Elastic Collisions - Online Tutor, Practice Problems & Exam Prep

In elastic collisions, both momentum and kinetic energy are conserved. The conservation of momentum is expressed as (m1+m2)(v1+v2)=(m1+m2)(v1+v2). For a moving object colliding with a stationary one, simplified equations for final velocities are (m_1 - m_2) / (m_1 + m_2) × ⁢v1 initial and (2⁢m_1 /(m1+m2))⁢ × v1initial. Understanding these principles is crucial for analyzing collisions in physics.

Intro To Elastic Collisions

Hey, guys. So in some problems, you'll see that objects collide, but they undergo what's called an elastic collision. Remember that this is one of the 2 sort of broad types of collisions that you could have. What I'm going to show you in this video is that like all collisions, we're going to solve these kinds of problems by using conservation of momentum. But for elastic collisions, we're also going to need a special extra equation to solve them. Let's go ahead and check this out. Remember that the whole idea is that momentum is always conserved regardless of the type of collision, whether it's elastic or inelastic. But inelastic collisions have a special sort of characteristic, which is that they also conserve kinetic energy, so the kinetic energy is also conserved. So kinitial=kfinal for the system. So we're going to use this characteristic right here to actually help us solve our problems. Let's go ahead and work out this example together. We'll come back through this in just a second. So the idea behind this problem is that you have these two blocks that are smashing into each other. You can kind of think of this as like 2 billiard balls that are crashing into each other. So we have this one that's going to the right, this one's going to the left and what we want to do is we want to calculate the final velocities of both the blocks after the collision. So we're going to go ahead and stick to our steps. We're going to need diagrams for before and after, so let's go ahead and draw that. Right? So afterwards, after these things collide, we're going to have this 3-kilogram box, this 5-kilogram box, and we want to figure out their final velocities. So if I call this object 1 and object 2, what we're really looking for is vonefinal and then vtwofinal. So basically, these are our target variables like this and that brings us to the second step. We're going to have to write our conservation of momentum equation and we'll come back to this in just a second here. So our conservation of momentum is moneinitial+mtwoinitial=monefinal+mtwopostfinal. Right? So we have our masses and some of the speeds. Go ahead and start plugging in our numbers. So we have 5+3=5+3. And now we have the speeds, right? So this 5 kilogram block is initially going to the right at 2, so I'm going to plug in 2. This one is going to the left at 4, so I'm going to plug in negative 4. And then these two final velocities are actually our target variables. Remember, this is what we're looking for here. So I can simplify the left side because I have all the numbers. This is just 10 minus 12, so you get negative 2. So negative2=5vonefinal+3vtwofinal. So we still have 2 unknowns on the right side of this problem here. Right? And you can't assume that the vonefinal and vtwofinal are the same because that doesn't necessarily happen in elastic collisions. So what happens is whenever we end up with an equation with 2 unknowns, we're going to need another equation to solve it. And that's what's special about elastic collisions. For elastic collisions only, we often must use an extra equation, which I like to call the elastic collision equation. It goes like this, voneinitial+vonefinal=vtwoinitial+vtwofinal. This really is sort of like the mathematical consequence of this conceptual point right here which is that the kinetic energies are the same. Your textbooks will derive this, but you don't really need to know the details. So all you need to do is just memorize this equation. And there's a couple of really important things about this equation which I like to talk about here. So the first has to do with sort of like the order of the variables and how it looks similar to the conservation of momentum equation. Here's what I mean. So your conservation of momentum has the m's and it also has 1212. It goes initial, initial, final, and final. The elastic collision equation has no masses and if you look at the letters and the numbers, it goes 1 one and then 2 two, initial, final, initial, final. So the order of the variables is different. What I want you to remember here is that conservation of momentum goes 1212 and the elastic collision equation goes 1 1 22. That's a really easy way to remember it. Alright. So the second thing I want to point out here about this equation is that this equation actually has the same unknowns as the one that equation that we got stuck with. So what happens is this equation here also has vonefinal and vtwofinal here. So because we have 2 equations that have the same unknowns, we're going to end up with what's called a system of equations. Remember, we have the system of equations. There are a couple of different ways we can solve for them, and the easiest way to solve this is by using equation addition. So here's what I'm going to do. I'm going to write out my now my elastic collision equation over here. So I'm going to have voneinitial+vtwoinitial−sorry. it'svonefinal.vone<

