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Introduction to momentum

Momentum (P) is equal to mass (M) times velocity (v). But there are other ways to think about momentum! Force (F) is equal to the change in momentum (ΔP) over the change in time (Δt). And the change in momentum (ΔP) is also equal to the impulse (J). Impulse has the same units as momentum (kg*m/s or N*s). Created by Sal Khan.

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  • leaf green style avatar for user danhogan13
    The units of impulse are the same as those for momentum. Look back at Sal's original definition of impluse = force x time = (kg.m/s2)*s = kg.m/s . Impulse is just the change in momentum - it is not the change in momentum per unit time.
    (102 votes)
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  • blobby green style avatar for user André Löfgren
    I understand that the momentum is the same before and after the collision, but what happened to all the lost kinetic energy?
    (39 votes)
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  • blobby green style avatar for user crisfusco
    Is there a difference between Momentum and Impulse? Im confused :(
    Thank youu!
    (27 votes)
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    • blobby green style avatar for user Rutvik Shah
      Momentum is the mass of the object times the velocity of the object. The impulse is if that momentum changed, i.e. change in momentum. So in the video, because the car and the truck were stuck together post-crash, there was NO change in momentum, i.e. the impulse was 0.

      however if this was an elastic collision (i.e. a collision in which the two objects do not stick together after colliding) then the momentum of each would have changed, however the overall momentum would have been conserved. (Conservation of Energy). This would have brought about an impulse.
      (53 votes)
  • blobby green style avatar for user Jasmine Kaur
    why does initial momentum has to be equal to final momentum?
    (10 votes)
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  • leaf green style avatar for user Greg  Pailet
    When doing momentum problems to obtain the combined velocity, why is it that we do not use the concept of kinetic energy? We can use mv^2/2 to obtain this value, so why would it not work if when we assume v is a constant value after collision and that m is the sum of the two masses, we solve for a different value of v compared to solved with the momentum method?
    (8 votes)
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    • leafers tree style avatar for user Apoorv Agnihotri
      The main reason one cant apply "Work Energy Theorem (Kinetic Energy Conservation, as in this case)" on this problem, because some Kinetic Energy is stored as Potential Energy (Which is due to deformation caused by the collision at the surface of impact) thus the final Kinetic Energy is not equal to initial KE.
      NOTE: If this question proclaimed an elastic collision between the objects.
      Then KE(initial)=KE(final), because elastic collision means that the surfaces of the bodies AFTER (NOT DURING) the impact will reform fully to their original shapes, thus no extra PE is stored AFTER the collision.
      (7 votes)
  • leafers ultimate style avatar for user Ernesto Aguilar
    So do photons also have momentum?
    (6 votes)
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  • piceratops ultimate style avatar for user manvithn
    what if there is, say, a 10kg object moving at 10 m/s east and a 9 kg object moving at 10 m/s west. If they both crash, will they both move east? one east, one west? what if both of them have the same mass? will they crash and bounce off in opposite directions?

    In response to @Henningd's answer, if it is a perfectly elastic collision, will both move east or will one move west and one east?
    (6 votes)
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    • blobby green style avatar for user henningd
      Whether they bouce or stick together is a matter of the materials involved and can be quite complicated. In beginning levels of physics we only look at the extreme cases where 1. they are perfectly elastic in which case they would bounce off and 2. where they are perfectly inelastic and they would stick together. In real life situations most collisions fall somewhere in the middle and require much more complicated approaches.
      (8 votes)
  • blobby green style avatar for user Farhan Naufal
    At , why do we have to calculate the COMBINED MOMENTUM before the collision?
    (3 votes)
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  • starky sapling style avatar for user di
    Is momentum a type of energy?
    (2 votes)
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  • female robot ada style avatar for user Liza Kurtenbach
    I am still confused about what impulse and momentum really are, like their definition in words. Can anyone help me? Maybe with an example or something?
    (4 votes)
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Video transcript

