Completely Inelastic Collisions - Online Tutor, Practice Problems & Exam Prep

In a completely inelastic collision, two objects collide and stick together, moving with a common final velocity. For example, a 1 kg block colliding with a 9 kg block at 20 m/s results in both moving together at 2 m/s after the collision. This outcome is derived from the conservation of momentum, expressed as m_1 × v_1i + m_2 × v_2i = (m1 + m2)v_f. This illustrates how increased mass leads to decreased speed, maintaining momentum conservation.

Completely Inelastic Collilsions

Guys, in this video, we're going to take a look at a specific type of collision that you're going to see probably pretty often in your homework and tests called a completely inelastic collision. Let's go ahead and take a look, and we'll do a quick example. The whole idea is that in some problems, you're going to have two objects that collide and stick to each other. Some keywords that you might see are lodged, embedded, or sticks to. For example, in our example down here, we have a 1 kilogram block that's going to collide with and stick to a 9 kilogram block. This is going to be a completely inelastic collision. The main idea here is that, after the collision, if both objects are stuck together, they're going to move together with the same final velocity. That's what sort of defines these types of problems, that afterwards, the velocity of both objects is going to be the same.

Let's take a look at our example here. We have a 1 kilogram block that collides with a 9 kilogram block that's initially at rest. We have the initial speed, and basically, what we want to do is figure out the final speed of the system here. So, let's go ahead and go through the steps. We have our before and after diagrams. This is going to be before and after. Basically, what happens now is that you're going to have these two blocks, right? These two blocks are actually going to move together sort of as a system. We're just going to imagine that these two objects are stuck with Velcro or something like that, and they're going to move together with some final velocity, and that's what we want to calculate.

Now, what happens is we have this 1 and this 9 kilogram block that effectively just become one big block like this. Let's go ahead and take a look at our momentum conservation. We have mv1initial + mv2initial = mv1final + mv2final. We want to figure out basically what is the final speed of the system. Our mass one is traveling at 20 meters per second, and our 9 kilogram block is initially at rest. That means that the initial speed of this guy is actually equal to 0, so that term just goes away. What about the final velocity? Well, both objects, the 1 and the 9 kilogram block are both going to be moving together with that same speed. Since v1final equals v2final, we're actually going to simplify our conservation of momentum equation.

The general form is going to be mv1vm2vv2. On the right side, we can combine both the masses, and we say both masses m1+m2 are going to be moving at the same final speed like this. So, we simplify and solve for this. Now, when you go ahead and simplify, you'll get 20 over 10 which is vfinal=2 meters per second. What happens is initially, you have this 1 kilogram block moving at 20 meters per second. So, once it hits the 9 kilogram block, and they're both moving, effectively, the mass has increased by a factor of 10. So in order for momentum to remain conserved, if your mass increases by a factor of 10, your speed has to decrease by a factor of 10. Notice how this 10 kilogram block is now moving at 2 meters per second, and momentum is conserved. So that's basically it for this one guys. Let me go ahead and show you a couple more examples.

A 70-kg hockey player skates on frictionless ice at 5m/s. An opposing player of mass 110kg collides head-on with him at 4m/s. The two players become entangled after the collision and move together. Determine their final velocity.

• A. 0.5 m/s in the direction of the opposing player
• B. 0.5 m/s in the direction of the 70 kg player
• C. 4.39 m/s in the direction of the opposing player
• D. 4.39 m/s in the direction of the 70 kg player

Bullet Fired Into Block

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.

