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2-dimensional momentum problem

An example of conservation of momentum in two dimensions. Created by Sal Khan.

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  • blobby green style avatar for user priyadarshiniravi03
    Is momentum a vector????
    (37 votes)
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  • male robot donald style avatar for user Abhijay Bhatnagar
    at , how is the momentum in the y direction for B = 10? If you subtract 10 from both sides, you get -10, but then thats in the opposite direction.
    (19 votes)
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  • blobby green style avatar for user Nathan Fife
    How can the final velocity of Ball B be greater than the Velocity of ball A. That is impossible unless I misunderstand. I think that the only reason he got that answer is because he gave a velocity and angle for A after the impact instead of finding it. There is too much energy in the system I think.
    (3 votes)
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    • leaf green style avatar for user Sean
      The reason that this happens is due to the mass of ball A and the mass of ball B. Momentum is mass times velocity. Ball B is half of the mass of ball A, so the velocity of ball B needs to be greater than the velocity of ball A in this example in order to make up for its smaller mass due to conservation of momentum. Even though it sometimes feels like there should be, there is no conservation of velocity rule... Hope this helps!
      (19 votes)
  • leaf blue style avatar for user shreeyathussu98
    I don't understand how there was a velocity for the X and Y directions...does ball B move in one direction and then the other?
    (3 votes)
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    • blobby green style avatar for user pam.howie%ntlworld.com
      If you throw a ball straight up in the air it only has vertical or y velocity.
      If you throw a ball horizontally, it only has horizontal or x velocity initially ( though gravity will rapidly change that)
      in general I might pitch a ball which is neither horizontal nor vertical - I can represent the balls motion with horizontal and vertical components e.g. if i throw it up at 10 m/s at 30 degrees from the horizontal it has a horizontal component of 10 cos 30 and a vertical component of 10 sin 30.
      Why can't I just represent it as 10 m/s at an angle of 30 degrees?
      the answer to this is that on the good earth, gravity acts vertically in a well known and easy to calculate manner. Nothing changes to the horizontal component if I ignore air resistance while the vertical component is first retarded to zero then gravity has the nerve to pull the ball back down!
      So we divide the motion into vertical and horizontal components, calculate what happens to each using what my students fondly call the SUVAT equations of motion under constant acceleration. then to find out where the ball ends up after a give time, use vector maths to add the componenets back together.
      (7 votes)
  • leaf green style avatar for user Thomas06181992
    May I please ask what is the equation for 3D collision of any two objects
    (4 votes)
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    • primosaur ultimate style avatar for user saml.boyajian
      The same thing with 2D elastic collisions apply with 3D elastic collisions, you just need to solve for the conservation of momentum in each direction. (i.e. pxi=pxf, pyi=pyf, & pzi=pzf) Again we are assuming these are elastic collisions, otherwise these equations would not be applicable.
      (2 votes)
  • leaf green style avatar for user amartinencosbr
    Why do you write to Pay + Pby = 0?
    (4 votes)
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    • leafers ultimate style avatar for user hannah
      The initial momentums (momenta) in the y direction is zero, which is the sum of the initial momentum in y direction for A and the initial momentum in y direction of B. We know that momentum will be conserved therefore the momenta after the collision will add up to equal zero. Really there could be another subscript on those... (Pay + Pby)i = 0 = (Pay + Pab)f. I --> initial and f ---> final.
      (2 votes)
  • blobby green style avatar for user eckwak
    Does anyone know the difference between a conservation of momentum problem and a change of momentum problem? I want to look at every problem and say momentum is conserved before and after a collision. When do I know I've come across a problem where momentum is not conserved and have to use the Impulse equation? Do I still use conservation of momentum equations in an Impulse problem?

    For example if a car collides with a car at rest I can use conservation of momentum to find the speed of the cars after they have collided.

    But then I have to use Impulse to find the change in momentum. I'm not sure when I'm supposed to switch my logic from saying momentum is conserved in this case but is changing in this other case.
    (1 vote)
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    • leafers ultimate style avatar for user Ingo
      Linear momentum is a conserved quantity. In science, laws are laws, because they are obeyed in every single case, we know of. The linear momentum of any (closed) system is conserved - always!

