Guys, in the last couple of videos, we saw how to use momentum conservation to solve these kinds of push-away problems when you have objects that are pushing away from each other. But in some of these problems, you're actually going to be asked for the energy before or after a push-away event. For example, in the example we are going to work on down here, we have two boxes that are pushed up against this spring. In part b, we want to calculate how much energy is stored inside that spring. So, I'm going to show you quickly how to deal with these push-away problems with energy. The main idea here is that you're actually going to use two conservation equations. We're going to have to use momentum. We're always going to use momentum conservation with these problems, but we're also going to have to go back and use energy conservation as well. Let's go ahead and take a look at our example here. We've got these two blocks, 3 kg and 4 kg, and they're pushed up against this light spring, and you release it, and the spring basically is going to fire both of them off in opposite directions. The blocks are going to get released. So, we want to draw a diagram for before and after. This is the before, and afterwards, what happens is you sort of have these boxes that start pushing away from each other like this. So this 4 kg box goes like this, and the 3 kg box is now going to the left. We actually know what the final velocity of this 3 kg block is. It's launched at 10 meters per second. So what happens is I'm going to call this \( m_1 \) and \( m_2 \), which means that \( v_{1,\text{final}} \) is equal to -10 because it's going to the left. And then \( v_{2,\text{final}} \) here is actually what we want to solve for in part a, the recoil speed of the 4 kg block. So this is what we're going to look for in part a and basically, we're just going to use momentum conservation. We know how to solve these kinds of push-away problems. So once we have our before and after, we're just going to write our momentum conservation. So for part a, we're going to do \( m_1 v_{1,\text{initial}} + m_2 v_{2,\text{initial}} = m_1 v_{1,\text{final}} + m_2 v_{2,\text{final}} \). Now remember, what happens in these kinds of problems is that initially, the system is at rest. So right after you release your hand, the blocks are still actually at rest which means that the initial velocity is equal to 0. So what that means is that basically both of our left terms are actually going to go away and we have 0 initial momentum. So the system's initial momentum is equal to 0, and that means that our momentum conservation equation is going to simplify because we know that the final momentum has to be 0 as well. So basically, what happens is we're going to have \( -m_1 v_{1,\text{final}} = m_2 v_{2,\text{final}} \). If the right block gains 10 momentum this way, the left block has to gain -10 momentum that way. And so, they have to cancel each other. So we can we can basically solve for this \( v_{2,\text{final}} \) here, and we can start plugging in values. So I have -3 times -10, which gives \( 30 \) when divided by 4, what you end up getting here is that \( v_{2,\text{final}} = 7.5 \) meters per second. So notice how we got a positive number, and that should make some sense. If we actually got a negative number, so if we forgot one of these minus signs, we would have gotten a negative number, and that would have meant that this block is actually going to the left and that doesn't make any sense. So it's always a good idea to keep track of your numbers and make sure they make sense. So this is a positive answer, positive 7.5, and that's just using momentum conservation. Let's take a look at part b now. Part b is asking us to figure out how much energy is stored inside the spring. So what's going on here is that you had these blocks that are pushed up against the spring, and there's some elastic potential energy that is stored inside the spring because they're compressed. Once you release, what happens is that these two blocks start firing away from each other. They separate like this, and the spring decompresses, which means that the final potential energy is equal to 0. So what happens is there's some stored energy here and then this spring is no longer storing that energy. Where did that energy go? Well, basically, what happens is that you have these two blocks that have now mass and speed and therefore, there's some kinetic energy. This is \( k_{2,\text{final}} \) and \( k_{1,\text{final}} \). So basically, we're going to use energy conservation to figure out the energy instead. So you're going to use \( k_{\text{initial}} + u_{\text{initial}} + \text{work done by non-conservative} = k_{\text{final}} + u_{\text{final}} \). Now what we said here is that the initial velocity of the system is 0, so there's no kinetic energy. The initial energy that's stored here is going to be elastic potential energy. There's no gravitational energy because everything's sort of on the horizontal plane like this. What about work done by non-conservative? Once you release your hands and the boxes start flying this way, there's no work done by you, and there's no work done by friction as well. So we're on a smooth floor. And then finally, what happens is we have some kinetic energy because the boxes are moving, but we have no potential energy because the spring has decompressed. So what happens is we have \( u_{\text{elastic}} = k_{\text{final}} \). Now what happens is we have 2 blocks that are moving. So the kinetic energy is actually going to be \( \frac{1}{2} m_1 v_{1,\text{final}}^2 + \frac{1}{2} m_2 v_{2,\text{final}}^2 \). The kinetic energy is made up of 2 objects that are moving now. So we actually have all those numbers. We can go ahead and calculate this. So this \( u_{\text{elastic}} \), the stored energy inside the spring is \( \frac{1}{2} \times 3 \times (-10)^2 \), and it becomes positive once you square it, plus \( \frac{1}{2} \times 4 \times 7.5^2 \), which you end up getting is 262.5 joules. So that's the energy that was stored inside the spring. Let's take a look now at part c, which is now we want to figure out how much the spring was compressed before launching the blocks. What we just calculated was we just calculated the elastic potential energy, 262.5 joules. What we want to do now is we want to calculate the compression distance which is really just going to be \( x_{\text{initial}} \). So we can do is we can just relate the elastic potential energy to \( \frac{1}{2} k x^2 \). So what we're looking for here is \( x_{\text{initial}} \), and we have the elastic potential energy, and we also have the spring constant. It's 800, newtons per meter. So we can just go ahead and, I'll just go through this quickly. This is \( \sqrt{\frac{2 \times u_{\text{elastic}}}{k}} \), and that equals \( x_{\text{initial}} \). So when you plug this in, what you're going to get is \( \frac{525}{800} \), and you're going to get 0.81 meters. That's the final answer. So these are all your answers. Right? This is how you use momentum and energy conservation to solve these kinds of problems. That's it for this one, guys. Let me know if you have any questions.
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