For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Collisions & Linear Momentum

- If there is no net
__external__force that acts on a system, momentum is conserved:

This is called the→p

sys, start= →p

sys, end. *conservation of momentum*. - We can still apply the idea of conservation of momentum in collisions where there are external forces. Generally, collisions happen so quickly and external forces tend to be very small (compared to the forces in the collision). This means that the change in momentum caused by external forces is usually negligible if we look at the instant before and the instant after the collision.
- Conservation of energy and conservation of momentum are very different. Just because one is conserved does
__not__necessarily mean the other is as well. We categorize collisions based on this:*Elastic*: Energy__and__momentum are conserved:∆E _{sys}= 0 & ∆→p

sys= 0. *Inelastic*: Energy is__not__conserved, momentum is conserved:∆E _{sys}≠ 0 & ∆→p

sys= 0. *Completely inelastic*is just like inelastic above, except the objects stick together after the collision.

### Collisions & Linear Momentum

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Investigating Collisions 0:45
- Momentum
- Center of Mass
- Derivation 1:56
- Extending Idea of Momentum to a System
- Impulse
- Conservation of Linear Momentum 6:14
- Conservation of Linear Momentum
- Conservation and External Forces 7:56
- Conservation and External Forces
- Momentum Vs. Energy 9:52
- Momentum Vs. Energy
- Types of Collisions 12:33
- Elastic
- Inelastic
- Completely Inelastic
- Everyday Collisions and Atomic Collisions
- Example 1: Impact of Two Cars 14:07
- Example 2: Billiard Balls 16:59
- Example 3: Elastic Collision 23:52
- Example 4: Bullet's Velocity 33:35

### High School Physics Online Course

### Transcription: Collisions & Linear Momentum

*Hi, welcome back to educator.com, today we are going to be talking about collisions and linear momentum.*0000

*One really great importance of Physics is, being able to understand the interactions between the objects, the way things work between one another in the world.*0006

*One of the most basic forms of interaction is collision.*0015

*When two things hit together, it is useful to be able to know how things are going to play out at that moment.*0018

*If you can model it, we have done all sorts of useful things.*0024

*Think of all the things that we can do with it.*0026

*We can investigate a car crash, you could dock a shuttle with a space station, you could look at the collisions between atomic particles.*0027

*There is all sorts of stuff, stuff bounces off one another, where stuff hit one another and sticks together, you will be able to understand it a lot better if you have momentum and how it is conserved, how it works for momentum.*0034

*Investigating collisions!*0046

*If you wanted to talk more about collisions, what ideas would be useful?*0048

*We can all figure out at this point , momentum is going to be what we are going to be talking about.*0052

*We just introduced the idea of momentum, and the reason why we did that was to set ourselves up to be talking about conservation of momentum, and how it works with collisions.*0058

*If we are going to investigate collisions, if we are going to investigate how motion changes, and what is happening as things go on, we want to be able to talk about how motion is working.*0066

*One of the best things we have for talking about motion is momentum, p = mv.*0076

*If we got that under our belts, that is going to really help us.*0080

*But that is not quite enough, we need one more thing.*0084

*The other thing we need, is the centre of mass.*0087

*Talking about the centre of mass allows us to explore the motion of a system, how things work together, and that is what we are going to need to be able to talk about collision.*0090

*Two objects working together, hitting each other, that is the system, you look at those two objects and the way they interact together, they are going to have to interact as a system because we have decided to make them a system.*0097

*If you look at them as a system, we will be able to know things, the way their momenta are working, and the way their centre of mass is doing things.*0106

*We are going to use to use centre of mass to learn indirectly more about momentum.*0113

*We are going to derive this.*0117

*We can extend the idea of momentum to a system, by looking at the total of the momenta in the system, makes up the momentum of the system.*0118

*This symbol right here mean, ADD things together. (Might be a little confusing if you have not seen it before.)*0126

*That says, add up all the momenta in the system.*0135

*Say we are looking at a pretty basic system that has some p_{1}, and p_{2}, a two object system.*0139

*Then the momentum of that system is adding them together, and that makes sense, the final momentum of the system is just all of them added together.*0146

*Just like we said that the energy of the system is all of the energies involved, you look at the energy for, potential gravity, kinetic energy, spring energy, you added them all to find out what the energy of the system was.*0153

