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

## Discussion

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## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Momentum & Impulse

- Momentum (p) is a vector describing how difficult it is to stop a moving object.
- The total momentum of a system is the sum of the individual momenta of the particles comprising the system.
- Units of momentum are kg*m/s, or N*s.
- A change in momentum is known as an impulse (J).
- Impulse is change in momentum, as well as the product of the average force and the time it is applied.
- The area under a force-time graph is the impulse applied to an object.

### Momentum & Impulse

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
- Momentum
- Example I: Changing Momentum
- Impulse
- Example II: Impulse
- Relationship Between Force and ∆p (Impulse)
- Example III: Force from Momentum
- Impulse-Momentum Theorem
- Example IV: Impulse-Momentum
- Example V: Water Gun & Horizontal Force
- Impulse from F-t Graphs
- Example VI: Non-constant Forces
- Example VII: F-t Graph
- Example VIII: Impulse from Force

- Intro 0:00
- Objectives 0:07
- Momentum 0:39
- Definition of Momentum
- Total Momentum
- Formula for Momentum
- Units of Momentum
- Example I: Changing Momentum 1:18
- Impulse 2:27
- Impulse
- Example II: Impulse 2:41
- Relationship Between Force and ∆p (Impulse) 3:36
- Relationship Between Force and ∆p (Impulse)
- Example III: Force from Momentum 4:37
- Impulse-Momentum Theorem 5:14
- Impulse-Momentum Theorem
- Example IV: Impulse-Momentum 6:26
- Example V: Water Gun & Horizontal Force 7:56
- Impulse from F-t Graphs 8:53
- Impulse from F-t Graphs
- Example VI: Non-constant Forces 9:16
- Example VII: F-t Graph 10:01
- Example VIII: Impulse from Force 11:19

### 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: Momentum & Impulse

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

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

*Our objectives include relating mass philosophy in linear momentum for moving object and calculating the total linear momentum of the system of objects.*0008

*Relating impulse to the change in linear momentum and the average force acting on object.*0017

*Calculating the area under force vs. time graph.*0023

*Relating to change in momentum which is the impulse of an object.*0025

*Calculating the change in momentum of an object given a force function as a function of time acting on the object.*0030

*Let us start by defining momentum.*0038

*Momentum as a vector describing how difficult it is to stop moving object.*0041

*To think about how hard it is to stop a fly flying in your hands compared to a bus coming in to you, it takes a lot more momentum to stop that bus.*0046

*It has more mass and that typically has more velocity.*0057

*Total momentum is the sum of all the individual momentum when you are talking about a system.*0061

*Momentum given the symbol P it is a vector is equal the mass × the velocity.*0066

*Units are kg meters /s or Newton × seconds -- they are equivalent.*0072

*Momentum can change, a D3 bomber with the mass 3600 kg departs from its aircraft carrier with the velocity of 85 m / s due east.*0078

*Find its momentum.*0088

*Our momentum, initial momentum is mass × velocity which is 3600 kg × 85 m / s which is 36000 kg m/s.*0091

*After it drops its payload, its new mass is 3000 kg and obtains a cruising speed of 120 m / s, what is its momentum now?*0113

*Final momentum here is mass × velocity which could be 3000 kg × 120 m / s which is 36000 kg m/s.*0123

*As you can see momentum can change.*0144

*This change in momentum is what we call it impulse.*0148

*It gets the symbol capital J, its formula is change in momentum and its units of course must also be in kg m/s.*0151

*The D3 bomber which had the momentum of 3.6 × 10⁵ kg meters / s comes to a halt on the ground.*0162

*What impulse was applied?*0169

*Change in momentum is our final value of momentum - our initial value which is 0 -3.6 × 10⁵ kg m/s or -3.6 × 10⁵ kg m/s is our impulse.*0172

*What is that negative mean?*0196

*That again is telling this direction.*0198

*We are calling the 3.6, whenever this direction is, as positive and because the impulses applied*0201

*in the opposite direction to stop it that is how we get a negative sign.*0209

*Let us explore the relationship between force and change in momentum or impulse.*0216

*Force is mass × acceleration but we know that acceleration is the derivative of velocity with respect to time.*0221

*We can write this as force is equal to Mass × DVD T, this implies that however that force is equal to the derivative with respect to time of MV.*0235

*We know MV is momentum so we can write this as force is equal to the derivative of momentum*0258

*with respect to time, force in the derivative of momentum.*0266

*Let us do an example.*0275

*The momentum of an object as a function of time is given by momentum equals KT² where K is a constant, what is the equation for the force causing this motion?*0278

*Force is the derivative of momentum with respect to time which is going to be the derivative with respect to time of*0290

*KT², K is a constant so that is K × the derivative with respect time of T² which is going to be 2KT.*0298

*We can take a look at this relationship between impulse momentum and take a little bit further.*0313

*If force is the derivative of momentum with respect to time this implies then*0319