2D Collision

Hey, everybody. So in this problem, we have a car and a truck that's colliding. Now luckily, nobody's hurt, but we do have the cars that stick together and lock together during the collision. So this is a completely inelastic collision, and we're going to calculate the magnitude and the direction of the final velocity after colliding. So let's get into the steps. We're going to write some, we're going to draw some diagrams from before and after. So here's what's going on here. I've got this car like this and this car has a mass of 1,000 and it's traveling east at 20 meters per second. So in other words, its velocity kind of looks like this and I'm going to call this basically this is car A. So its v_a is going to equal 20 meters per second. Now what's happening here is that it slams into a truck that's moving north. So what I'm going to do is I'm going to draw a little x and y axis like this. Alright? So I've got this velocity here. This truck has a mass of 3,000, and it's moving totally, north. Right? So, basically, vertically up like this. That's north and that's east. So its velocity vector is going to look like this. I'm going to call this v_b and this v_b here is equal to 10 meters per second. So once they lock together, what's going to happen? Well, if you draw the after diagram, you can kind of sort of intuitively guess what's going to happen. Afterwards, the two cars, the trucks, whatever, are going to stick together. They're going to have a combined mass of 4,000. Right? 1,000 plus 3,000. And then they're going to move off in which direction? Well, if you imagine this, right, if you have a car that's moving like this and a truck that's moving like this, once they stick together, they don't move totally to the right or perfectly up. They move sort of a combination of the two directions. They sort of go off in a diagonal. So here's what happens: This truck is going to go off. These two cars are going to go off like this, and this is my v final. And that's what I want to calculate. I want to calculate the magnitude of that vector, but I also want to calculate the direction as well. Right? So I also want to calculate the angle with respect to the x-axis. Okay? That's really what we're trying to calculate here: v final and theta. So how do we do that? Well, we actually just go back to, you know, using, you know, normal vectors. If I want to calculate the magnitude of this vector, then I need its components. Right? I'm going to need, basically, the x components and I also need the y components. So the tricky thing here is that I want to calculate v final, but I have to first calculate what v_fx and v_fy are. How do we do that? Well, let's move on to the second step here, which is we're going to write momentum conservation equations. So what I've got here is I've got m_av_a initial plus m_bv_b initial equals, and then remember, these two things combine, so we can combine their masses and sort of use that shortcut equation, m_a plus m_b times v final. Now are we done here? Well, actually, no. Because for all the collisions that we've seen so far, we've only had them in one dimension, on one axis. We've had a car moving to the right and truck moving to the left or whatever. Two objects are basically just moving along the line. But in this problem, we have a two-dimensional collision because one is moving sort of left and right and the other object is moving up and down. And here's the idea. For a two-dimensional collision, we're still going to write momentum conservation equations, but now we're just going to do it for both the x and the y-axis. It's just like any other 2d problem that we've encountered. When we did motion in 2d, when we did forces in 2d, We basically just wrote a bunch of equations in the x and the y-axis. We broke them down and then solved them separately. That's exactly what we're going to do here with momentum. So what I'm going to do is I'm going to write equations for the x and y conservation of momentum. So this is going to be m_av_ainitial plus m_bv_binitial equals m_a plus m_bv final. But what the only thing that's different here is that I'm going to include this is the x-axis and this is the y-axis, and all the velocities are sort of implied that you're just going to look at the x and y axes respectively. Alright? So then how do we do this? Well, let's just jump right into the x-axis because we're going to go ahead and plug values and solve. So n_a, remember, that's just the mass of the car. That's going to be 1,000. Now v_ax initial, so in other words, this the velocity component of this car that lies along the x-axis. Well, this just lies like this, so it's basically just going to be 20 like that. Now the second term, this is going to be m_b which is 3,000, but what about the velocity in the x-axis? Remember, this truck is still moving at 10 meters per second.

