Momentum and Impulse
At this point, we have discussed how things move, and why things move. In this section, we discuss another aspect of motion: momentum.
We have probably heard this term used liberally in a non-scientific setting, often to describe the performance of a favorite sports team as they ascend or fall down the playoff roster. This, as with some other terms we use, is not the definition we will use. Instead, we define momentum:
momentum: the product of an object’s mass and velocity.
Or:
Where is the symbol for momentum. As we can infer from the formula, the larger an object’s mass, the greater its momentum; or, the larger an object’s velocity, the greater its momentum. As a result, even a bullet- which has a very small mass- can have a large momentum, because it travels very quickly.
As a thought exercise: is it possible for a bullet to have the same momentum as a truck? If the truck has a large mass with a small velocity, and the bullet has a large velocity with a small mass, in principle, it is possible. We can do some math to actually show this. Suppose the truck has a mass of kg, and is moving with a velocity of . NASA fires small objects at very
high speeds- as high as - to simulate orbital impact for their satellites. The momentum of the truck would be:
We can then show the mass of a possible NASA test projectile:
This would mean that this projectile has a weight on the order of , which- while a large mass for a projectile- would not be unheard of in some classes of weaponry.
The act of exerting a force on an object will cause it to accelerate, or change its velocity. If its velocity changes, we are going to change its momentum. This is summarized with a variation on an equation we have already learned:
You may recall Newton’s Second Law:
But, as it turns out, Newton did not write his law as ; instead, at the time, he was talking
in terms of the momenta of objects. Which means the previous form, , is the more general case. (This equation actually was the first used to describe the motion and forces involved with rocketry, though it was not until Tsiolkovsky that this was used in this fashion.)
If we rearrange this formula to compare the momentum of an object at the beginning and end of a problem, we find:
This results in a new quantity: impulse.
impulse: the change of an object’s momentum; the result of a force acting through a period of time
The formula above is called the impulse-momentum theorem, which states that a force exerted through a period of time will cause an object’s momentum to change.
We can try an example with this new theorem. Suppose that a car is accelerated from rest to over the course of . If its mass is , what is the change in the car’s momentum? What is the impulse delivered to the car?
This is the solution to both questions asked, because the impulse is defined as the change of momentum. To be fully correct, we would specify a direction with this magnitude, as impulse and momentum are both vector quantities.
This example problem was relatively simple in nature; the mass of the car did not change throughout the course of the problem, and there was only one force acting to change the car’s momentum. In more complicated situations, we would need to find the magnitude of the net force, and we may need to use product rule to differentiate the momentum function. We see such a situation in an awesome application of physics: rockets. However, we need to introduce one more concept before we can properly analyze the dynamics of rockets.
Conservation of Momentum
Imagine that we have two billiard balls, one rolling toward another, as shown in the diagram below:
When the two objects collide, the cue ball will exert a force on the 8-ball. By Newton’s Third Law, the 8-ball will exert a force of the same magnitude on the cue ball. The force diagrams for each will look like this:
The “ denotes that this is the force the 8-ball exerts on the cue ball, and vice-versa. By Newton’s Third Law:
So, according to Newton’s Second Law:
Here, the primes do not denote derivatives; instead, they denote “final” from “initial.” We do this because we wish to avoid confusion about multiple subscripts. Rearranging, we can find:
This is the Principle of Conservation of Momentum: as long as there are no external forces acting on a system, the momentum of all objects interacting within that system must be conserved. This is true regardless of what type of collision the objects experience. In fact, it is true regardless of how many objects are involved in such collisions. The true form of this principle is actually expressed as:
There are several types of interactions that we can quantify:
elastic collisions: objects collide and separate; there is no loss of kinetic energy.
inelastic collisions: objects collide and stick together; there are losses of energy by the objects due to effects such as deformation, heating, etc
explosions: one object separates into two or more objects
In all these situations- in which there are no external forces- the conservation of momentum will hold. We return to our billiard ball example: this is an elastic collision, so both kinetic energy and momentum must be conserved. As a result, we can find the velocity of the 8-ball after colliding with the cue ball. If we assume the 8-ball initially at rest, we find:
If the masses are the same for the objects, the masses cancel out. We find from energy that:
The masses cancel out in this equation, too. Solving the energy equation for the cue ball’s final velocity, we can solve for the velocity of the cue ball and 8-ball after collision:
If this is true, that must mean that:
Now, what if we allowed the masses to differ? In this case, we would follow precisely the same procedure, but there would be the added factors of mass floating around. The speed of the cue ball would not necessarily be zero- and, in general, will not be zero. Cue balls are usually
manufactured to be slightly heavier than the other billiard balls used, which is why when we play pool, the cue ball often goes ricocheting off. Often into another pocket, causing the unfortunate player to scratch.
