In this video, I go over the physics of a ball rolling off a table and determine how far it falls, the angle and speed it bounces with, and the distance it travels for a second bounce after losing ...
Transcript
The solution to problem three, and this one looks at ball rolls off table. Yeah, so basically a ball rolls off a table. So a ball rolls off a table with a speed of two feet per second. That's falling off like that fast. That's our initial V-0. The table is 3.5 feet high. and we're asked to determine the point at which the ball hits the floor and find its speed at the instant of impact. So at the point that it hits it, basically it's going to go like this, at a speed, initial speed, this is going to be V0 equals 2 feet per second, and then you want to know where it lands, basically the distance here, I assume. That'll be X like that, something like that. So you want to see where it lands, and then also the speed that it's hitting it down. It's going to be, so what's the speed that gravity has sped it up? Alright, and then Part B says, find the angle between the path of the ball and the vertical line drawn through the point of impact. See, figure below, so we're asked to find this impact angle there, impact and bounce angle. So that's 3.5 feet height. Yeah, and Part C says, suppose the ball rebounds from the floor at the same angle with which it hits the floor, but loses 20% of its speed due to energy absorbed by the ball on impact, and then where does the ball strike the floor on the second bounce? And now it's saying, it loses 20%, where's it going to go now? All right, interesting stuff. Alright, so now let's take a look at the solution to A. So A is, again, asking determine the point at which the ball hits the floor and find its speed at the instant of impact. So, let's take a look at solution. So since gravity is the only force acting on the ball when it is falling, we can derive the acceleration velocity in position of the ball. So basically, in its flight, as it's moving this way. So again, has initial velocity, it's going this way. Yeah, I'll just put it like this. Here's the ball here at the initial velocity. We know it's V0, but at the entire time, the only force acting on it is gravity. So we have the force, yeah, that's the force due to gravity. That's going to be, yeah, that's going to be basically F equals M, yeah, G times M. But again, we're not going to be focused on that. All we care about is the gravity there, because we're not to be looking at force and so on. So it's going to be falling down, and basically what we have is, so we know that the acceleration of this ball, is equal to, and so this equals to, this is going to be negative G, and then times if I'm going to use the J vector. Again, this is just in 2D, so we have the J, or the Y component, the standard basis vector. So now we take the integral, so we take the integral of this and so on, what we're going to get is velocity, so again, A is going to be, I'm going to put this as a vector, so A vector, is just going downwards, and then the I'll put it like this, actually, makes it room. So the V vector equals to the integral of the gravity, that just equals to, this is going to be, yeah, this is going to be G, actually, I'll put it for completeness, integral of gravity there. And I'll also put DT, et cetera. So this equals to, this is going to be negative GT, T, yeah, GT, J, and then plus, it's going to be plus the constant. And this cons, I'm going to just call the C1, well, we know that at, yeah, just for completing us, yeah, we're going to get V0 there. So at T equals 0, we have, this, the speed is just going to be V0. So V0, I'll put a vector like that, because it's just going to be horizontal, and this can be a horizontal vector, just going like that, equals, yeah, equals that, so, and this is going to be zero. So we're just left with, yeah, plus C, yeah, just going to go to 0 at T equals 0, let's just C1. Yeah, so then we have V, I'm not going to box this. There's acceleration, V is going to equal to v0 vector minus GTJ, like that. I'm just going to box this in, so that's that, and now the next one is going to be, well, the position vector, I'm going to call it S. that equals the integral of v vector d t, and that equals two, it's going to be integral of this setup, so we're going to have v0 t, yeah, I mean, yeah, just v0, yeah, v0 t, minus g, t, this is going to be a squared over 2, so just take the power over 2, and then this is j vector, and then we're going to have a plus c2, c2 there, and we know the initial height of this, And here I'm just going to put the vector sign above this vector. It's a constant vector. There's also a J is a vector like that. Alright, so we have all that. And then next setup is, well, we know the initial position. So what we're going to do is, well, we're starting off our coordinate setup. This is going to be our 0. And this is going to be, so we're launching it from here. So basically, this is at the point here. The initial position is going