Push-Away Problems - Online Tutor, Practice Problems & Exam Prep

In push away problems, conservation of momentum is crucial. When two objects, like a gun and bullet, push away from each other, their initial momentum is zero, leading to the equation: m_1 × v_1 × f + m_2 × v_2 × f = 0. Additionally, energy conservation relates elastic potential energy in springs to kinetic energy post-release, allowing calculations of stored energy and compression distance. Understanding these principles is essential for solving related physics problems effectively.

"Push-Away"Problems

Guys, in the last couple of videos, we saw how to use conservation of momentum to solve problems where objects are interacting. Interactions between objects usually fall into two broad problem types. We've already seen the first one, which is called a collision. In the time you have a collision, the basic setup of these problems is you have objects that are moving towards each other and they're going to collide. We saw a couple of different outcomes of this, and we'll talk about that in a later video.

What I want to talk about in this video is the second type of problem, which I like to call a push away problem. We're going to see a couple of examples of this. Some really common examples include objects that are thrown or like the recoil of a gun, which is actually the one we're going to solve below. We'll also see things that are pushing off of each other like ice skaters on a frozen lake, or astronauts in space, or something like that. You might even see some problems involving fireworks explosions. But the general setup, the reason I like to call these push away problems, is because objects are initially together. And then what happens is they're going to push away from each other, and then afterwards they're going to be moving in opposite directions. And really, this is just because of action-reaction. But what's fortunate for us is that regardless of the problem type, whether it's collisions or push away problems, we're going to solve all problems by using conservation of momentum. Any problems involving interacting objects can be solved by using conservation of momentum because momentum always has to be conserved. That's a sacred law of the universe. Whether it's collisions or pushaway, momentum has to stay the same from initial to final.

Alright. So, we have a 4-kilogram sniper rifle that's shooting a 5-gram bullet. Basically, what's happening here is I've got this little gun like this. So that's my little gun, and I've got the bullet inside. This is before, but then afterwards, what happens is that the sniper rifle actually fires the bullet. I've got my gun like this and then I've got the bullet going away like this. I've got the mass of the gun which is 4 kilograms and I've got the mass of the bullet which is 0.005 kilograms. Now, what happens is, we know that the initial velocity of the "before" case, before the bullet is actually fired out of the barrel, is actually going to be 0. The initial speed for both of these objects, the bullet sitting inside the gun and the gun not moving anywhere. So both of these objects have our initial speed of 0. So, we have

V v2 final = 600 meters per second

I want to figure out what is v1 final. What is the recoil speed of the gun? So that's the first step, drawing the before and after. Now we're at my enter my conservation of momentum:

m1 v1 initial + m2 v2 initial = m1 v1 final + m2 v2 final

Let's look at each of our terms. We know the masses of all the objects. Now I just need to know the speeds. Well, like we just said, the initial speeds for both of these objects, the bullet and the gun, are just equal to 0. What does that mean? It just means that for both of the terms on the left side, they're actually both going to equal 0. So I end up with 0 momentum on the left side. In fact, this is going to happen in a lot of your push away problems. In most of your push away problems, objects are going to be initially at rest, which just means that the total momentum of the system initial is equal to 0. So that actually means that the conservation of momentum equation is going to simplify for us. Let's check out how. So we have:

0 = m1 v1 final + m2 v2 final

If you start out with 0 momentum, you have to end with 0 momentum. So how do I solve for this v1 final? Basically, I can just move this to the other side like this. And what you'll end up with is:

- m1 v1 final = m2 v2 final

If you have zero momentum initial and final, and you have these things that are moving in opposite directions, it means that the magnitude of those real momentums have to be the same. They're just in opposite directions. That's what this negative sign means. So this is going to be the general setup. Negative m1 v1 final is negative m2 v2 final here. So we can solve for this v1 final. and we can say that v1 final is equal to negative. Now we're just going to plug in the numbers.

m2 = 0.005 , v v2 final = 600 , and you ' re gonna divide on the mass , which is 4

If you go ahead and work this out, you're going to get -0.75 meters per second. So as expected, what happens is that the gun has actually been recoiled to the left. The bullet was traveling to the right at 600 meters per second, and so it has some momentum in this way. So the gun has to recoil backward and its speed is 0.75 meters per second. Basically, what happens is if the gun or if the bullet has gained 10 momentum to the right, the gun has to recoil and gain 10 momentum to the left. I'm just using 10 as an example there. Right? So that's the speed. It's -0.75.