      What could change, in case you're looking at an inelastic collision, is the kinetic energy. But the total energy of the system is again conserved. So the kinetic energy would just get transformed into another kind of energy (i.e. heat) - or vice versa.
      (4 votes)
  • mr pink red style avatar for user Karen
    What is the formula for two dimensional momentum problems or a step by step method to follow?
    (2 votes)
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  • leaf green style avatar for user DJ  Morin
    What would happen if we were given a problem without the mass of each ball, but are told they are equal? Would we just assign a mass to them (like 1 kg)?
    (2 votes)
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  • blobby green style avatar for user Ajay Taneja
    Thanks for the video. Something nonintuitive here is: what makes the ball A and ball B go in the direction they do? Why does ball A move with the Y component upwards and ball B with Y component downwards?Any physical reason for that?
    (2 votes)
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Video transcript

Welcome back. We will now do a momentum problem in two dimensions. So let's see what we have here. So we have this ball A and we could maybe even think of it as maybe what's going on on the surface of a pool table. We have ball A and it's moving with its 10 kilograms. So these numbers are the mass of the balls. This is a 10 kilogram ball and it's moving to the right at 3 meters per second. And then it hits this ball, B, which is a 5 kilogram ball. And then we know that ball A, ball A kind of ricochets off of ball B and gets set onto this new trajectory. Now, instead of going due right, it's going at a 30 degree angle to, I guess we could say, horizontal. It's going in a 30 degree angle at 2 meters per second. And the question is, what is the velocity of ball B? So velocity is both magnitude and direction. So we need to figure out essentially, what is ball B doing? Ball B is going to be going-- We can just think about it. If you ever played pool, we could guess that ball B is going to go roughly in that direction. But we need to figure out exactly what the angle is and exactly what its velocity is. So let's do this problem. So at first you're saying, oh, Sal, this looks confusing. You know, I know momentum should be conserved and all that. But now we have these vectors and there's two dimensions and how do I do that? And the key here is that there's just really one more step when you're working on it in two dimensions or really three dimensions or an arbitrary number of dimensions. When we did one dimension, you made sure that momentum was conserved in that one dimension. So when you do two dimensions, what you do is you figure out the initial momentum in each of the dimensions. So you break it up into the x and y components. And then you say the final momentum of both objects are going to equal the initial x momentum and are going to equal the initial y momentum. So let's figure out the initial x momentum. So P for momentum. Because the m is for mass. So let's say the initial momentum in the x direction-- And we don't have to write initial or final because really, the total momentum in the x direction is always going to be the same. So let's say what the initial-- Actually, let me write initial just so it hits the point home that initial and final don't change. So the initial momentum in the x direction. So i for initial, x-- I should do something better than keep writing these subscripts --is equal to what? Well ball B has no initial velocity, so it has no momentum. Ball A is 10 kilograms. And what is its velocity in the x direction? Well all of its velocity is in the x direction. So it's 3. I mean, this is only moving in the x direction. So the momentum in the x direction is 30 kilogram meters per second. Mass times velocity, kilogram meters per second. And what's the initial momentum in the y direction? Well B isn't moving at all, so it has no momentum in any direction. And A, all of A's movement is in the x direction. It's not moving at an angle or up at all, so it has no momentum in the y direction. So we immediately know that after the collision, the combined momentum of both of these balls in the x direction has to be 30, and the combined momentum of both of these balls in the y direction has to be 0. So let's figure out what A's momentum in the x and y directions are. So to figure out what A's momentum is, we just have to figure out what A's velocity in the x and y directions are and then multiply that times the mass. Because mass doesn't have any direction. So let's figure out the x and y components of this velocity. Let's do the x component of the vector first. So the x is just this vector. Change colors to keep things interesting. The y is this vector. That is the y component. And so, what are those? And this hopefully, is going to be almost second