Head-On Elastic Collision

Hey, everyone. So, hopefully, you got a chance to work this problem out on your own. So you've got these two blocks of equal mass. They're going to undergo a head-on elastic collision. We want to calculate the magnitude and direction of the final velocities after colliding. So let's go ahead and work this out step by step. So the first thing we're going to do is just draw a quick little sketch of what's going on. So we've got these two blocks that are heading toward each other. So I'm going to call this one block A and this one block B. We've got that \( v_{A_{\text{initial}}} \) is equal to 5, and then we've got \( v_{B_{\text{initial}}} \) like this is equal to negative 3. Right? It's to the left, so it picks up a negative sign. And, ultimately, they're going to collide. And then afterward, we want to figure out, basically, where are they going and how fast. So after the collision, block A and B, they're going to have sort of unknown speeds. Right? They could be both going to the right or they could be going off in different directions or to the left. So we can't draw any arrows, but, basically, we want to calculate what are the final velocities of these two blocks. So let's move on to the next step. We're going to write our equations for conservation of momentum and elastic collisions. Remember, ultimately, we want to solve a system of equations. We're going to have to write both of these. So for the conservation of momentum, this is going to look a little bit different than this because instead of \( m_1v_1 \), we're really going to have \( m_A v_A \). Right? That's the object. \( m_A v_{A_{\text{initial}}} + m_B v_{B_{\text{initial}}} = m_A v_{A_{\text{final}}} + m_A m_B v_{B_{\text{final}}} \). Right? And then, over here, what we've got is we've got our elastic collision equation. Remember, this is the only one that we can use. We can only use this equation for elastic collisions, and it's basically that \( v_{A_{\text{initial}}} + v_{B_{\text{initial}}} = v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Right? Remember, they look different. And so on and so forth. Okay. So we're basically just going to go ahead and start solving the system of equations by plugging in some numbers. So what happens is in the first problem here, usually, we would start plugging in the masses of each of these objects, but we actually don't know what they are. But that's fine because the mass of A is actually equal to the mass of B because they're equal mass. Alright. So they're equal mass, so that means that \( m_A \) is equal to \( m_B \). What that really means here is we can actually cancel out the \( m \) term from the entire equation. Remember, if you have the same number that goes through all of your terms, you can just cancel it out completely. So in other words, the mass actually really doesn't matter in this problem or at least in the equation. So, then let's go ahead and start plugging in some initial values. So at \( v_{A_{\text{initial}}} \) is going to be 5. The \( v_{B_{\text{initial}}} \) is going to be negative 3. This equals \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). And when you simplify this, what you're going to get here is you're going to get 2 equals \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Alright? This is the first equation that we're going to need solve our system of equations because, remember, this is the one that has 2 unknowns. So let's look at the other equation, the elastic collision equation. Remember, if we start plugging in our values, \( v_{A_{\text{initial}}} \) is 5, \( v_{A_{\text{final}}} \) is unknown, \( v_{B_{\text{initial}}} \) is negative 3, and then \( v_{B_{\text{final}}} \) is unknown. Okay? So we also have these same 2 unknowns in this problem. And so what we want to do is, again, we want to sort of add something to this first equation so that one of the terms will cancel out, and then, basically, you're left with one unknown. Alright? So we want to sort of stick another equation down here, so that we can cancel out one of the terms. Okay. So what happens is when you bring the negative 3 over to the other side, it becomes a positive, and you end up with 8. And then when you move the \( v_{A_{\text{final}}} \) to the other side, it picks up a negative sign. So in other words, you get 8 equals negative \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Alright? Now, again, this is where we would have to either multiply this equation to get one of these terms to cancel out, but we actually don't have to do that here because in this equation, we have a \( v_{A_{\text{final}}} \), and we also have a negative \( v_{A_{\text{final}}} \) in this equation. So we don't actually have to multiply this by anything just to make the numbers line up. So we can actually just go ahead and stick this equation right down in this box. So we're going to add this to, let's see. This is 8 equals negative \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Notice how now when you stack these two things on top of each other and then you add them down, the \( v_{A_{\text{final}}} \) will just cancel out. And what you'll end up with here is you'll end up with 10 on one side, and you'll end up with 2 \( v_{B_{\text{final}}} \) on the other. Alright. So now that we sort of eliminated one of the equations, we're just going to go ahead and solve. So this final velocity here, this \( v_{B_{\text{final}}} \) for B is going to be 5. So now let's move on to the last step here, which is this is one of our target variables. This \( v_{B_{\text{final}}} \) here is actually going to be going off to the right like this. So this \( v_{B_{\text{final}}} \) equals 5, and we got a positive number, so it points to the right. The last thing we have to do is we just have to plug the first target variable into any of the other equations to then solve for the other missing variable. So in other words, we can stick this \( v_{B_{\text{final}}} \) into either one of these two equations to solve for the other one. It really just your preference. They're both pretty much the same. I'm just going to go ahead and go with the first one here. So then, basically, if we rewrite equation number 1, what you're going to get is the 2 is equal to, \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \), but we actually know that that's 5 already. So in other words, this is just going to be 5 like this. Okay? So, we just go ahead and solve, and we're just going to get that \( v_{A_{\text{final}}} \) is equal to negative 3 meters per second, and that is your other final answer. So in other words, \( v_{A_{\text{final}}} \) is going to be moving to the left at negative 3. Alright? So now I want you to notice that something interesting happened in this problem. So we actually had these two blocks of equal mass. They collided, and what happens is the \( v_{A_{\text{initial}}} \) for object A was 5, and then the \( v_{B_{\text{initial}}} \) for object B was negative 3. But afterward, their velocities basically switched. So now, \( v_{A_{\text{final}}} \) is negative 3, whereas \( v_{B_{\text{final}}} \) is 5. So in other words, the \( v_A \) for object A became the \( v_B \), the final velocity for object B and vice versa for the for the second block. So in other words, when you have these two objects of equal mass and they undergo elastic collisions, one pro tip that you can use is they actually trade or exchange velocities. So they trade or exchange. Basically, they just swap the initial and final velocity between both objects. Alright, guys. So that's it for this one.