*Exact same thing here: To find out the momentum of a system, you just add all the momenta together.*0167

*How do we know for sure though, that the mass of the system, (all masses summed up) times the velocity of the centre of mass, is equal to the momentum of the system (all added together)?*0173

*We can actually prove that.*0185

*M _{sys}v_{CM} = (m_{1}+m_{2})(m_{1}v_{1}+m_{2}v_{2})/(m_{1}+m_{2}) (in 2 particle system, what happens when two objects interact with one another, this works with an arbitrary number of things in our system, but we are going to be looking at the collision between just two things, to understand how this came to be, also note that velocity of centre of mass is the weighted average.)*0186

*We have got something that cancels off pretty easily, and from there, (m _{1}v_{1}+m_{2}v_{2}) = p_{1}+p_{2} = Σp.*0255

*So, all the momenta in the system added up together, is equal to the momentum of the system, which turns out to be, = M _{sys}v_{CM}.*0290

*We look at the thing as a whole, it is the mass of the system times the movement of the system of a whole, or we look at the pieces that make it up, it is all those pieces added together, great intuition, works out perfectly.*0299

*Similarly, we are not going to show this mathematically, but we can do this with impulse.*0312

*The change in momentum of the system, Δ p_{sys} = j_{sys} = F_{ext}t.*0316

*That makes sense, if you are going to have a change in the system, you are going to put an impulse on the system, just like we did when we were working with single object at a time, but it is also going to be the case that the force on it is external, because remember, form our work with centre of mass, internal forces, because of newton's third law, every action has an equal and opposite reactionary force,*0322

*So, if it is an internal force in the system, sure, stuff is going to happen inside of the system, but the ultimate change to the centre of mass, is going to get a force going one way, but it is going to get the exact same force going the other way, they will cancel out.*0343

*Centre of mass is going to experience nothing.*0355

*So, form our work in centre of mass, we know that internal forces have no effect on the system's centre of mass, so no effect on the system's momentum.*0356

*So, Δ p_{sys} = j_{sys} = F_{ext}t.*0362

*Now we are ready to make something really useful.*0372

*Conservation of linear momentum: If there are no external forces acting on the system, there is no net external force acting on the system, and there should not be external forces acting on the system, but as long as they are going to cancel out, at static equilibrium, we are going to have the change in momentum of the system = zero.*0375

*That means, system's momentum at the beginning = system's momentum at the end.*0394

*If we know that there is no net external force, we know that we are going to have a constant momentum, a conserved linear momentum.*0401

*That is really handy, because that is going to give us a whole bunch of power, when we are looking at new kinds of problems.*0409

*Once again, notice that there can be internal forces, it does not matter if there are internal forces, they will have no effect on the system's momentum as a whole.*0416

*For example, if we had a bomb, or some explosive things, and we blew it up into a bunch of pieces.*0422

*Each one of those pieces would fly away.*0436

*But, since it is all internal forces in that system, the system's momentum is going to wind up being that same steady state when it was at rest, it is going to be zero.*0439

*Of course, there is also gas in there, so it is pretty complicated thing to model, you might have to deal with the gas moving in every which way as well, but the total system's momentum, when we take into account the pieces flying every way, the gas in the centre expanding out rapidly, the end of it is going to wind up also being zero momentum.*0450

*All sorts of things are going in opposite directions, but when we add them all up together, they ultimately cancel each other, and we get back to rest.*0470

*Conservation and external force:*0477

*This means, if we have an external force acting on our system, we require to have no external force to allow the conservation of momentum, so fi there are external forces, we fling up our hands, and go, 'Oh, we cannot do this problem'?*0479

*No, we do not have to do that, because we can still examine a collision, even when there is an external force. Why?*0491

*External forces are normally pretty small compared to collision forces.*0496

*Think about, if a car was crashing into another car, there is going to be little bit of friction on the ground, but for the amount of time that the crash occurs, friction on the road, like small amount of rolling resistance, small amount of internal friction inside the engine, it is going to be very little compared to the massive change in forces involved in two cars crashing together.*0501

*Those massive forces are going to be so large compared to all the other forces in effect, that we can basically forget about those other forces, we can forget about those external forces, because we are going to be looking at a brief period of time.*0527

*We can take a snap shot just before the incident of impact, a snap shot just after the incident of impact, because the impact is going to be more than an instant, but it will be something very brief, order of (1/10) of a second or less.*0542

