00:01
Okay, so we're doing in chapter 9 problem 21 here.
00:04
So we have this 224 kilogram projectile, and it was fired with the speed of 116 meters per second, at an angle 60 degrees with the vertical or horizontal.
00:17
And it breaks into three pieces of equal mass.
00:21
So we have this into three pieces of one third mass each.
00:30
Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion.
00:40
One of these moves vertically downward and the other horizontally.
00:44
And we want to determine the velocity of the third fragment.
00:49
Okay.
00:50
So for the initial projectile motion, we should know that the horizontal velocity in the x direction is constant because the only force acting is just gravity, and that's all in the y direction.
01:05
So the velocity at the highest point, right when the explosion happens, is the entire velocity, it's all horizontal at that point.
01:16
So that means the entire velocity there is just vx times, or vx equals the initial speed times cos theta.
01:26
Because that's just the x component of the initial velocity and the initial velocity.
01:31
And the horizontal component stays constant.
01:37
We know that the momentum before and after the explosion needs to be conserved.
01:44
So we know that p before equals p after.
01:51
Okay.
01:52
So if we let v3 be the velocity of the third fragment, this is what we're trying to find.
02:00
Okay.
02:01
So let's start writing out our momentum equation.
02:05
So initially, we know right before we are only moving horizontally with this as our speed.
02:12
So the momentum at that point is given as p before, and this is m v .0, cost, theta, and this is in the x direction of momentum.
02:26
So this is in the i -hat direction.
02:28
So this is our p before.
02:30
So now let's start writing out what our p after is.
02:35
So we have one fragment that says moving at the same.
02:38
Speed with one third mass.
02:41
So that's one third mass, oops, times v0 times cos theta.
02:51
And we have two of these.
02:52
It says two of the projectiles are moving with the same speed, but one's moving horizontally and one's moving vertically downward.
03:01
So for the horizontal one, this is in the i -th direction.
03:06
And now we can add another one to the same mass and speed, but they said this is moving vertically downward, so that's the negative j hat direction.
03:19
So this is two of our three components, and our last one is just one -third mass times whatever this v -3 velocity vector is.
03:28
So now that we have this equation, we can just solve for what v -3 is.
03:33
So if we rearrange things and solve for v3, we get 2v -0 -cos theta.
03:42
This is in the ihat direction and we have v not cos theta in the j hat direction.
03:55
So now we can plug stuff in and figure out what this is.
03:58
So this is 2 times 116 meters per second times coast 60 degrees in the ihat plus 116 meters per second times coast of 60 degrees in the j hat direction.
04:18
Awesome...