Week 4: Systems of Particles,

Momentum and Collisions

Summary

• The center of mass of a system of particles is given by:

One can di erentiate this expression once or twice with respect to time to get the two corollary

P

All three expressions may be summed up in the useful forms:

The center of mass coordinates are truly weighted averages of the coordinates – weighted with the actual weights of the particles91.

•The mass density of a solid object in one, two, or three dimensions is traditionally written in physics as:

91Near the Earth’s surface where the weight only depends on the mass, of course. Really they are weighted with the mass.

181

function of the coordinates92. However we will usually work only with very simple mass distributions that we can easily integrate/sum over in this class. When doing so we are likely to use these definitions backwards:

Use the following ritual incantation (which will be useful to you repeatedly for both semesters of this course!) when working with mass (or later, charge) density distributions:

The mass of the chunk is the mass per unit (length, area, volume) times the (length, area, volume) of the chunk!

• The Center of Mass of a solid object (continuous mass distribution) is given by:

This can be evaluated one component at a time, e.g.:

(and similarly for ycm and zcm).

It also can be written (componentwise) for mass distributions in one and two dimensions:

• The Momentum of a particle is defined to be:

p~ = m~v

The momentum of a system of particles is the sum of the momenta of the individual particles:

XX

p~tot = mi~vi = mi~vcm = Mtot~vcm i

where the last expression follows from the expression for the velocity of the center of mass above.

92Think about how mass is distributed in the human body! Or, for that matter, think about the Universe itself,

which can be thought of at least partially as a great big mass density distribution ρ(~x)...

These forms are very useful in collision problems where momentum is known and conserved; they will often save you a step or two in the algebra if you express kinetic energies in terms of momenta from the beginning.

•Newton’s Second Law for a single particle can be expressed (and was so expressed, originally, by Newton) as:

In this expression the internal forces directed along the lines between particles of the system cancel (due to Newton’s Third Law) and:

where the total force in this expression is the sum of only the total external forces acting on the various particles of the system.

• The Law of Conservation of Momentum states (following the previous result) that:

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

or (in equationspeak):

~

If and only if F tot = 0 then p~tot = p~i = p~f , a constant vector

where p~i and p~f are the initial and final momenta across some intervening process or time interval where no external forces acted. Momentum conservation is especially useful in collision problems because the collision force is internal and hence does not change the total momentum.

•The Center of Mass Reference Frame is a convenient frame for solving collision problems. It is the frame whose origin lies at the center of mass and that moves at the constant velocity (relative to “the lab frame”) of the center of mass. That is, it is the frame wherein:

which is why it is so very useful. The total momentum is the constant value 0 in the center of mass frame of a system of particles with no external forces acting on it!

• The Impulse of a collision is defined to be the total momentum transferred during the collision, where a collision is an event where a very large force is exerted over a very short time

~

interval t. Recalling that F = dp~/dt, it’s magnitude is:

•An Elastic Collision is by definition a collision in which both the momentum and the total kinetic energy of the particles is conserved across the collision. That is:

p~i = p~f

Ki = Kf

This is actually four independent conservation equations (three components of momentum and kinetic energy).

In general we will be given six “initial values” for a three-dimensional collision – the three components of the initial velocity for each particle. Our goal is to find the six final values – the three components of the final velocity of each particle. However, we don’t have enough simultaneous equations to accomplish this and therefore have to be given two more pieces of information in order to solve a general elastic collision problem in three dimensions.

In two dimensional collisions we are a bit better o – we have three conservation equations (two momenta, one energy) and four unknowns (four components of the final velocity) and can solve the collision if we know one more number, say the angle at which one of the particles emerges or the impact parameter of the collision93 , but it is still pretty di cult.

In one dimension we have two conservation equations – one momentum, one energy, and two unknowns (the two final velocities) and we can (almost) uniquely solve for the final velocities given the initial ones. In this latter case only, when the initial state of the two particles is given by m1, v1i, m2, v2i then the final state is given by:

•An Inelastic Collision is by definition not an elastic collision, that is, a collision where kinetic energy is not conserved. Note well that the term “elastic” therefore refers to conservation of energy which may or may not be present in a collision, but that MOMENTUM IS ALWAYS CONSERVED IN A COLLISION in the impact approximation, which we will universally make in this course.

A fully inelastic collision is one where the two particles collide and stick together to move as one after the collision. In three dimensions we therefore have three conserved quantities (the components of the momentum) and three unknown quantities (the three components of the final velocity and therefore fully inelastic collisions are trivial to solve! The solution is simply to find:

~ ~

P tot = P i = m1~v1,i + m2~v2,i

~

and set it equal to P f :

m1~v1,i + m2~v2,i = (m1 + m2)~vf = (m1 + m2)~vcm

93Wikipedia: http://www.wikipedia.org/wiki/impact parameter.