Collisions & Motion (Momentum & Energy)

Concept: Intro to Collisions with Motion

18m
Video Transcript

Hey guys, so now that you've learned how to use conservation of energy to solve motion problems and you've learned how to use conservation of momentum to solve either push away problems or collision problems we're going to put the two of them together into more elaborate sort of longer questions and these tend to be pretty popular especially if your professor is doing long answers or full response types of questions so let's check it out. Alright, so it says here some collusion problems have more than one part so rather than just having a collision we're going to have some motion either before or after the collision, so one or both objects will be moving either before or after the collision probably the most common example is a bullet that gets shot into a block and this block moves and it eventually stops. So, in these problems we will use as I mentioned both conservation of energy and conservation of momentum and the order might change depending on what's happening, energy will deal with motion the motion part of the problem and momentum will deal with the collision or the push away two objects pushing away from each other part of the problem so let's jump straight into the example and show how this works. A 20 Kg crate slides on a smooth horizontal surface so let's draw a horizontal surface here, crate has mass 20 Kgs and it slides this way with 40 meters per second, Ok? Smooth meaning the friction is zero, no friction, OK? The box then collides and sticks to a 30-kilogram crate that is initially at rest so there's a second crate so over here of mass 30 that is initially at rest, OK? So, I'm going to put here 0 meters per second not moving. After the collision, the crates stick together and go up in a frictionless again frictionless incline that makes an angle of 37 with the horizontal so somewhere after the collision point there's an incline like this makes an angle of 37 with the horizontal and I'm going to draw the 30 and the 40 sort of going up together, 30 and 40 going up together, this is after the collision eventually they will stop, right? Eventually they will stop that's common sense but also because it tells me how high up the hill the crates reach meaning they do eventually stop and then they come back down, right? They don't like fall over the hill over the incline and flyovers, so at some point they stop and I want to know what is the final heights over here? So, for now I'm going to call this H and I'm also going to want to know what is the length of the plane here? The length of how far up the plane do you go, OK? So just to be clear how high up the plane is H it's a vertical and how far up the plane is L it's this distance here versus this distance here, they're difference, OK? Slight difference in wording and it means a big difference, I want to really quick remind you before we jump into this that there is a relationship however between L and H which is when I have L here and I have theta here and I have H here, H is L sine of theta, so later on what we'll see is that if you find one you can get the other. Alright so let's go how do we find H? Well I want to first point out that there's two parts to this problem, the first one is from here to right about the first thing that happens here is the collision. First thing happens is the collision, so yes, this object is moving there's some motion before the collision but that motion doesn't change it's just moving with forty all the time so we don't worry about that, the first thing that happens the first change that happens is there's a collision somewhere here. After the collision, the objects will move up the hill there's some sort of motion here, OK? Because there's a collision here we're going to use the conservation of momentum equation (MOM) and I'm going to write that as a shorthand for momentum and here we're going to use the energy equation because there's motion. Now there's two parts, I'm going to go a little slower because I'm first introducing this but towards the beginning here of this piece which is what happens before the collision is the initial part after the collision is final, OK? Now, when you go to the second part of the problem the initial part of motion is here which is initial and the final part of motion is here which is final, now there's a problem here which is I have two initials and two finals and the final velocity of the first part of the problem is the same as the initial velocity of the second part, so here we can run into some complication so what I like to do and this is what I want to show you is instead of using If I'm going to use A, B and C and I'm going to draw a little diagram to help us organizer our thoughts, I like to draw this little diagram to show this is point A we call point eight in this case there is a collision here, point A actually let me put the collision up here, there is a collision, point A is the point or the velocity before the collision so I'm going to write pre-collision initial, the final point here is after the collision so I'm going to write post-collision, this is Point B. Point B is the end of the collision and the beginning of motion and point C is the end of the motion so I'm going to call this the end of motion where the object stops. OK so instead of saying in initial, final and initial and final and having to remember that the final of first the same as the initial second, I'm going to refer to these points as A, B and C, OK? I'm going to refer to this mass as the first mass just because it's on the left and the second mass as masses 1 and 2. So I have two equations to write because I have two conservations, I have the conservation there's a collision here so I'm going to write......Sorry I meant to write momentum here because there's a collision and I meant to write energy here because there's conservation of energy there, OK? so I'm going to write the momentum equation from points A to B. So, between A and B I have a momentum equation so it's going to look like this object one and two so it's M1, V1 plus M2, V2. Again, I'm setting this up for you because once you get a nice system it'll be easy to solve this. Now I'm going from A to B so it's initial and I'm going to call it not initial but we're going to call it A and instead of final we're going to call it B, that's the first equation. The second equation I can write is an equation from points B to C, it's the second part of the problem and the objects are moving so I'm going to write a conservation of energy equation, K+U+Work (Non-conservative) = K+U, now I'm going from B to C so it's B-B, C-C, alright? So, the first step is going to draw