to be 0, and then the height is 3.5 feet. Yeah, so this is going to be our Y, that's our X, and this is going to be our 3.5 feet, like that. I'll just put 3.5, so yes, we're going to be dealing with feet. So that's the initial position there. So what that means is at, at t equals to zero, what we get is, yeah, let's put it like this, the same thing as we did before, the position at t equals to 0 is equal to s not uh yes i'll put s not the initial position and that's going to be equals to and this equals to well it's zero x position it's again as a vector so this is this is going to be equal to let's put it as zero i plus 3.5 j and now this whole thing is equal to this whole part here, when you put in 0, yeah, yeah, when you put in the t-go-0, that just, it vanishes, we just love with C2. Yeah, it's going to be C2 vector. So, what that means is we're going to have the s vector, uh, like that as a function of t is equal to, yeah, just equal to s not vector, that's the c2, and then plus v knot of t minus gt squared. j vector and then when we plug this inside this is a j vectors we just add the j vectors the i is just uh the zeros we don't do anything with it so this we have s is equal to v not well actually before that i'm gonna put everything in in first yeah we have that i'm just gonna put 3.5 j vector plus v not is going to equal to well we know that's going to be two feet per second so two t Yeah, I'm just going to go back up here. So we know that that's the initial one. This is going to be just 2. Yeah, 2 feet per second. So that's just going to be launching it at 2 feet per second, like that. So let's put it like that. And then the next one's going to be, well, minus G. We know the juice gravity, just leave that like that, and then 2 divided by 2. And this is going to be iVector. It's going to be j vector. All right, here, I just fixed that up. So now we can move that inside. So we're going to get a. s like that equals 2, 2Ti minus, or actually plus, now we have a 3.5 minus gt squared, and this is i vector, divided by 2, and this is going to be j vector, like that, fascinating stuff there. Alright, erase this. Alright, so I moved it over here, I'm going to box this, like that. Alright, so now that we have all this, so now the next setup is, well, we want to know, again, we want to know the point at impact, like the distance, the x coordinate going here, and as well as the speed hitting the ground. So let's find the value of T when it hits the ground. So therefore, therefore right here, at this point right here, so that's at y equals zero. we get at, let's put it like here, actually I'll put it just here, so at y equals to 0, yeah, we're going to get, well, the y component is going to be 0, so that's this j setup there, so 3.5, this whole thing has equal to 0. So what we're going to get is, well, 3.5 minus g, t squared over 2 is equal to 0. Yeah, so what we could do now is just multiply by 2 on both sides and then move it over to this side here, and then we'll have that positive, just because it's on the negative. So we're going to get, now, we're going to get a G, t squared, and also, yeah, we're going to also divide this by G on both sides. So we're going to divide it out, move it over, so we're going to get a t squared is equal to 3.5, so then we're going to multiply by 2, divide by G. which equals to well 7, yeah, just 3 .5 times 2 is 7. So we're going to get 7 over G. And now take the square root of it, so we're going to get t is equal to well plus or minus. square root 7 over G, so we'll just take the positive. I'll just put the plus or minus, just because we can have a minus it for some reason. Yeah, so, but it'll be, but it's not going to make physical sense in our case because we're not using negative time. Because again, the graph can go backwards, forward, and time, and so. So we're just going to use positive like that. But I always put that plus or minus just to always keep it on your toes. So we have that, and now we want to know the position. Well, we know that. Now, we know that this part right here, the X component is just this, x equals to 2T. So we have, thus, x equals to 2T, which equals to, at the point of impact, 2 times square root 7 over G. And if you plug this in the calculator, this roughly equals 2.94 feet, which makes sense. Basically, 2 feet per second, falls down. and about 0.9, almost one foot away. So it doesn't go, it's not a giant number, you're not getting some sort of error of 10 to 3 feet and so on. So anyways, so that seems reasonable. All right, so now we'll do a quick calculation check, just go over here, so 2 square root 7 over 32.2, again, that's feet per second squared in the United States or imperial units for gravity. So 9.8 meters, per second squared or this 32.2 feet per second squared. And if you plug down to the OneNote calculator, what I ended up getting equals to this, 935. So it's close enough to 94, but it is 935. So I don't know if my calculus book had an error or not, but it is that. All right here, I'm