https://youtu.be/pVvK4wze13I
PHYS 1101: Lecture Nineteen, Part Three
Okay, this next example is going to lead in very nicely to our last subject for a lecture, which is the idea of an elastic collision. What does that mean? Let’s start out with just the tools that we know about. Let’s start out trying to solve this problem with what we already know. Because again for any abrupt event you usually can easily justify applying momentum conservation between these two objects involved in the abrupt event. So let’s start out in that direction.
Here we have a point 1 kilogram cart traveling in the positive X direction at 10 meters per second. And it collides with a larger cart. I’m going to draw my before and after pictures. Here’s before, cart 1 is coming in moving to the right at 10 meters per second. This is object 1. Here’s object 2. It initially is at rest. So it’s V20 is 0. Let me label this carefully this is V10.
That’s just before the collision just after the collision which is elastic we’re going to investigate that in a minute. I need to figure out the velocity of the point 1 kilogram cart. Now these two collide and I don’t know what happens to them. One could bounce back. Likely cart number 2 is going to start moving forward because of the collision. I’m going to draw a little picture of the two after.
So now I’m going to give cart 2 some final velocity which I don’t know. And I don’t even know the direction of cart 1. Let me just draw it. I’m just going to assume the thing bounces back. I have no idea. When we solve for the value of V1 final, if that number turns out to be negative, then we’ll know this velocity is to the left. If it’s positive, we’ll know the velocity is to the right.
So off to the right here I’m going to do my knowns. I’ve got M1 is .1 kilograms M2 is .3 V10 is plus 10 meters per second. I’m going to go I’ll jot it here to the left with the indication of positive direction is to the right. I do know that V20 is 0.
What do I want? I want the velocity of the .1 kilogram cart that is V1 final. I want this variable here. I want to know what the consequences are for this incoming cart that bounces off of, or collides with cart 2.
It does want the velocity. I’m going to specify the direction at the end of the problem. In WebAssigned for example you’ll probably have to pull down a menu and choose to the right or to the left or something. But this is the variable I have to go after.
What equation am I going to use to focus on that variable or go after it? I’m going to apply momentum conservation. This means I’m going to do M1 V10 plus M2 V20 has to add up to M1 V1 final plus M2 V2 final. I know some of these are 0, I know this one is 0, I have a number for this, I have a number for that, but I don’t know two variables.
Here’s the variable I want to solve for but I don’t know a value for V2 final. I’ve got no more information in terms of numbers to bring into the problem so there’s more information I need. I need to . . . there’s got to be more assumptions that I can make. I don’t have enough information to solve the problem.
That’s where this idea of it being an elastic collision comes into play and it’s useful. What information is that telling me what is an elastic collision mean? Well here’s a little schematic to try to convey the essence of it to you.
Let’s do a separate little thought process here off to the side. Let’s say I’ve got a cart and very practically to this cart is attached a spring with a little bumper. This cart is coming in with some initial speed, it collides with this hard wall, and of course your intuition is right in that you picture this spring compressing, but then the spring wants to stretch back out again so it pushes off, and this cart is going to bounce back off. This compressing and then stretching back out of this spring is another conversion of kinetic energy to potential energy.
We’ll learn a little bit about that later in the class. But in terms of what you visually see for this collision, this cart comes in, as the spring compresses, it stops ever so briefly, but then it starts speeding up, and as it bounces off and comes back out. If this collision is what’s called perfectly elastic, then the incoming speed is equal to the final speed. Let me write that a couple of ways. The momentum coming in is equal to the negative of the momentum coming out. Negative just because of the direction is opposite.
This means if the velocity coming in is to the right with a certain speed, then the final speed will be the same. The velocity is just directed to the left. So the speeds are the same. If the speed coming in and out is the same, then I know that the kinetic energy is the same on both sides.
That’s the meaning of elastic collision kinetic energy is conserved. The speed is conserved. The momentum is conserved in general though for a collection, a group of objects when it’s more than just one object bouncing off something. The total kinetic before has to equal to total kinetic after. You don’t lose any energy in this collision.
So that concept here I’ve got highlighted in red. When you see the words elastic collision and it’ll specify it that clearly it means that the total kinetic energy is conserved. So add up the kinetic of both objects before the collision. And you have to get the same number as adding up the kinetic energy after that collision.
Okay, so here is that stated in a mathematical equation when you go to write it out. Here is the kinetic energy of object 1 and object 2 before. Here’s the kinetic energy of object 1 and object 2 after. And this equation says that the sum before and after have to be the same. They have to be equal.
You get to use this idea of energy conservation only if it’s an elastic collision. If it’s any kind of collision, 99% of the time you can justify that momentum is conserved. To be 100% confident just ask yourself is your dominant force involved during this collision only the force between these two objects that they each exert on each other. If the answer is yes, in that direction of the collision, the direction of those forces, you can apply momentum conservation.
