For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

### Conservation of Linear Momentum

- In an isolated system, where no external forces act, linear momentum is always conserved. In any closed system, the total linear momentum of the system remains constant. This is a direct outcome of Newton’s 3rd Law of Motion.
- In the case of a collision or explosion, if you add up the individual momentum vectors of all the objects before the event, you’ll find that they are equal to the sum of the momentum vectors of the objects after the event.
- Kinetic Energy is conserved in an elastic collision.
- Momentum tables can be used to solve for unknown quantities in collisions.

### Conservation of 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
- Objectives
- Conservation of Linear Momentum
- Collisions and Explosions
- Solving Momentum Problems
- Types of Collisions
- Example I: Traffic Collision
- Example II: Collision of Two Moving Objects
- Example III: Recoil Velocity
- Example IV: Atomic Collision
- Example V: Collision in Multiple Dimensions
- Example VI: AP-C 2001 FR1
- Example VII: AP-C 2002 FR1
- Example VIII: AP-C 2014 FR1

- Intro 0:00
- Objectives 0:08
- Conservation of Linear Momentum 0:28
- In an Isolated System
- In Any Closed System
- Direct Outcome of Newton's 3rd Law of Motion
- Collisions and Explosions 1:07
- Collisions and Explosions
- The Law of Conservation of Linear Momentum
- Solving Momentum Problems 1:35
- Solving Momentum Problems
- Types of Collisions 2:08
- Elastic Collision
- Inelastic Collision
- Example I: Traffic Collision 3:00
- Example II: Collision of Two Moving Objects 6:55
- Example III: Recoil Velocity 9:47
- Example IV: Atomic Collision 12:12
- Example V: Collision in Multiple Dimensions 18:11
- Example VI: AP-C 2001 FR1 25:16
- Example VI: Part A
- Example VI: Part B
- Example VI: Part C
- Example VI: Part D
- Example VII: AP-C 2002 FR1 30:10
- Example VII: Part A
- Example VII: Part B
- Example VII: Part C
- Example VII: Part D
- Example VIII: AP-C 2014 FR1 38:55
- Example VIII: Part A
- Example VIII: Part B
- Example VIII: Part C
- Example VIII: Part D

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: Conservation of Linear Momentum

*Hello, everyone, and welcome back to www.educator.com. *0000

*I am Dan Fullerton and in this lesson we are going to talk about conservation of linear momentum.*0003

*Our objectives include explaining how linear momentum conservation follows a consequence of Newton’s 3rd law for an isolated system.*0009

*Identifying situations in which linear momentum is conserved.*0016

*Applying linear momentum conservation to situations in which two or more objects interact to solve variety of problems.*0020

*Let us talk about conservation of linear momentum.*0027

*In an isolated system where no external forces act, the total linear momentum of that system stays the same, it must be conserved.*0030

*In that close system, that means total linear momentum is constant whenever you have before the event*0039

*must be exactly the same as the total linear momentum you have after the event.*0044

*That is a direct outcome of Newton’s 3rd law of motion when you have a force of object 1 and object 2,*0050

*object 2 inserts the exact opposite force, same in magnitude opposite direction on force 1 *0057

*which maintains that amount of linear momentum in the close system.*0063

*Collisions and explosions, in the case of a collision or explosion if you add up the individual momentum vectors of all the objects before the event.*0068

*Then, you compare that to the sum of all the momentum vectors, of all of the objects after the event you will find that they are the exact same.*0077

*Mathematically, the law of conservation of linear momentum states that the initial total momentum = the final total momentum.*0085

*How do we solve momentum problems?*0096

*First, identify all the objects in the system then determine the moment of the objects before the event *0098

*and use variables for any thing or characteristics you do not know.*0104

*Determine the moment of the objects after the event using variables for anything you do not know.*0109

*Add up all the momentum from before the event and add up all the moment that after the event.*0114