Three-Body Collisions

Hey, everybody. So let's take a look at our problem here. This one's kind of interesting because we have three different objects, a bullet and two blocks, and two different collisions that are happening. Basically, what we want to do is we want to calculate the speed of the second block after the bullet embeds itself into the second one. So let's go ahead and stick to the steps. We're going to draw a diagram from before and after. This is what's going on in this problem. We have a bullet like this. It's flying and it's going to hit towards a block. It's going to head towards a block. So I'm going to call this \( m_1 \). This has a mass of 0.01 kg and the block has \( m_2 \), which I'm going to call 0.5 kg. What happens is the bullet actually keeps on going after it's passed through the first block. It keeps on going through the first block and actually embeds itself in a second block. I'm going to call the \( m_3 \), and this equals 0.5 kilograms as well. And, basically, once it embeds itself in there, the whole entire system is going to be moving off with some velocity. I'm going to call this \( v \) at this point. Alright? And that's really what we want to calculate. The tricky thing here is that we have three different objects and two different collisions. So if you're not careful, what happens is that if you use \( m_1v_1{initial} \) and \( m_2v_2{initial} \), the final will be the initial for the second collision, and that can be really confusing. So what we're going to do is we're going to do something that we've used in other, sort of chapters like motion and forces, which is we're going to label the different events that are happening. So here's what I'm going to do. I'm going to call this a, this one is b, and this one is c. So a is actually going to be so a is the point where it's before the first collision. So this a, b, and c do not refer to different objects. They refer to different sort of events or different points in time. I want to be really clear about that. So it's before the first collision. Now point b is after the first collision. Right? It's when the bullet exits the first block and keeps on going towards the second one. So it's after the first, but it's before the second collision. And then finally, point c is when the bullet embeds itself into the second block and so that's after the second collision. So, really, what we want to calculate is this final velocity here, and I'm going to call this velocity c. That's really what we're interested in. So how do we do that? Well, we're going to stick we're going to, again, stick to the steps, move on to the second thing, and we're going to write the conservation of momentum equation. If I want to calculate the \( v_c \), then I'm going to pick the interval that basically includes c. So in other words, I'm going to include the I'm going to have the interval from b to c. Right? That's the interval after the first collision but before the second one, and then when the bullet finally embeds itself. So let's set up our equation here for b to c. So what I'm going to do is, again, keep, just keep in mind that we have three different objects. So, really, what's going on is that the bullet, which is \( m_1 \), is going to collide with the third object, \( m_3 \). So here's how our equation's going to get set up. It's going to be \( m_1 v_1b \). Right? That's the initial from b to c. Plus \( m_3v_{3b} \) equals now what happens is when the bullet embeds itself, that is a completely inelastic collision. So what happens here is that from b to c is completely inelastic. Whereas, the first collision from a to b is just an inelastic one because nothing gets stuck together. Right? The bullet exits and both of these things sort of like you know they go apart. Right? They don't actually stick together. Okay. So this is going to be, again, using our shortcut, \( m_1 + m_3 \), and this is going to be \( v_c \). So this is really my target variable is what is the final velocity of the bullet plus block combo? So that's what I'm looking for here. So if you look through your variables, which what you're going to see is that all these masses are given to us. So can we actually just solve this by using this interval? Well, if you look at this, what happens is we're going to hit 0.01 and then \( v_{1b} \) is going to be the velocity of the bullet after it exits the first block. Right? So this is going to be \( v_{1b} \). Now what is that velocity? If you look through the problem, you might see that this 430 meters per second, but remember, that's the initial speed of the bullet. In other words, the \( v_{1a} \) is going to be 430, but what happens is after it exits the first block, it's going to change. Right? There's going to be an exchange of velocity there. There's going to be some exchange of momentum, and \( v_{1b} \) is definitely not going to be 430 meters per second. But we actually don't know what that is. So I don't know what this is yet, and I'm just going to leave it blank for here. Now, what about \( m_3v_{3b} \)? Well, basically, this is just the initial velocity of the second block before the collision. In that case, the third the second block is stationary. So in other words, that term actually just goes away because that term is 0. So we have 0 on the left side and then on the right side we have two velocities the two masses combined, so 0.01, 0.5, \( v_c \). So there's only one variable that we really need, and that's this \( v_{1b} \) over here. It's the velocity of the block after it exits the first block or, sorry, the velocity of the bullet after it exits the first block. Now how do we go find that? Remember, we in this problem, we have two different collisions, two different sorts of intervals. And so to go and find this \( v_{1b} \), I'm going to have to go and use the other interval, which is going to be the a to b interval. Alright? So let's set up that equation now. Now that equation is going to look like \( m_1 v_{1a} + m_2 v_{2a} = m_1 v_{1b} + m_2v_{2b} \). Now remember, we can't use the shortcut like we did over here because this collision is not completely inelastic. The two masses will not stick together. Alright? So that's the important part. So, really, this \( v_{1b} \) here is actually going to be this \( v_{1b} \), and that's, again, why we didn't use initials and finals because it can get really confusing. So let's go ahead and start plugging in the values and solving. So this is going to be 0.01. What about this \( v_{1a} \)? We actually know what that is. That's just the initial velocity of the bullet, which is 430 meters per second. So 430 plus \( m_2v_{2a} \). So, basically, that's the mass and velocity of the first block of this one. Now, remember, both blocks are at rest. So what happens here is you're just going to get 0.5, but then this cancels out because this is equal to 0. So, basically, the whole term goes away. Then we have a 0.01 and then \( v_{1b} \), that's exactly what we're trying to find here, plus. And then we've got 0.5. And then what about \( v_{2b} \)? It's basically the velocity of the block after the collision. Now what number is that? Well, if you look through your variables, what's gonna happen here is that the speed of the first block immediately afterward is 5.6 meters per second. So that's basically \( v_{2b} \). Alright? That's what's going to go over here, the 5.6. So now what we can do is we can solve this equation for \( v_{1b} \), and that's exactly what we were missing in the b to c interval. Okay? So if you plug all the stuff in, what you're going to get is 4.3 on the left side. This is going to equal 0.01\( v_{1b} \)+, and then this works out to 2.8. So when you bring this over to the other side, what you're going to get here is you're going to get 1.5 is equal to 0.01\( v_{1b} \). And then finally, \( v_{1b} \) is equal to 150 meters per second. Alright? So now that we have this 150, that's what we stick into this equation over here. So this becomes the 150, and now we can go ahead and solve for \( v_c \). This is going to be 0 point, we work this out. This is going to be 1.5 again equals 0.51\( v_1 \) or, sorry, \( v_c \). And now if you work this out, what you're going to get is the \( v_c \) is equal to 2.94 meters per second. Alright? So it's a little bit of, you know, sort of upfront work, getting organized, labeling everything, but once you do that, the rest of the math becomes pretty straightforward. Alright? So that's your final answer. It's 2.94 meters per second. Let me know if you have any questions.