this stuff, the entire situation I like to quickly draw a little diagram that summarizes what's happening and then I write my two equations and then you just plug numbers into the equations and you solve now, now there's a lot of algebra obviously but let's check it out. Alright, so what are we looking for? We're looking for H final, OK? So, I want you to look at these two equations and tell me where do you see an H final? Now by final It's the second part, It's the final of the second piece so this is A, this mill here is B, this is C, so I want to Hc, do you see an Hc in either one of these equations? And the answer is no there is no Hc. However, it is here, somewhere right? I got as over here, Bs over here, I got to B here and C here, C is here, do you see an Hc in here? And I hope you saw or you realize that there is an Hc here because the gravitational potential energy of point at point C is MGH at point C, so when I expand that equation the H is going to show up, now you don't necessarily have to do this you don't have to go look for the variable and figure out where it is you can just start expanding this and then look for it, OK? You don't have to know exactly where it's going to come from but it's helpful sometimes. Alright so I'm going to start here just to make the H show up kinetic energy, is the kinetic energy at point B? Point B, let's refer right here point B is post-collision immediately after the collision is this object moving? And the answer is yes in fact there will always be the case that post-collision the object is moving, that's the whole point of these questions so I have kinetic energy. What about potential energy? Remember there's two types of potential energy one is gravitational if you have a height and one is elastic if you have a spring, there is no springs here and right after the collision both boxes are still on the floor so there is no gravitational energy of any type. What about work non-conservative? Work non-conservative is the work done by you if you're doing anything here if you are not, you're just watching plus the work done by friction, there is no friction both this surface and this surface are smooth they're frictionless therefore there is no work done by you or no work done by friction. What about kinetic energy at point C? Point C is at the end where it stops and by definition since it stops and there is no kinetic energy at point C, what about potential energy at point C? Well again potentially there is no springs so there is no elastic but the object did gain some height so we are going to have potential energy. So now I can expand this equation here, I can expand the K into 1/2MVb squared equals MGHc and that's my variable right there that's what I need to get, OK? Before we go to the other equation let me talk about the mass, OK? Now while this object is going from right after the collision to up the plane which mass do you use? Is it the 20? Is it the 30? And I hope you're thinking that it's actually....By the way I wrote the masses wrong here, I hope you caught this I'm so sorry this is 20 and 30, I hope you thought that the mass you should use is actually both and that's true, it's both because they stick together so mass here is going to M1+M2 which in this case is 50, that being said that's the correct mass even if you got the wrong mass in this particular case it wouldn't have mattered because the Ms cancel anyway, OK? So, you wouldn't really write what the M is because it's going to cancel, sometimes the M isn't going to cancel so it's important it's important that you know how to figure out which M to use and in this case, Hc if I move the G over to the over side will be this. Now look what happens, I'm ready to plug stuff in or I'm almost ready to plug stuff the problem is I don't know what Vb is, I know what 2 and I know what G are but I don't know Vb, so what do I do? Well that's why there's two equations, I'm going to go over here and find Vb, okay? And that's what you're going to do in these problems, you're going to start from one equation going to the other if you don't know which equation to start from it doesn't matter just pick one, and if you hustle through it you're going to get it. Alright, now I need Vb notice that there's V1b and then there's V2b, well what we know about this problem? The objects are stuck together so these two velocities are actually going to be the same, OK? So, I can write this as a M1Vb+M2Vb, I don't have to write V1B and V2B I just write Vb because it's the same and what that means is that I'm going to be able to combine the two like this? Vb(M1+M2), what about on the left side here? The first mass is a 20, first mass......Let me actually write M1V1a, this velocity is a 40 but this velocity here is a 0, A is before the collision and before the collision the 30 kilogram isn't even moving, OK? So, this is all I have, it simplifies here like this, OK? Notice that I'm trying to avoid I'm trying to delay plugging numbers until the end just so you can see can see how to do this with the letters only, Vb equals M1V1a divided by M1+M2, I have all these numbers therefore I can just plug stuff in, M1 is 20, the velocity is 40, the masses are 20+30, so this is 80 divided by 50, do I have this here? I do, it going to be 16 meters per second. So Vb and B remember is post-collision, the velocity of post collision is 16, it drops from 40 to 16 makes sense because the object got heavier, it went from being 20 Kg block moving to now 20 with a 30 moving together so they're going to move slower, OK? Now that I know that Vb is 16 I can finally go back and plug this in here, OK? And I'm going to have 16 squared divided by 2 times 9.8 and if you plug this into the calculator you get 13.1 meters. Now that's the final answer, this thing is going to go up a distance a height of 18.1 meters. Now all of this is just for part A but luckily part is B pretty easy so all of this is part A, I'm going to part B in a little corner here. In fact, we had just talked about Part B, if I want to know how far up I want to know L I can use this equation right here, L=H/ (sine of theta), H we just got over here it's 13.1 divided by the sine of 37 and this means that L is 21.8 meters, cool? Again, took a little bit longer then what you would usually take to do or to explain one of these questions but I wanted to introduce the idea of using A, B and C and slowly show you what the equations are, we're going to do a good amount of practice because these are very popular and you should be able to do a bunch of these. Luckily, they're also very predictable so we're going to do a bunch of them and you're going to be ready, hopefully this makes sense let me know if you guys have any questions.

