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# 2-dimensional momentum problem

## 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 this is maybe what's going on on the surface of a pool table we have ball a and it's moving with its ten 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 and gets and gets set on to this new trajectory it's now instead of going to do right it's going at a 33 30 degree angle to I guess we could say horizontal we're going at 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 be doing ball B is going to be going you know we can just think about it just you know if you ever played pool we could guess that ball B is going to go roughly in that direction we need to figure out exactly what the angle is and exactly what its velocity is so so let's do this problem so at first you were saying well this is salad this looks confusing you know I know momentum should be conserved in all that but now we have these vectors and there's two dimensions and and how do I do that and and the key here is is that there's just really one more step when you're working on it in two dimensions really three dimensions or an arbitrary Mynt number number of dimensions when we need 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 let's say let's call this so P for momentum 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's 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 ten kilograms ten kilograms and what is its velocity in the x-direction well all of its velocities in the x-direction right so it's three I mean this is only moving in the X direction so the momentum in the X direction is 30 kilogram meters per second right 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 AIDS 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 zero so let's figure out what a smell in the X and y directions are so to figure out what a is momentum is we just have to figure out what a is velocity in the X and y directions are and then multiply that times the mass right 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 vector for the X component of the vector first so the X is just this vector and the Y change colors to keep things interesting the Y is this vector that is the Y component and so what what are those and this hopefully is is going to be almost second nature to you if you've been watching all of the other videos and Newton's laws this is just our trigonometry and we can write out our sohcahtoa again so Toa and I reassure you this is the hardest part of any of these multi-dimensional trig problems of a 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 so we know V a sub X or the velocity of a in the X direction over the hypotenuse over two meters per second is equal to what adjacent over hypotenuse adjacent over hypotenuse cosine is equal to cosine of 30 degrees or the velocity of a in the X Direction is equal to two cosine of 30 degrees and what's cosine of 30 degrees all right 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 this is 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 is second nature to you as well but since opposite over hypotenuse is equal to the sine of 30 so V a in the Y Direction is equal to 2 times the sine of 30 degrees sine of 30 degrees is 1/2 so this is one half one half times two is equal to one meter per second so after the collision a is moving at 1 meter per second up one meter sorry one meter per second in the upwards direction and it's moving at square root of 3 meters per second in the right worst direction so what is going to be a z' momentum in each of the directions well we figured out its velocity so we just multiply each of the velocities times the mass so the a has a mass of 10 kilograms so we could say the and this is going to be the final momentum the momentum of a in the X Direction is going to equal square root of 3 times 10 right square root of 3 is a 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 opposites it's positive it's 1 m/s is 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 zero so we know that 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 the momentum of B in the Y Direction have to equal what what was the initial momentum in the Y direction all right it was zero right there's no movement in the Y direction initially we know the momentum of a in the Y direction it's 1010 kilogram meters per second plus the momentum of B in the Y Direction is equal to zero so solving for this to subtract 10 from both sides so the momentum of B in the Y Direction is equal to 10 kilogram meters per second kilogram meters per second you know the unit's so if its momentum is 10 in in the Y direction what is its velocity in the Y direction well momentum is equal to mass times velocity right so we know that 5 times 5 times the velocity in the Y direction that's its mass is equal to 10 alright 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 Bea's velocity and so let's say this is B's velocity vector in the Y Direction is and this is a minus right because this was equal to minus 10 all right I just write this is going down right it was positive it was a velocity of positive 1 going up and then the minuses carried through 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 sea the momentum of a in the x-direction which is 10 square roots of 3 10 square roots of 3 plus the 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 12 point let's just say twelve point seven so we know that the momentum of B in the X Direction is equal to twelve point seven twelve point seven 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 twelve point seven divided by five so the velocity of B in the X Direction is twelve point seven divided by five twelve point seven divided by five is equal to two point five four meters per second so two point five four so its velocity in the X Direction is 2.54 meters per second so what's going faster in both directions and if you 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 the arctan well i won't go into that's a complexity right now actually I'll do that in the next video just so that I won't leave you hanging but we know what the x and y components of Bees velocity is see in the next video