*Set them equal to each other, the law of conservation of linear momentum says they have to be = and you can solve for any unknowns.*0119

*When we talk about collisions, there are two main types of collisions.*0127

*Elastic collision is collision which kinetic energy is conserved.*0131

*The kinetic energy before the collision is = to the kinetic energy after the collision.*0135

*That is a bouncy type of collision for example.*0139

*There is no law that says the kinetic energy before collision has to = the kinetic energy after the collision.*0143

*That is not always the case, however, it is the case we call that an elastic collision.*0149

*An inelastic collision is a collision which kinetic energy is not conserved.*0155

*A sticky collision, the two objects stick together.*0159

*You have various levels of things in between, partially inelastic collisions things like that.*0163

*The key thing you need to know, if something says it is an elastic collision the kinetic energy before = the kinetic energy after.*0168

*Let us do example of a traffic collision.*0180

*We have a 2000 kg car traveling 20 m/s and it collides with 8000 kg car which is at rest.*0182

*If the 2000 kg car has a velocity of 6.67 m/s after the collision, find the velocity of the 1000 kg car after the collision.*0190

*The way I like to solve these is with a momentum table.*0200

*Let us start by labeling our objects. *0204

*This 2000 kg car, let us call it A and the 1000 kg car let us call it B.*0206

*What we can do is we make a table organizer information.*0213

*We will have a column for our objects, we will have a column for the momentum before the collision in kg m/s, the momentum. *0217

*And a column for the momentum after the collision P after and kg m/s.*0228

*Since it is in 1 dimension, we do not have to worry about X ,Y and Z components, it is all 1 dimension.*0234

*Let us make ourselves in the nice, happy, little table.*0241

*Our objects, we have car A, we have car B, and we make row for total.*0252

*Just filling in lines on a table to help us stay nice and organized.*0265

*For car B and that should work.*0271

*Filling in what we know, car A is 2000 kg traveling at 20 m/s before the collision. *0283

*Its momentum is mass × velocity or 2000 × 20 which will be 40,000 kg meters/s.*0289

*Car B, before the collision is at rest so its momentum is 0.*0303

*Our total momentum before the collision is 40,000 + 0 or just 40,000 kg meters/s.*0307

*After the collision, car A still has the same mass but its velocity is 6.67 m/s *0317

*so its momentum after is 2000 × 6.67 or 13,340 kg meters/s.*0326

*The momentum of car B after the collision, its mass is still 1000 but we do not know its velocity that is why we want the variable in there, Vb for its velocity.*0336

*Its total momentum after is just 1000 Vb.*0347

*If I find that total by adding those up I come up with 13,340 + 1000 Vb.*0350

*We apply the law of conservation of momentum which says that the total momentum *0361

*before the collision has to = the total momentum after the collision.*0365

*We do that by just stating that the totals here must be equal.*0369

*A little bit of algebra, 40,000 = 13,340 + 1000 Vb - 13,340 from both sides.*0376

*I have the 26,660 must = 1000 Vb.*0389

*Therefore, Vb must = 26.7 meters/s.*0396

*There is the velocity of car B after the collision, applying law of conservation of linear momentum.*0404

*Take a look at another one here.*0414

*On a snow covered road, a car with the mass of 1100 kg collides on a van having a mass of 2500 kg traveling at 8 m/s.*0417

*Note, if they are traveling head on, the velocity of 1 with respect to the other in opposite directions, therefore, one of the velocity must be negative.*0426

*As a result of the collision, the vehicles locked together and immediately come to rest.*0437

*Calculate the speed of the car immediately before the collision.*0442

*They must be going this way, they hit, they stop right there.*0445

*Let us take a look at our momentum table.*0451

*We will have a column for objects, we will have a column for momentum before, and the column for momentum after the collision.*0454

*Our objects, we will have a car, a van and a ρ for total.*0462

*Let us start filling in our table.*0497

*Car A, its initial momentum, it is going to 1100 kg and we do not know its velocity.*0500

