Week 4. Impulse and momentum, and work and energy

Important new topic: Conservation laws
In physics, "conservation" doesn't mean "try to recycle whenever possible," it is indentifying quantities that are not allowed to change as long as certain assumptions are in place. Hence a "conservation law" is a kind of bean-counting rule that can be extremely convenient for achieving a shortcut to a solution. In everyday life, a conservation law might be that when you play a game of poker, the total amount that the winners take from the table must equal the total amount that the losers leave behind. In physics, there are three main quantities that are conserved when certain conditions are met: momentum, energy, and angular momentum. We will start with the first two this week, and angular momentum will come much later.

All the conservation laws in the motion of objects come from F=ma and the law of action/reaction, because that is all of mechanics right there. So we are not really learning about new physical principles, we are repackaging what we already know into a convenient shortcut that is good for solving certain types of problems. It's not always easy to know in advance what kinds of problems they are good for solving, so we often simply see if we can use conservation laws first, and if they don't solve our problem, we go back to the full analysis involving the forces and writing F=ma as usual.

Momentum and impulse
Let us first consider a system of fixed mass M. Then write F=Ma like F = M*dv/dt = d(Mv)/dt, and we can see that forces produce a rate of change of momentum. So if we apply a net force F for a time t, we can see that the momentum Mv will change by the amount Ft. This is called the impulse, the force times the time it is applied. If the force is a function of time, like F(t), we simply integrate F(t)dt to get the impulse. Takeaway messages: a force applied over a time causes a change in momentum called an impulse. If there are multiple forces, associate with each its own impulse contribution, and just add up the impulse contributions. So this is what motivates the concepts of momentum and impulse-- they are all about what is changing when you apply a given force F for a given time t, and multiply F*t (or integrate if F varies). That's a property that seems important enough to be worthy of a name (momentum), and to also name its changes (impulse), but what really makes the quantity useful is that we can know a wide range of situations where it must stay the same.

Conservation of momentum
Now let us see if we can use the center of mass concept to find a conserved quantity, using F=ma and the law of action/reaction. The latter point is what leads to conservation of momentum any time we are dealing with a system with no external forces on it, because the internal forces can produce no net force. This is the "no bootstrapping" principle of action/reaction-- you cannot pull your bootstraps up without them pulling you back down in return. Whenever you have forces applied to a system, you can add them all up to get a total net force, and since it doesn't matter on what part of the system they are applied, all internal forces cancel out and the system as a whole has no net force if it is not from something external to the system. Thus its acceleration is zero, thus its rate of change of Mv is zero, thus its momentum is constant.

This powerful principle means that if you isolate some system from external forces in some direction, then the total momentum in that direction, summed over all the parts of the system, must be kept the same. For example, if a stationary person on a frictionless surface pushes on another such person, both will move away from each other in such a way as to keep their total momentum the same, in this case zero. So if one person has mass M and ends up with velocity +V, and the other has mass m, then MV = -mv, so v = -MV/m.

The principle extends to much more complicated systems. For example, if a firework shell is shot from a cannon, then its center of mass will follow the usual trajectory you can calculate for a cannonball, regardless of whether or not the fireworks explode. That's because an explosion is internal forces only, so does not affect the motion of the center of mass since there is no net force produced by the explosion. The only way any system can have a net force on it is if those forces are external to the system, this is the basis of conservation of momentum by internal forces.

Work and changes in kinetic energy
The above analysis applied when you know about the time the forces are exerted for.Also, many types of forces are given as force as a function of position, not a function of time. For all these reasons, we need to understand what a force does when it is exerted over a distance instead of over a time.

As shown in class and in the book, if you take a force exerted over some distance, and multiply that force times that distance, you get a change in the quantity mv^2 / 2. This important fact motivates giving names to these quantities, so we call the force times the distance "work" and the mv^2 / 2 the "kinetic energy" of a system. If you have more than one force, just add up all the work done by each force on the system, and if you have more than one object in the system, just add up their kinetic energies. What is always true is the work done equals the change in total kinetic energy, so we have the concept that work causes a change in kinetiConservation of mechanical energy
This important aspect of work, that it involves the force times the displacement along the force, means that it responds to internal forces in a completely different way than momentum does. Since action/reaction pairs have opposite signs, and apply for the same positive time, they cancel when their impulses are added. But work is not necessarily like that, because the displacements of the parts of a system do not need to be the same, so the action/reaction pair work does not cancel to zero. In fact, it is common for the displacements of the parts of a system, say two masses colliding and sticking together (or the pieces of an exploding firework as above), to point in opposite directions-- just like the forces do. So the work of opposite forces on opposite displacements don't cancel, they add up. Hence, internal forces can very easily increase (as for fireworks) or decrease (as for masses sticking together) the total kinetic energy of a system. We cannot get a conservation law simply by adding up the kinetic energy in an isolated system, as we did for momentum. Nevertheless, there is a way to build a conservation law that involves kinetic energy, we merely have to track how it changes by inventing a notion of "potential energy", adding that to the kinetic energy, and getting "mechanical energy." Mechanical energy is conserved in situations where there is no friction and nothing sticking together.

