Collisions & Motion (Momentum & Energy) - Online Tutor, Practice Problems & Exam Prep

In problems involving collisions and subsequent motion, apply conservation of momentum during the collision and conservation of energy for motion. For a completely inelastic collision, where two objects stick together, use the equation ${\frac{m}{1}}^2=\frac{m}{g}y$. The final height can be calculated as $y=\frac{v^2}{2g}$, emphasizing the importance of energy conservation post-collision.

Collision Problems with Motion/Energy

Hey, guys. So in previous videos, we've seen how to solve motion problems by using conservation of energy. And in the last few videos, we've seen how to solve collision problems by using conservation of momentum. We're going to combine those two things in this video because some problems are going to involve a collision and then they'll involve motion with changing speeds or heights after the collision. So what I have, what I mean by that is that in this problem we're going to work out down here, we have a crate that collides with another crate. So that's a collision in the first part. And then what happens is they're going to stick together and they're going to both rise up this inclined plane. So they're going to change some speeds and heights after the collision. So this is going to be the motion parts. Now we know how to solve these problems individually, but really what happens is that to solve these problems, we're actually going to use both conservation equations. You're going to use conservation of momentum during the collision parts and conservation of energy during the motion parts. These problems are super important, and there's a lot of different variations that you're likely to see on a homework or test, and they can get very messy if you're not careful. But what I'm going to do in this video is I'm going to show you a step by step processing system so you can avoid any confusion and get the right answer. So let's go ahead and check out our problem here. So we have, like we said, a collision and then followed by motion. So let's go ahead and stick to the steps here. What we're going to do is we're going to draw diagrams for our problems, but then we're going to label points of interest. Here's what I mean by this. In the collision part of our problem, we have an initial and a final. In the momentum equation, we have m1v1im2v2f, things like that, but you also have an initial and final during the motion parts. In the energy equation, we have like KiUf and things like that. So what happens in these problems is that the final part of the collision part of the problem is actually the same as the initial of the motion part of the problem. So if you're not careful, you're going to have i's and f's floating around anywhere everywhere, and that can get very messy and potentially confusing. So to avoid this problem, I'm going to take a similar approach to what I did with projectile motion. I'm basically going to call these things these points of interest a's, b's, c's, and so on instead of i's and f's. So what happens is the different points of interest of our problem are right before the collision. It's right before the 20 hits the 30. So I'm going to call that point a. The next point of interest is what happens after the collision and they stick to each other. So I'm going to call that point b here. So point b is right after the collision. What's interesting about point b is that it's also the same point as when the motion part starts. It's basically where they are at the bottom of the incline and they're going to start moving up. So point b is after the collision, but it's also where the motion part starts. And then eventually, they're going to rise up and eventually stop once they hit the top of the incline. So the last point is where the motion part ends, and I'm going to call that point c. So what's our target variable? We're looking for how high the crates travel before they stop. So basically, they're going to travel some vertical distance or some height. I'm going to call that y, but that's going to be y at point c here. So this is what we're looking for in our problem. How do we solve for that? We've already drawn the diagrams and labeled the points of interest, but now we're going to go ahead and start writing some equations. So the second step is you're going to write out both your momentum and energy conservation equations here. So we're going to use the conservation of momentum for the collision parts and then we're going to use the conservation of energy for the motion parts. So let's go ahead and do that. So what I'm going to do here is I'm going to use m1v1am2v2a equals m1v1bm2v2b. And now for the energy equation, I'm going to use KbUiplusworkdonenonconservativeequalsKfUf here. So that's both of the equations. Now we just have to figure out which one to start off with. So hopefully, you guys realize that if we're trying to figure out y c, then we're going to start with the equation that includes point c, which is our energy conservation equation. So let's go ahead and start expanding out the terms here. Do we have any put kinetic energy at b? Remember, b is after the collision. So after these two things collide, they're both moving up the inclined plane with some speed so they have kinetic energy. However, do they have potential energy? Well, actually, no, because this is the point where they were still at the bottom of the incline. So here, what I'm going to do is I'm going to call this the ground level, y equals 0, and therefore, there's no potential energy here. There's also no work done by non-conservative forces because you're not doing anything. There's no friction. What about k c k final? What happens is when the blocks go up the incline, they're eventually going to stop. That's where their maximum height is. That's where to be the speed is 0 and therefore there's no kinetic energy. So basically, all of it gets converted to gravitational potential which is where our y c comes from. So let's go ahead and expand out the terms. So for k b, we're going to have 1/2mvb2. But hopefully you guys realize that this is actually a completely inelastic collision because one crate sticks to another one and they both move at the same speed. So what happens is this is a completely inelastic collision. So what happens is I'm not going to use little m. I'm actually going to use big M. So here what happens is that big M is equal to basically both the masses. They're going to stick together as 1. So one f big m vb2 is equal to big Mgy_c. So that's our target variable and we can actually see that the masses are going to cancel. So even if you didn't know that they stick together, that would have been okay because they cancel out. So let's go ahead and write an expression for our y c. I'm going to move the g to the other side and basically with this one half, y c just becomes vb2/2g. So this is our target variable y_c, and I'm almost ready to start plugging everything in. The problem is I don't know what this v_b is, this velocity which is after the collision. So how do we figure this out? Well, remember, point b is the part where our or is the point where our motion starts, but it's also the same point as where the collision ends. So if I'm stuck here solving for v_b, hopefully, you guys realize I can actually solve for this by using our conservation of momentum equation because I have v_b's on this equation as well. In fact, this will very commonly happen in your problems. You're going to start off by using one of these equations, but then you're going to have to go to the other one to fully solve the problem. So that's what we're going to do here. We're just going to use our conservation of momentum equation now. So what I'm going to do is I'm going to call this object 12, and we'll start plugging in all the numbers that we have. So this is going to be 20 times the initial speed of 40 plus 30 and then times the initial speed of 0. So this v a here, this v two a is equal to 0 because this block is initially at rest. So there's no momentum here. So then what happens afterwards? We already said this is a completely inelastic collision. So basically, what happens is that these masses combine and these two velocities have to be the same. So what I'm going to do is I'm going to add the masses 20 plus 30, and then this is just going to be v_b. So this is what I'm looking for here. If I can figure out this v_b, I can plug it back into this equation. So to solve for this, you're going to do 800 divided by the 50 on the other side. And you're going to get v_b equals 16 meters per second. Alright. So now we're just going to finish off the calculation here. We're going to plug this because now we know v_b. This is 16 squared divided by 2 times 9.8. And if you go ahead and work this out, you're going to get 13.1 meters. So what happens is these two things stick to each other and they're both going to rise a distance, a height of 13.1 meters above the incline. Notice how in this problem, the angle of the incline never factored into the problem and that's because in conservation of energy, this potential energy doesn't depend on the angle, it just depends on the height. Alright. So I have like That's really what there is. That's really all there is to the problem. I have a couple more conceptual points to make here. So some of you may be wondering why I was able to use conservation of energy even though this was a completely inelastic collision. And the idea here is that energy isn't conserved. You're going to lose some energy during the collision. That's in this interval from a to b. But once they stick together, energy is conserved afterwards. So basically it travels 1 object and energy is going to be conserved in the interval from b to c. So we're totally okay with using conservation of energy here. Alright. So only if the work done by non-conservative forces is 0. If you have no friction or any work done by you or something like that. Alright. So the last thing I want to point out is that we actually did one sort of specific example of this type of problem. We saw a collision where they both travel up an inclined plane afterwards. But a lot of these problems are going to have a couple of different variations or situations that you might see, but you're always going to use this system whenever you see these kinds of problems. You could have a collision where they go up against a spring like this, or you could have a collision in which both objects encounter some friction afterwards, or you could have a collision and then both objects basically swing up like this as a pendulum. Notice how what's common about these is in all of these situations, you have changing speeds or heights after the collision. So we're going to use all the steps that we just discussed here. You're going to write out your conservation equations and you're going to start that working down what your target variable is. Alright. So that's it for this one, guys. Let me know if you have any questions.

