https://youtu.be/-qPREtcLwsQ
PHYS 1101: Lecture Five, Part Eleven
We’re having this problem of a log that’s floating on water and a stone is dropped from rest, from a 75-meter high bridge, and it lands on the log as the log passes under the bridge. If the log moves with a constant speed of 5.0 meters per second, what’s the horizontal distance between the log and the bridge when the stone is released?
Okay, first point here. These problem solving steps in these equations can only apply to a single object. In this problem we have clearly two objects that are moving. The stone is falling while the log is moving along, left to right. We’re going to have to use these problem solving steps for each of these objects. They’re each going to have their set of variables and their set of equations.
The key to these two sets of equations is that the time, the variable t that shows up in both sets, is the same because this motion is happening at the same time for both of these objects. Starting out, the problem very similar to what we’ve done before, but we’re just going to have to sketch the motion and think about these variables for each of these two objects.
So off to the right-hand side here, I’m going to sketch my two objects. Let me first sketch the log. I’m going to sketch it down low, and the motion diagram for the log I know is just constant velocity. It’s moving along at a steady pace. Here’s the log at the start, and here’s going to be my log at the end. I’m adding some velocity vectors here. Right away I have a nice visual that my a from my log is 0.
I’m going to do log variables in pink to match the color of my log just to try to keep it clear and separate. Let’s use orange for the stone. Let me add to the sketch of the picture to what’s happening to the stone during the same time interval. It starts out up high at the same instant that the log is to the left. As that log is drifting, floating along to the right, the stone is falling. It’s in free fall. It starts from rest, but I know that it’s accelerating under the influence of gravity. And so its motion diagram looks like that.
Let me sketch in some velocity vectors for the stone. What’s a for the stone? It has to be my downward vector of -9.8 meters per second squared. And by choosing the negative there, I by default have already decided that my axis . . . I’ll take this back here for just a second. Let me tidy this up a minute because I want these to line up a little bit better than they are. Okay, I just tidied that up a little bit for us.
Before I draw my axis, I realized I wanted to be sure and sketch my rock at the end of the problem too. That stone lands on the log at this instant as the log floated to the right and the stone has undergone this type of motion falling.
So let’s draw our axis. I’m going to draw them in black fairly thick, and our x-axis, I’m going to pick the standard positive to the right. For the y-axis I’m going to pick the standard positive up. And I didn’t have to make this choice, but my inclination was to just draw the origin right there where the two cross, where the stone and the log are going to meet at the end of the problem. This location was correspond to the (0, 0) values for x and y.
Having drawn my axes, I’ve got each of these objects sketched at the start and the finish, start and the finish. Let me go in and think about these variables now for each one. Grabbing my pink pen, I know for the log what’s the t0, time, position, and velocity. At the start, what’s the time, position, and velocity at the end and what’s the acceleration?
For my other object, my orange stone, at the start, what’s the time, position, and velocity initially and then at the end here what’s my — let me drawn down below here — my time, my position, and my velocity for the stone right at the end?
Okay, actually I’ve realized I’ve kind of skipped over a step. Let’s ask ourselves if it would help to sketch in more velocity vectors here. My motion diagram does a good job of that, I think, for the log, and maybe I’m okay here for the stone, too. Right before that stone hits the log is the ending scope for my stone, and it will have still a large velocity here right before it makes contact.
My initial velocity, I know, is 0 for the stone. Okay, let’s go through the variables now for each one, and see what values we know or don’t know. We’re going to start the clock for both of them at 0. We have to do that to use our equations. Do we know the initial coordinate of the log when it’s right here? No, in fact, that’s what I want.
What’s the horizontal distance between the log and the bridge when the stone is released? Where the stone is released is the bridge. This distance, this is what I want. So if I were to know this coordinate relative to my origin, 0, I would have a value then, that distance. So this is what I want. This is the variable that I’m going to have to focus on.
Do I know the initial velocity? Let’ see. The log moves with a constant speed of 5.0 meters per second. Constant speed means at the beginning it’s positive 5.0 meters per second for the velocity because it’s to the right, and that’s actually what it has to be at the end, too, because the acceleration is 0.
Do I know the final time? No, that’s the time from when the clock starts that it takes the log to float to this location, and the same time that it takes the stone to fall this distance. I don’t know it. Do I know the final x-coordinate? Yes, it’s 0. That’s where I’ve put my origin. That’s where the log will be at the end of the problem.
Grabbing my orange pen, let’s go through the same exercise for the stone. Do I know my initial y-coordinate? Yeah, I do. The bridge is 75 meters high. At this initial instant it will be up here at this location. That’s coming to be at +75, that coordinate, because here’s my origin. My initial velocity is 0 I concluded. I’ve already also concluded that my final time, I don’t know.
Do I know my final y-coordinate? Yeah, it’s 0 because it’s also at the origin. Do I know my final y velocity? No, I don’t.