Alright. So that's it for the problem. I have one last sort of conceptual point to make, which is that momentum is conserved like we had in this example here only if your system is isolated. So we've seen that word before. Isolated just means that all the forces that are acting in your system or in your problem are internal. So remember, we defined this system as the gun plus the bullet. And if you look at the forces, what's happening is that the gun exerts a force on the bullet to the right. That's what shoots it out of the barrel. But because of action-reaction, the bullet has to exert a force backward on the gun. Now even though there are two forces here if you've defined your bubble to be the gun and the bullet, then these forces are internal, which means that momentum is going to be conserved. If I had an external force, if I actually put my hand against the gun with the bullet inside and pushed it, that's an external force that's coming from outside of the system. And therefore, momentum is not going to be conserved.

In your problems where you define your system as both of these interacting objects, these forces are usually going to be internal, which means your momentum is going to be conserved. That's what we can use our conservation of momentum equation.

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

An 80-kg astronaut is stranded floating in space is 30 m away from his spaceship. He wants to return to his spaceship in 20 s. How fast must he throw his 2-kg space hammer, directly away from the spaceship, to accomplish this?

• A. 40 m/s
• B. 120 m/s
• C. 240 m/s
• D. 60 m/s

Sliding Football Player

Guys, let's take a look at this problem here. Hopefully, you got a chance to work it out on your own. If not, here's a little bit of help. So the idea here is that we have a football player on frictionless ice who has a football. So they're both moving along together. He's got this football like this, and they're both moving at 2 meters per second. Afterwards, the player is actually going to throw the football ahead at about 15 meters per second. So this is going to be a push away problem. Both objects are together, and now they're moving separately afterwards. Let's go ahead and draw our before and after. So this is our before. What does our after look like? Well, basically, now what happens is you draw the football player like this, and now he's thrown the football ahead of him. We can actually figure out the speed. So what I'm going to do is I'm going to call this m1, which is 70, and I'm going to call this m2, which is 0.45, and we're told that the player is going to throw the ball with 15 meters per second relative to the player. What does that mean? Well, this is going to be my v₂ final here. This means that they are both initially, already moving at 2 meters per second, then the football player throws it an additional 15. So what actually happens is v=2+15=17. That's the new final velocity. We want to figure out what the player's velocity is after that happens, after this push away. Alright. So let's go ahead and set up our momentum conservation equation. This is going to be m1v1initial+m2v2initial=m1v1final+m2v2final. Okay. So, basically, I've got this 70 times something plus 0.45 times something equals 70 times something plus 0.45 times something. Alright. So what goes inside each of these parentheses? Well, remember, initially, they're actually both moving together. So these speeds are actually the same, and they're both moving with initial 2 meters per second. So that's what goes inside here. So there's 2. Afterwards, we're looking for the final velocity of this v1. So this is actually what goes inside here. That's my target variable. What about the football? Remember, the football we just calculated is moving at 17 meters per second forwards, so that's what goes inside here. Now we just go ahead and simplify the left and right sides. So what this becomes here is, on the left side, I get 140.9, and on the right side, I'm going to get 70 times v₁ final plus, and then this is going to be 7.65. So all I do is just move and subtract the 7.65, and I end up with 133.25 equals 70 times v₁ final, and I can just go ahead and divide by the 70. What you get is v₁ final is equal to 1.9 meters per second. So let's see. Basically, what happens is that the football player is still moving to the right, just slightly slower than he was initially. Initially, both objects were moving at 2 meters per second, and then because the football is a lot lighter, the fact that it's going at 17 meters per second now means there hasn't been a lot of recoil in the much heavier football player. So the football player has a final velocity of 1.9, so he's only slowed down a little bit, but he's given a lot of momentum to the football. Alright? So that's the idea here, guys. Let me know if you have any questions, and I'll see you in the next video.

