Linear Momentum and Collisions

Now lets extend our understanding of translational motion by
defining a new term, "linear momentum". Linear momentum
of a particle is the product of a particle's mass and its
velocity. For some reason which escapes me at the moment, the
symbol *p* is customarily used to designate momentum.
Since it is the product of a vector and a scalar, it is a vector
quantity.

The rate of change of momentum with respect to time, taking mass to be constant, is

where

In fact when Newton published his second law it was in the form

For a system of particles, the linear momentum is just the vector sum of that of each of the particles individually. As we worked out in the previous lesson,

m * *v*_{cm} =
m_{1}**v*_{1} +
m_{2}**v*_{2}...+
m_{n}**v*_{n} ,

so for a system of particles
In words this says that for a system of particles, the rate of change of the total momentum is equal to the applied force. Run the Two Particle Center of Mass display again and look at the center of mass track. Since the objects are moving with constant velocity we know that there is no force applied to the two object system. The momentum of the whole system must be constant, meaning that the center of mass travels in a straight line at constant velocity.

This idea that in the absence of external forces the momentum remains constant is called the principle of the "conservation of linear momentum". This principle, like the conservation of energy or work-energy theorem, seems to be near the heart of the laws of nature. It applies in all kinds of situations, even those outside the realm of classical dynamics.

So far we have developed two distinct modeling schemes for building mathematical models of dynamical systems. In the first instance we used Newton's second law. The implicit assumption in using that modeling scheme is that we know what forces are applied to the particle and how that force varies as a function of time. The other modeling scheme involved the first fundamental conservation law we discovered, the conservation of mechanical energy. In applying that model we need to know the way the potential energy varies as a function of position. Now that we have the conservation of linear momentum as another basic principle we can model some situations in which we know neither the forces nor the potential field.

One instance of a system where the detail of forces and fields is obscured is in collisions. It is theoretically possible to know in detail the way force varies with time in a collision but as a practical matter that may be very difficult to determine. By "collision" , I mean an interaction between two objects which is over is a very short time, compared to the time of observation. With this as a working definition, we may divide time up into three regions, before collision, during collision and after collision. Using the conservation of linear momentum and the conservation of mechanical energy, we may determine from the before collision conditions what the conditions will be after the collision without ever knowing exactly what went on during the collision.

You may recall one example of a collision already illustrated in the Billiards display. It was used in the discussion of Newton's third law. I cheated a little bit and used the conservation of linear momentum in that display before we understood the concept. Perhaps you would like to review that display with the idea of conservation of momentum in mind.

The model behind this display operates on Newton's first law during the times before and after the collision. We assume there are no forces on the balls before and after the collision so the equations of motion are the trivial case of no change in velocity. During the collision the two balls compress one another slightly exerting huge forces. Even though we will not work out in detail the variation of force with time, we can get some notion of the total force involved.

We know now that *f*=*p*'. Lets estimate the
forces involved in a head on, dead center collision between the
two balls of equal mass. The cue ball stops in that case and the
red ball goes off at the velocity which the cue ball had
initially. Verify this with the model or go down to your local
tavern and try it. The way to get a head on collision with the
model is to place the cursor right on the center of the red ball
for the shot. You might want to start with the cue ball at the
far right for maximum shot speed. Let's assume you achieve a
cue ball speed of 2 meters per second. The change in velocity of
the cue ball then is 2m/s since it comes to a complete stop.

A normal cue ball has a mass of about .25kg so the momentum
change is .25*2kgm/s = 0.5kgm/s. The question now becomes, how
long does it take for the cue ball to stop. Here is where we have
to estimate a number. Let's say the balls remain in contact
for 1e-3 seconds. The average force magnitude |f| then during the
1e-3 seconds is |f| = *p*' = .5/1e-3 = 500 Newtons

This is considerable force, perhaps enough to lift a small adult off the ground. Forces of the sort we are talking about here, large but short are called "impulse" forces. In general they are estimated by dividing a change in momentum by the time to make the change. Conversely the average force if known, multiplied by the time the collision lasts is called the impulse of the force and is equal to the change in momentum.

The basis for our collision model will be the two conservation principles, conservation of kinetic energy and conservation of linear momentum. We can work with conservation of kinetic energy only, because in studying collisions we want to isolate the effect of the collision from the effect of other forces. So we will set up our conditions such that other forces like those arising from potential energy changes do not apply. Actually collisions are classified as to whether or not kinetic energy is conserved. If it is, the collisions are called "elastic" collisions, if not they are called inelastic collisions. We are going to limit ourselves to elastic collisions.