Problem: A 40-kg crate is released from rest from the top of a 10-m long, smooth inclined plane that makes an angle of 53° with the horizontal. At the bottom of the plane, a 50-kg crate is initially at rest on a smooth horizontal surface. After hitting the second crate, the first crate stands still. Calculate the speed of the second crate after the collision.

11m

Concept: Collisions & Motion: Finding Initial Velocity

11m
Video Transcript

Hey guys, so in this video we're going to look at a slight variation of these collision with motion problems, now when you have both collision and motion or motion and collision, you're going to use both conservation of energy and conservation of momentum equation, in the standard problem something gets hit and the objects that gets hit moves maybe goes up a hill and you might want to know how far up the hill does it go? Or how much does the spring compress given the initial velocity? In other types of problems which is what we're going to do now, we're going to work backwards and I will tell you how far the block goes and ask for the initial velocity in the beginning of the problem so let's check it out. Alright so it says here in some collision motion problems you're going to be asked to calculate the initial velocity, so the problem works backwards, right? It's basically kind of like forensics if the block moves this much how fast must the bullet be moving before it hits the block, OK? So here I have a 300-gram bullet that strikes a 10 Kg block that is at rest so let me draw this real quick, here is the bullet I'm going to call it mass 1 and it's 0.3 kilograms and it's going to strike 10 kilograms block so mass 2 is 10 kilograms, the bullet has some sort of initial velocity the block is initially at rest, OK? It's on a horizontal surface, the bullet is going to stay inside of the block so later on the bullet it's going to stay inside of the block and it's going to cause the system to move after, right? it's as if the coefficient of friction between the block and the surface is 0.6 , so it's a coefficient of friction between the two so what I'm going to do is I'm going to draw the floor like this the surface like this just indicate that there's friction and the block slides a total distance of 35 meters so after being shot the block goes from being at rest and then it's slides 35 meters so let's say 35 meters ends right here, so this distance I'm going to call it delta X, this displacement is 35 meters. I want to know the speed with which the bullet hit the block, in other words I want to know the pre-collision speed of the bullet, OK? Now if it travels a total distance of 35 it's implying that the final velocity here is 0, that the block stops Ok? And as always, we're going to use conservation of energy and conservation of momentum, the first that happens in this problem is a collision so from A to B I have a collision so therefore we're going to use the conservation of momentum equation, and from B to C is where the block starts moving and then it stops so at point C it stops and we're going to use the conservation of energy equation here. I want to remind you that before the collision is A in this case because it's to the left here and then after the collision is B so this is pre-collision, we call B post-collision and we call C the end where the object stops. So, when I wrote here that V Final is 0 just to be more specific this is Vc, OK? Vc of both of block and the bullet, alright? So let's write these equations from A to B and B. to C. from A to B I have as it says here a momentum equation so I'm going to write a M1V1, M2V2, M1V1 M2V2 and this is from A to B, OK? I'm going to write the B to C equation already, it's going to be different color from B to C, I have the energy equation so it's K+U+ work (non-conservative) = K + U, in this case we're looking at B-B C-C, Alright? And I want to know the initial velocity of the bullet, bullet is one and before the collision is A so I know V1A, OK? [00:32:06.18] And I really think it helps to know exactly which variable with all the little subscript and what not you want once you identify that correctly you can just look at the equations here and you know exactly what you need and what you need is this guy right here, OK? So, because this is what we're looking for I'm going to start there, I'm going to start with that equation.

