 Let's do another example where we use linear momentum, this time we're going to do a two-dimensional collision We have initially one object, which is not moving at all and we have another object that's moving in horizontal direction and It's going to hit this object and then after the collision We have object one that moves up at an angle of 21 degrees and we have object two that moves down at 30 degrees Clockwise from x at the speed of 8 meters per second and the question is what was the v1 initial? What was the v1 initial and what was the v1 final? So how do we solve this? Well, I already hinted it. We're going to apply conservation of linear momentum So we have the final momentum of the system is equal to the initial momentum of the system Now initially I had mass one moving at v1 and I have mass two Moving at v2 we want initial and v2 initial Now we know that v2 is zero so we can cross out that part here being zero At the end I have two objects moving so I have to consider two momentum again So mass one times v one final plus mass two times v2 final now this is a vector loss actually there are two Equations hidden in there, which is good because you have two unknowns So we have x equation and we have a y equation. So let me rewrite this By splitting up the velocities in its x and y component for the initial side This is very easy. We have m1 and then we only have v Initial in x and we have nothing in y direction. So the initial part is very simple Now let's do the same thing for the final part. So final part. We had m1 times the vectors now here We have both x and y components. So we have v1 final and x v1 final the magnitude Now we have to determine cosine sign as we did when we worked with Newton's laws of motion So the x component here Is cosine of theta 1 So cosine of 21 degrees while the y component is sine of 21 degrees and Then for m2 We have v2 Final v2 final and now what do we have here? Here we have in x we have cosine of theta 2 so cosine of 30 and in y direction I have sine of 30 and negative because we're going downwards so negative making the negative here sine of 30 So here are my two equations here is the x equation and here is the y equation Now let's look where my unknowns are my unknowns are here my v1 final v1 final and my v1 initial Everything else is known. We know the v2 final. So this is all good So which equation is easier to solve? Well, it looks like the y equation is easier to solve So I'm going to start with the y equation so y equation Which says that m1 times v1 final times sine 21 degree Minus m2 times v2 final v2 final. We know 8 meters per second v2 final times sine 30 degrees is equal to zero So if I solve this for v1 final I get v1 final is equal to m2 times v2 final times sine 30 degrees divided by m1 times sine 21 degrees and what I get is 21.5 meters per second So I have found that the one final is 21.5 meters per second So now I'm going to solve the x equation I'm going to plug in the number I found from the y equation into the x equation So let me erase this So let's do this in x directions or in x direction. I have m1 times v1 final which I found from the y direction times cosine 21 plus m2 v2 final which I know times cosine 30 is m1 times v1 initial So this is very simple I just divide by m1 So I have all of this divided by m1 gives me v 1 initial which is I calculate this 30 1.1 meters per second So v1 initial is 31 0.1 meters per second So the problem is asked is solved but let now but now let's ask us a different question Is this an elastic or an inelastic collision? Well, we know that the two objects separate so we know it's not a hundred percent inelastic collision now the question is is it 100% elastic collision If it's a hundred percent elastic collision, what does that mean? That would mean that the kinetic energy is conserved So let's check if with these speeds here. We have conservation of Kinetic energy, so I'm going to need to make some space again is the collision 100% elastic It's the big question if it was elastic, we know that kinetic energy should be conserved So let's calculate our total initial kinetic energy. We know that we calculate kinetic energy That's one half and v squared. So my kinetic energy initial is One half and one times v one initial squared Plus one half and two v two initial squared v two was zero So that's all the energy we have where I get Around 1210 joules Plugging in my v1 initial and my m1 my kinetic energy final Is one half and one v one final squared plus one half and two v two Final squared Gives me around 580 joules for this part and If I use my eight meters per second for my v2, I get plus around 134 joules for that one gives me a total of around seven hundred and Ten joules So if I look at the two numbers, I see clearly that kinetic energy was lost therefore not 100% elastic The fact that kinetic energy was lost actually tells us that we had indeed a collision and not some hidden Explosive inside there that would have added energy only if there was something that adds energy kinetic energy could go up This case can again she was lost where did the energy go probably? There was some little deformation that was done on one of the objects or there was something transformed into heat or into sound But all in all can again she was clearly lost Therefore the collision was not 100% elastic and the fact that we have two objects traveling in separate directions also mean it was not in elastic So it's a realistic collision That's somewhere in between 100% elastic and 100% in elastic