Two objects collide and bounce apart. FIGURE EX11.31 shows the initial momenta of both and the final momentum of object 2. What is the final momentum of object 1? Write your answer using unit vectors.

Video transcript

Everyone in this problem, we have an image that shows the momentum of toys A and B before the collision as well as momentum of toy B after the collision, the toys separate after the collision. And we're asked to determine the momentum of toy A after the collision giving the result in unit vectors. Now, if we look at the image we are given and we're told that the momentum of toy A before the collision. OK. Initially is a vector pointing from the origin up to the 0.22, the momentum of Toy B initially is a vector pointing from the origin to the point negative 11, the momentum of Toy B after the collision is a vector pointing from the origin to the 0.2 comal where we have the X component of the momentum in the X on the X axis and the Y component on the Y axis. We're given four answer choices. Option A negative I, option B negative I plus J, option C I minus J and option D negative I plus two J. With all of those answer choices have the unit of kilogram meter per second. Now we have a collision. OK. We can think about the conservation of momentum, we know that momentum will be conserved. So we can look in both the X direction and the Y direction. If we look in the Y direction, we get that the initial momentum in the X direction P I X is going to be equal to the final momentum in the extraction P FX. Now we have two objects in our system toy A and toy B. So our momentum has to consider both of those. So the initial momentum in the extraction, it's gonna be the momentum of toy A initially in the extra action plus the momentum of toy B initially in the action. And this is gonna equal the momentum of toy A after the collision in the extraction plus the momentum of Toy B after the collision in the extraction. Now filling out what we know based on the diagram we were given, OK. The momentum of A again the vector points to two comma two. And so the X component of that vector is going to be two kilogram meter per second, the initial momentum of toy B in the X direction. OK. That vector points to negative one comma one. So the X component is negative one. So we have negative one kg meter per second. And this is equal to on the right hand side, we have the final momentum of toy A in the direction we don't know that. But we wanna calculate that. OK. That's gonna be part of that final or that total momentum of toy A after the collision that we're looking for the momentum of toy B after the collision in the X direction while that vector points to two comma one. So the X component is going to be two. So what we have now is two kg meter per second minus one kg meter per second. It's equal to the final momentum of toy A in the X direction plus two kg meter per second. We can move that two kg meter per second to the left hand side by subtracting it. We can simplify, we get that the momentum of toy A after the collision in the X direction is going to be two minus one minus two which gives us negative one kilogram meters per second. All right. So we're looking for the momentum of toy A. After the collision, we found the X component. We're gonna do the exact same for the Y component and then we can put them together. So we have the initial momentum in the Y direction is equal to the final momentum in the Y direction. And the momentum is made up of the momentum of two things. The two objects in our system, the momentum of joy A and the momentum of toy B, we have P A I Y plus PB I Y is equal to P A F Y plus PB F Y. The initial momentum of toy A in the Y direction. OK. Again, that vector points to two comma two. So the Y component is two, we get two kg meters per second for toy B in the Y direction. Initially that vector points to negative one comma one. So the Y component is positive one. So we add one kg meter per second. So the left hand side we're done with now we have two kg meter per second plus one kg meter per second. On the right hand side, we have P A F Y. OK. Got a Y component of the final momentum of toy A which we're looking for plus the final momentum of toy B in the right direction which we can get off of our graph. Now that vector points to two comma one. So the Y component is positive one. So we're adding one kg meter per second in order to isolate this momentum P A F Y, we can subtract one kg meter per second from both sides. And that is going to leave us with the Y component of this momentum P A F Y is equal to two kg meter per second. OK. So we have the X component and we have the Y component. So we wanna put them together. OK. So that final momentum P A F of toy A is going to be equal to. Now we have negative one in the X component or the X direction. Remember that the X direction corresponds with the I direction. And so putting this together, we get negative I in the Y direction we have two. So we get plus two J in our unit kilogram meter per second. And that is the momentum we were looking for the momentum of toy A after the collision. This corresponds with answer choice D negative I plus two J kilogram meters per second. Thanks everyone for watching. I hope this video helped see you in the next one.