal 111.1 kg x m/s directly after the bullet is fired. The velocity will be really small since the earth's mass is enormous (5.9736 × 10^24 kg) it will then be 111.1/[5.9736 × 10^24] = v and will decrease according to the acceleration above I'm still not sure how conservation of momentum would occur in the air particles that have gained momentum due to air friction. These will also have to rise and fall due to gravity. Does this mean that every time I blow air upwards air will come back down? Its hard to picture this in a fluid. As part of the liquid is forced upwards another part will be forced down - so that no void is created. If these two are equal momentum will cancel, no? We can agree that the shockwave will travel with a limited velocity. Lets say it is travelling from the bottom of an aquarium. In the time it takes to travel from the bottom to the surface, momentum will vanish, no? Its not like the surface will instantaneously bulge upwards, there will always be a time lag. During this time a certain volume of water will travel upwards while an equal volume will travel downwards - the two cancel. ... It seems that the momentum in the liquid is angular - since the liquid is circulating.

Can angular momentum be created from linear momentum? It seems like it by looking at a system such as a turbine - steam flowing in one direction turns a wheel. The wheel gains angular momentum and the steam loses linear momentum. Of course, this could be misleading - perhaps the turbine receives this momentum and transfers it to the ground (creating an infinitesimally small acceleration). But I don't see how the angular momentum could be offset - before steam is applied angular momentum = 0 so it should = 0 at the end.. This is a good thread on the subject - http://www.bautforum.com/archive/index.php/t-32846.html

How about this - Imagine a very tall vacuum chamber. Suspended from the ceiling of this vacuum chamber is a windmill vane, suspended using a string attached to its center so that it can spin freely. Pointed at one vane of the windmill is a laser beam. Either the windmill starts spinning (angular momentum) or the entire windmill gains linear momentum (from the photons) - however, it cannot really gain linear momentum since it is attached to a string and will only move in an arc. Any movement of the windmill results in angular momentum. So, linear momentum can be transformed into angular momentum.

Okay, I see where I'm wrong. If momentum was 'scattered' or lost in a liquid then liquids wouldn't have any weight . . . which is clearly false. This means that all the velocities of different particles do cancel out on average but the overall gravitational force exists to give them weight. Which means that a shockwave travelling through the liquid will also conserve momentum. I wonder how this works out in very small systems consisting of maybe 1000 particles. Its fun to think about at the very least Please Register or Log in to view the hidden image!

If an object has non-zero linear momentum then its moving. Angular momentum is then definable about any axis which doesn't coincide with the instantaneous velocity vector of the object. Gravity and weight have nothing to do with it. The material which makes up a fluid has inertial mass (inertial mass is not gravitational mass and gravitational mass is not weight) and thus can carry momentum, which is provided when the bullet enters the fluid. Generally such a container of the fluid is large, while the bullet is small, so it might seem like the liquid doesn't pick up any momentum but really it spreads the momentum transfer over a larger region. To be more specific suppose you have a cuboid tank of water which is the size of an average bath tub and got plenty of water in it. You put the tank on top of a table with 'smooth' wheels on it, like a well oiled trolley used on aeroplanes to dispense over priced drinks.Someone stands at one end and fires a bullet into the tank (like they do on CSI) and the bullet is 'captured' by the water (ie doesn't go straight through). The tank absorbs the momentum of the bullet, which is partly transferred into the trolley too and the entire thing moves in the direction the bullet was originally moving in. The momentum they have will be equal to the momentum of the bullet originally but obviously move slower due to having more inertial mass. Suppose someone stands at the other end from the person with the gun and 'braces' their hands against the tank so as to try to keep it still when the other person fires the gun. They'll manage to do that easily. Why? Because the pressure their hands exert will be over a large area. As a result although the total impulse must be the same as would be needed to stop a bullet with their bare hands the lower pressure is not going to cause them injury. Stopping a bullet bare handed would mean applying a lot of force in a small area, which you can't do (and that's why the bullet would enter your hand) but the water spreads the force out, even though the momentum change overall is the same. Another example to do with spreading out an impulse is a pin. No matter how hard you try you can't press one of your thumbs into the palm of the opposite hand (ie breaking the skin) but a pin easily breaks the skin with minimal effort. You can't do fluid mechanics on such small amounts of particles, there aren't enough to make a 'smooth' fluid, you have to include quantum effects and weird things in fluids which arise only at very short distances and, as I've already told you, quantum mechanics conserves momentum. Fluid mechanics does too, its built into the equations of motion. If your equations of motion satisfy the conditions needed to conserve momentum (ie use of Noether's theorem etc) then no solution of them will violate conservation of momentum and shockwaves are a particular kind of fluid mechanics solutions. Its the same reasoning as I explained in QED, the overreaching expression, from which all equations etc are derived, is invariant under spacial translation, thus conserves momentum and thus all predictions and solutions will too. If they don't then you have an anomaly and your model is inconsistent. The reason why physicists love doing all this complicated mathematics (ie variational principles, Lagrangians, Hamiltonians etc) is that once you know the basic machinery you can very quickly deduce results which are extremely hard to do otherwise. For instance, you can obtain the motion of a spinning top without having to solve some horrific specific case of F=ma. If you put in the time needed to understand variational principles you'll understand the basic methods of pretty much every area of physics. Classical mechanics, relativity, quantum field theory, all of them use it and if you know how one of them uses it you can pick up the other ones very quickly. By saying "This maths is too complicated" you're making life harder for yourself, you're only able to consider specific systems one at a time. Variational methods allow you to consider entire classes of problems at the same time.

I always figured it was different in liquids because liquid is more dense than air therefore it interferes with a bullets trajectory more so than if it was shot in open air. Since there are more particles the bullet needs to pass through in order to reach its destination, it loses more of its momentum to friction than it would to air resistance.