A 300g bullet is fired horizontally into a 10-kg wooden block initially at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.6. The bullet remains embedded in the block, which then slides 35 m along the surface before stopping. What was the initial speed of the bullet?

• A. 492.5 m/s
• B. 20.3 m/s
• C. 677 m//s
• D. 697 m/s

Bullet-Block with Height

Welcome back, everybody. So in this problem, we have a bullet that is being fired into the bottom of a wooden block. It's fired upwards. It basically keeps on going after it passes through, so the bullet continues rising. But having transferred some of its momentum to the block, the block now starts rising as well. Basically, we want to calculate how high the block reaches. So this is going to be a maximum height. What I'm going to do here is I'm going to just call this some ymax value. That's really what we're interested in. In this problem, we have a collision that is followed by some motion. So we're going to go ahead and stick to our steps for solving these kinds of problems, these conservation problems, with energy. We're going to have our diagrams. We're going to label our points of interest. Before the collision, you have the bullet that's being fired upwards, and then it hits the block, and that's when the block starts moving, and then finally, it goes upwards and reaches its maximum height. There are basically three points of interest here: a, b, and c. The only thing that's different is that most of our problems up until now have been mostly horizontal, but in this problem, we have basically purely vertical motion. That's the only thing that's really different here. Let me go ahead and draw some stuff out. Remember that point a is going to be before the collision. Point b is going to be after the collision. This is also where the motion of the block starts. And then, finally, when it rises up to point c, that's where the block ends its motion. Just so it's kind of consistent with the other diagrams or sort of timelines that we've set up.