just going to quickly double check live. So we have Gravity and Imperial Unis just Googled it or a search engine. I'm using Brave. browser search, yeah, so it's actually 32.174, so let's see if that's closer. So if we get 2 times square root of, yeah, I mean, 2 times square root 7, divided by G, which is going to be this, that equals 2 equals what? Okay, 9-328, yeah, so it's still not. it's still not the setup there, so I'm going to go 9, 2, 3, 3, so yeah, so that's pretty interesting. So this is actually, roughly, I'm just going to put this 3-3, yeah, 9-3, yeah, so this is Kaggle's book is wrong, so that's close enough, yeah, so 9-3. Yeah, so its coordinates are. So I don't know why the solution's manual had a 9-4 there, I'm pretty sure it's, yeah, it's not wrong. Okay, so anyways, we're gonna go with this. So at that instant, yeah, so basically at that instant, the ball is approximately 0.93 feet to the right of the tabletop, and its coordinates relative to an origin of the floor directly under the table's edge, so that's this little graph we did right here. that's going to be, yeah, the coordinates are .93 and 0, and these are all in feet. And at impact, the velocity is, so we basically take the velocity now, so we have the velocity here, and just plug in T at impact. So we just take that. So we'll get v0. I'll just copy this and paste it, so it looks like we're easier to follow that. So this is going to be copy this. All right, so here I just copied it, and I'm going to move all the down here. So at impact, we're going to get, here so v equals to v0 minus g tj and that's going to equal to well we know v not is just two feet per second just put it like that yeah but we're dealing with vectors so we need to put the component this is going horizontal so the i center basis and then minus g and instead of plugging this 0.93 or 0.94 so let's just make the exact value so screw it's 7 over g So we're going to get a, and actually I'm going to, what I'm doing, so I'm going to put the square root a bit lower, so it doesn't look like we're dividing G, dividing the square root by G. So, like that. Now we're going to have a, yeah, square root 7 over G. Well, what I'm going to do is, yeah, and this is going to be a J standard-based vector. What I'm doing is I'm going to multiply this inside. So it's the same thing as writing, well, square root 2 like that. Yeah, square root of G squared, like that. Yeah, so what that means is, yeah, we basically can move that inside. So this is going to equal to move that inside, cancels one of the Gs, so we're just going to have square root, yeah, square root 7g, yeah, so 7g's there. So thus our velocity is equal to 2i, minus velocity vector, and then this is going to be 7, yeah, square root. this is just going to be square root 7g, like that, and then j vector. And then the magnitude of, because we were asked to find the speed, not just the velocity vector. Because number of velocity is a vector as magnitude and direction, but we're just asked for the speed of impact, and you can see right here, and it's speed. And we can just use the... the distance of that, or the magnitude equation, or basically the distance in 3D or 2D, so we're going to get thus speed equals to the magnitude of the velocity at impact, and that equals to, if you recall, there's a square root, then you square the component, so this one's two squared, and then the next one's going to be plus, well, this whole set of them, and this is, the G is squared as well. like that, so we're going to have, well, I'm just going to put it for a complete, a 7G squared, like that, and then what happens here is, well, we're going to get a square root 4 plus 7g squared. Actually, whoops, it's not that G squared again. I got confused by this dish. Yeah, the 7's gone, the square root's gone, so just 7G. So that's what it is, and if you plug this into the calculator, you're going to get roughly 15 feet. per second, and I'll show you a calculations check soon. Alright, so now we have that, let's do a quick calculation check, so right, yeah, that's the, again, the speed of impact. So, calculation check, square root 4 plus 7 times 32.2, that's the feet per second squared in imperial units of gravity, equals to 15.1459, so a good calculus book got it. right this time so 15, yeah, it doesn't round up higher than it should, so 15 .14, 5, etc, that's close enough. Yeah, so recall now, the distance formula in 3D and 2D that I use, basically the magnitude between two points right here is going to be, well, the difference between them, but in this case we're starting from origin, so that's going to be 0, 0, these all just vanish to 0, and so on, so basically you square the difference and the component distance, and then you, yeah, you square that, and then square root the whole thing, and you get the distance. That's going to be the, and then same thing with the velocity, the magnitude is the speed. Alright, so now let's take a look at the solution to Part B, so if we scroll back up. So part B