00:01
In this question, we are pulling aside a pendulum bob that has the same length as a fellow pendulum, but only half the mass.
00:15
We're gonna pull aside to an angle theta.
00:17
They're gonna collide elastically, and we are gonna find the velocities of the bobs after that collision and how high they go.
00:27
So one thing we wanna note here is that m1, we were told, has half the mass of the one that's stationary, and i don't like dealing with fractions, so we're gonna rewrite it as 2m1 equals m2.
00:48
Let us start with that.
00:52
And now we're gonna have three phases of this question.
00:57
First of all, we have the pulling m1 aside and figuring out its velocity.
01:03
In fact, you know, let's write this down.
01:04
Let's record this.
01:05
Three phases.
01:08
Phase one, we have conservation of energy to get v1 before the collision.
01:27
And then second phase is going to be what goes on during the elastic collision.
01:41
And then the third phase of the question is going to be conservation of energy again to get the final heights.
01:59
And so actually, it'll be elastic collision to get those velocities, and then it'll be conservation of energy to get those final heights.
02:10
So one of the things we're gonna need to do is dive a little deeper into this pulling aside of mass m1 to an angle theta.
02:27
And i'm gonna just measure these initial positions from the top of the bob just for convenience.
02:36
So we're gonna lift that bob to a height h.
02:41
L is of course the hypotenuse over here.
02:44
And then l is also this total distance for the vertical direction.
02:50
But i'm gonna call this kind of extra distance between that top dashed line and the anchor point, i'm gonna call it d.
02:58
So that we can say that the length l is equal to d plus h.
03:02
Now, why did i do that? because the hypotenuse l and the side d and the angle theta make a right triangle.
03:13
So i can then express the height h in terms of l and theta.
03:19
And so in fact, i can say h is equal to l minus d, and d is gonna be equal to l cosine theta.
03:30
And so i can say that my height h is equal to l times the quantity one minus cosine theta.
03:39
And so now we can do a conservation of energy to figure out the speed v1 before m1 collides with m2.
03:48
So we have gravitational potential energy equals kinetic energy just before the collision.
03:57
So that's m1gh equals one half m1 v1 squared.
04:08
And we're gonna notice that m1 cancels.
04:14
And then i'm gonna multiply, well, i am gonna rearrange it also, but i'm gonna multiply both sides by two and then take the square root.
04:25
V1 equals the square root of 2gh.
04:28
So the square root of 2gl times the quantity of one minus cosine theta.
04:36
And what i'm gonna show you for the rest of the question will allow us not to have to write it that way until we get to the end of doing some, you know, other algebra and calculations.
04:49
And then at the very last step for each of these x2 parts, we can substitute in this square root phrase, just because, you know, it's just faster and easier and it keeps us from a lot of writing and a lot of extra effort that we kind of don't have to do.
05:07
Okay, so this elastic collision, we are gonna tackle, so by the way, in case it wasn't clear, this was phase one.
05:20
So now for this elastic collision, we have two things that are gonna happen.
05:25
One is conservation of kinetic energy, one half m1 times v1 squared plus one half m2 times v2 squared equals one half m1 v1 final squared plus one half m2 v2 final squared.
05:55
And then for conservation of momentum over here, we can say m1 v1, and we're gonna develop, so here's what we're gonna do, we're gonna develop a general relationship that we can use for honestly any elastic collision with any masses and any initial velocities.
06:16
And then we'll substitute in the zero for v2.
06:21
This has been derived this clearly in precisely one textbook that i've ever worked with in, you know, 25 years of being a student and a teacher and tutor.
06:35
And so that's why i'm gonna go ahead and show you how this works.
06:41
So m1 v1 plus m2 v2 equals m1 v1 final plus m2 v2 final.
06:50
So let's deal with this energy thing first.
06:55
We're gonna multiply, actually we don't have to multiply everything through by two.
06:58
Since there's a one half in every term, we are simply gonna cancel out the one half.
07:04
Like let's not deal with fractions.
07:07
The next thing i'm gonna do is collect my m1 terms on one side.
07:14
So m1 v1 squared minus m1 v1 final squared.
07:19
And then i'm gonna collect the m2 terms on the other side, m2 v2 final squared minus m2 v2 squared.
07:30
And let's factor out our masses on each side.
07:35
So m1 times the quantity v1 squared minus v1 final squared equals m2 times the quantity v2 final squared minus v2 squared.
07:49
And now we're gonna use a little piece of algebra here.
07:53
You know, when we have two equations with this many unknowns, one technique that often happens is dividing one equation into the other.
08:08
And so we wanna be able to get rid of these squares.
08:12
And we can do that legitimately, algebraically, remembering the fact that when you have a minus b times the quantity a plus b, that's how you get the difference of squares, right? the difference of two quantities squared is equal to the two quantities added multiplied by the two quantities subtracted.
08:35
So we're now setting ourselves up to be able to divide our momentum equation into our energy equation, which isn't something that you might think to do if you didn't have somebody telling you that it's possible.
08:48
So m1 times the quantity of v1 minus v1 final times the quantity v1 plus v1 final will be equal to m2 times the quantity v2 final minus, oops, it should be v2.
09:04
I'm gonna make that two final a little bit neater, times the quantity v2 final plus v2.
09:15
Now let's go rearrange our momentum equation similarly so that all of our m1 terms are on one side and all of our m2 terms are on the other.
09:31
So m1 v1 minus m1 v1 final equals m2 v2 final minus m2 v2.
09:38
Once again, similarly, we're gonna factor out those masses, m1 times the quantity v1 minus v1 final equals m2 times the quantity v2 final minus v2.
09:53
And ooh, now you might have noticed that this is starting to have some terms on both sides that look an awful lot like that energy equation.
10:04
So we're gonna divide one equation by the other.
10:08
Again, not a technique that you might think of if you didn't have someone telling you that this is a way that you can solve a system of equations.
10:20
So notice m1 cancels out with m1 and v1 minus v1 final cancels out with v1 minus v1 final...