nature to you if you've been watching all of the other videos on Newton's laws. This is just our trigonometry and we can write out our SOH-CAH-TOA again. And I reassure you, this is the hardest part of any of these multi-dimensional trig problems-- Multi-dimensional physics problems, which really are just trig problems. So if we want to figure out the x component, so the velocity of A in the x direction, what is it equal to? Well this is adjacent to the angle. We know the hypotenuse. So we know VA sub x or the velocity of A in the x direction over the hypotenuse, over 2 meters per second, is equal to what? Adjacent over hypotenuse. Cosine. Is equal to cosine of 30 degrees. Or the velocity of A in the x direction is equal to 2 cosine of 30 degrees. What's cosine of 30 degrees? It's square root of 3 over 2. This is square root of 3 over 2. And square root of 3 over 2 times 2 is equal to square root of 3. So this is equal to the square root of 3 meters per second. And what is the velocity of A in the y direction? Well hopefully, this second nature to you as well. But since opposite over hypotenuse is equal to the sine of 30. So VA in the y direction is equal to 2 times the sine of 30 degrees. The sine of 30 degrees is 1/2. So this is 1/2. 1/2 times 2 is equal to 1 meter per second. So after the collision, A is moving at 1 meter per second up. One meter per second in the upwards direction. And it's moving at square root of 3 meters per second in the rightwards direction. So what is going to be A's momentum in each of the directions? Well, we figured out its velocity, so we just multiply each of the velocities times the mass. So A has a mass of 10 kilograms. And this is going to be the final momentum. Momentum of A in the x direction is going to equal square root of 3 times 10. Square root of 3 is the velocity, 10 is the mass. So it's 10 square roots of 3 kilogram meters per second. And the momentum of A in the y direction is going to be-- and since it's going up, we'll say its positive --it's 1 meters per second is the velocity times the mass. So 10 times 1 is 10 kilogram meter per second. So now let's figure out B. Let's do the y direction first because they add up to 0. I'm going to switch colors. We know that the momentum of-- and this is after the collision. The momentum of A in the y direction plus momentum of B in the y direction have to equal what? What was the initial momentum in the y direction? Right, it was 0. There was no movement in the y direction initially. We know the momentum of A in the y direction. It's 10. 10 kilogram meters per second plus the momentum of B in the y direction is equal to 0. So solving for this, just subtract 10 from both sides. So the momentum of B in the y direction is equal to 10 kilogram meters per second. You know the units. So if its momentum is 10 in the y direction, what is its velocity in the y direction? Well, momentum is equal to mass times velocity. So we know that 5 times the velocity in the y direction-- that's its mass --is equal to 10. 10 is its momentum. So the velocity of the y direction of B must be 2 meters per second. So there we go. We figured out B's velocity. And so let's say this is B's velocity vector in the y direction is-- And this is a minus because this is equal to minus 10. So it's going down. It was a velocity of positive 1 going up and then the minuses carry through and this is a velocity of minus 2 meters per second for B in the y direction. So now let's figure out the velocity of B in the x direction. And I'm running out of space and it's getting messy. But we just have to remember that the momentum of B in the-- The momentum of A in the x direction, which is 10 square root of 3, plus momentum of B in the x direction has to equal what? It has to equal the initial momentum in the x direction, which is 30. So to figure out the momentum of B in the x direction, we just subtract 10 square root of 3 from 30. And let's do that. So let's figure out 3 square root times 10 equals. And then subtract that from 30. And we get let's just say 12.7. So we know that the momentum of B in the x direction is equal to 12.7, 12.7 kilogram meters per second. And we know the momentum, so we just divide by the mass and we get its velocity in the x direction. So 12.7 divided by 5. So velocity of B in the x direction is 12.7 divided by 5. 12.7 divided by 5 is equal to 2.54 meters per second. So its velocity in the x direction is 2.54 meters per second. So it's going faster in both directions. I'm not going to figure out the angle here because I've actually run out of time. But if you were to add these two vectors, you'd get an angle something like this. And you could figure out the angle by taking the arc tan. Well, I won't go into the-- that's a complexity right now. Actually, I'll do that in the next video just so I won't leave you hanging. But we know what the x and y components of B's velocity is. See you in the next video.