Push-Away Problems with Energy

Guys, in the last couple of videos, we saw how to use momentum conservation to solve these kinds of push-away problems when you have objects that are pushing away from each other. But in some of these problems, you're actually going to be asked for the energy before or after a push-away event. For example, in the example we are going to work on down here, we have two boxes that are pushed up against this spring. In part b, we want to calculate how much energy is stored inside that spring. So, I'm going to show you quickly how to deal with these push-away problems with energy. The main idea here is that you're actually going to use two conservation equations. We're going to have to use momentum. We're always going to use momentum conservation with these problems, but we're also going to have to go back and use energy conservation as well. Let's go ahead and take a look at our example here. We've got these two blocks, 3 kg and 4 kg, and they're pushed up against this light spring, and you release it, and the spring basically is going to fire both of them off in opposite directions. The blocks are going to get released. So, we want to draw a diagram for before and after. This is the before, and afterwards, what happens is you sort of have these boxes that start pushing away from each other like this. So this 4 kg box goes like this, and the 3 kg box is now going to the left. We actually know what the final velocity of this 3 kg block is. It's launched at 10 meters per second. So what happens is I'm going to call this \( m_1 \) and \( m_2 \), which means that \( v_{1,\text{final}} \) is equal to -10 because it's going to the left. And then \( v_{2,\text{final}} \) here is actually what we want to solve for in part a, the recoil speed of the 4 kg block. So this is what we're going to look for in part a and basically, we're just going to use momentum conservation. We know how to solve these kinds of push-away problems. So once we have our before and after, we're just going to write our momentum conservation. So for part a, we're going to do \( m_1 v_{1,\text{initial}} + m_2 v_{2,\text{initial}} = m_1 v_{1,\text{final}} + m_2 v_{2,\text{final}} \). Now remember, what happens in these kinds of problems is that initially, the system is at rest. So right after you release your hand, the blocks are still actually at rest which means that the initial velocity is equal to 0. So what that means is that basically both of our left terms are actually going to go away and we have 0 initial momentum. So the system's initial momentum is equal to 0, and that means that our momentum conservation equation is going to simplify because we know that the final momentum has to be 0 as well. So basically, what happens is we're going to have \( -m_1 v_{1,\text{final}} = m_2 v_{2,\text{final}} \). If the right block gains 10 momentum this way, the left block has to gain -10 momentum that way. And so, they have to cancel each other. So we can we can basically solve for this \( v_{2,\text{final}} \) here, and we can start plugging in values. So I have -3 times -10, which gives \( 30 \) when divided by 4, what you end up getting here is that \( v_{2,\text{final}} = 7.5 \) meters per second. So notice how we got a positive number, and that should make some sense. If we actually got a negative number, so if we forgot one of these minus signs, we would have gotten a negative number, and that would have meant that this block is actually going to the left and that doesn't make any sense. So it's always a good idea to keep track of your numbers and make sure they make sense. So this is a positive answer, positive 7.5, and that's just using momentum conservation. Let's take a look at part b now. Part b is asking us to figure out how much energy is stored inside the spring. So what's going on here is that you had these blocks that are pushed up against the spring, and there's some elastic potential energy that is stored inside the spring because they're compressed. Once you release, what happens is that these two blocks start firing away from each other. They separate like this, and the spring decompresses, which means that the final potential energy is equal to 0. So what happens is there's some stored energy here and then this spring is no longer storing that energy. Where did that energy go? Well, basically, what happens is that you have these two blocks that have now mass and speed and therefore, there's some kinetic energy. This is \( k_{2,\text{final}} \) and \( k_{1,\text{final}} \). So basically, we're going to use energy conservation to figure out the energy instead. So you're going to use \( k_{\text{initial}} + u_{\text{initial}} + \text{work done by non-conservative} = k_{\text{final}} + u_{\text{final}} \). Now what we said here is that the initial velocity of the system is 0, so there's no kinetic energy. The initial energy that's stored here is going to be elastic potential energy. There's no gravitational energy because everything's sort of on the horizontal plane like this. What about work done by non-conservative? Once you release your hands and the boxes start flying this way, there's no work done by you, and there's no work done by friction as well. So we're on a smooth floor. And then finally, what happens is we have some kinetic energy because the boxes are moving, but we have no potential energy because the spring has decompressed. So what happens is we have \( u_{\text{elastic}} = k_{\text{final}} \). Now what happens is we have 2 blocks that are moving. So the kinetic energy is actually going to be \( \frac{1}{2} m_1 v_{1,\text{final}}^2 + \frac{1}{2} m_2 v_{2,\text{final}}^2 \). The kinetic energy is made up of 2 objects that are moving now. So we actually have all those numbers. We can go ahead and calculate this. So this \( u_{\text{elastic}} \), the stored energy inside the spring is \( \frac{1}{2} \times 3 \times (-10)^2 \), and it becomes positive once you square it, plus \( \frac{1}{2} \times 4 \times 7.5^2 \), which you end up getting is 262.5 joules. So that's the energy that was stored inside the spring. Let's take a look now at part c, which is now we want to figure out how much the spring was compressed before launching the blocks. What we just calculated was we just calculated the elastic potential energy, 262.5 joules. What we want to do now is we want to calculate the compression distance which is really just going to be \( x_{\text{initial}} \). So we can do is we can just relate the elastic potential energy to \( \frac{1}{2} k x^2 \). So what we're looking for here is \( x_{\text{initial}} \), and we have the elastic potential energy, and we also have the spring constant. It's 800, newtons per meter. So we can just go ahead and, I'll just go through this quickly. This is \( \sqrt{\frac{2 \times u_{\text{elastic}}}{k}} \), and that equals \( x_{\text{initial}} \). So when you plug this in, what you're going to get is \( \frac{525}{800} \), and you're going to get 0.81 meters. That's the final answer. So these are all your answers. Right? This is how you use momentum and energy conservation to solve these kinds of problems. That's it for this one, guys. Let me know if you have any questions.