Now let's see what I have I have all the masses, the initial velocity of the second object before the collision, right? Is zero the block doesn't move so I can second scratch this whole thing and do I have these final velocities? I check this off but I actually don't have this I don't have V1B I don't have V2B however the bullet stays lodged this is a completely elastic collision and even though I don't have these velocities I do know that at least they are the same which means I can combine them so instead of writing V1B and V2B, I'm going to write just Vb and I'm able to combine the masses and M1+M2 because they are moving together and then here I have M1V1a, M1V1a is what I want so what I can do is just divide both sides by M1 I'm going to go ahead and remove this M1 here and we are done with this part of the problem with this equation here where as long as I have all these numbers to the right of the equal sign I'm good, I have everything except Vb I don't have Vb so what do I do I go over here I go to the other equation to find Vb and Vb by the way if you look here you should know that Vb will show up inside of Kb kinetic energy is half MV squared and I'm looking for the B one so it's going to be here but you don't necessarily have to look for it before expanding you can just go and expand it and it's going to show up, cool? So even if you're not sure if it's there sometimes it happens you expand the whole equation it's going to show up, so is there kinetic energy at point B? Kinetic energy a point B, Point B is post-collision and in post-collision you're always going to have kinetic energy so yes. There is no potential energy at point B because and no moments this thing either go up a hill or compresses as spring so no there's no potential energy in the beginning or in the end, is the work non-conservative? Work non-conservative is the work done by you, you're not doing anything you're just watching so you don't do that there's no work you but there is friction there is work done by friction we'll get back to that in a second and is there kinetic at point C? The answer is no because a point C you stop, alright? So, let's expand this, this is 1/2MVb squared Vb is what I want and then plus work non-conservative, I'm going to go off to the side here and I'm going to derive work non-conservative, OK? So, this is sort of to the side, work non-conservative is just the work done by friction and remember that the work done by friction is -Fd remember that friction is mu normal and remember that normal in this case here is because there are no other forces acting on the Y axis normal equal to MG. So now we're going to work our way back if normal equal to MG then friction is mu MG and if friction is mu MG then the work done by friction is - mu MGD which is also the work done by non-conservative forces, so you got to do all that, you have to be good at doing that and this is what goes right here negative mu GD equals zero I'm solving for Vb so I'm going to move this guy to the other side and you see that the masses will cancel check it out then MGD the masses cancel, now if the masses didn't cancel it will be important for you to know which mass to use and in this case mass would have been M1 plus M2 because the system moves together, the two objects move together but you could screw that up because the masses are going to cancel anyway so even if you had the wrong number there you'll be fine, Vb is what I'm looking for it's going to be the square root of 2 mu GD and if I plug in these numbers mu is 0.6 gravity is 9.8 and the distance D or delta x doesn't matter is 35 if you take the square root of all this you get I have here 20.3 as the velocity.

Now 20.3 is the Vb which is the post-collision velocity of both of these things but that's not what we want what we what is the initial velocity so what I got to do is plug this number into here, OK? But now we're basically done just got to plug it in, so V1A= 20.3 the masses are 0.3 and 10 and this is 0.3 and let me see do I have this here, yup I got it so is going to be 697 meters per second and this is the final answer. The initial velocity of the bullet is 697 so we were able to work backwards to get the velocity, it's a very simple, it's very very similar to what we did before we're just asking for a different variable the process is pretty much still the same which is you draw the collision, I like to draw this to help organize our thoughts when you write the two equations and plug stuff in, so that's it let me know if you have any questions.

Example: Collisions with Pendulums

13m
Video Transcript

So, some collision motion problems involve pendulums. Yay pendulums alright so let me show you the pendulum equation real quick so pendulums are made of a bob (a mass at the bottom) let's put a little M there and then there is a string, this string is always going to be massless and it has a length L and then the pendulum is then going to be hit by something and go up ends up over here somewhere and this distance here is the height gained by the pendulum H and what we want to do is we want to have a relationship between L the length of the pendulum, H the height gained and the maximum angle that it makes with the vertical when you have pendulum problems the angle that matters is the angle with the vertical because the pendulum moves around the vertical axis like this, so to do that remember we draw a line here and that so that we form a little triangle, Ok? And then I'm going to write two expressions for this little segment here, one of those two expressions is going to be in terms of angles in terms of trig functions now this is the adjacent side of the angle so this is going to be L cosine of theta and you can just remember that on the pendulum equation it's cosine that you have, another expression I can write is if this is H just this piece and the whole thing is L then this piece here is going to be L-H so these are the two expressions, L-H and L cosine of theta so let's put them together, it's the same it's two expressions for the same piece so I can say that L-H equals L cosine of theta this is the pendulum equation this equation can be written in different ways I'm going to write it where we have the H by itself, I'm going to move it over here so that it becomes a positive so it's going to be H equals L-Lcosine of theta, here's another version it's the same exact thing except just written a little bit different but it's the same equation, Ok? So that's the pendulum equation you need to know for some of these problems let's do an example here real quick the rest is pretty similar to what we've already done in terms of using momentum and energy equation.

