AP Physics C: Mechanics 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 applied0201

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 momentum0258

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 of0290

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 then0319

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 surface0386

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 × Velocity0493

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 / s0509

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 integral0714

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