https://youtu.be/-ItJqm5QKX8
PHYS 1101: Lecture Eighteen, Part Three
Now let’s move on to that second tool. This tool is really great for when you don’t have enough information or you don’t care or need to know about the duration of the collision or the size of the forces in the collision. All you want to know is the end result. What’s the consequence of that collision on these objects involved?
Usually there are just going to be two objects, and what I’m going to show you is that the implications of forces causing momentum to change leads you to the conclusion that if you simply add up the momentum of both objects at the end, that’ll be the same as the total momentum of the objects at the beginning if there were no forces from the outside playing a role.
Said another way, during this collision if the only forces that are significant is this third-law pair, the action-reaction forces between these two objects, then you’ll be able to draw this conclusion.
Think of this as a new, separate starting equation for many problems for which this applies. Again, it’s an abrupt event that I want you to concentrate on or consider when you want to apply this tool, and this is for the case where we don’t care at all about the force or the time during the event. All we want to know about is the final consequence.
Here’s the logic of why this is true, and that following through will help us see or appreciate what we need to plug into that equation to have it work for us.
Consider a scenario where I just have two objects, and they’re about to collide. Here’s a snapshot of them before. They’re going to collide. Say these are two billiard balls. And then after the collision they head off in different directions.
Now Newton’s Third Law tells us that during this collision they each experience a force that’s equal and opposite to the one that the other experiences. There’s a third law action-reaction pair between these objects when they’re in contact.
The quiz questions you thought through above hopefully had you appreciating the fact that the impulse that this object feels or undergoes because of this collision has to be the same magnitude, opposite direction, that this impulse feels because the forces are a third-law pair. If the forces are equal and opposite between these two objects–the collision of course lasts the same time for the two, they’re both in contact for the same amount of time–then the product is going to have the same magnitude for both but negative direction. The impulse is same magnitude, opposite direction.
Remember the impulse is nothing more than a shorthand way of saying the change in momentum, pfinal – pinitial. So we come to this necessary conclusion if during this collision the only interaction I had, the only forces that are significant are between these objects, then whatever momentum change one object feels it has to be at the expense or equivalent to the opposite-sign same-amount-of-magnitude change that the other one experiences. This has to be true.
Let me write this equation out here again for you, and let’s just rearrange it ever so slightly. What’s delta p1 again? That was p1f minus p1i. That change has to be equal to the negative of p2f minus p2i.
If I multiply through by this negative sign, all I want to do now is just collect all of these terms so that I end up with the momentum at an instant, for example at the end or at the beginning, being on the same side of the equation.
So let’s move this final momentum to the left by adding it and let’s move this initial momentum to the right by adding it to both sides. So I’m going to end up with p1f + p2f is equal to the sum of the initial momentums, p10 + p20. Adding up everybody’s momentum at the end, I better add up to the same number as adding up all the momentum at the beginning.
Some of these might be negative, meaning they could be headed to the left versus the right, but in the end the total sum can’t change. This is only a consequence. It logically has to follow given that the only force was between these objects and they have to experience equal and opposite forces. That puts a constraint then on how the momentum can change. It’s not arbitrary.
If one loses some, the other object has to gain that momentum. So here is this equation translated into words. Let’s read it. The left side of the equation, it says one object’s final momentum plus the other one’s final momentum, that’s the total final momentum, has to equal the equivalent to the total initial momentum, the initial sum of the ps for the two objects, and what I’ve got in red points out the only assumption we made in that analysis was that the force was between these objects that caused the momentum to change, and that’s always going to be true if there are no outside forces, if you will.
There’s nothing outside that’s in contact with one of these objects or doing something that in itself is going to change the momentum. All I have are these third-law pairs in this interaction between the two objects.
Here is the bare useful logical consequence of that statement, or I’ll call it a corollary. Because of the vector nature of that equation, this translates to if I have a scenario where there are no outside horizontal forces then my horizontal momentum has to be conserved. If there are no outside vertical forces, then my vertical momentum has to be conserved.
We’re going to see some problems where I may have forces vertically so I can’t apply vertical momentum conservation, but perhaps for that problem horizontally there’s nothing going on except this collision in terms of horizontal forces. In that case, I can apply my momentum conservation equation to the horizontal momentum initially and the momentum finally. This as an aside, let me point out, is what call conservation of momentum.
Notice by being able to use this equation now I’ve been able to completely bypass all details of the collision. I don’t care what the average forces are between these two objects. I don’t care how long the collision lasted. All I want to know is given an object came in with a certain velocity, perhaps the object was at rest, where did they end up or what were their final velocities?
So this is how you often will see it written. You’ll see these capital ps, so momentum conservation is just that the momentum doesn’t change. The total momentum at the beginning is equal to the total at the end.