So a pendulum is constructed from a 40 kilogram block, so let me draw a little string here and there is a 40 Kg block and a light 2-meter long string that's the length and the length equals 2 meters, now light means that its massless the strings will always be massless in these problems, the pendulum is at rest so it has initial velocity V pendulum Let's call pendulum the second object because there's going to be a bullet here that's the first object so V2 before the collision so V2a equals 0. And then I have it's at rest and in its equilibrium position, the equilibrium position of the pendulum is down here so the picture looks like that, there's a bullet Mass 1 500 grams so 0.5 Kg goes this way with a velocity V1a of 700 meters per second, OK? The beginning of the problem is A it hits the pendulum and the pendulum is going to swing up this way and we want to know what is the maximum height that the pendulum will obtain? This height is always relative to the lowest point and we want to know what is this theta right here? So, we want to know H and we want to know theta. Alright, so what I'm going to do is I'm going to draw the little diagram that I like to draw showing that this problem has two parts, first from points A to B there is a collision so we're going to use the momentum equation and then from B to C this piece here there is a motion so we're going to use the energy equation, the initial part of momentum is before the collision so A pre-collision, B is after the momentum the final part of the momentum equation which is the velocity after the collision so post-collision and C is the end, the end of the motion where the object stops. So, I'm going to write an equation from A to B, I'm going to write the momentum equation M1V1, M2V2, M1V1, M2V2, this is from A to B, OK? So that's the momentum equation. Alright and the second equation is going to be from B to C and from B to C there is motion so I'm going to write an energy equation which is going to be K+U plus work non-conservative=K+U and this is from B so B-B to C C-C, the final height I want is for the motion part it's at the end of the second part of the problem so what I'm looking for is actually Hc and this angle here is theta C, OK? So, if I would Hc, Hc is here because it has to do with this the potential energy to height MGH So that's where we're going to find it, OK? So, I'm going to start here, now is there a kinetic energy at Point B? Point B is post-collision and there's always kinetic energy right after the collision so yes, what about potential energy? Potential energy will be 0 and that's because the pendulum is at its lowest point, remember with pendulums even though the pendulum is not on the floor what we do is we don't say that this is where 0, we move the 0 up because potential energy is relative to a reference point and we make the reference point here so the lowest point of a pendulum is where we say H equals 0 so at the lowest point there is no potential energy. Now B is post-collision the bob just started moving but it hasn't really gaining any significant height yet which is why say the potential is 0, there's no work because there's no friction and you're not doing anything you're just watching, the kinetic energy at point C is 0 because the block stops at the end, right? So there's no kinetic energy because the highest point it stops and there is potential because you've gained some heights, so let's expand this is going to be 1/2MVbsquare and this is MGHc, now Hc is what we're looking for notice that the masses cancel, so because the mass cancels since the mass cancels you don't have to know what the mass was but let's talk about it because sometimes it won't cancel and I want to point out to you that something different about this question is that the bullet comes in at 700 but the bullet actually emerges out of the block with 400 so the bullet sort of what keeps going here so the velocity of objects 1 after the collision is B is four hundred the bullet doesn't stay lodged inside, so for the second part which is the motion of the pendulum the bullet is not in there so you're only looking at the pendulum so the mass here is only M1 I'm sorry the mass here is only M2 which is 40, OK? However, it doesn't really matter because they cancel anyway, Ok? Alright so I want to find Hc, Hc is Vb squared/2G and at this point we can't make this any simpler we're ready to plug in the numbers except that we don't have Vb, right? You don't have Vb so we're going to have to go get Vb and then come back so obviously if we're stuck on the second equation then it's because we have to go to the first equation and in fact you see that there is Vbs right here.

So let's look at this first equation, I have all the masses, the initial velocity of the first bullet I have it is 700, the initial velocity of the block before the collision is 0 it's not moving and these final velocities here is what I want, now they're not.....actually I want one of them and I have the other, the velocity of the bullet after the collision is 400 so I have it and this is actually what I want, I want that Vb, OK? I want Vb of the second object don't confuse the two that's what I want, I have everything else so I can solve so let's start plugging stuff in. So the first mass.....Actually let me just derive this with the letters and then we'll plug it in at the end sometimes the professor might want to do that so M1V1a I want this guy by itself so I'm going to move everything out of the way -M1V1b=M2V2b, notice here that I have M1 and M1 so I can factor it out, I have a M1(V1a-V1b)=M2V2b, V2b is what I want so I'm going to divide both sides by M2 And V2b which is what I wanted right here I can find by plugging in these numbers. Cool? So let's do that now we're ready to plug we got all the numbers V2b is the first mass and M1 is the bullet which is 0.500 and M2 is the block which is 40 and the difference of the velocities, the first velocity of the bullet is 700, it comes out with 400 and if you do this I have it you get that this velocity is 3.75 meters per second, so the block gets hit and moves with 3.75 I can now put this in here and Hc will be 3.75 squared/2x9.8 and again if you do this you get to the height of C this 0.72 meters, that is the first answer part A, so all of this was for part A however luckily Part B is very easy, part B I want to theta C and we're just going to use the pendulum equation that I derived earlier OK? So, H=...Actually the first version of the equation's a little bit faster to work with for this particular question L cosine of theta=L-H, I want theta so I'm going to move L over here, OK? So, I'm going to have cosine of theta= (L-H)/L and theta becomes the arc cosine of the (L-H)/L make sure your calculator is in degrees and it's going to be the arc cosine of L is (2-0.72)/2, right and if you put this into the calculator you get that the angle is 50.2 degrees, right? So that's it for this one hopefully makes sense let me know if you guys have any questions.

Example: Collisions with Pendulums

17m
Video Transcript

Hey guys, so here's a practice problem of collision with motion involving two pendulums Let's check it out.

