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Week 4: Systems of Particles, Momentum and Collisions |
the collision time t they are negligible, in the specific sense that:
p~ = p~c + p~b ≈ p~c |
(395) |
(for just one mass) over that time only. Since the collision force is an internal force between the two colliding objects, it cancels for the system making the momentum change of the system during the collision approximately zero.
We call this approximation p~ ≈ p~c (neglecting the change of momentum resulting from background external forces during the collision) the impulse approximation and we will always assume that it is valid in the problems we solve in this course. It justifies treating the center of mass reference frame (discussed in the next section) as an inertial reference frame even when technically it is not for the purpose of analyzing a collision or explosion.
It is, however, useful to have an understanding of when this approximation might fail. In a nutshell, it will fail for collisions that take place over a long enough time t that the external forces produce a change of momentum that is not negligibly small compared to the momentum exchange between the colliding particles, so that the total momentum before the collision is not approximately equal to the total momentum after the collision.
This can happen because the external forces are unusually large (comparable to the collision force), or because the collision force is unusually small (comparable to the external force), or because the collision force acts over a long time t so that the external forces have time to build up a significant p~ for the system. None of these circumstances are typical, however, although we can imagine setting up an problem where it is true – a collision between two masses sliding on a rough table during the collision where the collision force is caused by a weak spring (a variant of a homework problem, in other words). We will consider this sort of problem (which is considerably more di cult to solve) to be beyond the scope of this course, although it is not beyond the scope of what the concepts of this course would permit you to set up and solve if your life or job depended on it.
4.3.2: Impulse, Fluids, and Pressure
Another valuable use of impulse is when we have many objects colliding with something – so many that even though each collision takes only a short time t, there are so many collisions that they exert a nearly continuous force on the object. This is critical to understanding the notion of pressure exerted by a fluid, because microscopically the fluid is just a lot of very small particles that are constantly colliding with a surface and thereby transferring momentum to it, so many that they exert a nearly continuous and smooth force on it that is the average force exerted per particle times the number of particles that collide. In this case t is conveniently considered to be the inverse of the rate (number per second) with which the fluid particles collide with a section of the surface.
To give you a very crude idea of how this works, let’s review a small piece of the kinetic theory of gases. Suppose you have a cube with sides of length L containing N molecules of a gas. We’ll imagine that all of the molecules have a mass m and an average speed in the x direction of vx, with (on average) one half going left and one half going right at any given time.
In order to be in equilibrium (so vx doesn’t change) the change in momentum of any molecule that hits, say, the right hand wall perpendicular to x is px = 2mvx. This is the impulse transmitted
to the wall per molecular collision. To find the total impulse in the time |
t, one must multiply this |
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by one half the number of molecules in in a volume L2vx |
t. That is, |
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ptot = 2 |
µ L3 |
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t (2mvx) |
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Let’s call the volume of the box L3 = V and the area of the wall receiving the impulse L2 = A. We
Week 4: Systems of Particles, Momentum and Collisions |
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combine the pieces to get: |
= A t |
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µ V |
¶ µ |
2 mvx2 |
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µ V ¶ Kx,avg |
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P = A |
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Favg |
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ptot |
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N |
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where the average force per unit area applied to the wall is the pressure, which has SI units of Newtons/meter2 or Pascals.
If we add a result called the equipartition theorem100 :
Kx,avg = |
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mvx2 |
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kbT 2 |
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where kb is Boltzmann’s constant and T is the temperature in degrees absolute, one gets:
P V = N kT |
(399) |
which is the Ideal Gas Law101 .
This all rather amazing and useful, and is generally covered and/or derived in a thermodynamics course, but is a bit beyond our scope for this semester. It’s an excellent use of impulse, though, and the homework problem involving bouncing of a stream of beads o of the pan of a scale is intended to be “practice” for doing it then, or at least reinforcing the understanding of how pressure arises for later on in this course when we treat fluids.
In the meantime, the impulse approximation reduces a potentially complicated force of interaction during a collision to its most basic parameters – the change in momentum it causes and the (short) time over which it occurs. Life is simple, life is good. Momentum conservation (as an equation or set of equations) will yield one or more relations between the various momentum components of the initial and final state in a collision, and with luck and enough additional data in the problem description will enable us to solve them simultaneously for one or more unknowns. Let’s see how this works.
100Wikipedia: http://www.wikipedia.org/wiki/Equipartition Theorem.
101Wikipedia: http://www.wikipedia.org/wiki/Ideal Gas Law. The physicist version of it anyway. Chemists have the
pesky habit of converting the number of molecules into the number of moles using Avogadro’s number N = 6 × 1023 and expressing it as P V = nRT instead, where R = kB NA . then using truly horrendous units such as liter-atmospheres in