Dot Physics

Pre Reqs: momentum principle, forces,energy, vectors

Really, there is not much new here. This is an introduction to objects that interact. To describe this, I will need to pull several different ideas together (that you have probably already looked at). Let me start with a simple case. Suppose I have two objects, maybe they are two asteroids in space. I will call them asteroid A and B:

i-dc0d0631de7b5ac39162b134b649d679-interaction-1.jpg

In this situation, the two objects have different momentums but one interaction between them. Notice that the gravitational force on asteroid A is the same magnitude but opposite direction as the force on asteroid B. This is because there is just one interaction, the gravitational force between the two asteroids. (notice that I am using the notation FAB to indicate the force from object A acting on object B) Anyway, what does this force do to the two asteroids? Well, forces change the momentum. If I want to look at the momentum of the two asteroids after some very short time, I can write:

i-4617904937fc3e2f791c43b9aba9ff68-deltap-1.jpg

However, I can write a relationship between the two forces (since they are really the same force):

i-179261bdd2535ac607754e96ac1bc802-fa-fb.jpg

I didn’t say it earlier, but the time that the force acts on asteroid A is the same as the time the force acts on asteroid B. So, I can re-write the two momentum equations as:

i-fdde993db9000c2e39415dece040b8d6-deltap-2.jpg

So, now I can relate the two changes in momentum as:

i-2f5a04f334010bba25b47f505937411e-comparep.jpg

If you think about this, since they have the same magnitude of force acting on them (because they are interacting) for the same time, they have the same change in momentum. If I think of the two asteroids as one system, then I can write:

i-4962786f4a0e7e976298c2ef911aac71-conservation-p.jpg

This is what people usually think of when they think of collisions – conservation of momentum. In this case, this essentially means that the total momentum is the same before and after something happens. Is this always true? No. This is true if there is no external forces on the system. Suppose I had these two asteroids near a star, but the star was not in my system (why would I do that? I don’t know, just pretend). Then both of these asteroids would have another force on them such that after some time delta t, the total final momentum would be different than the total initial momentum.

Another important thing to realize is that this does not have to be a gravitational force acting between the two objects. It could be anything. Suppose these two asteroids collided. There would again be a force between them that would be something (probably the coulomb force). As long as this was an internal force, the total momentum would not change. Even if this force changed the internal structure of the two objects, momentum would still be conserved.

Now, for more details.

Really, the above stuff is all there is. All that is left is to look at a couple of special cases. I think it is important to NOT get bogged down in the particular solutions of particular cases, but instead realize that conservation of momentum is just an application of Newton’s laws.

In the following stuff, I am mostly going to talk about things that “collide”. These are interactions that are very quick with very large internal forces. For these cases, the external forces can usually be ignored. Take for instance two baseballs colliding in mid-air. If I look at the momentum of the two balls right before they collide and right after they collide, I can say that momentum is conserved. If I wait too long, clearly momentum will not be conserved because the balls will have an external gravitational force on them. So, I can cheat since the time for the collision is so short.

Special Case: Objects stick together

If the two objects are stuck together after the collision, this is called an inelastic collision (the name really isn’t that important). Here is a before and after picture of two objects in an inelastic collision:

i-4ff5e541650f789e29600e929d1c96a5-inelastic-1.jpg

So, here the initial momentum is due to two objects. After the collision, there is essentially just one object (I used the subscript T to indicate the total). Since momentum is still conserved, I can write:

i-6a72e3b9a2daddaf045a7030a337e7cd-inelastic-2.jpg

How do you deal with vector equations? One way is to break them up into components. If the collision takes place in just two dimensions, you could break this above equation into an x-direction and a y-direction. Not sure if it is worth given further details on this special case. If you know the masses and the two initial velocities (or momentums), you could easily find the vector value for the final velocity after the collision. Another interesting aspect to look at is the change in energy. What should be greater, the total kinetic energy before or the total kinetic energy after the collision? If this is just a plain-jane collision, the final kinetic energy should be less than the initial kinetic energy. This lost energy would go into deforming the two objects and increasing their thermal energy (plus maybe making some sound). If were some type of explosion, the kinetic energy afterwards would be greater. In either case, momentum is still conserved.

Special Case: One dimensional perfectly elastic collision between two similar objects where one is initially at rest.

Yes, this is a little specialized, but I want to show you an example where kinetic energy is conserved. Here is a before and after picture:

i-8ed23f9402c8c7da4755ad93c1e89d7f-samecollision-1.jpg

So, ball A and B have the same mass and ball B is initially at rest. After the collision, they do something. I just randomly put that they are both going to the right with a lower momentum. Also, for simplicity, I will call the right direction the positive x direction. This means that the following is true from the conservation of momentum:(since both balls have the same mass, I will use m for both ball A and B) (double-also: I left off the sub-x notation to indicate that these are the components of velocity in the x-direction)

i-cfa82b28c6f30e985fd6a21bd782bae5-collisionsame-2.jpg

Clearly, the masses cancel. However, I can not determine the velocity of the two balls after the collision because there are too many unknowns and only one equation. I can get another equation by looking at the kinetic energy. If KE is conserved then:

i-7434b48a4b6150dbdc277c5986d570ac-kesame.jpg

Again, the masses cancel. The “1/2″ cancels also. If I take the equation from conservation of momentum, I can solve for the final velocity of object A squared:

i-f2e71be06fd31b1caba2b37b03954ecc-sdfsdw.jpg

And now plugging this into the kinetic energy equation:

i-e9b8affea8f2f494c2878763c5513bef-vba2.jpg

So, the final velocity of the “target ball” that was originally stationary is the same as the initial velocity of “shooting” ball. What about the final velocity of the shooting ball (Ball A)? I can plug this solution for the final velocity of ball B into the other equation and I get:

i-2bc6af0788a538cc1ddecd9ff4402a25-piwer.jpg

I cheated. There is actually another valid solution. This works also:

i-c7b2a30c59c18fba581ced5307fd6168-possiblevelocty.jpg

So, the first solution has Ball A come in and hit ball B. Ball B leaves with the initial velocity of ball A and ball A stops. The other solution is that Ball A continues on with the same speed and ball B stays stationary. This second case would be what would happen if they “missed” but it would still satisfy the momentum and kinetic energy equations.

One application where you sort of see something like this example is in billiards (or pool, I don’t know which is the proper term). If you shoot the cue ball straight at a stationary ball, you COULD make the cue ball stop and the other ball continue. This doesn’t usually happen because the cue ball is rolling not just moving. If you could make a non-rolling collision between these two balls, you would achieve the result from this example.

Comments

  1. #1 tulcod
    December 12, 2008

    Hey, good explanations. could you also try and introduce a non-classical piece of theory some time? that would be really cool!

  2. #2 Uncle Al
    December 13, 2008

    A glancing blow will impart spin. Inelastic components (including internal disruption lessening binding energy). Descriptions are easy, instructions are hard.

    The Standard Model, arriving massless, suffers similar deficiencies. Assume mathematical elegance, demand SUSY. Curve-fit masses, diddle a Higgs mechanism.

    http://math.ucr.edu/home/baez/constants.html

    That empirical evidence supports neither SUSY or Higgs is met with more studies not different approaches. I’ve seen engineers optimizing a flat response surface. Good for monthly reports, bad for bottom lines.

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