AP Physics 1 and 2
Linear Momentum
Momentum is mass in motion, p = mv, and it is a vector. Almost all of the trouble in this unit comes out of that second half: momenta add as arrows rather than as sizes, so two large momenta can add to nothing, and a momentum reported without a direction is only half of an answer.
1. Mass in motion, with a direction attached
Unit 3 gave you one way to say how much motion a thing carries. Kinetic energy is a single number, it has no direction, and it grows as the square of the speed. Unit 4 gives you the other way. The linear momentum of an object is its mass multiplied by its velocity, p = mv, and it grows as the first power of the speed.
The unit has no special name. It is kg m/s, kilogram meters per second, and it is read exactly as it is written. If a quantity you have worked out does not come out in kilogram meters per second, whatever it is, it is not a momentum.
Momentum is a vector. It points the same way the velocity points, because mass is a positive number and multiplying a vector by a positive number stretches it without turning it. On a straight track, with right chosen as positive, a cart moving right has a positive momentum and a cart moving left has a negative one. That sign is part of the answer, not decoration on it.
2. The same speed is not the same momentum
Momentum answers a question that speed by itself cannot: how much motion is there to get rid of. A loaded shopping cart and an empty one rolling side by side at the same speed are not equally easy to bring to a halt, and the arithmetic says so before your arms do.
Both factors count once. Double the mass at the same velocity and the momentum doubles. Keep the mass and double the velocity and the momentum doubles again. That is where momentum parts company with kinetic energy, which doubles for the first change and quadruples for the second.
So a 4.0 kg object at 2.0 m/s and a 2.0 kg object at 4.0 m/s carry the same momentum, 8.0 kg m/s, while their kinetic energies are 8.0 J and 16 J. Equal momentum does not mean equal anything else.
Two carts on a level track, and the cart on the right has the larger mass. Move the slider to change the speed of the lighter cart, and watch the momentum arrow drawn under each one. The figure is here so you can find the setting where the two arrows match.
Two things are worth noticing while you move the slider. The arrows match while the speeds do not, and the lighter cart has to be going faster to get there. And once they match, nothing about the two carts is the same except the product mv, which is the only thing momentum ever asks about.
3. A system has one momentum, and it is a sum
Draw a boundary around any group of objects and call it a system. The momentum of that system is the vector sum of the momenta of its parts. There is nothing else to the definition, and there is no separate formula to learn: you find each mv, you keep its sign, and you add.
Adding the sizes instead is the most common way to lose this. Two carts carrying 6.0 kg m/s each, moving toward each other, give a system momentum of 0, not 12 kg m/s. The sizes do add to twelve. The sizes are not what the question asked for.
A system whose total momentum is zero can still be full of motion. The momenta cancel because they point opposite ways, and the kinetic energy of both objects is still sitting there, because kinetic energy has no direction to cancel with.
Two objects moving toward each other. The buttons choose what you are calling the system, the left object by itself or both objects together, and the slider changes the velocity of the left one; the figure is here to show that the total depends on the boundary you drew.
Switch between the two boundaries without touching the slider. Neither object has changed, and the total has, because the question changed. Then look for the slider setting where the total for both objects together falls to zero, and notice that the two objects are still moving as fast as they ever were.
4. Collisions and explosions, where momentum earns its keep
A collision is an interaction in which the forces the objects put on each other are much larger than any net force reaching them from outside. For the fraction of a second two carts are in contact, the push between them is far bigger than the friction of the track, which is what allows you to set the outside world aside for the length of the interaction.
Because only the state just before and the state just after are examined, each object can be treated as a single point carrying a mass. Its shape, its spin, and where on it the contact happened never enter the work. That is permitted here for one reason: nothing inside the interaction is ever looked at.
An explosion is the same idea pointed the other way. Forces inside the system drive its parts apart: a compressed spring released between two carts, a person stepping off a skateboard. Nothing is handed in from outside, so whatever momentum the system had before, it still has after. A 4.0 kg object at rest that comes apart into a 1.0 kg piece at 6.0 m/s and a 3.0 kg piece must send the heavier piece the other way at 2.0 m/s, because (1.0)(6.0) = (3.0)(2.0).