The way to define potential energy that works is to keep track of the work done on a system by some external force, like gravity, and attach a negative sign to that work. Then call the result a "change in potential energy." Since positive work done by the external force means a negative change in potential energy, and since it comes with an equal positive change in kinetic energy of the system, when we add kinetic and potential energies we get a conserved quantity, the mechanical energy. Since it is often easy to track the work done by an external force, it is easy to define a potential energy function. For example, with gravity we can arbitrarily define some y=0 level, like the ground or a tabletop, and then the gravitational potential energy of any object is just mgy, where y is the height above the y=0 level. The change in all the mgy terms in the potential energy then gives the negative of the change in total kinetic energy-- that's conservation of mechanical energy.

As mentioned, friction, or having masses stick together, does not conserve mechanical energy. This is because they generate heat, and heat is a form of energy not included in mechanical energy. So if there is friction, or if objects stick together, you will need to figure out what happens some other way, not by conserving mechanical energy. When you do it some other way, you can then see that mechanical energy will always drop, and the "missing" energy is the amount of heat generated.

Elastic collisions in two dimensions
An application of the power of the center of mass frame, and how the total momentum is zero in that frame and so stays zero whenever there are no external forces on the system, is elastic collisions in two dimensions. What I added to what is in Chapter 9 is that elastic collisions (collisions conserving total kinetic energy) look very simple from the center of mass frame, they just look like two objects with equal and opposite momentum p (and you have to calculate p from the initial data and knowledge of the center of mass velocity, which is the mass-weighted velocity of all the objects in the system, just like the center of mass location is the mass-weighted location of all the objects), which then collide and still look like two objects with the same equal and opposite momenta as before, only moving along a different direction. (The new direction is not easy to calculate, you would need to be told that, but what you can easily see is that the only way to keep the total momentum zero and keep a constant kinetic energy is to just keep the equal and opposite momenta of the two objects the same, only in a new direction). In other words, an elastic collision between any two objects is much simpler to understand in the center of mass frame, so we need to know how to get into that frame (just subtract the center of mass velocity from every velocity), and how to get back out of it (just add that velocity to every velocity).

The rocket equation
Another good application of the zero-total-momentum center-of-mass frame is the equation for acceleration of a rocket. The rocket equation tells you that the mass times acceleration of the rocket equals the rate the rocket is expelling mass out the back as rocket exhaust times the speed at which it is expelling it (in the rocket's own frame). That result is a simple consequence of being the rocket's center-of-mass frame where its momentum is zero, and then say after it expells a small mass dm, the backward momentum of that exhaust must equal the forward momentum the rocket acquires. That means dm*v_rel = m*dv where v_rel is the exhaust speed and dv is the speed the rocket acquires as a result. That equation might look weird to you, so just divide both sides by the time interval it happens in, which is dt, and you get v_rel * dm/dt = m*dv/dt. The left-hand side is the force on the rocket, the right-hand side is m*a.

Sample problems:
Three equal masses slide along a straight wire with no friction and no external forces. One mass is initially moving to the left at speed v, another is stationary, and the third is moving to the right at speed 2v. Some time later, two of the masses are stuck together and stationary. What is the change in kinetic energy of the system? (Yes there is enough information to answer this.)

Solution: Try it yourself first. If you get stuck, notice that although this question asks about energy, the conserved quantity here is total momentum. We don't expect kinetic energy to be conserved because the collisions are not elastic. But we know all the forces come in action/reation pairs, so the final momentum must equal the initial momentum. That means one mass has to end up moving to the right at speed v. That also means the final kinetic energy is 1/5 of what it was originally, as it started at 5/2 times m v squared, and ends at 1/2 m v squared. Put differently, the amount lost was 2 m v squared.