00:01
So this time we're given a new problem.
00:05
It's, we have two particles.
00:08
Think about the x axis.
00:13
So this happens to be the origin.
00:16
And there's a particle at the origin.
00:18
And also the center of mass of the two particles.
00:31
So, you know, this is one particle and then another one is here.
00:36
Going to call this one, a, the one at the origin, where x equals to zero, that's the position.
00:44
And then the second one, i'm going to call it b.
00:52
This is going to be b at x equals to 8 .0 meters.
01:00
This is just zero meters.
01:02
And then this is the center of mass that happens to be at x equals to 2 .0 meters.
01:18
And the second particle, you know, the b particle happens to have the mass of 0 .10 kilograms.
01:29
And then the a particle, we don't know what the mass is for that one.
01:34
So that's the information we've given.
01:36
We're also told that the velocity of the system at the center of mass happens to be 5 .0.
01:48
Meters per second ihat, you know, that's the velocity of the system.
01:56
So in part a, we want to find the mass of the particle at the origin.
02:04
We want to find m .a.
02:07
And then in b, we want to find the total momentum, p total of the system.
02:17
You know, that's the momentum of the system.
02:19
And then in part c, we want to find the velocity of the particle at the origin.
02:27
So, va, we're calling it va.
02:30
This is the velocity of the particle at origin.
02:42
And so, you know, we're going to do a stepwise process.
02:46
Remember, we'll use the center of mass to resolve this problem.
02:51
The center of mass happens when we're going to call that this should be called xc so xc is the center of mass so xc equals to m a plus x a plus mb times xb so so just to make sure the labels are right, this is xb and this is xa and this is xc.
03:47
And so this is divided by m, ma plus mb.
03:59
So that's what we're given.
04:02
And so our goal is to solve for the mass of a.
04:07
We'll multiply both sides by this value.
04:10
So we get xc, ma plus mb.
04:16
Equals to m .a plus xa plus mb xb.
04:25
And so m .a, we use distributive property right there.
04:29
M .a .x .c.
04:31
Plus m .b .x .c.
04:34
Equals to m .a.
04:37
Plus x .a.
04:39
Plus m .b.
04:41
Xb.
04:42
And then our goal again is to find m .a.
04:46
So we'll move this m .a.
04:49
To this side.
04:52
And then we'll move this to the right.
04:53
So we have m .a.
04:56
X .c.
04:58
Plus, or rather minus, because it's moving to the left.
05:03
So, you know, this is going to be minus.
05:10
This will be minus.
05:12
M .a.
05:13
Equals to x a plus mb xb minus m b xc so we pull out the m a and we're left with xc minus xc minus 1 equals to xa plus mb x b minus xc those are the positions we have um so we're getting closer to solving for for for m so we'll just shift to the next page and make sure that we have our numbers right so m a xc minus 1 equals to x a plus a plus mb xb minus xc recall that xa is the dimension at the origin right here so that means it's zero so x a is zero so we're left with you know we could divide both sides by xc minus one over xc minus one and so we end up with ma being equal to mb xb minus xc over xc minus one and so if we go back we see that xb is 8 .0 meters.
07:00
Sorry, mb.
07:01
Mb is 0 .10 kilograms.
07:05
So we need to change that real quick.
07:10
We need to change this one.
07:13
This is m .b.
07:14
Happens to be 0 .0 .0.
07:24
0 .10 kilograms.
07:29
And then xb is 8.
07:33
Meters and then xc is 2 .0 meters and then xc is 2 .0 meters minus 1.
07:51
2 .0 meters minus 1 and so if we simplify this problem we get 0 .1 times 6 so we get 0 .6 the mass being 0 .6 kilograms so, ma equals to 0 .6 kilograms.
08:20
So we just want to go back and make sure that we had our answers right.
08:24
So we had, we had, we had m1, oh, sorry, m -a, x -a.
08:47
So that would have made it x -a -0.
09:01
So this one, yeah, so we need to change some things here.
09:10
So, you know, we had a plus.
09:11
It shouldn't be a plus.
09:13
It should be a multiplication instead of a plus...