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Momentum conservation (2012)

Page history last edited by Joe Redish 7 years, 7 months ago

Class content > Newton's Laws > Newton's Laws as Foothold PrinciplesLinear momentum




A hyper-simplified example

Newton's third law tells us that when two objects interact, they exert forces on each other and those forces are equal and opposite:



for any type of force. When we combine this with the momentum form of Newton's 2nd law, something interesting happens. Let's consider the simplest case we can imagine just to see what happens.  Suppose that we have two objects, A and B, and they interact with nothing else except each other (with an unspecified type of force) as shown in the system schema shown at the right. Then Newton's 2nd law applied to each of the two objects reads:



We've put the momenta on the left because that's what we want to pay attention to. Now let's do something a little different. Since the only things in our system are the two objects, let's consider the AB pair of objects as a single system by adding these two equations together ( just as we did when we introduced Newton's third law.) The result is



But since the two forces are of the same type, equal and opposite, by Newton's 3rd law, the right hand Forces cancel! The result is interesting:

Now you might say, "So what?"  But when something has its derivative equal to 0 that means that it doesn't depend on the variable you are differentiating with respect to, in this case time. This says that the total momentum of the system doesn't change.  But the individual momenta might change dramatically! If the two objects are billiard balls that are colliding they might take off in totally different directions.  Each momentum changes a lot, but the equations above tell us, as long as the two billiard balls only interact with each other and nothing else, the total momentum doesn't! (This turns out to have important consequences for collisions of atoms and molecules that lead to chemical reactions. We will discuss this when we consider collisions, after we have discussed the issue of energy and energy conservation.)


A more realistic analysis

Let's try to make this a little more general. Suppose we have two objects, A and B, that we are going to consider as a "system".  Each will interact with each other by some (unspecified type of) force and with other objects in the world. We'll refer to those objects that are not A or B as "external" to the system as shown in the system schema at the right. There might be gravity acting on our objects or other forces. Then our N2 equations become



If we add these two together, taking into account that the forces that A and B exert on each other will cancel, we get the result


As a result we can now easily see the conditions for us to get the "total momentum of the system doesn't change" result again.


If we have a system of two interacting objects such that the net external force on the objects cancel, then the total momentum of the system doesn't change even though the individual momentum might change.


We refer to this as the Theorem of Momentum Conservation.  We can easily see that this can be generalized to any number of objects since for every pair, N3 will say that they will cancel when you add the N2 equations together.  The general result is


If we have a system consisting of any number of interacting objects such that the net external force on the objects cancel, then the total momentum of the system doesn't change even though the individual momenta might.


In the system schema the Theorem of momentum conservation can be applied in a direct way:  You can define the system of interest in any way you like by drawing a dashed line around the objects that you wish to be part of your system.  The theorem of momentum conservation then states that interactions between objects within the system do not change the total momentum of the system, only interactions that cross the (dashed line) system boundary change the total momentum. 


It's interesting to realize that whether momentum conservation hold depends on what system we are considering (where we draw the boundary of our system in the system schema). If we only consider the thrown ball, momentum is not conserved for the ball because there is an external force -- gravity.  But if we included the earth (and were able to measure its momentum to an incredible accuracy) then the momentum of the two would be conserved. If we could consider all the objects in the universe (?) it seems like the total momentum would be conserved. (This is not really legitimate since as we get to sizes comparable to the universe non-Newtonian effects become important.) What matters is what we are considering as inside our system and what as outside.


Total Momentum vs Change in Momentum

There are a number of different ways that to represent this result (momentum conservation).  Let's just consider two objects for whom the net external forces cancel.  We then have the result

We can express this in a variety of ways. We can say that if we look at the total momenta at two different times the result for the total is the same.  Or we could say that the change is 0.  For two items this becomes particularly useful since the total of the two stays the same says that they must change in equal but opposite ways.


So there are two ways we can express the fact that the momentum of a pair of objects is conserved:


  • The momenta added together at any (initial) time is equal to the sum of the momenta at a later (final) time.
  • That change in one object's momentum over any time interval is equal and opposite to the change in the other object's momentum.


Be careful! This is a place where if you are sloppy or are doing one-step thinking, you can easily mess up. There are examples we will do in which one of the object has  momentum 0 either initially or finally and for that special case the momentum and the change in the momentum are the same.  But that is not usually the case!  Be very careful to distinguish the momentum from its change and to state your momentum conservation carefully, paying attention to whether you are talking about a momentum or a change in momentum.



  • Perfectly inelastic collisions


Joe Redish 10/18/11


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