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Anderson Video - Weighing Lunch

Professor Anderson
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 >> Hello, class. Professor Anderson here. Let's try a problem where we're going to drop a mass onto a spring, and it's going to compress the spring, and we'd like to know how far that spring is going to get compressed. And let's do a completely inelastic collision between mass 1 and mass 2, which is attached to the spring. So here's our spring. We put a box of mass 2 on the end of the spring, and now we're going to drop mass 1 from a height h, where h is the height above mass 2. And now they're going to have a perfectly inelastic collision, which means we can't conserve energy, but we can, of course, conserve momentum. And so immediately after they collide, we have this sort of picture -- m1 connected to m2. Spring underneath them. And this whole thing is starting to move. And if we call this speed here -- let's make this arrow a little bit longer -- if we call this speed v1, the speed of the box m1 right before it hits m2, and we call this v2, then we can use conservation of momentum here to solve for v2. All right, so what happens in that collision is things get deformed. They stick together. However it happens, we're saying it's completely inelastic. So momentum initially has to be equal to momentum finally, where initially, we mean just before the collision, and finally, we mean just after the collision. So it's all box number one moving, so it's m1v1. Afterwards, it is m1 and m2 moving together at a speed v2. And so conservation of momentum tells us that. And let's go one more step. We know from freefall that the speed v1 is just going to be the square root of 2gh. So if you drop an object from a height h, that's the speed when it's at the bottom of that distance. And we can rewrite this and now solve for v2. So what do we get? We get v2 is equal to m1 over (m1 plus m2), all that times 2gh. All right. Let's now take it a step further and see if we can figure out what the amplitude of oscillation of this spring system is, this block spring system. So our initial picture now looks like this. We have the two blocks right on top of each other, and they are heading down at v2. Now, as soon as I put m1 on top of m2, there is in fact a new equilibrium position for this spring. And that equilibrium position is a little bit lower than it was before, and we'll say that it is a distance L lower than it was before. The spring system will get compressed, and the box will go down to a distance a. And now it's going to go up and down around that new equilibrium position L, and it will go from a on this side to a on that side of the new equilibrium position. So let's take a look at conservation of energy here and see what happens. So one thing that we said was on a vertical spring system, the new equilibrium position is going to be a little bit lower, and if we measure everything relative to that new equilibrium position, we can ignore gravity, all right. So what do we have? Initially in that picture, we have kinetic energy. We have two masses, m1 and m2, and they are moving at a speed v2. But in that picture, we also have a little bit of stretch to the spring, and that amount of stretch is 1/2, the amount of stretch is L, and so the energy is 1/2kL squared. Now, when it gets down to the bottom, it's all spring potential energy, and so it's 1/2ka squared. And now we're almost there, but we need to know something about this new equilibrium position L. And how do I figure out what that L is? Well, by adding this mass m1, we've added a little bit more force down due to gravity, and so we have m1g going down, and the spring is pushing up with kL. And so we get kL is equal to m1g. And now we can solve this for L and we can stick it into our equation. So what has our equation become? We have 1/2 (m1 plus m2) times v2 squared. And if we remember from just a minute ago, v2 was equal to m1 over (m1 plus m2), and we're going to square all that. And then we had a square root of 2gh that we're going to square, so that just becomes 2gh. We have 1/2kL squared. There is our L, so we get k times m1g over k, quantity squared. And all of that is equal to 1/2ka squared. And now the task is to solve this equation for the amplitude capital A. So I don't want to take away all your fun in doing the math and solving this for capital A, but I will tell you what I got so you can double-check. So I got the following. I got (2gh over k) (m1 squared over (m1 plus m2)) plus (m1g over k) squared, and then that entire thing square rooted. Now, let me give you a few numbers just as an example to try out, and you can try it with your own numbers. Let's say that m1 is a tenth of a kilogram, m2 is 4/10 of a kilogram. The height h is 25 centimeters. And the spring constant k is 200 newtons per meter. If you try all those numbers in this equation, you should end up with amplitude a is equal to 0.023 meters, okay. Try that with these numbers, and make sure you get the same answer, and then you can try it with your own numbers. All right? Hopefully that's clear. If not, come see me in my office. Cheers.