Momentum & Impulse in 2D

Concept: Momentum & Impulse in 2D

9m
Video Transcript

Hey guys. So, if you remember impulse and momentum are vectors. So, in some cases we're going to have collisions happening in two dimensions and we can use momentum and impulse in two dimensions to solve these problems, let's check it out. So, impulse is also a vector and its direction is the same as a direction of force and there was a really easy way to see this, impulse is Force times time, impulse is a vector, force is a vector they both have Direction, time has no direction. So, J has to get its direction from, from f, okay? So, the direction of J is the same as the direction of f, this is very similar to how, when you have p equals m, v, the P gets its direction from the vector, the other vector, which is V. Alright, so let's talk about two-dimensional momentum and impulse problems, here's the long form of the momentum impulse equation and we can write this in the x axis and the y axis and it would look something like this, you just put x is everywhere. So, J x equals f x Delta T, changing momentum in the x axis, mass. velocity final in the x, velocity final in the Y and you could do the same thing for the y axis would be exactly the same just replace X's with Y's, let's do an example here, an object of unknown mass. So, mass is unknown, initially moving with 200 kilograms meters per second momentum, if I didn't say momentum you would have known that this is momentum by the units. So, here is this object, it's moving this way with a momentum p, let's call this p initial of 200, this is similar to saying you have an object moving with a velocity of 200 except now it's a momentum of 200, okay? Which is velocity times the mass, in the positive x axis and has 100 Newtons second, this is impulse and says there, of impulse, delivered to it by a force directed along the y-axis. So, force, J equals f, t, if the force is in the y axis, negative y-axis this means that the impulse would be the negative Y axis. So, you can think of it this way, you're going like this and then a force pushes you down, what's going to happen? Well, if you're moving this way and something pushes you down you still keep moving that way but now you're going to face down a little bit and instead go this way and we can do this with vector analysis. So, if you're moving this way then there's a J here, of 100 and you now end up going sort of this way here, and this is going to be your p final. Alright, the idea here is that J is the change in p, that's part of J, right? So, I can expand this p, f minus p, I. and if I move stuff around look what I get, let's move the p, I to the other side, it's going to look like this, J plus p, I equals p, f, I want to rewrite this to look like this, p, f equals p, I plus J, I'm going to talk about this because that's what I just showed you with my arms your final momentum is whatever your initial momentum was plus J, J is a changing momentum. So, you're going this way I push you this. So, you're going to do this now, okay? So, this is your final momentum and we want to know the magnitude and direction of the objects final momentum, this is two dimensional. So, one of the ways that we can do this, is by calculating p, f, if we want to know P, F total, right? I didn't I didn't mention that I want the horizontal vertical. So, we just assume it's the total, I have to find p, FY and then I have to find p, FX, okay? And, that's because p, F is going to be the hypotenuse of the Pythagorean of its components, cool? I'm going to show you the sort of the long way to do it and then I'm going to show you how you could have done this a little bit faster, let me just block off some stuff here. So, let's try to find PF, okay? So, j, x, PF x i mean, j x is f x delta t x, delta t, time has no dimension, Delta px, okay? And I could also write m, V final x minus V final, the initial x but notice that I don't have the mass and I don't have the velocities, what I do have is I have the momentum and I have the impulse. So, we're going to stay just between these two, J Delta x is p final minus p initial in the x axis. Now, I'm being pushed down by an impulse of 100 not 10, sorry about that, 100, by the way, this is negative because it's down but there is no impulse in the x axis, if there's no impulse in the x axis and this is 0, the input. there's no impulse, there's no change in, there's no change in momentum and I can move this over to the other side and you simply get p I, let's do this over here, we simply get p FX equals p IX, right? Should make sense, the p's in the x axis, your momentum in the x axis it didn't change at all and you didn't have to go through all this, you can just look at the picture and you could have realized that well, you're going to be pushed down. So, you just keep going to the right there's nothing happening on the x axis you keep your same p I, which is 200, okay? Now, for the y axis I can write J Y, Delta py and let's expand this p Y or p final Y minus p initial Y, there is a J the y axis, which is negative 100, the p final in the y axis is what we're looking for and what about the p initial? Well, if you're moving straight in the right you're not moving up or down you have no velocity in the y-axis. So, you have no momentum in the y-axis, this is just 0. So, your final impulse in the x is 200 and your final impulse on the y axis is 100, okay? When you put those two numbers here, this is 200 and this is 100 not much space there but we're going to cram it in here, when you do that you get 224 kilograms meters per second is the unit for impulse, okay? So, that's it all you have to do is you, the fast way that I promised was going to explain to have done this is to first have to realize that p is going to be a combination of these two guys but then to get these two guys p x is 200, it doesn't change so it stays at 200, there is no p Y. So, when you add 100 this way you have 100 of p Y, okay? So, that's it, the answer is 224, there is one last part, which is asking for the direction of the final momentum, so this is just vector stuff, I'm going to the direction right here, theta is the arctangent of Y over x, remember that's how we find the direction for vectors, when we know the two components. So, it's going to be p Y over p x always Y over x as long as we're getting the angle with the x-axis right there, which is what we want, let's call that theta x and then this is theta x right here, if you do this you get an angle of, I have it here, negative 26.6 degrees and it makes sense that it's negative because it is below the positive x-axis right here, right? So, negative makes sense, cool? So, those are the two answers for this part just a quick example of how to, how we can combine components of momentum to find the total momentum and the angle of the total momentum, I have a second problem here, let's jump into that real quick.