So, two pendulums are set up as shown below right here, pendulum 1 has a mass of 3 Kgs M1=3 and is held at rest horizontally so imagine a pendulum is hung vertically usually we're going to pick it up and then move it this way until with this completely horizontal and then the other pendulum is here when you let it go this pendulum going to go hit this and then cause it to move, OK? So, this pendulum is going to do this over here and hit the other guy. Second pendulum I'm going write it over here mass of 2 this 4 Kg and both of them have the same length so I'm going to write that length 1= length 2, we're just going to call it L and they are both 5 long strings, they have to be the same length of string so that they actually match down there, so when you release number 1 it goes this way, it falls and strikes number 1 at the lowest point, so number 1 is going to look sort of like this and then it's going to hit number 2 and stays in place after the collision, OK? So, after the collision it's going to look like this, you have one here at 0 meters per second it stays in place and 2 is going to get thrown this way with some non-zero velocity which will cause it to then swing up, Ok? And what we want to know is we want to know.... Let me \draw number 2 over here, we want to know a bunch of stuff we want to know the speed of block 1 just before the collision, the speed of block 2 after the collision and so forth so let's work through it. So the first I want to talk about here is that this actually has we have three parts, this is a three part problem because first is a motion so I use energy then there's a collision so I use momentum and then there's a motion against going to energy so let me do my little ABC thing except now we're going to have ABC and D because there's three parts, so first I have motion from A to B, A to B right there and therefore we're going to use the energy equation there then there's a collision from B to C and then there's another motion from C to D, Ok? So, I'm going to use momentum here and I'm going to use energy here we're going to have to use three equations to solve this whole thing, OK? A is the start of the problem when everything is at rest it hasn't moved yet, B is the beginning of the collision so it's the pre-collision, C is the post-collision velocity and Dis the end where this thing stops, OK? It's where the second block goes all the up. So, I want to know Speed of block 1 so for part A I want to know V1 just before colliding, (pre-collision) so I want to know V1b, OK? I want to know V1B. Now there's a lot of parts to this since I want some information at point B, B is shared between energy and momentum, the first energy equation, the first moment equation so I'm going to find this either by using this energy equation from A to B or this momentum equation from B to C, OK? If you were to write out the two of them you would realize that you have to use the energy equation because you don't have enough information for the momentum equation, another way to go about this is just to start from the beginning so you first do this equation then you do this equation then you do this equation, so we're going to write the first equation first and it turns out that that's going to work if it didn't work that's fine you would go to the second equation and that one would have to work, OK? So, if you're not sure just get busy start writing, right so we're going to write the energy equation from A to B so it's K+U+work non-conservative=K+U this is from A to B and A to B. Let's do some color coding here when we're talking about this guy, so I'm going to actually do this OK, is there a kinetic energy at point A? Point A is the beginning of the problem it's at rest no kinetic energy, what about potential energy? The block in question here is block number 1, it's at a height so there is potential energy, there is no work because you're not doing anything and there is no friction, there is kinetic energy at point (this was I meant to make this a B) there is kinetic energy at point B because it felt, right? It went all the way down and it's about to hit the blocks so it still has some velocity so it has kinetic energy, there is no potential energy at point B because at this point the pendulum is at its lowest point and remember the lowest point of a pendulum we consider to be your 0 height so there's not potential energy, so all my potential energy in the beginning I'm so sorry I meant to make this an A, all my potential energy it's A-A B-B so all my potential at the beginning goes into kinetic energy at the bottom so let's expand this, MGHa=1/2MVb squared, OK? So, what I can do here is I can cancel the masses because I have M's on both sides if I weren't cancelling the mass it would be important for me to know which mass to use, here we're looking at the motion of the first pendulum so this would have been M1 and M1 is 3 Kg, OK? But it cancels, I'm looking for V1b all of this is about the first object so this is H1b, V1b, Ok? I'm looking for V1b so we can solve for V1b, V1b is 2GH1a square root of that, right? Square root of 2GH classic and we can plug in numbers 2, 9.8 and then the height, what's the initial height of this block? Let's see If you can figure this out what's the initial height of this block? I hope you are thinking that if this length here is 5 meters than that's it and that's actually correct so the initial height here is going to be 5 meters because this length is 5 and I went all the way up to the top of the it's parallel with the same height as the top of the rope or the string, right? Ok now if you plug all of this in you get a 9.9 meters for second, Ok? So that's the first part V1B is 9.9 meters per second. Now it's really just basic conservation of energy now I want to know the speed of block 2 So let's go here B, I want to know V2 after the collision, after the collision is point C after the collision is points C, Ok? And if I want to know that is where am I going to get that information, well since I want something at point C the two candidate equations are either the momentum equation from B to C because it has C in it or the energy equation from C to D because it has C in it as well and the one that's going to work is going to be the momentum equation, now you could have written both of them and picked one or you could have just gone with the next equation the sequence, you've written the first equation and I'm going to write the second equation if it fails at least you already wrote it and then later on you're going to need to use it, In these problems if you have three parts you're going to use all three equations so if you write the equation out of order it's not really a waste of time you're going to have to use it eventually, OK? So, I'm going to go with the second equation here let's make it blue and this is from B to C, B to C and we're going to use the momentum equation, OK? So M1V1, M2V2, M1V1, M2V2 this is from B so B-B to C C-C and we're looking for V2C which is this guy, OK? so the power of having all these different letters is that once you've interpreted that the post-collision velocity of the second pendulum means V2C once you know that it's really easy to find it on the equation, this is exactly what I want right here, OK? I have the masses, OK? Vb is the velocity just before the collision, right? Just before the collision and I have them as well just before the collision V1 is 9.9 we just got it right here, V2 is 0 because the second block is not moving at all so this is gone. After the collision, the first block doesn't move, right? After the collision the first block stays in place so this is 0 as well and the only thing I don't have is exactly what I want which is my target V2c so I can solve by plugging in numbers, OK? So, it's going to look like this, M1V1b= M2V2c therefore V2c is this is just algebra now, M1V1b/M2 and we're going to plug in all the numbers and get the answer, OK? so I have this here 3(9.9)/4 and this is 7.4 meters per second, So that's the final velocity right here, OK? So that's it for this part let's go to the third part.