says, find the angle between the path of the ball and the vertical line drawn through the point of impact C figure below. So it falls like this, but notice right here, so wants to know what this angle is. Well, this angle, this is actually a bit misleading, it should have been like, because it's going to go like this. but then the actual angle is going to be actually, would it be more vertical? I'm assuming this would be off it like that, yeah, just because, for example, this angle is less steep, yeah. So this is a bit misleading there. It should be like something like that. And we can see what the angle is because... It's just going to be the slope of the derivative, or just, yeah, it's just a derivative. So, a solution to be the slope of the curve when, basically when T is equal to, so when, or at T is equal to the impact, which is square root 7 over G. Okay, I've got to put a big square root here. I'm gonna, yeah, I'll just put a big square root, 7 over G. Again, that's the impact site we had in part A. I'm just going to go back. up there, so at this point 7 over G, so this is the point right here, so that impact, if we look at the slope, this is going to be, so at here what we have is, well, DY over DX, that's the slope of the curve, and basically at impact, yeah, so what we're going to have is, let's say it's falling like this, and then there's the impact, it's not going to fall vertically down like that, it's going to fall at an angle like this. and then there's the impact, but then if we zoom into this ball there, I'm going to go like this. Yeah, so this is the table, but, yeah, there's the table, etc. All right, so that's the table, and erase this, so there's the table like that, there's the origin there. Yeah, so it falls like that, but then if we zoom into this, let's zoom in, what we end up having is, Yeah, we're going to have, basically, here's the impact, it's going to, I go at an angle, so I can be vertically like that. So now it's going to be, well, the slope's going to be, there's going to be set up here, D, Y, and this is DX, like that. And here, I just drew this right angle there, so DX over D, let me just move it over. All right, so then this is the D, D, Y, over D, that's the slope, and there's the angle Theta. I'll put this bigger, so there's a theta angle. And now since everything's in terms of T, well, we could take the derivative of D, well, of Y in terms of T. So this is the same thing as writing, we could just write this as DY over DT over DX over DX. So we basically take the derivative of it, and again, the D over the T's cancel, the same thing as writing DX. So basically the derivative, of y in terms of y, divided by derivative of x in terms of y. So what we end up having is, yeah, we're going to have, and what I'll do is I'm going to write here, D over DT, just for completeness, we'll just make it clear. And this is going to be now, the Y component is just the Y component of the equation we had here. So that's just this whole setup there, but then is going to be at t is equal to 7 over g. Yeah, so now take a derivative of this. I'm just going to put the whole thing in there, just so we know what we're dealing with. So it's going to be derivative of 3 .5 minus gt squared over 2. That's the Y component divided by D over DT of, and this is going to be well in terms of T. go all the way back up where the x component is the x one is just two so the x one is just two i mean two t two t is the x component so that's the yeah two t so it's going to be two t so that's in terms of t said to take derivative it's going to be now so this is we're going to have d y or d x is equal to uh here's going to be uh 3.5 just goes to zero is constant we're going to have put the two down we're going to have, and then it cancels with this, we're going to have negative GT over and then 2. All right, and now we plug in this inside here. So we're going to get now negative g times it by square root 7g and over 2. Let's put this over 2 like that. All right, and then likewise as before, we just take this, put this square, and square, so we're not changing anything, just so we could bring this inside, and then we're going to bring that inside, those will cancel, so we're going to be left, but this is going to equal to, well, negative, and then negative is going to be, yeah, put this whole thing like that, negative, yeah, then put this inside, the squares cancel, we're just going to be left with 7g over 2. So that's the slope. Yeah, so this is our DY over DX. I'm just going to box this, but then now what this means is, yeah, so what this means is, well, now this DY or DX, so I'll make this bigger. So we're going to have now, this is going to be like this, and it's going downhill, hence the negative. So that's why, yeah, that's why it's negative like that. This is going to be, like this is Theta, it's going to be 7G going downhill, and now we're going to have, this is going to be DX. is just going to be, yeah, for, yeah, this is 7g over 2. I'll put this, um, yeah,…