*that we should be able to integrate from 0 to T, our force with respect to time.*0326

*If we integrate the right hand side from some corresponding initial momentum to find a momentum of our differential of momentum.*0332

*We find that, we have impulse is equal to F δ T, change in time is equal to our change in momentum.*0343

*Impulse is force applied for some time which is a change in momentum, they all work.*0360

*You can use any piece of that its most comfortable for you.*0366

*Apply a force for some of the time and change in objects momentum.*0369

*And that force applied for a time is what we call an impulse.*0372

*It all have units of kg m/s and just a very useful relationship to remember.*0376

*Let us do an example problem around that, a 6 kg block sliding to the east across a horizontal frictionless surface*0386

*with the momentum of 30 kg m/s strikes an obstacle.*0393

*The obstacle exerts an impulse of 10 N seconds to the west on the block.*0397

*Find the speed of the block after the collision.*0402

*Let us start, impulse is change in momentum which is our final momentum MV final - our initial momentum MV initial.*0405

*MV final is going to be our impulse + our initial momentum or V final is going to be M impulse + initial momentum divided by the mass.*0420

*Therefore, final velocity we can substitute in our values, momentum 10 N/s to the west that means it must be negative.*0440

*-10 N/s + 30 kg m/s ÷ 6 kg, which implies that our final velocity must be 3.33 m/s and since that is positive it must be to the east.*0449

*Let us take a look at another example.*0475

*A girl with a water gun shoots a stream of water that ejects 0.2 kg of water / s horizontally at the speed of 10 m/s.*0478

*What horizontal force must the girl apply on the gun in order to hold it in position?*0486

*Force is a derivative of momentum with respect to time which is the derivative with respect to time of Mass × Velocity*0493

*or we could write this as a DMDT × V.*0504

*Where we know the DMV T, it tells us that 0.2 kg of water / s to that is going to be 0.2 kg of water / s × the speed of 10 m / s*0509

*which implies that the force required to hold that gun in place must be 2 N.*0523

*We can also look at this graphically.*0533

*Impulse is the area under a forced time graph, it is equivalent to change in momentum.*0535

*We can also write impulse as the integral of FDT, the area under force time graph gives you the impulse.*0541

*Take a look at an example with the non constant force.*0556

*The area under the force time curve is the impulse to change a momentum.*0559

*We just mentioned that, determine the impulse applied here by calculating the area of the triangle under the curve.*0562

*Going to change force time graph so we can go find the area of that, impulse is going to be the area for triangle which is 1/2 base × height,*0569

*is going to be 1/2 our base is 10s our height is 5 N.*0584

*That is just going to be 25 N/s.*0591

*Let us take a look at another one.*0602

*The graph indicates the force on a truck of mass 2000 kg as a function of time.*0603

*The interval from 0 to 3s, determine the change in the trucks velocity.*0609

*I am going to do there is first is by finding the impulse.*0614

*The impulse is going to be the integral from T equal 0 to 3s of FDT which is really just looking at the area between 0 and 3s.*0618

*The area here 1000 N × 2s will be 2000 N/s - the area down here 1000 N × 1 s.*0638

*Or 1000 N/s and that has to be equal to the change in momentum or M δ V.*0649

*Therefore, 2000 -1000 N/s = the mass of our truck 2000 kg × δ V or a change in velocity is going to be 1000/2000 just 0.5 meters / s.*0657

*Let us take a look at one last example problem here.*0678

*A force FT is T³ is applied to a 10 kg mass.*0682

*What is the total impulse applied to the object between 1 and 3 s?*0687

*Let us start off with force as the derivative of momentum which implies then that DP or differential of momentum is forced DT.*0694

*Which implies that DP is going to be equal to, our force is T³ DT.*0706

*If we go and we integrate both sides then, the integral of DP from some P initial to P final must be equal to the integral*0714

*from T = 1 to T = 3 s of T³ DT.*0723

*Our left hand side becomes P final minus the P initial, our right hand side becomes T⁴/4 evaluated from 1 to 3 s,*0730

*which implies then P final - P initial, that is δ P must = 3⁴ is going to be 81 force -*0741

*and plug 1 in there, 1⁴ is 1/4 so that is going to be the 8/4 but we also know as 20.*0754

*Which implies the change in momentum is 20 kg meters / s and change in momentum is the impulse.*0764

*There is our answer, 20 kg m/ s.*0774

*Hopefully, that gets you a good start with momentum and impulse.*0781

*Thank you for taking the time and watching www.educator.com.*0784

*We will see you soon and make it a great day everyone.*0787

1 answer

Last reply by: Professor Dan Fullerton

Sat Jan 28, 2017 7:30 AM

Post by Duhoe Lee on January 27 at 09:28:00 PM

Why do we use momentum instead of kinetic energy?

1 answer

Last reply by: Professor Dan Fullerton

Sun Apr 19, 2015 1:46 PM

Post by Huijie Shen on April 19, 2015

Really helpful class!!! Thank you professor!