If we want to calculate the ymax, we're going to do that by writing out our momentum and energy conservation equations. Remember that from a to b, you have to use conservation of momentum because this is a collision. So we're going to use conservation of momentum here, and then we're going to use conservation of energy for the b to c interval. First, I want to label out these masses. The bullet, this is going to be my m1, which is going to be 6 grams. 6 grams are going to be 0.006 kilograms. The mass of the block, I'm going to call this m2, is equal to 1.2 kg. So I've got 1.2 here. So I'm going to write out my momentum conservation. So I've got m1v1a + m2v2a = m1v1b + m2v2b Right? That's the initial and final, the a and the b. One thing I also did here is I noticed that the blocks don't stick together, so we can't use the shortcut of combining everything into one object. Remember, the bullet, after it passes through the center of the block, it continues moving upwards. So these things are separate objects. They don't become 1.

Let's look at the b to c interval because now we're going to use conservation of energy. Remember, initial to final. So we have to use KEinitial + PEinitial + workdone by nonconservative forces = KEfinal + PEfinal. What are we looking for? We're looking for the maximum height. So in other words, we're looking for a variable that happens at point c where the block ends its motion. Naturally, the best equation to start with is by using the b to c interval. Let's expand out the terms. But first of all, what are we going to consider as our system? Remember, this is kinetic and potential, but first, for conservation problems, you have to define what your system is. Do we consider the bullet plus the block? Do we just consider the block only? Well, the best thing here is actually just to consider just the block. Write this out here. The system is only the block. The reason for that is, remember, in energy conservation problems, you can always pick your system as long as you're consistent with the terms. What happens is after the collision, and the bullet just becomes separate objects, we don't really care about it anymore. We're only really just considering or interested in the maximum height of the block. So we only have to worry about the energy conservation of the block. That's really important here. Make sure you keep that in mind. What kinds of terms can we cancel out here? Let's see. In the initial, is there any kinetic energy? Well, what happens here is at point b, this is after the collision between the bullet and the block. That's when the block starts moving. So there's definitely some kinetic energy here. What about any potential? Remember that spring or gravitational potential? We don't have any springs here, but we do have things moving in the vertical axis. This potential energy just depends on where we choose our 0 points. And remember, with energy problems, you always want to pick the lowest points to be your 0 points. So here, where the block is on the floor, I'm just going to consider that my y = 0. And so what that means is that there's no potential energy in my calculation. There's also no work done by nonconservative forces. There's no friction or anything like that. What about here at point c? Is there any kinetic energy? Well, remember, once the block travels from b to c and stops and ends its motion, that's when it's at the maximum height. So there is no kinetic energy because vc = 0. Once the block reaches its maximum height, there's no velocity. But now there's definitely some gravitational potential because the block has risen some distance, and this is really what we're trying to find here. This is really the ymax. That's really our target variable here. Then when we expand out our terms, what you're going to see here is you're going to have 12m2v2bsquared, right, 12bsquared, that's kinetic energy, equals gravitational potential energy, mgymax. So this is my target variable here. What you'll also notice is that the masses will cancel out because they're on both sides of the equation. So, really, I'm really close to just being able to plug in everything and solve. The problem is that I actually don't have what this v2b is. That's the velocity of this block right after the collision with the bullet. So right after the collision, the bullet, the block picks up some of the speed from the bullet, and it's going to go upwards like this with v2b. I don't know what that is yet, and so I can't go ahead and finish out this equation. But this is basically just going to be 12v2b, 12somethingsquared divided by 9.8 is going to equal your ymax. All I have to do is figure out what's going to go inside this equation. And to do that, when we get stuck in one of our intervals in these problems, we're always going to go to the other interval here, because remember, v2b is also in my momentum conservation. That's why I have to use both of them.