The immediate useful form to write that in is what I’ve got shown here. The only assumptions I’ve made is that these are just two objects that are interacting along a line. Each of these ps is just the momentum of that particular object, object two for example, and its velocity is at the final instant. The only perhaps tricky part to using this very powerful equation is only really just keeping track of what become kind of annoying little subscripts just so you can really sort out for the two objects what their momentum was each before and after the collision.
Okay, let’s apply that to an example. I’ll ask you some questions as we go along. Here’s our problem. We’ve got an 80-kilogram astronaut carrying a 20-kilogram toolkit, and she’s initially drifting toward a stationary space shuttle at a speed of 2 meters per second. If she throws the toolkit toward the shuttle with a speed of 6 meters per second as seen from the shuttle, what’s her final speed?
Great example of application of momentum conservation. Notice we have an abrupt event here, a throwing of the toolkit. It doesn’t ask us about the duration of the throw, the average force on the toolkit or on the astronaut during this throw. We just want to know the consequence, what’s her final speed?
So I’m going to apply momentum conservation. First question I ask myself is, what’s the group? What’s object one, object two? Do I have just two objects here? Do I need to consider three?
Think back to our derivation. What we’re focusing on is the two objects involved in the abrupt event. What two objects were in contact during this abrupt event? That’s your group.
Next question: is it true that during that event the only forces that you had were between these two objects? That would then mean that you have no outside forces, so you can apply momentum conservation to the problem. I can apply that equation p0t equals pft.
Okay, here’s my little cartoon of what’s going on. I have my space shuttle. I don’t know how to draw those, so I like my little house I’ve got drawn. I’m going to define an axis direction. I’m going to just make all this throwing and all these vectors happen horizontally, so I’m going to indicate positive direction to the right with a little axis that I’ve drawn.
And now I always want to draw a quick snapshot at the beginning and at the end. What are my objects? It’s the two that are involved in this abrupt event. That’s the astronaut and her toolkit. One of them I’m going to label one. The other one gets labeled two.
I’m told initially that both her and the toolkit are moving toward the space shuttle at 2 meters per second. This would mean my initial velocity is +2, my initial momentum is to the right.
My next picture represents the final snapshot, which is going to be the instant in time just after this abrupt event, just after she lets go of the toolbox. At that snapshot, what variables will I have?
She will have a final momentum p1f. She’s object one. The toolkit will have a momentum p2f. And what I’m going to apply to this problem is that the total momentum at the beginning has to add up to the total momentum at the end because only forces between these objects is relevant or occurs during the event. There’s nothing outside touching them or pushing or pulling on them, so momentum is conserved.
Here’s what it is written out. Let’s go in and put more specifically, write out what these are. It’s always the mass times then the velocity at that instant. This is object one. Then I have add m2v20, the velocity of object two at the start, and that better add up to m1 times its velocity at the end or after this abrupt event plus m2v2f. In the end, using this equation is pretty straightforward, but the notation does get cumbersome. You have to just slow down and take your time and be sure you’re representing the right velocities at the right instant for each of these objects.
Okay, next quiz question for you. Given the labels that I’ve chosen, we need to decide what variable we need to solve for. What do we need to go after? That’s your next question: pick which variable we need to focus on. Scroll back in the movie and reread the question if you need to. Let’s do our list of knowns.
We were given the mass of the astronaut and the toolkit, toolbox, as 80 kilograms and 20. We know that their initial velocities were both the same. She was holding onto the toolkit and they were to the right toward the space shuttle, so I’m making those both +2 meters per second.
And what I want is what? If you go back and read the problem, it asks for her final speed. Well, the equation that I’m going to work with has to do with momentum, and so it’s velocities that’ll show up in this equation, but if I pick the right velocity then the magnitude of it will be her speed.
I want her final velocity. That’s right in here, her final momentum. Her final momentum is her mass times her final velocity. So what do I want? I want v1f, and more specifically, I want the magnitude of this.
If I were to end up with a negative number, I would just use the positive value. The negative would just tell me she ended up heading to the left.
I just looked back. There’s one more known, and that is I’m told that the toolkit after the throw is headed toward the space shuttle with a velocity of 6 meters per second. That would be then the variable v2f is equal to +6 meters per second. See how careful you have to be with those subscripts? It’s easy to get, to make a little mistake.
Okay, we’re ready to work toward our solution, and of course we’re starting with our fundamental equation. This equation will ensure that we maintain conservation of momentum. What variables do we know in here? We know their masses, we know the final speed of the toolbox. We want her final velocity or her final speed, and we know the initial velocities were the same for both her and the toolbox.
I know everything except the thing that I want, so I’m ready to do my algebra. You can check me later, but the algebra ends up simplifying to this, and all I have to do now is plug numbers in.
When I work all that out, plug that into my calculator, I end up with her final velocity as being +1 meters per second. This tells me that she is still moving toward the space shuttle, fortunately. If she’d thrown it harder she might have even ended up moving away from the space shuttle. She slowed down, though. She was headed toward the shuttle with 2 meters per second velocity, and now it’s 1, and so her speed is going to be the 1 meter per second.