*We want to know that so we will call that V car.*0511

*The van has a momentum of 2500 × its velocity, we will call this negative compared to the velocity in the car *0515

*which we will assume to be a positive because it is going in the opposite direction.*0525

*-8 will be -20,000 so our total is going to be 1100 V car - 20,000.*0529

*After they stick together, so in reality, practically they become one object with the final momentum of 0 because they are not moving.*0542

*The total momentum after is 0 and the law of conservation momentum states that these two must be equal.*0554

*A little bit of algebra here, that means 1100 V car = 20,000 kg m/s or the velocity of the car is 20,000/1100, which will be 18.2 m/s.*0557

*The car must have been going 18.2 m/s.*0574

*One way it was going 8 m/s to the opposite direction, they hit, they stuck together, and they are immediately at rest.*0577

*Let us take a look at recoil.*0587

*Suppose we have a 4 kg rifle which fires a 20kg with the velocity of 300 m/s, find the recoil velocity of the rifle.*0589

*This time, we do not have two objects colliding.*0600

*Instead, we start as one object and they become 2 in explosion.*0602

*We will start objects, our momentum before, and our momentum after. *0607

*Our objects are a rifle, a bullet, and ρ for total.*0617

*Initially, the rifle and bullet are all one object so we can treat that is if they are just one item and their total momentum before the collision is 0, they are at rest.*0656

*Afterwards, we have the rifle with the mass of 4 kg and some recoil velocity.*0668

*We call that V recoil, we do not know what that is yet.*0676

*The bullet which has a mass 0.02kg, 20g and it is traveling at 300 m/s.*0680

*Its momentum after the explosion, after the event, is 6 kg meters/s.*0688

*When we add these up, we get 4 V recoil + 6, our total over here is 0 and *0694

*the law conservation of linear momentum says we can set this equal.*0702

*4 VR = -6 VR = -1.5 meters/s.*0708

*The recoil velocity of the rifle is 1.5 m/s.*0716

*The whole thing sitting there, you fire a bullet out one way at 300 m/s.*0721

*The rifle kicks back the other direction with the velocity of 1.5 m/s.*0725

*An atomic collision problem, a proton of mass M and a lithium nucleus mass 7M undergo an elastic collision.*0733

*Elastic collision, that could be important, that means the total kinetic energy before the collision = the total kinetic energy *0743

*after the collision, an extra piece of information.*0750

*Find the velocity of the lithium nucleus following the collision.*0753

*Let us start with our momentum tables.*0759

*Our objects, we have our proton, we have a lithium nucleus, and we have ρ for our total.*0761

*We have momentum before and momentum after.*0775

*What do we know to begin with?*0788

*Our proton that is traveling 1000 m/s with mass M so we call that 1000M for its initial momentum and lithium nucleus is at rest.*0790

*Its initial momentum is 0 so our total when we add this up is 1000M.*0802

*After the collision, the momentum of our proton is its mass × the velocity of the proton, we will call that VP.*0808

*The lithium nucleus, its mass is 7 M × at some velocity will call a VL.*0819

*After, we have 1000 M = MVP + 7MVL and those must be = by the law of conservation of linear momentum.*0828

*1000 M = MVP + 7MVL or we can divide the M out of all of that, 1000 = VP + 7VL or VP = 1000 -7 VL.*0845

*There is a start but we need some more information.*0863

*Now, we will have to go back to the fact that this is an elastic collision to get a little bit more info.*0865

*We know that the kinetic energy before = the kinetic energy after.*0871

*Before, our total kinetic energy ½ MV² is ½ M × its speed 1000².*0878

*Our lithium nucleus was not moving so that is 0.*0889

*After the collision, we have the kinetic energy of our proton ½ M and the velocity of our proton *0892