Okay, that’s an excellent sketch. I’ve thought through all of my variables, even the acceleration here for my stone. I have that written down. I’m ready to think about the equations and to solve them.
I often work these things from thinking about what I want and work backwards. What is it I want? I’ve decided I need to figure out what the variable of the value x0 is. So I need to look at my set of equations for the log and try to extract or figure out what x0 is. Then based on what I need in order to figure that out, I’m going to look at the other parts of the problem and see what’s going to help me out.
To do this though, and so we be sure we keep it straight, I have to have two sets of equations. On the left I’m going to do my equations for the log, and on the right I’m going to work with my set of equations for the stone.
Let’s think about what these equations are going to look like. Let me just copy it down here. I’m going to copy it over here for the stone, too. Okay, for the log I have 0 for the acceleration, so all terms that involve a drop out. This third equation as soon as you don’t have a really doesn’t have much meaning for you. It just confirms what you already know, that the final velocity equals the initial, so that’s not very helpful. And you see that same thing with the first equation. It also becomes not very useful when a is 0.
Equation two is what we have to work with so we’re going to have to make this equation work out for us. What do we need to make it work? What variables here do we know? We want this, and as soon as we have values for everything in the equation we’re set.
We know the coordinate at the end. We do know the velocity of the log, but we don’t know the time. How can we get the time? We’ve exhausted our equations and, therefore, our information from the log at this point that we can use. Whenever you get stuck like this, I suggest that you start thinking about the other motion that’s going on at the same time and see if it can help you get where you need to go.
Can we get the value of t from the stone’s information? Okay, for t, let’s see what we could use. Equation three doesn’t have time in it at all. That’s not going to be useful to us. Let me highlight, again, what we’re trying to focus on. We need a value for t. What else do we know? Of course, these become the variable y and y0 for my stone just because I am working with vertical motion.
I know acceleration. I know the initial velocity, and I know the two position coordinates. So, in fact, equation two looks great. Equation one is simpler. I know v0, but I don’t know the final v, so I can’t use equation one, so I’m going to have to go with equation two. So use two to get t. Let’s rewrite it, y equals y0 plus v0t plus one-half at squared.
I’m going to kill off my 0s first. I know my initial velocity is 0 for the stone. Which coordinate is 0? I put my origin at the bottom here so the final coordinate is 0. I need to do my algebra now to solve for t. This is the same algebra we need to do that we did in the last problem. I’m going to subtract y0 from both sides. That will remove that from the right-hand side and move it to the left. So I have –y0 is equal to one-half at squared.
I’m going to multiple by 2, and then I’m going to divide by a. And eventually then I’m going to take the square root. I have -2y0 divided by a is equaled to t squared, and I square root both sides. I end up with t being equal to the square root of minus 2y0 over a.
Let me keep going with my line down the middle of the page here because I’m almost done, at least, for getting a number for t. I’ve got y0 and a. t has to be equal to the square root of minus 2 times — what was the height of my bridge — I can’t remember. Ah, 75 meters. That was my initial y-coordinate, and I have to divide by minus 9.8 meters per second squared. We’ve been through the units. We know that those work out. I’m going to end up with seconds, and for a number I get 3.912 seconds.
As a rule of thumb when you’re doing intermediate calculations like this to get to our final answer, keep 4 digits and the end round to 3. If you do that when you type it into WebAssign, you’re most likely not going to get dinged for a rounding error. It should work out okay for you.
Okay, now that I have time I’m in good shape to go back to my equation for the log, and now I can work it through and solve for x0. So now I’m going to erase this question mark and put a little red check over t. I’ve got that variable now. Let’s do some algebra to isolate x0. I’m going to subtract v0t from both sides. I’m moving that to the left side, and now I just have to rewrite my equation. I’m going to flip it so that it reads left to right. x0, the thing that I want, is equal to x minus v0t.
I’m ready to plug in numbers. My final x-coordinate was 0 minus my initial velocity, which is my final velocity. What was that again? I can’t remember. Ah, +5.0 meters per second and then I need to multiply that by the time, 3.912 seconds.
Units are looking good. The seconds cancel. I’m left with meters, which is what a distance or a position coordinate should be in. When I multiply this out, I have negative 5 times that, and that gets me 19.6 meters.
Okay, let me remind you this negative sign here for this variable, a position coordinate just tells me the location relative to the origin. I’m to the left of the origin. Does that match the picture that we drew?
Ah, it does. At the initial position of the log, it was to the left of the origin. That’s the negative direction because to the right is what I called positive. So this initial coordinate is the minus 19.6 meters. The answer, though, that we want to report was the distance that that log floats from when the stone is dropped until it hits the log. So our final answer to type in is just going to be a positive 19.6 meters. It’s just the distance, so we’re using this variable that’s the coordinate information to tell us what the distance is.
And that brings us to the end of Lecture 5. I encourage you to look at these problem solving steps. Think about them as you work your homework problems. Go back over these examples to help you internalize and really understand what each of those steps is asking you to do. It’s guiding you through this problem solving process. There you go.