Special Equations in Elastic Collisions

Hey, everyone. So now that we've covered the basics of elastic collisions, there's a special type of problem that you'll see which has a very common setup. It's where one moving object, like this sliding block over here, is going to hit and collide with a stationary object like this one over here. And the basic idea of these problems is, instead of having to use a system of equations to solve for their final velocities, we're actually going to be able to use these special equations that I'm going to give you in just a second down here to solve them. Alright? So for these problems only, we can use these kinds of special equations. So let's go ahead and get started and we'll just jump right into a problem.

The basic idea of these problems is that this stationary object here isn't moving. This m₂ is never moving and because of that, the initial velocity of object 2 is equal to 0. And what that helps us do, is simplifies our equations. So what your textbooks are going to do, is they're going to do a derivation where they're going to take the momentum conservation and the elastic collision equations, and they're going to cancel out this term because it's 0. And then, basically through some substitutions and some algebra, you're going to get to these two expressions for the final velocities of both of the objects, and I'm just going to give them to you. The first one is you're going to do (m₁ minus m₂) over (m₁ plus m₂) times v₁ initial. And the second one is going to be (2 m₁ divided by (m₁ plus m₂)) times v₁ initial. So what you'll notice here is that these two equations actually look very similar. They both have the total mass in the denominator, and they're both multiplied only by the initial velocity of the first object. And then the numerators are a little bit different. Right? But that's basically it. You just use these two equations to find the final velocities of both of the objects. So let's just get right into our problem and see how we use these equations.