Consider two spherical objects one of mass m_{1} and
the other of mass m_{2}. Let's set things up so these
objects are approaching each other along the line joining their
centers, a recipe for a head on collision. Let these objects not
be subject to any forces and be not rotating or vibrating. The
motion is purely translation. Under these conditions we may
choose a reference frame which is one dimensional, with the
objects on the x axis.

Applying the conservation of linear momentum to this situation
we have,

m_{1} * v_{1i} +
m_{2} * v_{2i} = m_{1} * v_{1} +
m_{2} * v_{2} ,

where v

1/2 * m_{1} *
v_{1i}^{2} + 1/2 * m_{2} *
v_{2i}^{2} = 1/2 * m_{1} *
v_{1}^{2} + 1/2 * m_{2} *
v_{2}^{2} .

These may be rewritten as follows:

m_{1} * (v_{1i} -
v_{1}) = m_{2} * (v_{2} -
v_{2i})

for the momentum equation and
m_{1} * (v_{1i}^{2} -
v_{1}^{2}) = m_{2} *
(v_{2}^{2} - v_{2i}^{2})

for the energy.As long as the difference between final and initial velocities is not zero for either object (meaning a collision actually happens), we may divide the second equation by the first one which yields

v_{1i} + v_{1} =
v_{2} + v_{2i} ,

or
v_{1i} - v_{2i} =
v_{2} - v_{1} .

In other words in a one dimensional elastic collision, the relative velocity of approach before the collision equals the relative velocity of separation after collision.

Now here is one of those places where we are going to lean on that algebra
background I told you would need in this course. To get the final velocities
in terms of the initial velocities and the masses, you would solve the last
equation above for v_{2} and plug that into the momentum equation and
solve to get

v_{1} = v_{1i} *
(m_{1} - m_{2}) / (m_{1} + m_{2})
+ v_{2i} * (2 * m_{2}) / (m_{1} +
m_{2}) .

Likewise
v_{2} = v_{1i} * (2 *
m_{1}) / (m_{1} + m_{2}) + v_{2i}
* (m_{2} - m_{1}) / (m_{2} +
m_{1})

So we have the basis for a model of a one dimensional elastic
collision. For initial conditions v_{1i} and
v_{2i}, if a collision happens, the final velocities
depend on the masses as above.

One dimensional elastic collisions though leave a lot of naturally occurring collisions unaccounted for. We should extend this model to two dimensions to be more realistic. In three dimensions the models get too complicated.

In two dimensions the final velocities have two components each, an x and a y velocity, so here are four unknowns to be determined. Since momentum is a vector quantity, in two dimensions, conservation of momentum gives us two equations, one in x and one in y. Kinetic energy is a scalar quantity so the conservation of kinetic energy only gives us only one equation, leaving us one equation short of the number we need to solve for four unknowns.

What we need here is some additional information. To get that we have to look at the geometry of the collision itself. Since we are dealing with spherically symmetrical objects that geometry is not too hard to work out. The velocity of each ball is a vector which can be resolved into components one of which is on the line connecting the centers and the other perpendicular to this line. At the instant of the collision, the component in the line of sight, so to speak, will be the only component producing a force on the other ball. The perpendicular component can not change a ball's momentum because it does not act through the center of mass at all (Think about it.) Likewise the collision can not change the velocity across the line of sight.

So in a sense all collisions are one dimensional if the dimension is carefully chosen at the moment of impact. We can calculate the velocities in the line of sight at the instant of impact from the initial velocities of the balls. Then we can apply the one dimensional analysis to determine what the velocities in the line of sight will be immediately after the collision. Next we take the change in line of sight velocities from before to after the collision, and recognizing that no other component of the initial velocities will be affected by the collision, we add this change for each ball to the initial velocities to get the after collision velocities. Run the Collision Geometry display to see the situation. A bit later you will be able to experiment with collisions where you will control the masses and the velocities.

Let's Just to briefly revisit the notion of elasticity. So far we have been assuming perfect elasticity, meaning that kinetic energy was conserved. In real collisions, except for some of those between nuclear particles, the elasticity is less than 100%. For now we will just accept that there may be some loss of kinetic energy between the before collision and after collision system. Even a 0% elastic collision does not mean that the objects come to a stop, they just stick together. So the loss in kinetic energy can only be as much as is consistent with the conservation of momentum. I have not been able to write a good simple model for inelastic collisions. When I solve that problem, we may be able to go further with this discussion.

Now we can get a bit more precise with a billiard type display. Actually we are going to use two balls and a box in this model. Sort of like a billiard table with the possibility of using a variety of balls in a set up shot. We will not consider the effect of rotation. My computer is not fast enough to deal with the math involved in that. Run the Collision Model display to try out different values of the mass, speed and directness for two dimensional collisions.

Next we will deal with any rotational motion about the center of mass.