*We can take a look just before the impact happens, and just after the impact is complete, the external forces are likely to be pretty small compared to the forces involved in the collision, so we can still effectively use conservation of momentum.*0555

*It is not perfectly correct, but it is so close as to be plenty good.*0570

*We are going to be totally be able to use it, and it will still make a lot of sense.*0576

*As long as we consider the instant just before and just after the collision, we can treat it as the momentum is conserved during the collision.*0578

*It will not be perfectly conserved, but it will be such a small change, that it is as if it was conserved, so it is good enough for us.*0584

*Finally, momentum versus energy:*0593

*It is really important to notice that momentum and energy are totally different.*0596

*mv is very different from (1/2)mv ^{2}, conservation of momentum does not imply conservation of the other.*0599

*In fact, in almost all every day collisions, some energy is lost to heat via friction.*0607

*When, for example, you drop a ball, the ball hits the ground and rebounds.*0611

*If it had the same exact amount of energy that it started with, the amount of energy in that ball would have to be the same as what it was when it was dropped in the beginning.*0617

*That means, that it would be able to bounce and come up to the same height, again and again, but it does not happen.*0625

*For anybody who has dropped a ball, if you drop a ball, it does not keep bouncing forever, it eventually stops moving.*0632

*If it stops moving, the energy is being sapped.*0642

*Momentum is conserved, when you consider the earth and the ball interacting with one another, but the energy is not conserved in the ball-earth system.*0644

*The ball-earth system winds up dissipating energy.*0654

*That energy does get dissipated, conservation of energy does still work, but it winds up being dissipated into things that we cannot really watch, like heat energy, sound energy, so we do not really have a good way of doing it.*0658

*So from our point of view, mechanical energy is not conserved, the way that we know how to deal with energy, just is not conserved.*0669

*So we cannot use it, we can only keep momentum in that case.*0675

*The system's momentum will remain constant through a collision, but the energy of the system has no such restrictions.*0678

*When we are dealing with collisions, we know for sure, that momentum is going to stay the same.*0685

*But, energy might change.*0689

*Similarly, when we throw an object, we know that the energy in the object is going to remain the same through its motion.*0693

*However, since there are external forces, like gravity acting on the object, we know that the momentum of the object is not going to be the same.*0699

*Sometimes energy will be conserved, with momentum not being conserved, sometimes momentum will be conserved with energy not being conserved.*0706

*It depends on what you are looking at, you want to think about it as how short is the time frame, what external forces are acting on it, think in terms of collisions, means momentum you can trust, and no external forces means that we can trust in energy.*0712

*And external forces and energy, remember, gravity is not an external force for energy, because we have taken care of it within our energy formula.*0730

*But for momentum, there is no such formulae for these other things, like gravity or friction.*0737

*Still, finally, there are some collisions where energy is not lost.*0742

*So, we are going to have to categorize our collisions into different types.*0747

*When momentum is kept, when energy is kept; when momentum is kept, when energy is not kept.*0749

*Fist one, elastic: Collision is called elastic when energy is conserved in addition to momentum.*0754

*The change in the energy of the system = zero, and the change in the momentum of the system = zero.*0760

*These are really uncommon in everyday activities.*0764

*Stuff that we are used to in reality, in real life, it pretty much never happens, if not completely never.*0767

*Inelastic: Collisions that we do see in everyday life.*0774

*A collision is called elastic when energy is not conserved during the collisions.*0777

*So, we cannot trust the energy of the system to be equal to the energy of the system at the end.*0780

*Beginning system energy, not equal to ending system energy.*0784

*But, the change in the momentum of the system, is trustable, we can keep that.*0790

*The beginning of the collision, the end of the collision, they are going to wind up having the same momentum on both sides.*0794

*Those are called inelastic collisions.*0799

*When energy is not conserved, inelastic.*0802

*Finally, in addition, when object involved sticks together after the collision, when they hit each other and they just keep moving as just one thing, it is called completely inelastic.*0805

*They just do not hit each other, and lose some energy, they hit each other, and get turned into one object, or something that can be treated as one single object.*0814

*Finally, almost all everyday collisions are inelastic, sometimes it is useful to approximate some collisions as elastic.*0823

*However, in the atomic world, particles often have elastic collisions.*0830

*In the atomic world, elastic collisions are actually really common and standard.*0833