Part C asks for the maximum height that the second block will reach so I want to know this height here and this is at the very end of the problem so this is Hd, Ok? So, let's find Hd, let's color code that green So C I want to know what is Hd, now that's H of the second block because the first block stays at the bottom, right? So, I want to know H2d to be more precise, now if you look at your little set up here the only interval out of those three intervals that has the D in it is the third one so you have to use the energy equation which is also the only one we haven't used yet, OK? So, we have to use the energy equation from C to D which we haven't used yet, OK? Energy equation will be K+U+work non-conservative=K+U so Kc, Uc, Kd, alright so let's do this and by the way all this is for object 2 so if you want you can put a bunch of little 2s here everywhere, OK? Now is the kinetic at point C for the second object? That's a lot of stuff, right? The second object is this pendulum, the second pendulum here point C is post-collision, after the collision does the second object have velocity, it does, right? It does in fact we just calculated it so that we calculated it right here, Ok? So yes, after the collision you always have velocity for the second object in this case the first one actually doesn't, is there potential energy? Well right after the collision the second pendulum is about to go up but it hasn't really gone up yet so it has a tiny little height gain in the beginning so we're going to count that as negligible and say there's no potential, there's no work because you're not doing anything so the work done by you is 0 and there's no friction, there's no kinetic energy at the end because by definition of it being the highest point it means that you stop temporarily so the kinetic energy there is 0 there is no velocity and finally that second block gains some potential energy in the process of doing this so I have potential energy so all the kinetic becomes potential at the very end of the problem and let's expand this, so if I expand this I have 1/2MV2c squared, this is M2 equals potential energy which is M2GH2d, Ok? Put all the 2s everywhere so I don't screw up the masses, this block is moving by itself M2s cancel and we're looking for H2D so H2D is V2C squared/ 2G, I have this we just got it here look V2c we have it here so we're going to use that number 7.4 squared/ 2(9.8) and if you do this you get 2.8 meters and that is your final maximum heights, cool. The last part of the question there's one more piece asks you to be the angle that the second plane, the second pendulum rather will make with the horizontal this is just using the pendulum equation this is just using the pendulum equation so we're going to write L cosine of theta=L-H I derived this earlier and I want theta so theta will be move everything out of the way L-H divided by L and then take the arc cosine of both ,sides make sure the calculator is in degrees and if you do this you get 5 (which is the length of the pendulum)- 2.8 (which is the height gained)/ length 5 if you plug this in the calculator in degrees you get 63.9degrees and that's the end of this part so this is a really long question but I want to do one where you have two pendulums combining sometimes stuff like that shows up especially if your professor likes longer questions and I also will show you one has three parts as motion before collision then it has collision then it has more motion, OK? The last thing I want to talk about is how this block here which has mass 3 Kg starts here this block here is mass 4, Ok? This block goes all the way here at a height of 5 meters but after the collision the 4 only rises 2.8 meters, why is that? Well it's because it's heavier, OK? so the energy here gets transferred here and then the momentum gets transferred but because it's heavier it doesn't manage to go as far up if there were both 3 you could actually have immediately answered this question that because of conservation of energy and conservation momentum had this been a three it actually would have gone all the way up here and then he would have fallen back in these things we've been doing this but because it's heavier it doesn't make quite as high, let's say this was 2 kg this is 3 this is lighter so once it gets hit it's going to go farther up it would actually go somewhere here, OK? Because it's lighter so I just wanted to talk about that from a conceptual standpoint it should make sense that it doesn't go as far high as the first block that H2d is less than H1a at the beginning the problem, cool? I hope this makes sense let me know if you have any questions.

