This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Energy and Momentum

Topics include work, energy, conservation of energy, impulse, linear momentum, and conservation of linear momentum. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes.Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

In this video, we discuss momentum, or

Â what scientists sometimes refer to as a quantity of motion.

Â It is defined as the product of an object's mass and its velocity.

Â Momentum is a vector quantity and

Â is measured in units of kilograms multiplied by meters per second.

Â Since momentum is a vector,

Â direction is very important when describing this quantity.

Â Pay close attention to positive and negative signs.

Â An object traveling left will have a negative momentum

Â because of its negative velocity.

Â Recall in Newton's second law,

Â a net force acting on an object causes an acceleration.

Â As seen here, we can replace the acceleration term with its definition,

Â change in velocity over time.

Â Rearranging this expression, we can see that a force multiplied by time will

Â equal the change in an object's momentum.

Â We call this term impulse, and

Â it is the measure of the total change in momentum of an object.

Â Impulse is also a vector measured in kilograms multiplied by meters per second.

Â >> Let's look at an example.

Â A 1,500 kilogram car moved to the right with a speed of ten meters per second.

Â Calculate the momentum of the car as it moves.

Â The car then collides with another vehicle and

Â afterward is moving left with a speed of two meters per second.

Â If the cars were in contact for only 0.01 seconds,

Â calculate the average force exerted on the car during the collision.

Â When calculating momentum, don't forget that it is a vector, not a scalar.

Â So direction is important.

Â Our momentum equation for an object is mass times velocity.

Â And so for p equaling the mass of the car times velocity of

Â the car will give me the momentum of this vehicle as it moves.

Â So this first part is a pretty short problem for us.

Â The mass of the car, 1,500 kilograms.

Â The speed of the car, 10 meters per second.

Â And so it's this quantity of motion that we're used to referring to.

Â Really, a larger car moving at the same speed as a smaller car,

Â a larger car has more momentum.

Â It's very kind of a gut feeling when it turns into

Â imagining what momentum is, what impact it might have on a collision or

Â on a motion of a body, and we'll see that shortly here.

Â Our units, mass times velocity.

Â So kilograms times meters per second,

Â that'll also be written as Newtons times seconds.

Â Either are perfectly valid.

Â If you forget Newtons times seconds, make sure just to go back to your equation.

Â Let's say you can't remember the units.

Â And plug in the base units, and that's perfectly acceptable.

Â The units of mass are kilograms, which you know, and the mass of velocity

Â is meters divided by seconds, distance divided by time, which you know.

Â So it's perfectly acceptable to leave the units in its pieces rather than maybe some

Â larger unit that you may not remember the name of.

Â The question then goes on to say that the car then collides with another vehicle.

Â And is now moving to the left with a speed of 2 meters per second.

Â It hit and is now flying backwards in the opposite direction of its original motion.

Â They were in contact for only a short amount of time.

Â And it wants to know what is the average force.

Â This is where the topic of impulse comes up.

Â Impulse is represented by a J.

Â And it is a force times the change in time.

Â And equal to the change of momentum.

Â All of these things are equal, so you can pick and

Â choose what part of the equation you need to solve the particular problem.

Â The units of all three parts will be Newtons seconds, or

Â kilograms times meter divided by seconds, whichever you're more comfortable with.

Â In this problem since I'm looking for

Â a force, I'm going to keep that portion of the equation.

Â And I also know a lot about the car's momentum.

Â So m v final minus m of the initial,

Â remember a change is always a final minus initial value.

Â So I'm looking for this average force, which I don't know.

Â The time of the collision it didn't give me.

Â That's this 0.01 second value at the top of the board here.

Â The mass of the car, they did give me, 1,500 kilograms.

Â And now, this is where the whole vector nature of momentum

Â comes in as a critical point in our discussion.

Â The final velocity is not just 2.

Â The final velocity is a negative 2 because it's now moving left.

Â You forget this negative sign, it completely changes your answer.

Â It goes left.

Â That's more change than if at the end of the problem it had just been going to

Â the right with 2.

Â Minus the mass again, 1500.

Â And at the beginning of the problem this car was going to the right at

Â a speed of 10 meters per second.

Â That means that now when I solve for the force,

Â I'll actually get a larger force if, than if I'd forgotten a negative sign.

Â Very common mistake.

Â Solving for the average force then, notice that I end up with a negative answer.