*we said as 1000 -7VL + the kinetic energy of lithium nucleus, *0901

*which is going to be ½ × its mass 7M × its velocity VL².*0911

*As I look at this, it looks like I can do some simplifications here, we can ÷ 1/2M out of all of these.*0920

*I can get 1000² = ½, 1000 -7 VL² + we can pull ½ M out of that, we have got 7 VL².*0928

*Which implies then that 1000² is a million, so 10⁶ = if we expand this,*0949

*10⁶ - 7000VL- 14000VL + 49VL² + 7VL².*0958

*Or rearranging this a bit further, 10⁶ -10⁶ we can say is 0 = -14,000 VL + 56VL² ÷ 14,000VL at both sides may be so 56VL² = 14,000VL.*0974

*We can factor VL out of that to say that VL = 0 or 56VL = 14,000.*1003

*In this case, VL would be = to the 14,000 ÷ 56 is 250 m/s.*1012

*VL = 0 or VL = 250 m/s we will choose the 250 as the answer.*1020

*It is going to make sense here for a problem.*1028

*If VL, the velocity of our lithium nucleus is 250 m/s we can plug that in our equation for the velocity of our proton.*1031

*But over here for VL, MVP = 1000 -7 × VL which we just said was 250 m/s or 1000 -1750 VP is going to be = -750 m/s.*1040

*And negative, implying it is going back in the opposite direction.*1066

*Our little proton is bouncing into our lithium nucleus this one was moving away to the right at 250 m/s.*1070

*But our proton has bounced back in the opposite direction at 750 m/s.*1077

*We found the velocity of our lithium nucleus following the collision.*1084

*If we have to deal with things in multiple dimensions like playing billiards. *1091

*Bart strikes a cue ball of mass 0.17 kg giving the velocity of 3 m/s in the x direction, there it goes.*1096

*When the cue ball strikes the 8 ball of mass 0.16kg previously at rest, the 8 ball was deflected 45° from the cue ball’s previous path.*1102

*The cue balls deflected 40° in the opposite direction.*1113

*Find the velocity of the cue ball and the 8 ball after the collision.*1117

*Let us give ourselves some more room here.*1123

*What we are going to do is we are going to make a momentum tables. *1127

*We are going to make separate ones for the X direction and the Y direction and treat them separately.*1129

*Just like we did when we separated horizontal and vertical motion for projectile motion.*1134

*Let us start off by setting the VC, we will call that the velocity of our cue ball.*1140

*We will call V8 the velocity of our 8 ball.*1150

*If we make a momentum table for the X direction first, we have our objects, we will have the momentum before in the X direction*1156

*and the momentum after the collision in the X direction.*1166

*Our objects include our cue ball, our 8 ball, and we will also have ρ for total.*1169

*Before the collision in the X direction, the cue ball could have mass a 0.17 is traveling 3 m/s.*1191

*We have 0.17 × 3 which is 0.51 kg meters/s as its initial momentum in the X direction.*1199

*The 8 ball was at rest, so that is easy that 0.*1209

*our total momentum before the collision in the X direction must be 0.51.*1212

*After the collision, we still have the same mass 0.17 kg.*1218

*The velocity of our cue ball, however, we do not know yet.*1223

*What we know its variable VC, we defined, and we also know the X component of that is going to be the X component of these path,*1228

*which is going to bring in the component cos 40°.*1234

*Looking at the 8 ball after the collision, its mass is 0.16 kg. *1244

*We do not know its total velocity and we call that V8.*1250

*The X component, the horizontal component is going to be its mass × velocity × that cos 45°.*1255

*When I do this 0.17 × VC × cos 40 that is going to give me 0.13 VC and 0.16 V8 cos 45 is going to be 0.113 V8.*1266

*By the law of conservation of linear momentum, we can state that these must be equal.*1282

*Let us call that equation 1.*1288

*Let us look in the Y direction.*1292

*We make a table for momentum in the Y direction, our objects again are going to be our cue ball and the 8 ball will have ρ for total.*1295