Now, what about the case of an inelastic collision? Suppose that we replaced the billiard balls above with balls of clay. Instead of bouncing off of one another, they clay would bond together. How would this change our conclusions? We start again with the conservation of momentum, changing the subscripts to “1” and “2”, but keeping all else in the above problem the same:
If they collide and stick together, they effectively become one object, and move with the same velocity. In other words:
So the speed of the combined object is reduced from by a factor of ; in this case,
. How does the kinetic energy of the two change? We can see:
Which, in this example, simplifies to a reduction of equal to . Now, the catch: where does all of this energy go? We learned previously that energy cannot be spontaneously generated or removed from the universe, but only transformed into a new form of energy. In this case, the energy went to the objects; the surfaces of the balls of clay did work on one another,
compressing and deforming their surfaces and generating heat in the process. If we were to account for the deformation and the heat generated in this collision, we would find that the
of lost kinetic energy was transformed in this manner.
Finally, what about explosions? The conservation of momentum works perfectly well here, too. Take a bomb that has a mass of , place it on the ground at rest, and allow it to combust. In this next analysis, we will deal with only two pieces of a bomb, but the principle works the same for any number of pieces of shrapnel. The conservation law tells us that when the gases inside the bomb combust and do work/impart impulse by the work-energy and impulse-momentum theorems, the total momentum of the bomb’s parts must still be zero. If we let the pieces be exactly one half of the original mass, we find:
And should the masses differ, the velocities will differ by appropriate factors, but the total momentum will always be conserved. This leads to very interesting discussions on the center of mass of an object, when we reach the rotations section several chapters from now.
Collisions in 2-D
With an understanding of the conservation of momentum, we can apply the same in more than one dimension. After all, we live in 3-D, not 1-D!
For our purposes, collisions in two dimensions- at least for now- will be sufficient. We simply remember that momentum is a vector, not a scalar, so we can break the momentum conservation principle into constituent components. The examples above could have been treated in two dimensions, and in effect, were; the y-component of the momentum vectors was simply zero.
Solving such problems works exactly the same way as any other vector addition problem- we can use components to solve whatever unknowns we need!
We set up conservation of momentum in two dimensions:
Substituting in facts about their velocity, we solve:
,
If it were an elastic collision, instead, and we wanted to find the velocities of both objects, the problem becomes a little more complicated, but solvable:
Substituting in facts about their velocity and remembering that in elastic collisions, energy is conserved in both dimensions, we can solve:
Simplifying and writing the direction angle of one mass’s velocity as and the other as yields:
Here, we run into a problem. There are three equations and four unknowns; as a result, there is no unique solution. In other words, there is no way for a general two-dimensional collision to have a “closed form” solution, unless we are either given information about one of the object’s final velocity (either a magnitude or direction), or we do not consider the objects to be “point-like” objects. At this level, we must know something about the end of a two-dimensional elastic collision in order for the problem to have a particular solution. An inelastic collision
the same by allowing us to combine initial velocity terms. A problem involving two or more objects in an elastic collision, however, will instead have a dependent set of solutions, not an independent solution.
If we did want a closed-form solution, we would need to actually consider the geometry of the problem by seeing exactly where one object hit the other, and finding the angle the impact location makes with the center of mass of the object and our coordinate axes. As a general concept, this is well outside the scope of our course; there are simpler cases that can be considered later on in this course, but there are other ways of dealing with such situations discusses in the angular momentum unit.
Rocket Propulsion
In the case of rockets, gases are ignited and leave out the bottom in a huge ball of fire. As fuel is consumed, the mass of the rocket changes; further, as fuel is consumed, the force of the expelled gases continues to act on the rocket, by Newton’s Third Law. This causes the rocket to change its momentum at the same time that its mass is changes. It can be shown, using methods of calculus, that one of the more straightforward solutions of the impulse-momentum theorem gives:
Where stands for the mass of the rocket at any point in time, stands for the original mass of the rocket, corresponds to the change in the rocket’s velocity as a result of the thrust, and is the speed with which gases are expelled from the rocket (providing the thrust). This is the well-known “rocket equation,” as solved by Konstantin Tsiokovsky in 1896.
What is interesting is that when we graph as a function of time, the same can take place over any time period, suggesting that no matter the thrust, if we allow an appropriate time period, we will achieve the same change in velocity. We can also conclude that this change in velocity can happen in stages. What is also true is that this change in velocity need not be positive; instead, we could have a negative change of velocity, corresponding to a thrust opposite the direction of travel. This is because for a rocket (or similar system), any thrust consumes mass (fuel).
When we consider the solutions to these equations when we do not let the original force term be zero, but instead account for gravity, or drag, or both, we would find that more mass is needed to lift the rocket. We would also find a limiting relationship, where must be