*But, form the sort of thing that we see in everyday life, we are almost never going to see an elastic collision.*0836

*However there are certain things that we can approximate as elastic collisions, and learn some more stuff about.*0841

*On to the examples.*0847

*Two cars are being driven directly towards one another.*0849

*They are remote controlled, so nobody is hurt in the process of this example!*0851

*First car has mass, m _{1} = 1000 kg, and velocity, v_{1} = 30 m/s.*0855

*Second car has mass, m _{1} = 2000 kg, and traveling in the opposite direction, so, -20 m/s.*0861

*On impact, they stick together, so we are having a completely inelastic collision.*0866

*What is the velocity of the twisted hulk of metal they now make just after they collide?*0870

*We know that, p_{sys}(beginning) = p_{sys}(end).*0875

* p_{1}(initial) + p_{2}(initial) = p_{1}(final)+p_{2}(final).*0886

*What about the things on the right, we no more have two separate objects, we have one, stuck together.*0900

*In reality, this is not going to be that, it is going to the momentum of the twisted hulk of metal, momentum of the wreck, momentum of the whole.*0911

*So, m _{1}v_{1} (initial) + m_{2}v_{2}(initial) = (m_{1}+m_{2})v_{hulk}.*0930

*So, 1000×30 + 2000×(-20) = 3000v _{hulk}, so, -10000 kg m/s / 3000 = v_{hulk} = -3.33 m/s, approximately.*0957

*That is what is left, that is what the speed that is left, actually velocity, since we know that it is going to move in the direction of the heavier car, heavier car was going to slower, it had more momentum, so it winds out in the end.*1000

*We are going at a -3.33 m/s velocity.*1013

*Example 2: Two billiard balls of mass m _{1} and m_{2}, same mass of 0.16 kg, are on a pool table.*1020

*One of them is moving horizontally with speed of v _{1} = 5 m/s towards the other.*1029

*After the collision, the second ball is moving with a speed of v _{2} = 3 m/s, at 30 degrees below the horizontal.*1033

*What is the first's velocity?*1040

*In this case, we need to figure out, how much is going to the right, and how much is going up.*1042

*Now, we can figure out what the momentum is for the first ball, the ball that is moving.*1052

*We can figure out, the second one, we know what the momentum in the second one is.*1059

*We can figure out how much it is moving this way, and how much it is moving this way, and once we know that, we can go back and we can figure out, in this one, these two added up, are going to have to wind up equaling this one's vertical motion, which turns out to be nothing.*1062

*And we know that these two added up, are going to have to wind up equaling the horizontal motion.*1092

*We are now going to start working on it.*1104

*Velocity for the ball that is first moving before impact is 5 m/s to the right, o m/s up and down.*1111

*v _{2} = 3 m/s, 30 degrees below the horizontal.*1119

*Let us figure out what is v _{2}(final) in terms of component vector.*1124

*v _{2}(final) = cos(30)×3 m/s (since this side is adjacent), and the other side will be, sin(30)×3 m/s.*1132

*So, (2.598, -1.5).*1157

*That is an important thing that we have to catch specifically.*1166

*It is not just sin(30), it is actually, sin(-30), because it is below the horizontal, and we probably want to make going up positive, and going to the right positive, because that is what we are used to as standard.*1168

*So, it is not just sin(30), it is sin(-30) because it is below the horizontal, it is up to us to pay attention to the signs, we have to be understanding what we are saying with this.*1181

*So, (2.598,-1.5) m/s.*1194

*Now we know what the velocity of the second thing is.*1198

*From there, we can figure out what its momentum is.*1200

*First off, we know that the momentum at the beginning of the system is going to be equal to the momentum at the end of the system.*1203

*So, m _{1}v_{1} = m_{1}v_{1}(after impact) + m_{2}v_{2}(after impact).*1249

*Remember, m _{1} and m_{2} are the same thing, so divide everything by that m_{1}, m_{2} combination, since m_{1} = m_{2} = 0.16 kg.*1275

*So, velocity = v _{1}(final) + v_{2}(final), (5,0) = (v_{1}_{x},v_{1}_{y}) + (2.598,-1.5).*1292

*At this point, we can add things together, so, remember, we are dealing with vectors, each one of the pieces of vector, x component is totally separate from the y component, so now we can break this into two separate equations.*1335

