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194

Week 4: Systems of Particles, Momentum and Collisions

4.2: Momentum

Momentum is a useful idea that follows naturally from our decision to treat collections as objects. It is a way of combining the mass (which is a characteristic of the object) with the velocity of the object. We define the momentum to be:

p~ = m~v

(376)

Thus (since the mass of an object is generally constant):

~

 

d~v

=

d

(m~v) =

dp~

(377)

F

= m~a = m

dt

dt

dt

 

is another way of writing Newton’s second law. In fact, this is the way Newton actually wrote

~

Newton’s second law – he did not say “F = m~a” the way we have been reciting. We emphasize this connection because it makes the path to solving for the trajectories of constant mass particles a bit easier, not because things really make more sense that way.

Note that there exist systems (like rocket ships, cars, etc.) where the mass is not constant. As the rocket rises, its thrust (the force exerted by its exhaust) can be constant, but it continually gets

~

lighter as it burns fuel. Newton’s second law (expressed as F = m~a) does tell us what to do in this case – but only if we treat each little bit of burned and exhausted gas as a “particle”, which is

~

dp~

 

a pain. On the other hand, Newton’s second law expressed as F =

dt

 

still works fine and makes

perfect sense – it simultaneously describes the loss of mass and the increase of velocity as a function of the mass correctly.

Clearly we can repeat our previous argument for the sum of the momenta of a collection of

particles:

 

 

P~ tot = X p~i = X m~vi

(378)

i

i

 

so that

~

 

dp~i

 

~ ~

 

dP tot

X

X

 

 

 

 

 

 

 

=

 

=

 

F i = F tot

(379)

dt

dt

i

 

i

 

 

 

 

Di erentiating our expression for the position of the center of mass above, we also get:

d i mi~xi

X

d~xi

X

 

 

Pdt

=

mi

 

=

p~i = P~ tot = Mtot~vcm

(380)

i

dt

i

4.2.1: The Law of Conservation of Momentum

We are now in a position to state and trivially prove the Law of Conservation of Momentum. It reads94:

If and only if the total external force acting on a system is zero, then the total momentum of a system (of particles) is a constant vector.

You are welcome to learn this in its more succinct algebraic form:

 

~

~

~

~

(381)

If and only if F tot = 0 then P tot = P initial = P final = a constant vector.

 

 

 

 

~

 

Please learn this law exactly as it is written here. The condition F tot = 0 is essential – otherwise,

~

~

 

 

 

 

dP TOT

!

 

 

 

as you can see, F tot =

 

 

 

 

dt

 

 

 

 

 

 

 

 

94The “if and only if” bit, recall, means that if the total momentum of a system is a constant vector, it also implies that the total force acting on it is zero, there is no other way that this condition can come about.

Week 4: Systems of Particles, Momentum and Collisions

195

The proof is almost a one-liner at this point:

~

X

~

(382)

F tot =

 

F i = 0

i

implies

~

dP tot = 0 (383) dt

~

so that P tot is a constant if the forces all sum to zero. This is not quite enough. We need to note that for the internal forces (between the ith and jth particles in the system, for example) from Newton’s third law we get:

~

~

(384)

F ij = −F ji

so that

 

 

~

~

(385)

F ij + F ji = 0

pairwise, between every pair of particles in the system. That is, although internal forces may not be zero (and generally are not, in fact) the changes the cause in the momentum of the system cancel. We can thus subtract:

~

F internal =

~ ~ ~

from F tot = F external + F internal to get:

X

~

(386)

 

F ij = 0

i,j

~

~

 

 

dP tot

= 0

(387)

F external =

 

dt

 

 

 

and the total momentum must be a constant (vector).

This can be thought of as the “bootstrap law” – You cannot lift yourself up by your own bootstraps! No matter what force one part of you exerts on another, those internal forces can never alter the velocity of your center of mass or (equivalently) your total momentum, nor can they overcome or even alter any net external force (such as gravity) to lift you up.

As we shall see, the idea of momentum and its conservation greatly simplify doing a wide range of problems, just like energy and its conservation did in the last chapter. It is especially useful in understanding what happens when one object collides with another object.

Evaluating the dynamics and kinetics of microscopic collisions (between, e.g. electrons, protons, neutrons and targets such as atoms or nuclei) is a big part of contemporary physics – so big that we call it by a special name: Scattering Theory95 . The idea is to take some initial (presumed known) state of an about-to-collide “system”, to let it collide, and to either infer from the observed scattering something about the nature of the force that acted during the collision, or to predict, from the measured final state of some of the particles, the final state of the rest.

Sound confusing? It’s not, really, but it can be complicated because there are lots of things that might make up an initial and final state. In this class we have humbler goals – we will be content simply understanding what happens when macroscopic objects like cars or billiard96 balls collide, where (as we will see) momentum conservation plays an enormous role. This is still the first

95Wikipedia: http://www.wikipedia.org/wiki/Scattering Theory. This link is mostly for more advanced students, e.g. physics majors, but future radiologists might want to look it over as well as it is the basis for a whole lot of radiology...

96Wikipedia: http://www.wikipedia.org/wiki/Billiards. It is always dangerous to assume the every student has had

any given experience or knows the same games or was raised in the same culture as the author/teacher, especially nowadays when a significant fraction of my students, at least, come from other countries and cultures, and when this book is in use by students all over the world outside of my own classroom, so I provide this (and various other) links. In this case, as you will see, billiards or “pool” is a game played on a table where the players try to knock balls in holes by poking one ball (the “cue ball”) with a stick to drive another identically sized ball into a hole. Since the balls are very hard and perfectly spherical, the game is an excellent model for two-dimensional elastic collisions.