So, the basic setup of this problem is that we have a round boulder with a mass of 40 kilograms and a golf ball with a mass of 0.1. And we're going to calculate their final velocities of both of these objects, basically after they collide for these three cases where we're going to have the 2 masses that are the same, and then we're going to have the lighter one hitting the heavier one and then vice versa. So let's just get right to it. So, in part A, we have the boulder that hits another boulder. So, in other words, we want to calculate v₁ final and we're just going to write this out. A good way to remember this is that you subtract the masses and then you add the masses and then you multiply by the first the initial velocity of the first object. So in other words, we're going to take 40 minus 40 divided by 40 plus 40, right because they're the same. And then we're going to multiply this by the initial velocity of the boulder, which is 5 meters per second. And in fact, it's always going to be 5 throughout the entire problem. So what happens in this equation is you actually end up canceling out 40 minus 40 because it's 0, and then it doesn't matter what all this other stuff is because 0 in the numerator is going to make everything equal to 0. So in other words, the boulder basically stops.

What happens to the second boulder? What's the final velocity? Well, our equation again is (2m₁) over (m₁+m₂) times v₁ initial. So in other words, it's 2 times 40 divided by 40 + 40 and then times 5. So, really, 80 over 80 is just 1, and so this v₂ final here is equal to 5 meters per second. So those are our basically 2 answers. We have that this boulder basically just stops, right, the first one, but then the second one is going to go off at 5 meters per second. Now this should make some sense because elsewhere in our elastic collision videos, we said that if 2 objects of equal mass collide, they basically just trade velocities. The velocity of the first one becomes the final velocity of the second object, and that's exactly what happened here.

Let's take a look at part B. In part B, now we have that the golf ball, which is m₁, is going to hit the boulder, which is the 40. Alright. So v₁ final is going to equal. Now I'm just going to start plugging in the numbers here. So m₁ is going to be your 0.1, and it's really important that you plug them in in the correct order. The one that's moving is always going to be m₁. The one that's stationary is always going to be m₂. So when you plug this into your equations here, keep track of that. So 0.1 minus 40 divided by and this is going to be, 0.1 + 40 and then times 5. When you work this out, what you're going to get is you're going to get a negative number. You're going to get negative 4.98 meters per second. Alright? So that's the first one. The second one, v₂ final, is going to be it's going to be (2 \* 0.1) divided by (0.1 + 40) \* 5. So in other words, what you're going to get here is 0.02 meters per second. So, let's take a look at what happened. The golf ball, after hitting the boulder, basically just goes ricochets backwards at almost the same speed that it came in, but just negative. It can't go out with negative 49.8 meters per second. The boulder, on the other hand, picks up a tiny little bit of speed, and it goes off to the right at 0.02 meters per second. This should make some sense because when the golf ball hits the boulder, it transfers a very little amount of momentum. It's going fast, but it has very little mass, whereas the boulder has a lot of mass. So it only picks up a little bit of speed, but the golf ball basically just ricochets backwards at almost the same speed at which it came in with. Alright?

So let's take a look at now our final answer, our final part, which is where the boulder now hits the golf ball. Here, the situation is reversed because now what happens is that your golf ball is m₂ and your m₁ is going to be the boulders. That's the most important thing. So your v₁ final is going to look like m₁ minus m₂. So now it's going to be 40 minus 0.1. Notice how it's reversed from this, from part B over 40 plus 0.1 times 5. Now when you work this out, what you're going to get is that this is 4.98 meters per second. V₂ final is going to be (2 \times 40) divided by (40.0 + 0.1) and then times 5. When you work this out, what you're going to get is 9.98 meters per second. So now let's look at what happened here. Here what happens is that the boulder, when it hits the golf ball, it loses very little momentum, so it's still traveling to the right at almost the same speed at which it hits. But now the golf ball has picked up a ton of speed from the boulder. Once the boulder smacks into the golf ball, it goes off at a much higher speed, 9.98 meters per second. Alright?