*We got, 5 = v _{1}_{x} + 2.598, and, 0 = v_{1}_{y} + (-1.5).*1346

*So, 1.5 = v _{1}_{y}, and , 5-2.598 = 2.402 = v_{1}_{x}.*1362

*Putting these together, v_{1} = (2.402,1.5) m/s, and that makes good intuitive sense.*1380

*If you hit the ball, the other ball winds up moving away as well, you wind up getting something that is moving slower now.*1399

*If it is going down, if it was going horizontally first, and the other thing bounces down, it is going to have to get rebounded upwards.*1405

*So, it makes good intuitive sense, passes the sanity check, and works out.*1413

*We are able to start these things moving, because of conservation of momentum, and we can make it a little bit easier by canceling out the masses, because we know that they all have the same mass, so it is a little bit easier for us, and from there we break one of them into its component pieces, and from there we are able to work in factors, because vectors, we can just deal with the component pieces.*1415

*Example 3: Two rubber balls are on a flat surface.*1434

*The first one has a mass, m _{1} = 2 kg, and the second has a mass, m_{2} = 1 kg.*1437

*The first one is moving at a velocity 5 m/s.*1442

*The first ball collides with the second ball, and we are going to be able to treat this collision as elastic.*1445

*I know, normal everyday collisions do not wind up being elastic, they are almost all inelastic, but it is good for us to get some practice with an elastic one, and rubber balls do manage to keep a lot of their energy.*1449

*What is the velocity of each ball after the collision?*1459

*Notice, this ball is going to be moving, and this ball is going to be moving after the collision.*1462

*One thing to notice is, which ball has to be moving faster.*1467

*The ball on the right is moving slower than the ball on the left, that is not going to make sense, because that means that somehow collision happened and then this ball managed to move through the other ball.*1471

*That does not make sense, we know that these are solid objects that are moving in one direction, it cannot move through it, so that is going to guarantee the fact that the velocity of the second ball is going to have to be greater than the velocity of the first ball.*1483

*In this problem, we are also going to wind up ditching the vector notation on the top, because we are working only in one dimension.*1496

*It is little bit easier than normal , because we are only working in one dimension, and we got a whole lot of writing coming up to do in an elastic problem, so we are going to make it a little easier by ditching the arrows, but it is important to remember that we are working on the basis of the idea of vectors, we are still going to care about direction and all those things.*1503

*For the beginning, we know that v _{2} > v_{1}.*1525

*Now, let us look at, what is momentum going to tell us about this!*1530

*We know that, p _{sys}(beginning) = p_{1}(final) + p_{2}(final), (one dimension, so ditching arrows on top, but all the ideas of vectors are still there, we really care about direction of momentum.)*1532

*m _{1}v_{1} (initial) = m_{1}v_{1}(final) + m_{2}v_{2}(final).*1584

*So, 2 kg×5 m/s = 2×v _{1} + 1×v_{2}.*1617

*Sp, 10 = 2v _{1} + v_{2}.*1630

*Now, that i snot enough to solve, we got one equation and two unknowns, so we are not able to do it.*1637

*But luckily, this is an elastic collision, so now we can bring energy.*1642

*We know, Energy(initial) = Energy(final), and the final energy of the system, since one thing is moving at the beginning, it is just that one thing's motion.*1645

*The final energy of the system is going to be both the things moving together, so, E _{1} + E_{2}.*1655

*What is the energy at the beginning? (1/2)×2×(v _{i})^{2} = (1/2)m_{1}v_{1}^{2}+(1/2)m_{2}v_{2}^{2}.*1663

*We cancel (1/2)'s all the way across, substitute things in, 2×5 ^{2} = 2v_{1}^{1}+v_{2}^{2}.*1680

*So, 50 = 2v _{1}^{2}+v_{2}^{2}.*1699

*At this point, we have got two unknowns over in this one.*1708

*How do we do this? Go back, and substitute one equation into the other one.*1711

*We get, 10 - 2v _{1} = v_{2}.*1717

*Move that over here, so, 50 = 2v _{1}^{2}+(10-2v_{1})^{2}.*1721

*Simplify that out, 50 = 2v _{1}^{2}+100-40v_{1}+4v_{1}^{2}.*1736

*Rearranging, 6v _{1}^{2}-40v_{1}+50 = 0. (quadratic).*1750

*If you are really good at factoring, you might be able to factorize this, but, most of us are not that great at factoring, so what do we use now?*1770

