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Lecture Comments (11)

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!

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 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

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 event0039

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 direction1156

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 × V81395

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 T21617

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