11. Momentum & Impulse

Completely Inelastic Collisions

# Bullet Fired Into Block

Patrick Ford

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Hey 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 bullets. Now, what happens is the bullet is going to embed itself into the block. So we know this is a completely an elastic collision. Let's go ahead and write out some diagrams. So we've got, let's see um let's say, I've got this speeding bullet like this, that's speeding along this way. I have the mass of this bullet is 0.01. It's going to strike a block of wood, which is five kg. 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 that's stuck inside of it. And these things are both moving together with a final velocity of 0.6. That's what we're told right here, is that the bullet would system immediately is moving at 0.6. All right. So this is basically what the after case looks like we want to do is we want to figure out the initial speed of the bullets. So if this is M. One we really want to do is figure out what is V one initial. So I need to figure out what is this. V one initial right here. So to do that, I'm gonna go ahead and move onto the second step. Gonna write out my momentum conservation before I do that. I just want to point out that this V two initial is also going to be zero because the block of wood was stationary. So let's take a look at part A. So for part A we're gonna write our momentum conservation. M one V one initial plus M two V two initial equals M one V one final plus M two V two final. Now, I'm just gonna write all the masses. So again this is going to be 0.1 plus. This is gonna be five equals 0. plus five. Okay, What we can do here is we can actually use the shortcut because we know that if in a completely an elastic collision, these two velocities are going to be the same. So instead of writing it this way, I can actually say this is going to be 0.01-plus five. Right? Both of these things are going to combine together as a single mass and they're both gonna be moving at some final velocity, which I already know. Uh it's really this this initial velocity here that I'm actually interested in. So what's the initial velocity of the bullets? So what goes in here? What's the initial velocity of the five kg block? What's actually stationary? So, you know, it's going to be zero. So basically that's what we have here. We know the final velocity of the whole system is actually going to be 0.0 point six. So if we go ahead and simplify the left and right sides, this is just going to be 0.01 V1 initial equals 5.01 times 0.6. So if you multiply, you're gonna get 0.01, The initial equals, and then you're gonna get 343.006. So now what happens when you divide, you're gonna get viewing initial is equal to 301 m per second. And that sounds about right. For a speeding bullet, you know, it's about like 700 miles an hour or something like that. So that is the speed of the bullets. So it's traveling at 301 m per second, lodges in the wood and they both moved together at 3010.6. All right, So now what we're gonna do in part B is we want to figure out how much kinetic energy gets dispersed inside of this, an elastic collision. Remember that completely inelastic collisions, actually, all type of an elastic collisions do not conserve kinetic energy. So in order to figure out how much kinetic energy is lost, we want to figure out delta K delta case just K final minus K initial. Right? So it's how much energy you have now versus how much energy you had before. So let's go ahead and calculate those individually. What's the kinetic energy final? We're gonna look at the after case. Right. So, we basically have these two objects that are moving together with a combined speed of 0.6. So you're K final is going to be one half. And what we can do is we can say that this was someone and this is M. Two. But when they combined together their masses, big M. Is actually just going to be M one plus M two, which is just 5.1 That's what we calculated over here. So it's going to be one half of big M times V final squared, which is gonna be one half times 5. Times This is going to be 06 square. Which end up getting for K. Final is you end up getting 0. jewels. So just under one jewel. What about K. Initial? Okay. 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, has no kinetic energy, that doesn't contribute anything to the system. The only kinetic energy in the before cases, actually the kinetic energy of the speeding bullets. So here I'm going to use one half and I'm going to use M. One the little M. Notice how for the after case I use the big M. And then this one I'm using just little M. One uh times if you want initial, which is the speed we just calculated. So this is gonna be one half. This is going to be 0.1 And the initial speed that we just calculated was You're gonna square. That what you end up getting here is going to be getting about 453 jewels. So these are remember 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 0.902 minus jewels, which end up getting is negative 452. jules. So it loses, basically the system loses almost all of its kinetic energy that basically that energy goes in from the bullet that basically embeds itself into the wood gets dissipated as heat and sound and other things, but he loses a ton of energy during the collision. All right, so listen to this one, let me know if you have any questions.

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