Collisions & Motion (Momentum & Energy) Additional Practice Problems

A 13.9 g bullet is fired horizontally into a 0.518 kg wooden block resting on a horizontal surface (μ = 0.125). The bullet goes through the block and comes out with a speed of 233 m/s. If the block travels 7.29 m before coming to rest, what was the initial speed of the bullet? The acceleration of gravity is 9.8 m/s2.

1. 421.185

2. 462.083

3. 475.131

4. 428.393

5. 771.008

6. 608.548

7. 390.493

8. 415.722

9. 506.991

10. 636.296

Watch Solution

If all three collisions in the figure are totally inelastic, which cause(s) the most damage (deformation of objects, thermal energy increase, etc.)? Assume that the wall is stationary and the car is completely stopped by it in the first diagram.

1. I, III

2. I

3. all three

4. I, II

5. II, III

6. III

7. II

Watch Solution

In the figure, a block sitting on a frictionless horizontal surface is attached to a rigid wall on the right through a spring (whose axis is horizontal). A bullet is shot at the block from the left and gets embedded in it, causing the block to move to the right, thus compressing the spring. (Assume the bullet is travelling perfectly horizontally, along the axis of the spring, before hitting the block). Which of the following are true?

A. The initial kinetic energy of the bullet is completely converted to spring potential energy when the spring reaches its maximal compression.

B. The initial momentum of the bullet is equal to the momentum of the bullet+block system just after the bullet enters the block.

C. Part of the momentum of the bullet+block system is lost during the collision (i.e. before the spring-compression starts).

D. Part of the energy of the bullet+block system is ”lost” (no longer present as macroscopic kinetic energy) during the collision, before the spring-compression starts.

E. If we are given the masses of the block and the bullet, the initial speed of the bullet and the spring constant, it is possible to find the maximum compression of the spring.

1. A, E
2. A, B, D
3. B, D
4. A, B
5. A, C
6. D
7. B, D, E
8. A, C, E
9. A, B, E

Watch Solution

A test car of mass 740 kg is moving at a speed of 7.3 m/s when it crashes into a wall to test its bumper. If the car comes to rest in 0.36 s, how much average power is expended in the process?

1. 42820.9

2. 47940.6

3. 59118.6

4. 49662.2

5. 50789.2

6. 54770.3

7. 45825.2

8. 44990.3

9. 46540.4

10. 51904.1

Watch Solution

A truck is stopped at a stop light on a street in San Francisco with an 18% grade (a 10° incline), facing down the incline. A distracted driver driving a sedan down the same road does not notice the red light, nor the truck, and collides with the back of the truck at full speed. The car and truck stick together after the collision and slide to a stop down the road. The police investigating the accident want to know if the car was going above the speed limit, set at 25 miles per hour. The police determine that the car and truck slid a distance of 4 m together after the collision. Determine if the car was speeding before it collided with the truck. A typical mass for a sedan is 1400 kg, while a truck has a mass of 2000 kg. The coefficient of kinetic friction between car tires and dry pavement is 0.6.

Watch Solution

The ballistic pendulum is a system used to measure the speed of a fast-moving projectile, such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet embeds in the block, and the entire system swings through a height h since the collision is perfectly inelastic. Suppose that h = 5.0 cm, m1 = 0.5 g, and m2 = 1 kg. Find the initial speed of the bullet.

Watch Solution

A bullet of mass 24 g is fired into the bob of a ballistic pendulum of mass 2.85 kg. When the bob is at its maximum height, the strings make an angle of 40° with the vertical. The length of the pendulum is 3.75 m. Find the speed of the bullet (in units of m/s). The acceleration due to gravity is 9.81 m/s2

Watch Solution

A 3.0 kg block moving at 12.0 m/s along a horizontal frictionless surface collides with a 2.0 kg block that is initially at rest. After the collision, the two blocks stick together and then slide up a 53° frictionless inclined plane, as shown in the sketch. Calculate the maximum distance L that the two blocks travel up the incline.

Watch Solution

A bullet with mass 6.00 x 10-3 kg is fired horizontally with speed V0 into a 1.20 kg wood block that is initially at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.200, The bullet remains embedded in the block and after the collision with the bullet the block slides 0.120 m before coming to rest. What was the initial speed V0 of the bullet?

Watch Solution

Block A with mass 0.050 kg is released from rest at the rim of a frictionless hemispherical bowl that has radius R = 0.600 m. Block A slides down the side of the bowl and strikes block B, which has mass 0.150 kg and that is sitting at rest at the bottom of the bowl, and the two blocks stick together.

What maximum vertical distance above the bottom of the bowl will the combined object reach after the collision?

 

Watch Solution

A block with mass 4.95 kg is on a horizontal frictionless surface. It rest against a horizontal spring that has force constant = 300 N/m. Initially the spring is not compressed. A bullet with mass 0.0500 kg is travelling horizontally with speed v0 = 400 m/s. The bullet strikes the block and remains embedded in it. What is the maximum distance that the spring is compressed after the collision with the bullet?

Watch Solution