Example: Momentum & Impulse in 2D

9m
Video Transcript

Part B asks for the average force exerted on the ball, let's do that here, the average, I'm sorry, it asks for the horizontal and vertical component of the average force exerted by the ball, right? So, there's like a semicolon there, just in case, just to be clear, I would want to calculate the components of impulse and force, cool? So, basically I want to know what is Fx and what is Fy, and this is pretty easy now, because I can come from here and use the fact that I already know Jx, Jx equals Fx delta t, if you stop there and not write the rest you'll notice that you know Jx, you're looking for Fx, and you know t. So, it's pretty straightforward Fx is Jx over t, Jx is negative 10.9, t is 0.01, so this is going to be negative 1090 Newtons, for Fy it's the same thing, Fy will simply be Jy over delta t and that is 4.35 divided by 8.01 will be 435, it's a force. So, it's a Newtons, okay? So, that's part a, that's Part B. Part C is very straightforward as well in, Part C says, calculate the magnitude of the impulse and the force on the ball. So, now I'm not asking for horizontal and vertical I'm not asking for the components, I'm asking for the whole thing, this is just vector addition and if I want the magnitude of the total impulse in a two dimensional problem, that's just the Pythagorean theorem of this. So, it's the square root of the square of its component. So, Jx, Jy and we can just put those numbers here, the numbers that you would use are these Jx + Jy and if you do this, I have it here, actually I don't have it here. So, I didn't calculate these numbers ahead of time so this is roughly, I'm going to just kind of in my head real quick, this is roughly 121, this is roughly 20. So, it's going to be the square root of 141, which is roughly 12, this is probably really long, hopefully not and then if we do F it's also going to, please do this by yourselves and you may get a different number, this is Fx squared and Fy squared, at this point it doesn't really matter, this last calculation is the least important part of the problem, the point is to even get to this point. So, FX is 1090 squared 435 squared and then this will give you a some number, it's hard to tell, but yeah I'm just going to put a little tree here and then whatever that is just plug it in there I didn't calculate this number in advance but all you have to do is just do these two and get the answer, cool? So, let me know you have any questions I'll be glad to help.

Problem: A 200-g ball going up diagonally hits a vertical wall with 10 m/s at an angle of 30° with the wall. It then rebounds with the same speed and same angle. What is the magnitude of the impulse delivered to the ball?

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