Guys, got a classic example here for you where a bullet gets fired inside a block of wood, and we want to figure out the original speed of the bullet. Now, what happens is the bullet is going to embed itself into the block, so we know this is a completely inelastic collision. Let's go ahead and write out some diagrams. So we've got, let's see, a speeding bullet like this that's speeding along this way. I have the mass of this bullet, which is 0.01 kg. It's going to strike a block of wood which is 5 kilograms; that's my m₂. And then eventually, it's going to sort of embed itself inside. So this is the before, and in the after case, what happens is now we have the block that has the bullet stuck inside of it, and these things are both moving together with a final velocity of 0.6, as we're told right here. So this is basically what the after case looks like. What we want to do is figure out the initial speed of the bullet. So if this is m₁, we really want to figure out what is v₁ initial. So I need to figure out what is this v₁ initial right here. So to do that, I'm going to go ahead and move on to the second step. I'm going to write out my momentum conservation. Before I do that, I just want to point out that this v₂ initial is also going to be 0 because the block of wood was stationary. So let's take a look at part a. So for part a, we're going to write our momentum conservation.

m1v1 initial + m2v2 initial = m1v1 final + m2v2 final

Now I am just going to write out all the masses, which are going to be 0.01 and 5, respectively. What we can do here is use a shortcut because we know that in a completely inelastic collision, these two velocities are going to be the same. Instead of writing it out the complex way, I can actually say, 'this is going to be 0.01+5' for both mass values. These things are going to combine together as a single mass, and they're both going to be moving at some final velocity which I already know. It’s really just this initial velocity here that I'm actually interested in. What's the initial velocity of the 5 kilogram block? It's actually stationary, so we know it's going to be 0. So basically, that's what we have here. We know the final velocity of the whole system is going to be 0.6. So, if we simplify the left and right sides, we get 0.01v₁initial = 5.01 × 0.6. If you multiply, you're going to get 0.01 v₁ initial = 3.006. So, when you divide, you're going to get v₁ initial is equal to 301 meters per second. And that sounds about right for a speeding bullet, it's about like 700 miles per hour or something like that. So that is the speed of the bullet. It's traveling at 301 meters per second, lodges in the wood, and then they both move together at 0.6.

Now what we want to do in part b is to figure out how much kinetic energy is dispersed inside of this inelastic collision. Remember, in completely inelastic collisions, kinetic energy is not conserved. So, in order to figure out how much kinetic energy is lost, we want to calculate delta K. Delta K is just K final minus K_initial. This is the difference in kinetic energy. So let's go ahead and calculate those individually: What's the kinetic energy final? We're going to look at the after case. We basically have these 2 objects that are now moving together with a combined speed of 0.6. So your K final is going to be 1/2 and what we can do is we can say that this was m₁ and this was m₂, but when they combine together, their mass, big M, is just going to be m₁+ m₂ which is just 5.01. That's what we calculated over here. So it's going to be 1/2 of big M times v _final squared which is going to be 1/2 times 5.1 times 0.6 squared. What you end up getting for K final is about 0.0912 joules. So just under 1 joule.

What about K_initial? Well, K_initial, we're going to look at the before case. What's the only object that has kinetic energy here? Well, if the block is stationary, it has no kinetic energy, so it doesn't contribute anything to the system. The only kinetic energy in the before case is actually the kinetic energy of the speeding bullet. So here, I'm going to use one half. I'm going to use m₁, the little m. Notice how for the after case, I use the big M and then, in this case, I'm using just little m₁ times v₁ initial which is the speed we just calculated. So this is going to be 1/2 times 0.01 times 301 squared. What you end up getting here is about 453 joules. So these are really just the kinetic energy initial and final.

Now, what we have to do is subtract them. So the delta K, which is really what I'm interested in here, is going to be my final minus initial. So it's 0.0912-453 joules, which I end up getting as negative 452.9088 joules. So the system loses almost all of its kinetic energy. That energy goes from the bullet that embeds itself into the wood. It gets dissipated as heat and sound and other things, but it loses a ton of energy during the collision.

That's it for this one. Let me know if you have any questions.