Here’s what students ask on this topic:

What is a push away problem in physics?

A push away problem in physics involves two objects that start together and then push away from each other, moving in opposite directions. Common examples include the recoil of a gun when a bullet is fired, ice skaters pushing off each other, or astronauts in space. These problems are typically solved using the conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it. The initial momentum is usually zero, leading to the equation: $m_1\times v_1+m_2\times v_2=0$.

How do you solve a push away problem using conservation of momentum?

To solve a push away problem using conservation of momentum, follow these steps: 1) Draw a diagram of the initial and final states of the system. 2) Write the conservation of momentum equation: $m_1{v}_1+m_2{v}_2=0$. 3) Substitute the known values for masses and velocities. 4) Solve for the unknown velocity. For example, if a 4 kg gun fires a 0.005 kg bullet at 600 m/s, the recoil speed of the gun can be found using: $-m_1{v}_1=m_2{v}_2$.

What role does energy conservation play in push away problems?

Energy conservation is crucial in push away problems, especially when dealing with springs or other potential energy sources. The total mechanical energy (kinetic + potential) of the system remains constant if no external work is done. For example, if two blocks are pushed by a compressed spring, the elastic potential energy stored in the spring is converted into the kinetic energy of the blocks. The energy conservation equation is: $U_{\mathrm{elastic}}=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$.

How do you calculate the recoil speed of a gun in a push away problem?

To calculate the recoil speed of a gun in a push away problem, use the conservation of momentum. Assume the initial momentum is zero. The equation is: $m_1{v}_1+m_2{v}_2=0$. For a 4 kg gun firing a 0.005 kg bullet at 600 m/s, the recoil speed $v_1$ is found by solving: $-m_1{v}_1=m_2{v}_2$. Plugging in the values: $v_1=-\frac{0.005\times 600}{4}=-0.75 m/s$.

How do you determine the energy stored in a spring in a push away problem?

To determine the energy stored in a spring in a push away problem, use the conservation of energy. The elastic potential energy stored in the spring is converted into the kinetic energy of the objects. The equation is: $U_{\mathrm{elastic}}=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$. For example, if a 3 kg block moves at 10 m/s and a 4 kg block moves at 7.5 m/s, the stored energy is: $U_{\mathrm{elastic}}=\frac{1}{2}(3\times 10^2+4\times 7.5^2)=262.5 J$.

What is the importance of defining an isolated system in push away problems?

Defining an isolated system in push away problems is crucial because it ensures that the conservation of momentum applies. An isolated system means that all forces acting within the system are internal, and no external forces influence the system's momentum. For example, in a gun and bullet scenario, the forces between the gun and bullet are internal. If an external force, like a hand pushing the gun, is introduced, the system is no longer isolated, and momentum conservation does not hold. Therefore, correctly defining the system helps in accurately applying the conservation laws to solve the problem.