Â A negative 1.8 times 10 to the 6th Newtons.

Â A couple of things I want to point out.

Â The force is negative, which makes sense,

Â because if this car were originally moving to the right with a certain velocity.

Â But after the collision has been hit and is now moving left,

Â then of course the force in that collision pushed left on that car.

Â It changed its velocity, and is now making it move left.

Â So we should get a negative answer.

Â That's what that negative sign represents, a direction to the left.

Â The other thing I want to point out is our time frame was very short

Â while our force was very long, and that's very typical for collision problems.

Â They happen over very short amounts of time and the forces can be very large, so

Â don't be bothered by getting a large value for force or a short value for time.

Â >> One important piece of information that I can get from this question is that

Â the two cars are coupled together after the collision.

Â This is important because if they're coupled together or

Â attached together that means that this is an inelastic collision.

Â If it's inelastic, then what I also know is that kinetic energy is not conserved.

Â However, keep in mind no matter what type of collision you're looking at, all three

Â of the collision types have momentum, total momentum as being conserved.

Â Now let's go ahead and solve for this question.

Â It says calculate the speed of the two cars after the collision.

Â Well they're stuck together, so they're going to have the same speed.

Â The way that I'm going to solve for

Â this is going to allow me to use conservation of momentum.

Â The sum of my initial momentum equals the sum of my final momentum.

Â Initially I have m1 v1 initial,

Â plus m2 v2 initial.

Â Say I consider vehicle one to be the one that was in, that was

Â moving at 6 meters per second left, and vehicle two to be the one that's at rest.

Â This means that m2 v2 initial equalled 0.

Â Because these two cars have combined together after the collision,

Â they're now moving as one object, one mass.

Â I can write this as m1 plus m2 multiplied by final velocity.

Â Not a velocity 1 or a velocity 2,

Â just a final velocity because it's both of them combined.

Â From here, I can include my numerical values.

Â Notice that the initial velocity for vehicle one is plugged in as negative 6.

Â That's because the question told me that it was

Â moving with 6 meters per second, going left.

Â You have to be very careful when you read questions to figure out if your velocity's

Â going to be positive or negative.

Â That's because momentum is a vector and so is velocity.

Â [NOISE] When I solve for this final velocity of the two vehicles moving

Â together, I end up with negative 2.4 meters per second or

Â 2.4 meters per second going left.

Â The next part asks me to solve for

Â the percentage of kinetic energy lost during this collision.

Â Well I know that kinetic energy is not conserved.

Â So I can go ahead and solve for

Â the sum of my initial kinetic energies,

Â knowing that they will not equal the sum of my final kinetic energy.

Â You can do this separately, or as I am about to do over here, next to each other.

Â So initially the only energy, the only kinetic energy that I had was for

Â vehicle one.

Â So I have one-half m1v1, initial, squared.

Â [SOUND] In final, I had the two of them moving together, so

Â I still have just that one total combined kinetic energy.

Â So I have one-half m1 plus m2 multiplied by

Â the final velocity that I just solved for, squared.

Â And plug in these values.

Â Notice even though these velocities are negative because it is squared, my answer,

Â since it's energy's going to come out to a positive value.

Â So my total initial kinetic energy is 18,000 joules,

Â which as suspected, does not equal to my total final

Â kinetic energy, which is 7,200 joules.

Â When I take the difference in my kinetic energy,

Â I end up with a loss [SOUND] of 1,000 or

Â 10,800 Joules.

Â And if I want to solve for the percentage of kinetic energy lost,

Â I'm going to take this value, divided by the original,

Â the initial kinetic energy that I had, which was the 18,000.

Â This gives me 0.6.

Â I want the percentage so I multiply it by 100.

Â Then I end up with 60%

Â of kinetic energy lost.

Â >> Now, we need to take care of momentum conservation in two dimensions.

Â In this case, the object will move in the x direction and

Â in the y direction during the problem.

Â So we're going to have to make sure we keep track of both.

Â Let's draw a quick sketch.

Â The first marble, we'll call it m1, is moving to the right.

Â m1 initial speed of 2 meters per second.

Â Here's another marble of the same mass, we'll call it m2, it was at rest.

Â So its initial velocity v2i is 0 meters per second.

Â After the collision, though, now they have different behaviors.

Â M1 is now going Southeast.

Â Not perfectly Southeast which would be a 45 degree angle, but

Â instead 38 degrees South of the Eastward direction.

