consider this problem a four kilogram ball moving east at a speed of five meters per second strikes a two kilogram ball at rest calculate the velocities of the two balls assuming a perfectly elastic collision now if you want to try this problem yourself go ahead and pause the video so what equations do we need now notice that we need to calculate the velocity of the first ball in the second ball so we have two missing variables therefore we need two equations if you only have one missing variable you need to use the conservation of momentum for any collision be it inelastic or elastic momentum is always conserved so m1 v1 plus m2 v2 is equal to m1 v1 prime plus m2 v2 prime so v1 is the initial velocity of the first ball v1 prime is the final velocity of the first ball v2 is the initial velocity of the second ball and v2 prime is the final velocity of the second ball so what this equation is saying is that the total momentum before the collision that is the momentum of both objects is equal to the final momentum after the collision now if you have just one missing variable this equation will be enough to solve for the missing variable but in this problem we have two missing variables and so we need to use a second equation for an elastic collision kinetic energy is also conserved however it is not conserved for an inelastic collision but it's always conserved for a perfectly elastic collision now you don't want to use the kinetic energy formula because it's going to be a lot of work instead you want to use the simplified version of the conservation of kinetic energy for these problems that equation simplifies to this it's the velocity of the first ball that is the initial velocity of the first ball plus the final velocity of the first ball and that's equal to the initial velocity of the second ball plus the final velocity of the second ball so for elastic collisions these are the two equations that you want to use if you need to find two minutes and variables so let's start with the first equation so we have a four kilogram ball moving east so m1 is four and it's moving east at a speed of five meters per second so the velocity is positive five if it was moving west the velocity would be negative and you got to put in that negative value otherwise you won't get the right answer now the second ball is at rest so v2 is 0 which means 2 times 0 is going to be 0. now the mass of the first ball is 4. now we don't know the final velocity of the first ball after the collision that's what we're looking for m2 the second mass or the mass of the second ball that's two kilograms and we don't know the final velocity of the second ball so four times five is 20 and that's equal to four v 1 prime plus 2 v 2 prime so let's save this equation now let's focus on the second equation v 1 plus v 1 prime is equal to v2 plus v2 prime so v1 is still 5 that's the velocity of the first ball v1 prime we don't know what that is v2 the initial velocity of the second ball is zero and we don't know what v2 prime is so i'm going to take this term and move it to this side so 5 is equal to negative v1 prime plus v2 prime so what i'm going to do now is line up these two equations we need to solve for v1 prime and v2 prime using a system of equations so this is going to be 4 v 1 prime plus 2 v 2 prime is equal to 20 and negative v 1 prime plus v 2 prime is equal to 5. so now what we have is an algebra problem how can we solve this system of equations so we need to use elimination so we have positive 4 v 1 prime what i need is negative 4 v 1 prime so that those two will cancel so i'm going to multiply the second equation by 4. so i'm going to rewrite the first equation it's still the same it doesn't change and the second equation is going to be negative 4 v 1 prime plus 4 v 2 prime and 5 times 4 is 20. so now let's add these two equations four negative four adds up to zero two plus four is six and twenty plus twenty is forty so now what we need to do is take forty and divided by six so v2 prime is 6.67 meters per second because it's positive that tells us that the second ball is moving to the right with that speed so now let's calculate v1 prime using this equation so if negative v1 prime plus v2 prime is equal to 5 what i'm going to do is take this move it to this side so it's going to be v1 prime but positive i'm going to take this number and move it to that side so it's going to turn to negative 5. so it's a v2 prime minus 5. so v1 prime is going to be v2 prime which is 6.67 minus 5. so that's 1.67 meters per second so for the sake of space i'm going to rewrite it here now how do we know if we have the right answer but we need to make sure that momentum and kinetic energy are conserved in this problem so let's start with momentum m1 v1 plus m2 v2 has to equal m one v one prime plus m two v two prime so the mass of the first ball is four and it has a velocity of positive five the second ball is at rest the mass of the first ball is still four but its final velocity is 1.67 and the second ball has a mass of 2 with a final velocity of 6.67 now 4 times 5 is 20 and 4 times 1.67 is 6.68 and 2 times 6.67 is 13.34 now if we add 13.34 with 6.68 that's going to be about 20.02 which is approximately equal to 20. so that's good enough there's going to be some rounded error but if it's close enough then you know you did it correctly so momentum has been conserved that's a good sign so now let's make sure kinetic energy is conserved we're going to do it two ways the most simplest way is to use this equation v1 plus v1 prime has to equal v2 plus v2 prime so v1 the initial velocity of the first ball is 5. v1 prime is that value that's 1.67 the initial velocity of the second ball because it was at rest is zero and v2 prime is 6.67 now 5 plus 1.67 is 6.67 so kinetic energy is conserved now let's confirm it using a more familiar equation let's use this one the kinetic energy of the first object the first ball is one-half m1v1 squared and the kinetic energy of the second ball is well supposed to be m2 v2 squared and this is going to be equal to the final kinetic energy of the first ball which is one half m1 v1 prime square plus the kinetic energy the final kinetic energy of the second ball which is one half m2 v2 prime squared so this is gonna be one half times a mass of four times an initial speed of five now once you square v the speed doesn't the velocity doesn't matter in terms of the sign if you use a negative or positive five when you square it is still 25 so kinetic energy is always positive the mass of the second ball is zero but its initial speed is zero so its kinetic energy on the left side is zero now the final speed of the first ball we have 1.67 and the final speed of the second ball is 6.67 so 5 squared is 25 times 4 that's 100 times 0.5 so this is going to be 50 and this is that's going to be zero and then 1.67 squared times 4 times 0.5 that's 5.5778 and then we have half times 2 times 6.67 squared which is 44.4889 so if we add that to 5.5778 you should get 50.06 so that's approximately equal to 50. so therefore we can see that kinetic energy is indeed conserved which means these two answers are correct so those are the final velocities of the two balls mentioned in this problem you