Welcome back. I will now introduce you to the concept of momentum. And the letter for momentum is, in physics, or at least in mechanics, it's the letter P. P for momentum. And I assume that's because the letter M has already been used for mass, which is I guess an even more fundamental idea. So P for momentum. So what is momentum? Well, you probably have a general idea of it. If you see a big guy running really fast, they'll say, he has a lot of momentum. And if there's a big guy running really fast and a small guy running really fast, most people would say, well, the big guy has more momentum. Maybe they don't have a quantitative sense of why they're saying that, but they just feel that that must be true. And if we look at the definition of momentum, it'll make sense. The definition of momentum is equal to mass times velocity. So something with, say, a medium mass and a huge velocity is going to have a big momentum. Or something with maybe a medium mass, but-- the other way around. I forgot what I just said. So medium mass and big velocity, huge momentum, or the other way around. Huge mass, medium velocity would have maybe the same momentum, but it would still have a big momentum. Or another way of doing momentum is how little you would like to be in the way of that object as it passes by. How unpleasant would it be to be hit by that object? That's a good way of thinking about momentum. So momentum is mass times velocity. So how does it relate to everything we've been learning so far? So we know that force is equal to mass times acceleration. And what's acceleration? Well acceleration is just change in velocity. So we also know that force is equal to mass times change in velocity per unit of time, right? Per change in time. T for time. So force is also equal to-- well, mass times change in velocity. Mass, let's assume that mass doesn't change. So that could also be viewed as the change in mass times velocity in the unit amount of time. And this is a little tricky here, I said, you know, the mass times the change in velocity, that's the same thing as the change in the mass times the velocity, assuming the mass doesn't change. And here we have mass times velocity, which is momentum. So force can also be viewed as change in momentum per unit of time. And I'll introduce you to another concept called impulse. And impulse kind of means that you think it means. An impulse is defined as force times time. And I just want to introduce this to you just in case you see it on the exam or whatever, show you it's not a difficult concept. So force times change in time, or time, if you assume time starts at time 0. But force times change in time is equal to impulse. I actually don't know-- I should look up what letters they use for impulse. But another way of viewing impulse is force times change in time. Well that's the same thing as change in momentum over change in time times change in time. Right? Because this is just the same thing as force. And that's just change in momentum, so that's impulse as well. And the unit of impulse is the joule. And we'll go more into the joule when we do work in all of that. And if this confuses you, don't worry about it too much. The main thing about momentum is that you realize it's mass times velocity. And since force is change in momentum per unit of time, if you don't have any external forces on a system or, on say, on a set of objects, their combined, or their net momentum won't change. And that comes from Newton's Laws. The only way you can get a combined change in momentum is if you have some type of net force acting on the system. So with that in mind, let's do some momentum problems. Whoops. Invert colors. OK. So let's say we have a car. Say it's a car. Let me do some more interesting colors. A car with a magenta bottom. And it is, let's see, what does this problem say? It's 1,000 kilograms. So a little over a ton. And it's moving at 9 meters per second east. So its velocity is equal to 9 meters per second east, or to the right in this example. And it strikes a stationary 2, 000 kilogram truck. So here's my truck. Here's my truck and this is a 2,000 kilogram truck. And it's stationary, so the velocity is 0. And when the car hits the truck, let's just say that it somehow gets stuck in the truck and they just both keep moving together. So they get stuck together. The question is, what is the resulting speed of the combination truck and car after the collision? Well, all we have to do is think about what is the combined momentum before the collision? Well let's see. The momentum of the car is going to be the mass times the car-- mass of the car. Well the total momentum is going to the mass of the car times the velocity of the car plus the mass of the truck times the velocity of the truck. And this is before they hit each other. So what's the mass of the car? That's 1,000. What's the velocity of the car? It's 9 meters per second. So as you can imagine, a unit of momentum would be kilogram meters per second. So it's 1,000 times 9 kilogram meters per second, but I won't write that right now just to keep things simple, or so I save space. And then the mass of the truck is 2,000. And what's its velocity? Well, it's 0. It's stationary initially. So the initial momentum of the system-- this is 2,000 times 0-- is 9,000 plus 0, which equals 9,000 kilogram meters per second. That's the momentum before the car hits the back of the truck. Now what happens after the car hits the back of the truck? So let's go to that situation. So we have the truck. I'll draw it a little less neatly. And then you have the car and it's probably a little bit-- well, I won't go into whether it's banged up and whether it released heat and all of that. Let's assume that there was nothing-- if this is a simple problem that we can do. So if we assume that, there would be no change in momentum. Because we're saying that there's no net forces acting on the system. And when I say system, I mean the combination of the car and the truck. So what we're saying is, is this combination, this new vehicle called a car truck, its momentum will have to be the same as the car and the truck's momentum when they were separate. So what do we know about this car truck object? Well we know its new mass. The car truck object, it will be the combined mass of the two. So it's 1,000 kilograms plus 2,000 kilograms. So it's 3,000 kilograms. And now we can use that information to figure out its velocity. How? Well, its momentum-- this 3,000 kilogram object's momentum-- has to be the same as the momentum of the two objects before the collision. So it still has to be 9,000 kilogram meters per second. So once again, mass times velocity. So mass is 3,000 times the new velocity. So we could call that, I don't know, new velocity, v sub n. That will equal 9,000. Because momentum is conserved. That's what you always have to remember. Momentum doesn't change unless there's a net force acting on the system. Because we saw a force is change in momentum per time. So if you have no force in it, you have no change in momentum. So let's just solve. Divide both sides of this by 3,000 and you get the new velocity is 3 meters per second. And that kind of makes sense. You have a relatively light car moving at 9 meters per second and a stationary truck. Then it smacks the truck and they move together. The combined object-- and it's going to be to the east. And we'll do more later, but we assume that a positive velocity is east. If somehow we ended up with a negative, it would have been west. But it makes sense because we have a light object and a stationery, heavy object. And when the light object hits the stationery, heavy object, the combined objects still keeps moving to the right, but it moves at a relatively slower speed. So hopefully that gives you a little bit of intuition for momentum, and that was not too confusing of a problem. And in the next couple of videos, I'll do more momentum problems and then I'll introduce you to momentum problems in two dimensions. I will see you soon.