Â It tells me that the speed of that particle,

Â that marble, is now 0.5 meters per second.

Â We don't know yet what this final speed of marble two is.

Â In fact, that's what it wants us to find out.

Â To do so, I need to break this problem up in the horizontal and the vertical.

Â Let's do horizontal first.

Â So I know this is a collision.

Â Happens over a very short amount of time.

Â This is an excellent candidate for momentum conservation,

Â which is conserved in every collision.

Â Starting off then.

Â Taking a close look at my picture.

Â I noticed that initially on the left hand side here

Â there's only one object with a momentum in the x direction.

Â The second marble's at rest.

Â After the collision though both objects will be moving in the x direction or

Â at least that what it seems to me.

Â If not, we might find out that that second marble has no momentum in the x direction.

Â And perhaps flies directly upward.

Â Let's see what we find.

Â So, always start off with your fundamental principle.

Â Sum of momentum initially in the x direction

Â will equal your sum of momentum's at the end of the problem in the x direction.

Â That's telling your grader or your professor that you know momentum

Â is conserved and that's the approach you're taking in this problem.

Â Setting it up then,

Â I notice that my mass one was moving at the beginning directly horizontally.

Â So it's positive momentum.

Â And I don't have to break it up.

Â It's entire velocity there is in the x direction.

Â Then, after the collision, I notice that, this only has this much horizontal speed.

Â I'm doing it only in the horizontal direction.

Â I need to break up this velocity vector into its horizontal and

Â vertical components.

Â That means that when I do m1 down here, I'm then going to take

Â v1 final and break it up by using the cosine of theta.

Â That gives me the horizontal components, the adjacent side of that triangle.

Â It's positive because it goes to the right.

Â I don't know what direction m2 goes yet.

Â So I'm going to leave this positive and if it's supposed to come on negative and

Â maybe go left afterward, it will let us know and

Â actually come up negative at the end of my problem.

Â So our mass 2, I'm going to then solve for v2 final in the x direction.

Â Since this is all in the x direction, solving for

Â v there will give me the x component of my speed.

Â Giving myself some more room so we can sub in values,

Â I know each marble had a mass of 0.02.

Â The initial one was traveling with a positive 2.

Â I've already set up all of my positive negative signs, so

Â I'm not going to worry about subbing anything in a negative.

Â We've already taken care of anything that was going left.

Â In this case, nothing seems to be going left yet.

Â 0.02 was the mass, that final speed of that first marble, 0.5 meters per second.

Â And the angle to give me the adjacent side, the portion in the x direction,

Â cosine of 38 plus 0.02.

Â And I'm solving for the final x component of that second marble.

Â Solving, I get v2fx to be 1.6 meters per second.

Â That's just the x component, we're not done yet.

Â We need to also do this in the y direction to find the y component of the velocity.

Â And then we're going to have to combine those using vectors.

Â And the distance formula or

Â Pythagorean theorem however you like to think about it, to solve for the total.

Â So setting it up then.

Â Sum momentum initial in the y direction.

Â Always starting off with your fundamental principle.

Â I notice looking up at my diagram again, nothing is moving in the,

Â y, direction at the beginning of this problem.

Â There is no momentum in the, y, direction, that's okay.

Â That just means that on the left hand side of our momentum equation

Â that will equal zero.

Â That means that the sum of momentum's over here on the right hand side,

Â needs to also be zero.

Â Coming down then,

Â no momentum in the y direction to begin with at the beginning of the problem.

Â My m2, I don't know what direction it is, so I'm just going to make it positive.

Â If it needs to come out negative, it will at the end of the problem.

Â So it's got some momentum in the y direction.

Â Plus, but this second marble, or sorry the first marble in this case is moving down.

Â It's moving south of east.

Â So in this case, I need to make this negative.

Â My m1, my v1 final, but I need the vertical component.

Â And looking up at my diagram again, that vertical component is going

Â to be 0.5 of v1 final times the sine of that 38 to get me the vertical component.

Â And so that's what I'm going to put here.

Â Those are my two.

Â I'm going to sub in my numbers.

Â We have 0.

Â The mass again for both, 0.02.

Â We're looking for this final y component of the second marble, 0.02.

Â The final speed of the first marble,

Â 0.5 meters per second, sine of 38.

Â Solving for that, I have v2 final in

Â the y direction was 0.31 meters per second.

