00:01
So for part two of this problem, we have two equations.
00:05
We have the observed fractional energy loss.
00:10
So f observed, and this is equal to initial kinetic energy minus final kinetic energy divided by initial kinetic energy.
00:25
And we have f predicted, which is equal to the ratio of the masses of the two cards.
00:34
That we are using.
00:38
And so we have this system where we have one cart, we'll call that mass 1, going at some speed v, and it is crashing into some cart of mass m2, which is at rest at first.
00:59
And then, so this is before the crash, and then after the crash, we know that these two masses are now moving together.
01:15
So we have some mass m1 plus m2 and it is moving at some speed v.
01:22
F.
01:23
And so we want to show that the plot, we want to show the plot of f observed versus f predicted.
01:32
And we're assuming that our observations match the predictions of the theory.
01:37
And so we just want to show that f observed and f predicted are equal to one another and so first let's do this so we know that f observed is equal to initial kinetic energy so one half m1 v i squared and we are subtracting final kinetic energy and so that would be one half times m1 plus plus m2 times vf squared, and this is divided by initial kinetic energy.
02:18
So 1ā2, vi squared, right? so immediately what we can see is that we can get rid of these one -haves.
02:29
Those are gone.
02:31
We want to somehow get vf in terms of vi, and we can do that using conservation of momentum.
02:39
So we know that momentum initially will equal momentum the final momentum for our system.
02:49
And so this implies that we have m1, vi, because we don't care about the momentum of mass 2, because it has a speed of 0, is equal to the total mass of the system at the end, m1 plus m2 times vf.
03:10
And so we see that vf can be equal to this.
03:20
Vf can be equal.
03:21
To m1 divided by m1 plus m2 times initial velocity and so we can plug that in for vf so what we end up getting is this we have m1 v i squared minus m1 plus m2 times the square of this and so that gives us m1 divided by m1 plus m2 squared times v initial squared and that's all divided by m1 times v initial squared so now we can see we can get rid of these v initials because we have them in every term of our expression on the in the numerator and the denominator now we just want to simplify this right and so we can get rid of the m1 and the numerator and the numerator and the numberator and the exponent in the denominator.
04:35
And so now we want to simplify, though.
04:41
And so in order to simplify, we want to add up all the terms in the numerator, and to do that, we have to get this expression in the denominator of this expression.
04:56
So to do that, we have m1 times m1 plus m2, my.
05:04
Minus m1 squared divided by m1 plus m2.
05:12
And this is all the stuff in our numerator, but then don't forget in the denominator of our total expression, we have divided by m1...