Here’s what students ask on this topic:

What is a completely inelastic collision?

A completely inelastic collision is a type of collision where two objects collide and stick together, moving with a common final velocity. This means that after the collision, the two objects move as a single combined mass. The defining characteristic of a completely inelastic collision is that the objects do not bounce off each other but instead remain attached. The final velocity of the combined mass can be determined using the conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

How do you calculate the final velocity in a completely inelastic collision?

To calculate the final velocity in a completely inelastic collision, you use the conservation of momentum principle. The equation is:

$m_1{v}_{1}i+m_2{v}_{2}i=\; (m_1+m_2)v_f$

Here, m1 and m2 are the masses of the two objects, v1i and v2i are their initial velocities, and vf is the final velocity of the combined mass. By solving for vf, you can find the final velocity after the collision.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved. The objects bounce off each other without any loss of kinetic energy. In contrast, in an inelastic collision, only momentum is conserved, and some kinetic energy is converted into other forms of energy, such as heat or sound. A completely inelastic collision is a special case where the colliding objects stick together and move with a common final velocity, resulting in the maximum possible loss of kinetic energy.

Why is kinetic energy not conserved in a completely inelastic collision?

Kinetic energy is not conserved in a completely inelastic collision because some of the kinetic energy is transformed into other forms of energy, such as heat, sound, or deformation energy. When the objects stick together, the internal forces during the collision cause energy dissipation. While momentum is conserved, the loss of kinetic energy is a characteristic feature of inelastic collisions, especially completely inelastic ones where the objects move together as a single mass after the collision.

Can you provide an example of a completely inelastic collision?

Sure! Consider a 1 kg block moving at 20 m/s that collides with a 9 kg block initially at rest. After the collision, the two blocks stick together and move as a single mass. Using the conservation of momentum:

$1\times 20+9\times 0=\; (1+9)v_f$

Simplifying, we get:

$20=10v_f$

Solving for vf, we find:

$v_f=2\text{m/s}$

So, the combined mass moves at 2 m/s after the collision.