So I just want to sort of summarize these cases. See sort of limiting cases where the masses are equal and much less and much greater than. So this is basically what happens. Whenever you have the masses that are equal, the final velocity of the first one is going to be 0, and the final velocity of the second object is basically going to be whatever the initial velocity of the first one was. That's exactly what we saw here with the boulder to boulder case. You might remember if you ever played a ball a game of billiards, when you hit the cue ball into another ball, the cue ball basically stops, and then the other one just goes off with the same speed at which you shot the white ball with or the cue ball. Alright? So that should make some sense if you've ever played the game of pool. So for the second option for the second case where one object is much much less massive than the other one, in other words, you have a very massive target, what happens is that the v final of the first one is going to be the negative v final, v initial, of the first object. Right? It basically just goes back with as much speed, but just in the opposite direction. Whereas the second object, if it's much much much more massive, what you'll see is that this is basically just equal to 0. It picks up a very very little amount of speed. Now for the final, basically, what happens is that the v₁ final is going to be pretty much whatever it hit withm right, because the massive object, massive projectile isn't going to lose a whole lot of speed, whereas the final of the second is actually going to be 2 times v₁ initial. That's exactly what we saw with the golf ball. It picks up basically a ton of speed that's almost double of the 5 meters per second that it came in with. So the very last point I want to make is that after the collision, the second mass is always going to move forward. Right? Whatever is being hit is always going to move forward just a little bit, but this m₁ might move forward or backward depending on its mass. What we saw here is that one could either stop, it could also go backwards or forwards, and it really all depends on the mass. Alright?<\/p>

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

Here’s what students ask on this topic:

What is an elastic collision in physics?

An elastic collision is a type of collision where both momentum and kinetic energy are conserved. This means that the total momentum and the total kinetic energy of the system remain constant before and after the collision. In mathematical terms, for two colliding objects, the conservation of momentum is expressed as:

$(m_1+m_2)(v_1+v_2)=(m_1+m_2)(v_1+v_2)$

Additionally, the conservation of kinetic energy is given by:

$K_{\mathrm{initial}}=K_{\mathrm{final}}$

Elastic collisions are often exemplified by collisions between billiard balls or gas molecules.

How do you solve elastic collision problems using conservation of momentum and kinetic energy?

To solve elastic collision problems, you need to use both the conservation of momentum and the conservation of kinetic energy. The steps are as follows:

1. Write the conservation of momentum equation:

$m_1{v}_1+m_2{v}_2=m_1{v}_1\text{'}+m_2{v}_2\text{'}$

2. Write the conservation of kinetic energy equation:

$0.5m_1{v}_1^2+0.5m_2{v}_2^2=0.5m_1{v}_1\text{'}^2+0.5m_2{v}_2\text{'}^2$

3. Solve the system of equations to find the final velocities $v_1\text{'}$ and $v_2\text{'}$.

These steps will help you determine the final velocities of the colliding objects.

What is the elastic collision equation and how is it used?

The elastic collision equation is a special equation used to solve problems involving elastic collisions. It is given by:

$v_1+v_1\text{'}=v_2+v_2\text{'}$

This equation is derived from the conservation of kinetic energy and is used alongside the conservation of momentum equation:

$m_1{v}_1+m_2{v}_2=m_1{v}_1\text{'}+m_2{v}_2\text{'}$

By using these two equations together, you can solve for the final velocities of the colliding objects. This method is particularly useful when dealing with two unknowns in elastic collision problems.

How do you solve elastic collision problems when one object is stationary?

When one object is stationary in an elastic collision, the problem simplifies significantly. The initial velocity of the stationary object is zero, which allows us to use simplified equations for the final velocities:

1. For the moving object:

$(m_1-m_2)/(m_1+m_2)\times v_{1}initial$

2. For the stationary object:

$(2m_1/(m_1+m_2)\times v_{1}initial$

These equations allow you to directly calculate the final velocities without solving a system of equations. This method is particularly useful for problems involving a moving object colliding with a stationary one, such as a sliding block hitting a stationary block.

What happens to the velocities of two objects of equal mass in an elastic collision?

In an elastic collision between two objects of equal mass, the objects essentially exchange their velocities. If object 1 with mass $m_1$ and initial velocity $v_1$ collides with object 2 with mass $m_2$ and initial velocity $v_2$, the final velocities will be:

$v_1\text{'}=v_2$

$v_2\text{'}=v_1$

This means that after the collision, object 1 will have the initial velocity of object 2, and object 2 will have the initial velocity of object 1. This behavior is often observed in billiard balls or other similar systems where the masses are equal.