*We will have the momentum before and will have the momentum after.*1307

*In the Y direction, things get a little bit simpler because our cue ball initially is not moving in the Y direction.*1322

*Its momentum in the Y direction is 0, the 8 ball is at rest 0 so the total 0.*1329

*After the collision, the mass of our cue ball 0.17 × its velocity VC but now we need the Y component of its velocity which is going to be sin -40°.*1336

*For the 8 ball, its mass still 0.16, its velocity V8, but we need to Y component of its motion which is going to be sin 45°.*1351

*When I add these together, 0.17 sin -40 is about -0.109 VC + 0.16945 is going to be a 0.113 V8 and we will call that equation 2.*1363

*When we set these = 0 of course because of the law of conservation of linear momentum.*1383

*Let us see if we can solve this for VC.*1391

*-0.109 VC + 0.113 V8 + 0.109VC to both sides I have 0.109VC = 0.113 V8 or VC = 0.113/0.109 × V8*1395

*which is going to be 1.04 V8 so that is from equation 2.*1413

*Let us take a look at equation 1 if we pull that over here.*1421

*We have a 0.51 = 0.13 VC, let us replace VC with 1.04 V8 + 0.113 V8.*1425

*An equation that has just a single variable V8 should be able to solve that.*1445

*0.51 = we are going to have 0.13 × 1.04 + 0.113 or 0.248 V8. *1450

*And then just solving for V8, it is going to be 0.51 ÷ 0.248 or right around 2.06 m/s.*1465

*Now that we have V8, we can figure out VC = 1.04 × 2.06 m/s.*1478

*VC is going to be = 2.14 m/s.*1494

*How we can deal with that in two dimensions?*1504

*Break up your momentum into X and Y components.*1507

*Let us take a look at a couple of free response problems from old AP exams.*1512

*Let us start off with a 2001 exam free response question number 1.*1516

*You can find the link up here or google search it, you can find these online,*1520

*pause the video, and take another to look at it, over see if you can solve it, and come back *1525

*and check your solution or where you got stuck as we go through here.*1529

*The 2001 free response exam, it looks like we have a cart with a couple motion sensors. *1534

*A couple of graphs of velocity and force as they collide.*1541

*Part A says, determine the cart average acceleration between 0.33 and 0.37 s. *1546

*Let us take a look, average acceleration.*1553

*Average is going to be a change in velocity ÷ the time interval, which is your final velocity minus your initial over your time interval.*1560

*It looks like a 0.37s where what may be -0.18 m/s - the initial.*1572

*In the initial at 1.33 looks like that is what about 0.22 m/s and all of that happens over the time interval at 0.04 s *1584

*which is going to be -0.4 m/s over 0.04 s which is -10 m/s².*1591

*Alright, taking a look at part B.*1606

*Determine the magnitude of the change in the carts momentum during the collision.*1610

*Change in momentum is the integral of the force time graph F.DT from T1 to T2*1617

*or in this case, we are given that FT graph it is just going to be the area under the graph.*1628

*It looks like we can break that up into a couple triangles and a rectangle. *1634

*If I do that and you can break it up by the way you wanted to.*1638

*To find the area, I am going to take the area of that first triangle, I see I have a tall triangle, 2 little triangles on the side and 1 rectangle in the middle.*1642

*Area of the triangle ½ × our base 0.01 × our height 10 N +*1651

*our rectangle length 0.02, height 10 N + our big tall triangle ½ × its base 0.02 × its height 30 N +*1660

*We got 1 more triangle there on the right ½ × 0.01 × 10.*1678

*I come up with a total of about 0.6 kg-m/s.*1684

*If you brought that up using other shapes that would certainly work as well.*1691

*Find the area under that graph time integral.*1694

*Alright, part C.*1699

*Determine the mass of the cart.*1701

*Change in momentum, assuming mass is constant is mass × change in velocity.*1704

