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, p = m * v. The rate of change of momentum with respect to time, taking mass to be constant, is p' = m * v', where v' is the vector acceleration of the particle. Remember that we are using the ' symbol to represent the rate of change with respect to time. From our expression of Newton's second law, we know that m * v' is the force on a particle so p' = f. In fact when Newton published his second law it was in the form p' = f, rather than as we first expressed it.

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 = m₁*v₁ + m₂*v₂...+ m_n*v_n so for a system of particles p = m * v_cm, and p' = m * a_cm = f_a, where f_a is the force resultant external force applied to the 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.

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.

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 that 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. Impulse forces even though of short duration can do considerable damage to the objects involved. 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.

Consider two spherical objects one of mass m₁ and the other of mass m₂. 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, m₁ * v_1i + m₂ * v_2i = m₁ * v₁ + m₂ * v₂, where v_i is the initial velocity of each object and v is the final velocity of each object. The conservation of kinetic energy gives us the equation 1/2 * m₁ * v_1i² + 1/2 * m₂ * v_2i² = 1/2 * m₁ * v₁² + 1/2 * m₂ * v₂². These may be rewritten as follows:
m₁ * (v_1i - v₁) = m₂ * (v₂ - v_2i) for the momentum equation and m₁ * (v_1i² - v₁²) = m₂ * (v₂² - v_2i²) 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 + v₁ = v₂ + v_2i , or v_1i - v_2i = v₂ - v₁. 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 v₂ and plug that into the momentum equation and solve to get v₁ = v_1i * (m₁ - m₂) / (m₁ + m₂) + v_2i * (2 * m₂) / (m₁ + m₂). Likewise v₂ = v_1i * (2 * m₁) / (m₁ + m₂) + v_2i * (m₂ - m₁) / (m₂ + m₁).

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. In the special case where the objects have the same mass and object 2 was initially at rest, you can easily see from the equations that object 1 comes to a stop and object 2 takes off with the same velocity object 1 originally had. The objects just exchange velocities.

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 cannot 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.

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.

Are there any questions?