11. Momentum & Impulse
Collisions with Springs
Collisions with Springs
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Hey guys, so for this video, I'm gonna show you another type of collision with motion and energy type problem, which is where you have a collision and then some interaction with the spring. So in our example that we're gonna work out down here, we have a crate that collides with another crates, that's the collision parts. And then afterwards these crates are gonna stick together and they're gonna compress up against the spring. So that's the motion or the energy part of the problem. There's really nothing new to this. We're just gonna use conservation of momentum for the collision. That's what we always do. And then conservation of energy for the spring parts. Alright, so there's nothing new, we've seen how to do these kinds of problems before let's go ahead and take a look. So we have the initial speeds of the crates and we want to go ahead and draw the diagrams and then label the points of interest. Right? So before the crates actually collide, that's before the collision, that's gonna be part A. After the collision it's going to be part B. And that's also right where the motion starts. That's right. When the blocks start compressing up against the spring and then finally, when they reach that maximum compression distance, that's where the motion ends. That's gonna be part C. So what we really want to find here is we want to find the maximum compression distance of that spring. So before the collision, what happens is that the spring is just relaxed like this and then afterwards when these when these two creates combined, they compress against this spring. Now the tip of the spring is over here? We want to find is this distance over here? What is this? X. How far did it actually compress? Alright, so let's go ahead and take a look at the second step. We're gonna write our momentum energy conservation equations. So, I'm gonna go ahead and do that over here and we're gonna use conservation of momentum for the collision and then conservation of energy for the motion parts. So, I've got M one V one A plus M two V two A equals M. One V one B plus M two V two B. And for the energy part from B to C. I've got K. B plus U. B plus work non conservative equals K. C. Plus, you see. All right. So, now we have our two conservation equations. I just figure out which one I'm going to start with. And hopefully you guys realize that what I'm trying to actually find here is what's the maximum compression distance, which is basically what is the distance when you reach part, see where the motion ends. So, we're trying to find X. C. That's gonna come from the potential energy term inside of my conservation of energy equation. So, that's what I'm gonna start with. So, let's go ahead and expand out each of the terms figure out what I can cancel. So, do we have any kinetic energy initial? At B. Well, yes, because these two crates are gonna stick together and they're both gonna start compressing the springs? They have to have some kinetic energy. Is there any potential energy? Or remember that this potential energy is gravitational plus elastic potential energy. And right here at point B, that's where the spring is still relaxed. It hasn't started compressing yet. So there's no gravitational potential and there's also no elastic potential energy. Remember your elastic potential energy is one F. K. X. Squared, but it hasn't started compressing yet. There's also no work done by non conservative forces because you're not doing anything and there's no friction. What about K. Final? So what happens? I'm sorry. That's there's no potential energy here. So what about K. C. What will hopefully you guys realize that the maximum compression distance of the spring is gonna happen where the blocks have stopped moving. So when the spring is maximally compressed, the velocity here for both the crates is gonna be zero. And so therefore there's no kinetic energy. And so therefore all of this sort of potential energy happens because you have some you have some compression distance like this, right? So we're gonna go ahead and expand, expand our terms. So our kinetic energy at B. Is gonna be one half and we're gonna use a little M. Or we're gonna use big M. What happens is these two crates are going to combine together? They're gonna stick to each other. So this is a completely an elastic collision. And so what happens here for both the crates is that your big M. Is equal to M one plus M. Two, they stick together. So we're gonna use one half big M V B squared equals and then this is gonna be one half K times X c square. So there's our target variable, that's what we're looking for here. And unlike what happens in other problems are mass is not going to cancel, we have em on one side and then K on the other, but we can cancel out the one half and then we can write an expression for this XC term here. So let's go ahead and do that. So X. C. When I basically move the K over and I take the square roots, I'm gonna get the square roots of M over K times X. Symes V b Squared. So this is gonna be the square roots of, I've got the big mask, which is really just the combination of 10 and 30, so it's gonna be 40 here divided by K. My compression of my spring constant was equal to 500. That's what I was given here. So all I have to do is just figure out what is my velocity. So notice how I don't have the velocity after the collision and where the motion starts. So what happens here is I need to go figure out this VB by looking at my conservation of energy equation because I have VBS over here. Alright, so once I can figure that out, that's what I'm going to plug into this this equation right here. So let's go ahead and and do this conservation of momentum. So I'm gonna have 10 times 20 if I call this object one and object to so 10 and 20 plus this 30 kg block is actually initially at rest, so V two is equal to zero up here, so I have zero over here cancels out and this is gonna equal well remember if this is a completely an elastic collision, then both of these velocities here are actually gonna be the same. So I can group them together. Um and I can basically use my my completely an elastic collision shortcut. So I'm gonna use 10 plus 30 and they're both gonna have the same speed of VB. So that's what I'm looking for here. And so this is gonna end up being 200 divided by, I'm gonna move this 40 to the other side, that's gonna equal V. B. And you're gonna get five m per second. So that's the number that we plug into this equation over here. So if you just plug in this square root of 40 times whatever, You're gonna get the maximum compression distance of 1.41 m. And that's the final answer. So again, you're just using conservation of momentum and energy to solve your problem. That's it for this one guys, let's move on!
An 8g piece of sticky clay strikes and embeds itself in a 0.992kg block at rest on a frictionless, horizontal surface. The block is attached to a spring with a spring constant of 5N/m. The impact compresses the spring 75.0 cm. What was the initial speed of the clay?