*Therefore, mass is going to be change in momentum over change in velocity.*1710

*We just found change in momentum is 0.6 kg-m/s and our change in velocity was 0.4 m/s.*1715

*We found that right up there so that is just going to be 1.5 kg.*1728

*Part D, determine the energy lost in the collision between the force sensor in the cart.*1740

*That is going to be our change in kinetic energy which is our final kinetic energy - initial kinetic energy,*1750

*which will be 1/2 MV final² – 1/2 MV initial².*1758

*It is going to be M/2 × RV² - the initial² or mass 1.5 kg/2, *1767

*our final velocity 0.22 m/s² - our initial velocity 0.18m/s²,*1778

*which gives me a change in kinetic energy of 0.012 joules.*1789

*Alright, relatively straightforward there.*1802

*Let us take a look at another free response problem, this time going to the 2002 exam free response number 1.*1805

*Take a minute, download it, look it over, print it out, give it a shot, comeback when you have a chance to look it over.*1815

*As I look at this one, we got a crash test cars at some mass moving it a speed colliding completely *1826

*inelastically with an object of mass M at time T = 0, it was initially at rest.*1832

*The speed V after the collision is given by that function.*1839

*Now A, says calculate the mass of the object.*1844

*The way I would go about doing that is I look at conservation of linear momentum.*1849

*The total momentum before must = the total momentum after.*1854

*Our momentum before, we have 1000 kg, the mass of our car × its velocity 12 m/s + the mass of our second object *1859

*which is not moving so 0 must = collide and it is a completely inelastic collision.*1871

*They are sticking together, our new combined mass is going to be 1000 + M × their velocity at T = 0.*1879

*The left hand side becomes 12,000 = 1000 + M and then we find the velocity at T = 0 using the function they give us which is going to be 8/1 or 8.*1891

*12,000 = 8000 + 8M = 4000 which implies then that the mass capital M must be 500kg.*1910

*Let us move on to part B.*1921

*Assuming an initial position of X = 0 find the expression for the position of the system after the collision is a function of time.*1938

*We are given the velocity function so we should be able to find the position function as the integral from 0 to T of our velocity function of 8 DT/1 + 5T.*1946

*Or we can do it indefinitely and use that boundary condition that we know that the initial position at × 0 = 0.*1963

*That is going to be 8 integral from 0 to T of DT/1 + 5T.*1973

*And that looks like it is in the form of DU/U in order to do that if our U is = 1 + 5T. *1982

*DU must = 5DT and we only have a DT.*1991

*I am going to multiply this by 5 to make it fit that form which means I have to ÷ that side by 5.*1998

*This implies then that our position function must = we have 8/5 the integral of something of a form DU/U is natural log of U.*2005

*That is going to be the natural log of 1 + 5T evaluated from 0 to T which is going to be 8/5 × the log of 1 + 5T *2018

*substituting T in for my variable - the log of 1 + 0,*2032

*Which would just be natural log of 1, log of 1 is 0.*2039

*What we are going to end up with then is the position as a function of time is 8/5 log of 1 + 5T.*2044

*Alright, moving on to part C, let us give ourselves some more room here.*2057

*Determine an expression for the resisting force on the car object system after the collision is a function of time.*2066

*When I’m thinking about the resisting force we know the velocity function should be able *2073

*to find the acceleration, we know the mass so we can use Newton’s second law F = MA to find that force.*2078

*Let us start off with, acceleration is the time rate of change of velocity which is the derivative with respect to time of 8/1 + 5T.*2086

*Which is going to be 8 × the derivative of 1 + 5T to -1.*2098

*I have 8 the derivative of that -1 is going to be -1 + 5T⁻² × DT 5 which is going to be -40/1 + 5T².*2110