*We got two options: (1) calculator, graph it and see the roots, because we will be able to treat this as a single function and when that function equals zero, you would have found the answer to this equation.*1776

*Alternately, we could just throw down with the quadratic formula, and we will be able to figure out what it is.*1789

*What is the quadratic formula? (if you want to do advanced mathematics, or even physics, you could buy a graphing calculator, but it is not necessary.)*1794

*[-b +/- sqrt(b ^{2} - 4ac)]/2a.*1821

*Plug things in, [40 +/- sqrt(40 ^{2} - 4×6×50)]/(2×6) = (40 +/- 20)/12, gives us two possible answers.*1829

*The two answers are, 60/12 = 5 m/s, and 20/12 = 1.67 m/s, approximately.*1861

*We have got two answers, so it is up to us to figure out which one makes sense, which one is going to allow v _{2}to be greater than v_{1}.*1881

*If it is 5 m/s for v _{1}, was it not 5 m/s beforehand?*1887

*That means that it hits the ball and then it keeps going at the same speed, that does not make any sense.*1889

*It manages to just go through the ball, remember, math is your friend, but it is not something that just spits out your answer.*1905

*It is up to you to keep it tamed, you have to understand what you are doing with it.*1911

*We brought this quadratic equation to bare, but it is up to us to understand how to interpret the results that it gives us.*1915

*It gives us two answers that work mathematically, the energy will be conserved, and the momentum will be conserved if that other ball was never there, it just manages to pass through it.*1921

*But, we know that it cannot be the truth, it did not just pass through it, it has to contact that ball.*1929

*So, the answer that has got to be, is this one tight here.*1934

*So we have got, v _{1} = 1.67 m/s.*1938

*If that is the case, we can then plug this back in, and we get, 10 - 2×(1.67) = v _{2}, so, 6.66 m/s = v_{2}, approximately.*1943

*There we go, we are able to figure out what the other velocity has got to be, because we are able to bring all these equations bare.*1964

*We bring what we know about energy formula, what we know about momentum, and we are able to do both of them, because we got two unknowns and two equations, we are able to solve it.*1971

*And there are our answers.*1979

*We know that v _{1} is 1.67 m/s, still in the positive direction, and v_{2} is positive at 6.66 m/s, and it turns out it is going to be faster than the initial object actually was.*1980

*That is the stuff that happens in elastic collisions, things do not behave necessarily quite like you expected, because energy is conserved, so we got that v ^{2} thing going on, so some weird stuff can happen sometimes.*1997

*Well, not weird, because it is physics, it is reality, so it cannot be weird in that way, but it is certainly not what we necessarily expect.*2008

*Example 4: A ballistic pendulum is one way to measure the velocity of a bullet fired from a gun.*2028

*Ballistic pendulum is a big heavy thing sitting on a rod, and there is this thing here, and a bullet will come in here, and it is going to impart its momentum, so the thing will swing up.*2032

*It winds swinging up, just like we got here, bullet goes in here, launches itself in, and put some momentum into it, and the ballistic pendulum will continue to swing up.*2050

*It is one way to measure the velocity of a bullet fired form a gun.*2061

*A bullet is fired into a large mass at the end of a pendulum where it launched itself.*2062

*By measuring how high the pendulum rises, we will be able to figure out the velocity of the bullet before impact. How do we do that?*2068

*Denote the bullet's mass as m _{b}, the pendulum's mass as m_{p}, assume the pendulum's rod's mass is negligible, so all we have to worry about is the mass of the block.*2074

*The height that the block achieves will be 'h'.*2086

*Using all these things, (in real life, we could build something like this, and using a pencil we could mark how high it went up, and we will be able to weigh the mass of the bullet and the pendulum, we will be able to find a fairly light weight strong rod that was able to hold it in place without being much compared to the mass of the pendulum, and we will actually be able to do this, this is what people did in the 1700's).*2091

*What is the trick that we will be able to use to solve for this?*2123

*Momentum is not going to be conserved throughout the entire thing.*2126

*The momentum of the bullet, as it goes up, we have got gravity that is dealing against it.*2130

*So, gravity is an external force from the point of view of the momentum, if it is going up, it is going to have to have some force moving it up, so we have got external forces on the bullet.*2135