Â I now know how fast this particle is moving in the x direction,

Â and how fast the marble is moving in the y direction.

Â It moves to the right.

Â And it moves up.

Â It moves east and north in this scenario.

Â So, let's solve for the total because that's what it asks us to do.

Â My v2f in the x direction, 1.6 meters per second.

Â My v2f in the y direction,

Â 0.31 meters per second.

Â You add your vectors tip to tail like this, so

Â I'm combining the components to get a total final velocity here.

Â I also, if I'm looking for velocity, need the angle.

Â Remember, you need magnitude and direction.

Â So this is being a right triangle, I am going to use my distance formula.

Â V2f would equal the square root of v2f in the x direction squared plus

Â v2f in the y direction squared.

Â That's my a squared plus b squared equals c squared.

Â V2f, subbing in my numbers,

Â 1.6 squared, 0.31 squared.

Â I get a final velocity of this object

Â of 1.63 meters per second.

Â Now that's just the speed.

Â Don't forget that we also need the direction.

Â We can do that by using triangle trig, using the sides of this triangle.

Â That means that since I already know the horizontal and

Â vertical components, why don't we use tangent.

Â Inverse tangent of opposite, in this case the vertical component

Â over the adjacent, the horizontal component, should give me the angle.

Â Solving them, I get 10.97 degrees.

Â And I have to look at my picture at this point, and notice that it went east, and

Â it went north.

Â It went to the right and up.

Â And so my angle then is that many degrees North of East.

Â And I'm trying to reference those directions in the same way they do in

Â the problem.

Â So, now that we've found the final speed of this problem,

Â of this marble and its direction.

Â The problem goes on to ask one last question,

Â what type of collision are we looking at?

Â Remember that there are three types, elastic collisions where energy,

Â kinetic energy is conserved, which means it's the same before and after.

Â You have inelastic collisions where they stick together,

Â that's a perfectly inelastic collision.

Â They stick together, and you know energy is not conserved.

Â But then there are just regular, inelastic collisions.

Â In that case, energy is not conserved, but they also don't stick together.

Â The only way to tell for

Â sure if the objects don't stick together is by doing the math.

Â We're going to find kinetic energy initial of the problem.

Â We're going to find the kinetic energy final of the problem.

Â If those two are equal, then I know this was an elastic collision.

Â Kinetic energy was conserved.

Â If they are not equal and we've lost energy, well,

Â that's what should happen in most realistic problems.

Â That's going to be inelastic.

Â They didn't stick together so it's not perfectly inelastic,

Â but they still lost energy.

Â And you should never see in a realistic problem the kinetic energy increase.

Â Something strange would be going on there.

Â There's some potential energy that was released that you didn't know existed in

Â the problem to begin with.

Â So let's try this out.

Â Remember at the beginning of the problem, there was only one object moving.

Â So I'm going to calculate the kinetic energy for that one object.

Â There was the first marble with its initial speed.

Â So one-half, 0.02.

Â It had a speed at the beginning of the problem of 2 meters per second.

Â So I'm going to make sure to square that.

Â And I calculate that initially, the system had 0.04 joules of kinetic energy.

Â After the collision there are two objects moving now.

Â You have the first object with its final kinetic energy.

Â And you have the second marble with its final kinetic energy there.

Â Which means I'm going to have a one-half m1 v1 final squared,

Â and a one-half m2 v2 final squared.

Â And we know all those numbers now that we've done the problem using momentum and

Â found the speed.

Â So, let's do that.

Â K final equal one-half, 0.02.

Â Now notice that term shows up in both.

Â They have the same math.

Â So I'm going to pull that out.

Â The speed of the first marble after the collision was 0.05.

Â And we calculated the speed of the second marble, 1.63 after the collision.

Â So that's my calculation now.

Â I've pulled out the one-half and the m.

Â Solving, I get a final kinetic energy of

Â 0.029 joules.

Â Notice what that tells me.

Â Notice that here we have more energy than we do at the end.

Â The beginning of the problem we have 0.04 joules.

Â Now we have more something like 0.029, about 0.03.

Â We've lost energy, that's what we expect.

Â But they didn't stick together.

Â That tells me that we have an inelastic collision.

Â Remember, if they don't stick together, it could be elastic, it could be inelastic,

Â you have to compare the kinetic energies to tell for sure.

Â [SOUND]

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