00:01
Okay, in this problem, we have a wagon with a person in it, and the person's holding a rock.
00:07
So i'll have the person.
00:09
You can sort of see their top half, and the person's holding a rock.
00:13
The total mass of the system is 95 kilograms, and they're moving to the horizontally, we'll call it to the right, with a speed of 0 .5 meters per second.
00:29
Then the person throws the rock.
00:31
In one case, the person throws the rack forward, and in one case, the person throws the rock backwards.
00:36
So we'll consider the forward throw first.
00:39
If the velocity of the rock is 16 meters per second, and the mass of the rock is 0 .3 kilograms, what is going to be the velocity of the wagon after the throw? so we're going to use momentum.
00:59
The initial momentum is equal to the final momentum, so there's no external forces on the system.
01:04
Initially, it's the total mass times the initial velocity.
01:09
So that's everything.
01:10
It's all one object, one system.
01:13
Then after the throw, we have the momentum of the rock.
01:19
And then the momentum of the wagon, which is the person and the wagon together.
01:24
I'll just call it m -wagon and v -wagon, but that includes the person.
01:28
So plugging in our values, we have 95 kilograms times 0 .5 meters per second.
01:39
Equals the mass of the rock, so 0 .3 kilograms, that should be 0 .300 times the velocity of the rock, plus the mass of the wagon.
01:52
So if the wagon and the person in the rock were 95 kilograms, now they are 94 .7, because we've lost a little bit of mass going through the rock.
02:03
And we don't know the speed of the wagon.
02:06
So 95 times 0 .5 is 47 .5 kilogram meters per second and that's equal to 4 .8 kilogram meter per second plus 94 .7 kilograms times the velocity of the wagon.
02:27
If we move our 4 .8 over to the side, subtracting it on both sides, we have 42 .7 kilogram meters per second is equal to 94 .7 kilograms times the velocity of the wagon, dividing both sides by the mass of the person in wagon...