*But at the same time, energy is not going to be conserved.*2147

*We got this impact, we never said it was elastic, this is an inelastic collision, this is definitely an inelastic collision.*2149

*We cannot conserve energy throughout, we cannot conserve momentum throughout, how do we do this?*2160

*We can conserve them one piece at a time.*2165

*Notice, the instant, if we break this into three things, the bullet goes in, and at this moment, when it is just after the collision occurs, before it has the chance to swing up, momentum is conserved.*2168

*So, p_{i} = p_{f}, then from here, once it is swinging up, there is no more collisions, it is just a basic pendulum rod going up, so we can use energy.*2187

*So, for here, E _{i} = E_{f}.*2199

*We are able to break this into two pieces: the momentum is going to be conserved for just the collision, and after the collision, energy will be conserved.*2203

*We break it into two pieces: beforehand and afterwards.*2212

*If p_{i} = p_{f}, what is in the initial momentum?*2215

*m _{b}v_{b}, there is nothing else moving, we got the pendulum at rest.*2222

*Once again, we are going to nick those vectors, since we are dealing with linear quantities.*2227

*Wait a second! I hear you ask, is it not going to move up and to the right when it swings up!*2237

*Yes, that is true, but we will be dealing with that as energy, so when we get to that point, we only have to worry about speed.*2242

*For the collision, as far as we are concerned, it hits and goes in, and it is still basically moving in just one dimension, because it launches itself very quickly, and the collision happens very quickly compared to the movement of the swinging.*2247

*So, m _{b}v_{b} = (m_{b}+m_{v})v_{sys}, for ease, we will denote, v_{sys} after collision as, v_{p}.*2258

*That makes it a little bit easier to write, because we are going to have to be talking about two very different world, the world of the impact, and the world after impact where it is swinging.*2297

*This will make it a little bit easier, v _{p}, it is important to pay attention to what wee are doing here.*2303

*We can figure out what the velocity of the bullet is, in terms of, (m _{b}+m_{p})/m_{b}, but are not able to solve for this, we still need to figure out what the velocity of the pendulum is, we do not know what the velocity of the pendulum is.*2309

*Remember, we only have the height.*2328

*Now, start looking in terms of energy.*2330

*E _{i} = E_{f}.*2332

*The energy of the system at the beginning of the swing, does it have any height?*2336

*No, we can consider that as its base line height. Is it moving?*2340

*Yes, it is definitely moving, there are no spring involved, and the only thing that we got there is the kinetic mechanical energy, kinetic energy.*2342

*(1/2)(m _{b}+m_{p})v_{p}^{2} = Mgh (at the top of its arc, it has got to just finish moving, then the only thing it has is stored gravitational potential energy.)*2349

*'M' in this case, is not just 'M', it is the mass of the whole thing, m _{b}+m_{p}, so, m_{b}+m_{p}gh.*2356

*Cancel out m _{b}+m_{p} on both sides, so, v_{p}^{2} = 2gh, v_{p} = sqrt(2gh).*2387

*Substitute that back in, v _{b} = (m_{b}+m_{p})/(m_{b}×sqrt(2gh)), there we are.*2402

*By just measuring the height that the thing goes up to, by already knowing the constant of gravity, by weighing the mass of the bullet, weighing the block, we are able to figure out the velocity of the bullet, which is pretty great.*2427

*These are all the things that we were able to do in the 1700's.*2437

*Even without fancy technologies like LASER, and all these other cool stuff, we can actually come up with a really solid way to measure the velocity of a bullet by what we know about momentum and what we know about energy.*2440

*Hope you enjoyed this.*2452

1 answer

Last reply by: Professor Selhorst-Jones

Sun Nov 8, 2015 4:33 PM

Post by Gautham Padmakumar on October 30, 2015

Why are we assuming velocity of pendulum and system are equal ?

1 answer

Last reply by: Professor Selhorst-Jones

Sun Mar 10, 2013 7:20 PM

Post by Norman Cervantes on March 9, 2013

29:05 where does -40v1 come from?

2 answers

Last reply by: Tanveer Sehgal

Sat Nov 24, 2012 6:13 AM

Post by Tanveer Sehgal on November 22, 2012

Hey,

Which calculator would you recommend I purchase? I am planning to study physics and math at a higher level.