So momentum conservation alone can get you really far in solutions to problems. As you’ve seen, we’ve done a few examples now. If it’s an elastic collision though with even less initial information, you can determine what some final speeds are, what the initial speed are. So this gives you an extra tool to use if it’s this particular kind of a collision.
Here’s the rub though. I don’t know if rubs the right word for it, but here’s what you have to be careful of if you set out and you just write down as two starting equations for this collision momentum and energy conservation and you try to do your algebra, you try to perhaps combine these two equations to solve for the variable you want, you soon run into an algebra nightmare. It becomes really complicated with all these terms and these speeds being squared. If you don’t simplify it in just the right way, you just end up with more of a mess than you started out with.
So people have worked with these equations and they have already derived for us a smart way of algebraically simplifying this that captures the essence of both energy and momentum conservation. So instead of using and starting out with this basic equation to capture energy conservation you can use these two equations. Okay these two equations tell you what the final speed is. Sorry, the final velocity for object 1 as it depends on the mass of the two objects and the incoming velocity of object 1. The second one here tells you what the final velocity is of object 2 depending on what the masses are and again the incoming velocity.
Very important point though, these equations are derived only for the special case where the second object M2 was just sitting there minding its own business when object 1 plowed into it. Okay let me emphasize that. I have to have object 2 being at rest in order to use those equations I just highlighted. And I have that for this problem. Here’s object 2 so I’ve got to . . . it’s the object that’s at rest so I’ve got to call it object 2. It has a V20 that’s equal to 0. It’s an elastic collision therefore I can use these two equations. Notice that V20 doesn’t appear anywhere in these equations and that’s because it’s assumed that object 2 is initially at rest.
Okay, so any elastic collision that you’re given will only give you a problem where one of the objects is initially at rest. You have to make that object number 2 if you’re going to use these equations. Okay, we’re going to get back to that problem in just a minute and apply these equations to getting our answer.
In the meantime though I want to say something about an elastic collision, an inelastic collision. It’s very few collisions are perfectly elastic, meaning that kinetic energy is exactly conserved. In real life all collisions usually fall somewhere between perfectly elastic and what’s called perfectly inelastic or perfectly non-elastic.
Two objects collide, they come and they collide and they stick together as a result of this collision. That’s called a perfectly inelastic collision. The bullet in the block, the bullet block were stuck together after the collision perfectly inelastic. A lot of this incoming kinetic energy was lost and it goes into heat. It goes into friction. In the case of the bullet hitting the block that energy goes into breaking up and splintering the wood.
Billiard balls are probably the best example I can think of, of a nearly perfect elastic collision, right? Billiard balls, really, if you hit a cue ball straight into an object ball, if you don’t put any spin on that ball, if you hit it dead straight, the cue ball will come to a stop and the object ball will take off with the same speed.
Okay, real life is usually somewhere in between so if it’s not a perfect elastic collision then it’s inelastic. If it’s inelastic collision, the only tool you get to use is momentum conservation and here it is highlighted. Okay quiz question 7 before we get back to solving our example with those carts. Remember the apple example that we did in Lecture 18, what kind of collision was that?
Okay now back to our problem. I’ve copied the pictures here for us of just before just after the collision with my two carts. I was after remember the velocity of cart 1, after the collision V1F, and I had started out applying momentum conservation. The total momentum at the beginning equals total momentum just after.
I’ve copied here to the side the picture or the equations that I also can use because this is an elastic collision. I want to solve for V1f so my eye is drawn to this top equation. And I see that yes in fact it will work for me. I do know the masses of each of these objects and I know the speed of the incoming cart. Remember very important in order to use these two equations, it has to be an elastic collision with one of the objects that you label number 2 initially at rest. It has to be or else these are not logically consistent or valid. These were derived based on that assumption.
Okay, so let’s go in and use then that top equation directly. If it satisfies my criteria for using it, I can use it. M1 minus M2 over the sum times my incoming velocity. Important point this is a velocity equation so the sign is important. If V01 is positive that means I’m assuming velocity is to the right or positive. That’s how I drew it this object moving to the right.
So let’s plug in some numbers here and see what we get. M1 was the light cart. It was .1 kilograms and then I have to subtract in M2 which is .3. And then I have to divide by the sum which is going to be .4 kilograms. And that’s then times plus 10 meters per second. Well .1 minus .3 is minus .2 kilograms divided by 0.4 times my 10 meters per second. And I see that I can simplify this to ½ right .2 divided by .4 is going to be ½. I end up with minus ½ of 10 meters per second. Which is minus 5 meters per second.
So cart number 1 it recoils from this collision and bounces back, not with its full speed, but bounces back with half of its speed and it’s now headed to the left. Cart 2 has now picked up some velocity to the right. We could calculate that using this equation. I would put in 10 meters per second here, 2 times .1 kilograms over .4 would tell me what the consequence is for cart 2.