00:01
Problem 8 .67.
00:03
We have two blocks moving towards each other on a frictionless table.
00:08
They have these spring bumper things that make their collisions perfectly elastic.
00:21
What we want to know is what is the maximum potential energy that gets stored into springs.
00:27
And also we want to know what is the final speed of each block and direction.
00:38
So the net momentum of this system is not zero.
00:55
So when the blocks are in contact with each other, they're still going to be moving.
01:02
So the potential energy is going to be the kinetic energy at the beginning, minus the kinetic energy that the system still has while the blocks are moving.
01:26
Now we need to know how fast they're going.
01:32
So when they're so the center of mass of this is moving at some speed and when they're stuck together they're going to be moving at that speed basically.
01:48
So and you can see how if you rearrange what i'm about to write that you could get there from conservation of momentum just as easily.
02:08
So yeah the initial momentum of a plus the initial momentum b divided by the sum of their masses.
02:29
So you can think of this either as conservation of momentum or as asking what is the center of mass speed.
02:42
They're the equivalent questions.
02:47
In any event, this works out to be negative 112 meter per second or 0 .0 .0.
02:57
08 -33.
03:03
Pretend that looks like a 3 at the end there.
03:08
It's not really something we're looking for, but yeah.
03:12
So we need that.
03:17
And then umax is just a simple matter of plugging in all of our kinetic energies.
03:28
It's going to take the one half out front so we don't have to keep writing it.
03:36
May be a squared plus mb b b b initial squared squared minus their kinetic energy when they're moving together which is an object with the combined mass of the two moving at this center of mass speed and so if you put the numbers into that you get 5 .21 joules now, there's an equation in the book, direct, it gives you the speed of each object when you have a collision like this, or one of them is stationary, but when both are moving, we need to go, we need to start with this equation that relates the relative speeds before and after the collision, and of course the conservation of momentum.
05:24
So m -a -v -a -final plus m -b -b -b -final equals m -a -v -b -b -i -nual, now you have two equations and two unknowns, and it's not terribly, difficult although it is rather tedious to work through solving all of these so i'm not going to walk through all of that here i don't think it is it is good practice so you know you might as well do it at least once one of those sort of things but here we've got the we end up with the final speed of a being equal actually let me write this in a different order yeah so we have the difference of mass between a and b times a's initial speed plus twice b's initial momentum and this is just divided by the total mass yeah and then just to make sure what i did when i was working this out is actually you you can solve this for vbf in terms of b a f put that in here for vbf solve for vaaf and then let's see you have an expression for vaa if you can put that back in here and get vbf and you know, i've done it that way but also what you can do and i did it that way just to check to make sure that you get the same result is notice that there's not really anything special about a and b so if you swap a for b everywhere in this equation you get the equation for vbf, so so let's go ahead and write that down just sort of by i...