Let's go ahead and plug in some values and then just solve. So m1 is going to be the mass of the bullet, that's 0.006 times v1a. That's the initial speed of the bullet. Remember, the initial speed of the bullet is going to be 800 meters per second. That's what it's fired at, so it's 800. Now what's m2? Mass 2 is 1.2. What's the initial speed of the block? That's v2a. Well, it's actually just nothing because, again, at point a, when the bullet is still being fired upwards, it's before the collision, there is no initial velocity for this block. It's equal to 0, so that term goes away. So then what we have is we have 0.006 and then v1b. That's going to be the speed of the bullet after the collision here. So what happens here is that this bullet exits and it keeps on going upwards. I'm going to call this v1b, and it's emerging moving upwards at 150 meters per second. So that's v1b over here. It's 150. So you plug that in, it's going to be 150, and this is going to be a mass 2, which is going to be 1.2 times v2b. So remember, I came over here because I needed what this v2b is. If you look through your variables, this is the only one that's missing. So I'm just going to go ahead and simplify some stuff. This is going to be 4.8 when you work this out. This is going to be 0.9 + 1.2v2b. So pretty complicated equations, but they work out and they basically simplify to just a couple of numbers. So then when you work this out and you move the 0.9 over to the other side, what you're going to get is 3.9. And then when you divide by the 1.2, that's going to be v2b. And when you work this out, this is going to be 3.25 meters per second. Remember, we're not done yet. This is not our final answer because this is just the velocity of the block right after the collision. Now we just plug it back into this equation over here. So this is going to be my 3.25. Now when you work this out, this is going to be the final answer because this is your ymax, and this is going to be a final maximum height of 0.54 meters. That's pretty reasonable. Right? A bullet was really, really small mass, but it has a lot of momentum because it's being fired very, very quickly at this 800 meters per second. When it hits something like a block, which is 20 times heavier, it gives a little bit of speed and it rises a little bit of distance, but nothing too crazy. That makes sense. Let me know if you have any questions.

Here’s what students ask on this topic:

How do you solve problems involving both collisions and subsequent motion?

To solve problems involving both collisions and subsequent motion, you need to apply the conservation of momentum during the collision and the conservation of energy for the motion. First, identify the points of interest: before the collision (point A), after the collision (point B), and the final state of motion (point C). Use the conservation of momentum equation for the collision: $m_1{v}_1{A}_{}+m_2{v}_2{A}_{}=m_1{v}_1{B}_{}+m_2{v}_2{B}_{}$. Then, use the conservation of energy equation for the motion: $K_B+U_B=K_C+U_C$. This approach helps you systematically solve for unknowns like final height or speed.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved. The objects collide and bounce off each other without any loss of kinetic energy. The equation for an elastic collision is: $m_1{v}_1{i}_{}+m_2{v}_2{i}_{}=m_1{v}_1{f}_{}+m_2{v}_2{f}_{}$. In an inelastic collision, momentum is conserved, but kinetic energy is not. The objects may stick together, and some kinetic energy is converted into other forms of energy, such as heat or sound. The equation for a completely inelastic collision is: $m_1{v}_1{i}_{}+m_2{v}_2{i}_{}=\; (m_1+m_2)v_{\mathrm{f}}$.

How do you calculate the final height in a collision problem involving an inclined plane?

To calculate the final height in a collision problem involving an inclined plane, first use the conservation of momentum to find the velocity after the collision: $m_1{v}_1{A}_{}+m_2{v}_2{A}_{}=\; (m_1+m_2)v_{\mathrm{B}}$. Then, apply the conservation of energy to find the height: $\frac{v^2}{2}$. The final height $y$ is given by $y=\frac{v^2}{2}$, where $v$ is the velocity after the collision and $g$ is the acceleration due to gravity.

Why is energy not conserved in a completely inelastic collision?

In a completely inelastic collision, energy is not conserved because some of the kinetic energy is converted into other forms of energy, such as heat, sound, or deformation energy. While momentum is conserved, the loss of kinetic energy occurs due to the internal forces acting during the collision. The equation for a completely inelastic collision is: $m_1{v}_1{i}_{}+m_2{v}_2{i}_{}=\; (m_1+m_2)v_{\mathrm{f}}$. The kinetic energy before and after the collision is different, indicating that some energy has been transformed into other forms.

How do you use conservation of momentum and energy in a problem involving a collision and an inclined plane?

To use conservation of momentum and energy in a problem involving a collision and an inclined plane, follow these steps: 1) Apply the conservation of momentum to the collision to find the velocity after the collision: $m_1{v}_1{A}_{}+m_2{v}_2{A}_{}=\; (m_1+m_2)v_{\mathrm{B}}$. 2) Use the conservation of energy to find the height the objects reach on the inclined plane: $K_B+U_B=K_C+U_C$. The final height $y$ can be calculated using $y=\frac{v^2}{2}$, where $v$ is the velocity after the collision and $g$ is the acceleration due to gravity.