*If we want our force, that is our mass × acceleration.*2133

*Our mass 1500 × acceleration -40/1 + 5T², it is going to be -60,000/1 + 5T².*2139

*Part D, determine the impulse delivered to the car object system from time T = 0 to T = 2 s.*2167

*Let us give ourselves some more room for part D here.*2175

*As we look at part D, find the impulse.*2178

*Impulse is the integral of force with respect to time.*2181

*That is going to be the integral from T = 0 to 2s of our force function -60,000/1 + 5T² DT.*2187

*We will pull out our constants.*2203

*Let us see, that will be -60,000 integral from 0 to 2 of 5 T + 1⁻² DT.*2207

*But if I'm going to do this, I’m going to integrate it, U to some power, I need to have the du here which means we need a 5 DT.*2222

*I'm going to put a 5 here which means I am going to divide by 5 there.*2231

*That implies then that we have a-60,000/5 × the integral of that is going to be that 1 + 5 T⁻¹/-1 *2236

*all evaluate from T = 0 to 2s or -60,000/5 that is -12,000.*2258

*We will have -1/1 + 5 T, just rewriting that right hand side.*2267

*Evaluated from 0 to 2 which implies then that our impulse is going to be -12,000 × we will plug our 2 in first.*2274

*-1/1 + 10 is going to be -1/11 - -1/1 + 5 × 0 that is just going to be - 1.*2286

*This would be -12,000 × -1/11 + 1.*2299

*Which when I run through all that math there, -1/11 + 1 I end up with our impulse = -10,909kg m/s.*2309

*I think that finishes up the 2002 question 1.*2327

*Alright, let us do one more here.*2332

*Let us go to the 2014 exam Mechanics question number 1.*2336

*As we look at this one, we have a cart and some photo gates in the spring and *2343

*the experiment the student uses a spring to accelerate the cart along the track.*2349

*The restoring force is given FS = as² + bs, it is not a linear restoring force.*2353

*It is given by the function.*2359

*First, where are asked to find an expression for the potential energy as a function of that compression s.*2361

*If we take a look at that A, remember force = -DUDL which implies then that DU=-FDL.*2369

*Or if we integrate both sides to potential energy is minus the integral of force DL as long as we are dealing with the conservative force and we are.*2385

*We can say that our potential energy function U is the integral from S = 0 to some final value s of -our force function as² + bs all with respect to s.*2395

*Which will be -as³/3 + bs²/2, all evaluate from 0 to s which is just going to be -as³/3 - bs²/2.*2412

*That is the work done by the force.*2432

*The work done by the force in compressing the spring that means that the work that potential energy stored in the spring *2435

*must be the opposite of that which is going to be as³/3 + bs²/2.*2442

*Let us take a look at part B.*2454

*Part B, talks a little bit more about the experiment and we are to calculate the speed of the cart immediately after it loses contact with the spring.*2460

*We can use an energy argument there that the initial energy = the final energy which implies that *2469

*the initial energy stored in the spring must = the final kinetic energy in the cart.*2477

*as³/3 + bs²/2 = ½ MV².*2486

*Plugging in with what we know, substituting in, we have 200 a × 0.04³/3 + 150 b × 0.04²/2 = ½ × mass 0.3 V².*2495

*Or the left hand side becomes 0.1243 = ½ × 0.3 × V² complies then that V² = 0.828 or V is about 0.91 m/s.*2521

*Let us take a look at B2.*2547

*The impulse given to the cart by the spring.*2550

*Impulse is just change in momentum which is going to MV final - MV initial is going to be our mass is 0.3 × our final 0.91 - initial velocity 0 *2554

*or about 0.273kg m/s.*2569

*There is B.*2576

*For C, it looks like we are making a graph so let us give ourselves more room here on the next page.*2578

*For C, we have a graph, we are set up with axis that is a nice detail graph to help as plot points very accurately.*2583