The two momenta in that last example are equal in size and opposite in direction, and they are not free to be anything else. Topics 4.2 and 4.3 are about why.
5. Three ways this goes wrong
Adding magnitudes instead of vectors. Two momenta of equal size pointing opposite ways total zero. Write each one with its sign before you add anything, and the error cannot happen.
Reporting a momentum with no direction. A momentum of 2.0 kg m/s is an unfinished answer. East, west, up the ramp, or a sign attached to a stated positive direction: pick one and say it every time.
Treating momentum and kinetic energy as the same news. They disagree on purpose. Equal momenta usually mean unequal kinetic energies, and a total momentum of zero tells you nothing at all about how much kinetic energy a system holds.
For this course the arithmetic stays small. Two objects, sometimes three, on a straight line, with every velocity measured from the same chosen positive direction.
Check yourself
1. A 0.50 kg ball moves east at 4.0 m/s. Give its momentum, and say what is still missing if a student stops after writing 2.0 kg m/s.
p = mv = (0.50)(4.0) = 2.0 kg m/s, directed east. The number by itself is half of the quantity, because momentum is a vector: the direction belongs in the answer, either named or carried as a sign against a stated positive direction.
2. Cart A has mass 2.0 kg and moves right at 3.0 m/s. Cart B has mass 3.0 kg and moves left at 2.0 m/s. Find the momentum of the system, and say where the answer 12 kg m/s comes from.
Taking right as positive, the momenta are +6.0 kg m/s and -6.0 kg m/s, so the system momentum is zero. Both carts are moving and the system as a whole is not. The answer twelve is the sum of the two sizes, which is the one sum the definition does not ask for.
3. Object X has mass 4.0 kg and speed 2.0 m/s. Object Y has mass 2.0 kg and speed 4.0 m/s. Compare their momenta and their kinetic energies.
The momenta are equal in size, 8.0 kg m/s each. The kinetic energies are not: 8.0 J for X and 16 J for Y. Momentum counts the speed once and kinetic energy counts it twice, so the faster, lighter object wins on energy while tying on momentum.
4. A 4.0 kg object sits at rest and then comes apart into a 1.0 kg piece moving at 6.0 m/s and a 3.0 kg piece. Find the speed of the heavier piece and say which idea from this topic you used.
The system momentum was zero before, and nothing from outside was added, so the two momenta have to cancel: (1.0)(6.0) = (3.0)v gives v = 2.0 m/s, in the direction opposite the lighter piece. The idea used is that the momentum of a system is the vector sum of the momenta of its parts, so a zero total forces the two pieces to match in size and oppose in direction.
Topic 4.2, Change in Momentum and Impulse. This topic said what a momentum is. The next one asks what changes one, and the answer turns out to be a force acting for a length of time. That product carries its own name and has a graph of its own.
Two papers on Units 1 and 2
Thirty multiple choice questions and four free-response questions in each. Version A is the printed sheet. Version B is a second set, harder, on the site only. Socrates sits beside every question.
Open the practice testEvery force situation, one diagram at a time
Draw the free body diagram, name every arrow, choose the axes, then write the equations. Blocks on a floor, blocks on a ramp, Atwood machines, pulleys, cars on curves, banked curves, loops. It marks all of it, and it names the forces you drew that are not there. Socrates sits beside every situation.
Open the Force BoardEvery simulation on the site
All 109 interactive simulations in pictures, by topic and course. Click any one and it opens ready to use. Open to anyone, with no sign-in.
Open the LabThe one-page sheet
The whole unit on one page. Sign in with your class code and it opens at printing size.
The sheet for this topic
The same material as the topic page, laid out to write on in class. This is Topic 4.1, Linear Momentum. Every other topic has its own sheet, and its own answer key, on its own page.
The worked answers, for checking your own after you have tried them.