WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to educator.com.
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I am Dan Fullerton and in this lesson we are going to talk about momentum and impulse.
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Our objectives include relating mass philosophy in linear momentum for moving object and calculating the total linear momentum of the system of objects.
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Relating impulse to the change in linear momentum and the average force acting on object.
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Calculating the area under force vs. time graph.
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Relating to change in momentum which is the impulse of an object.
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Calculating the change in momentum of an object given a force function as a function of time acting on the object.
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Let us start by defining momentum.
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Momentum as a vector describing how difficult it is to stop moving object.
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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.
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It has more mass and that typically has more velocity.
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Total momentum is the sum of all the individual momentum when you are talking about a system.
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Momentum given the symbol P it is a vector is equal the mass × the velocity.
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Units are kg meters /s or Newton × seconds -- they are equivalent.
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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.
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Find its momentum.
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Our momentum, initial momentum is mass × velocity which is 3600 kg × 85 m / s which is 36000 kg m/s.
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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?
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Final momentum here is mass × velocity which could be 3000 kg × 120 m / s which is 36000 kg m/s.
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As you can see momentum can change.
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This change in momentum is what we call it impulse.
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It gets the symbol capital J, its formula is change in momentum and its units of course must also be in kg m/s.
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The D3 bomber which had the momentum of 3.6 × 10⁵ kg meters / s comes to a halt on the ground.
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What impulse was applied?
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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.
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What is that negative mean?
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That again is telling this direction.
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We are calling the 3.6, whenever this direction is, as positive and because the impulses applied
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in the opposite direction to stop it that is how we get a negative sign.
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Let us explore the relationship between force and change in momentum or impulse.
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Force is mass × acceleration but we know that acceleration is the derivative of velocity with respect to time.
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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.
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We know MV is momentum so we can write this as force is equal to the derivative of momentum
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with respect to time, force in the derivative of momentum.
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Let us do an example.
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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?
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Force is the derivative of momentum with respect to time which is going to be the derivative with respect to time of
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KT², K is a constant so that is K × the derivative with respect time of T² which is going to be 2KT.
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We can take a look at this relationship between impulse momentum and take a little bit further.
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If force is the derivative of momentum with respect to time this implies then
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that we should be able to integrate from 0 to T, our force with respect to time.
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If we integrate the right hand side from some corresponding initial momentum to find a momentum of our differential of momentum.
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We find that, we have impulse is equal to F δ T, change in time is equal to our change in momentum.
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Impulse is force applied for some time which is a change in momentum, they all work.
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You can use any piece of that its most comfortable for you.
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Apply a force for some of the time and change in objects momentum.
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And that force applied for a time is what we call an impulse.
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It all have units of kg m/s and just a very useful relationship to remember.
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Let us do an example problem around that, a 6 kg block sliding to the east across a horizontal frictionless surface
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with the momentum of 30 kg m/s strikes an obstacle.
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The obstacle exerts an impulse of 10 N seconds to the west on the block.
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Find the speed of the block after the collision.
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Let us start, impulse is change in momentum which is our final momentum MV final - our initial momentum MV initial.
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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.
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Therefore, final velocity we can substitute in our values, momentum 10 N/s to the west that means it must be negative.
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-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.
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Let us take a look at another example.
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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.
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What horizontal force must the girl apply on the gun in order to hold it in position?
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Force is a derivative of momentum with respect to time which is the derivative with respect to time of Mass × Velocity
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or we could write this as a DMDT × V.
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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
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which implies that the force required to hold that gun in place must be 2 N.
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We can also look at this graphically.
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Impulse is the area under a forced time graph, it is equivalent to change in momentum.
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We can also write impulse as the integral of FDT, the area under force time graph gives you the impulse.
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Take a look at an example with the non constant force.
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The area under the force time curve is the impulse to change a momentum.
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We just mentioned that, determine the impulse applied here by calculating the area of the triangle under the curve.
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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,
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is going to be 1/2 our base is 10s our height is 5 N.
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That is just going to be 25 N/s.
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Let us take a look at another one.
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The graph indicates the force on a truck of mass 2000 kg as a function of time.
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The interval from 0 to 3s, determine the change in the trucks velocity.
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I am going to do there is first is by finding the impulse.
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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.
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The area here 1000 N × 2s will be 2000 N/s - the area down here 1000 N × 1 s.
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Or 1000 N/s and that has to be equal to the change in momentum or M δ V.
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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.
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Let us take a look at one last example problem here.
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A force FT is T³ is applied to a 10 kg mass.
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What is the total impulse applied to the object between 1 and 3 s?
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Let us start off with force as the derivative of momentum which implies then that DP or differential of momentum is forced DT.
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Which implies that DP is going to be equal to, our force is T³ DT.
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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
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from T = 1 to T = 3 s of T³ DT.
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Our left hand side becomes P final minus the P initial, our right hand side becomes T⁴/4 evaluated from 1 to 3 s,
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which implies then P final - P initial, that is δ P must = 3⁴ is going to be 81 force -
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and plug 1 in there, 1⁴ is 1/4 so that is going to be the 8/4 but we also know as 20.
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Which implies the change in momentum is 20 kg meters / s and change in momentum is the impulse.
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There is our answer, 20 kg m/ s.
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Hopefully, that gets you a good start with momentum and impulse.
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Thank you for taking the time and watching www.educator.com.
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We will see you soon and make it a great day everyone.