*A lot of grid lines.*2596

*Label this I would have velocity here in m/s.*2601

*Our x would be in meters position.*2609

*Plot the data points for the speed of the car as a function of position, scale and label all axis as appropriate.*2615

*We might have something like 0.2, .4, .6, .8, 1, 0.1, .2, 0.3, 0.4.*2620

*When I plot my points, I needed up with something that looks like we are *2637

*just above the point for here at 0.2 or pretty close to that over here at 0.4.*2642

*We are starting to tail off a little bit more quickly as we get over to the right.*2648

*Something like that.*2655

*That would be part C.*2659

*For part D, compare the speed of the cart measured by photo gate 1 the predicted value *2662

*of the speed of a cart just after it loses contact with the spring.*2669

*The velocity is measured by the photo gate 1 was 0.412 m/s.*2674

*If we predicted that, compared to the predicted value, I would say that is about 0.320 m/s.*2684

*How did that happen?*2702

*Why are we off?*2705

*You could look at this from a bunch of different perspectives and there are bunch of different right answers.*2706

*The one that I chose was stating that the uncertainty of the position measurement of + or -0.005m coupled *2711

*with the uncertainty in the speed measurements of + or -0.002 m/s.*2718

*That could account for this difference.*2723

*It does not take a whole lot of difference in that position measurement coupled with the speed measurement to account for this.*2726

*That would be one way you could explain that.*2732

*For D2, from the measured speed values of the cart as it rolls down the track,*2736

*give a physical explanation for any trend you observe.*2741

*The key thing that I see here is, as we go further, it looks like the rate at which our speed is dropping off is actually increasing.*2744

*I said something to the effect of friction from the carts axis and air resistance are doing some sort of negative work on the cart.*2753

*Those resistance forces must be increasing which can be absorbed by noticing the increasing negative slope of the graph.*2760

*The cart is losing more speed in the last 0.2m that was in the first 0.2m.*2767

*Some sort of explanation like that where you give an explanation for why we are falling off more quickly there is what they are really looking for.*2772

*With that, I think we are all finish up with conservation of linear momentum.*2781

*Thank you so much for watching www.educator.com and make it a great day everyone.*2786

1 answer

Last reply by: Professor Dan Fullerton

Sat Dec 26, 2015 8:34 AM

Post by Shehryar Khursheed on December 25, 2015

I have a question concerning part a on the 2014 FRQ1. When I was solving it on my own, I first used U=-W and since W=integral(F dr), then U= -integral(F dr). Doing so, I got the same expression for potential energy except with negative signs. Looking at the scoring guidelines for this question, I'd still get credit for this expression since they are only concerned with the magnitude. However, when I use this for part b, I obviously get the wrong answer. My question is how do you know when to put the negative sign or leave an expression positive when dealing with work and energy? Also, why did my rationale for part a give me incorrect signs?

1 answer

Last reply by: Professor Dan Fullerton

Sun Apr 19, 2015 1:47 PM

Post by Huijie Shen on April 19, 2015

Hi professor,

I have some questions here:

When should we use the conservation of KE to solve question? Is it only when the question mentions that it's an elastic collision?

Is example V an elastic collision? Why don't we use the conservation of KE? Or is it that when we deal with any collisions in multiple dimension with given initial velocity and final directions , we don't have to use conservation of KE?

Sincere, Alina

1 answer

Last reply by: Professor Dan Fullerton

Sat Dec 13, 2014 5:12 PM

Post by Thadeus McNamara on December 13, 2014

at 29:45 did you switch the initial and final velocities?

2 answers

Last reply by: Michael Sramek

Sat Apr 11, 2015 1:44 PM

Post by zihni kaleci on October 31, 2014

i got it to work on google chrome as safari kept giving me Network error! [Error #2032]! great lectures so far

1 answer

Last reply by: Professor Dan Fullerton

Thu Oct 30, 2014 6:02 AM

Post by zihni kaleci on October 29, 2014

this lecture won't work!