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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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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University of Houston System

35 ratings

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

Â by time will 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 10 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 2 meters per second.

Â If the cars went 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 the velocity of the car,

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

Â So, this first part's a pretty short problem for us.

Â The mass of the car, 1500 kilograms, 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 really,

Â a larger car moving at the same speed as a smaller car.

Â The 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, so

Â kilograms times meters per second.

Â That will 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, and plug in the base units and

Â 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

Â maybe some 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 meter 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, and

Â 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, 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 a particular problem.

Â The units of all three parts will be Newton seconds or

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

Â In this problem, since I'm looking for a force, I'm going to keep that portion of

Â the equation, and I also know a lot about the car's momentum.

Â So m v final- m v 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 did 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 comes in as a critical

Â point in our discussion.

Â The final velocity is not just 2.

Â The final velocity is a -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

Â to the right with 2.

Â Minus the mass again, 1500, and at the beginning of the problem

Â this car was going to the right at a speed of 10 meters per second.

Â That means, that now when I solve for the force, I'll actually get a larger force

Â than if I had forgotten that negative sign, very common mistake.

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

Â a -1.8 x 10 to the 6th Newtons.

Â A couple of things I want to point out, the force is negative, which makes sense.

Â Because if this car were originally moving to the right with a certain velocity but

Â 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.

Â So don't be bothered by getting a large value for force, or

Â 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 attached together,

Â 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 m1v1

Â initial + m2v2 initial.

Â Say I consider vehicle 1 to be the one that was moving at 6 meters per second

Â left, and vehicle 2 to be the one that's at rest.

Â This means that m2v2 initial equaled 0.

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

Â they are now moving as one object, one mass.

Â I can write this as m1 + m2, multiplied by a final velocity, not a velocity 1,

Â or a velocity 2, nust 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 1 is plugged in as -6.

Â That's because the question told me

Â that it was 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

Â is going to be positive or negative.

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

Â When I solve for this final velocity of the two vehicles moving together,

Â I end up with -2.4 meters per second, or 2.4 meters for 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 this sum of my initial kinetic energies.

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

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

Â So initially, the only kinetic energy I had was for beagle one.

Â So I have 1/2, M1, V1 initial squared.

Â In final, I have the two of them moving together.

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

Â So I have 1/2 m1 + m2 multiplied

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

Â And plug in these values.

Â Notice, even though these velocities are negative,

Â because it is squared, my answer, since it's energy,

Â is going to come out to a positive value.

Â So my total initial kinetic energy is 18,000

Â joules which as suspected does not equal to my total

Â final kinetic energy which is 7,200 joules.

Â When I take the difference in my kinetic energy I end up with a loss.

Â Of 1,000 or 10,800 joules.

Â And if I want to solve for the percentage of the kinetic energy loss.

Â I'm going to take this value, divide it by the original, the initial kinetic energy

Â that I had which was the 18,000 this gives me .6.

Â I want the percentage so I multiply by 100.

Â And I end up with 60% of kinetic energy.

Â 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 have to make sure we keep track of both.

Â Let's draw a quick sketch.

Â The first marble, we'll call in M1, is moving to the right.

Â M1 initial speed of two meters per second.

Â When it hits another marble of the same mass, we'll call is M2, that was at rest.

Â So it's initial velocity, V2i is zero 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.

Â But instead, 38 degrees south of the eastward direction.

Â It tells me that the speed of that particle, that marble,

Â is now 0.5 meters per second.

Â We don't know yet what this final speed of marble 2 is.

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

Â To do so, I need to break this problem up.

Â Let's do horizontal first.

Â Since I know this is a collision, happens over a very short amount of time.

Â This is an excellent candidate for

Â moment of conservation which is conserved in every collision.

Â Starting off then 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 x stretch.

Â The second marble is at risk.

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 the second marble has that momentum in x stretch and

Â perhaps flies directly upward.

Â Let's see what we find.

Â So always start off with your fundamental principle,

Â the sum of momentum initially in the x direction

Â will equal your sum of momentums at the end of the problem in the x direction.

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

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

Â Setting it up then,

Â I noticed that my mass 1 was moving at the beginning directly horizontally.

Â So it's positive momentum.

Â And I don't have to break it up.

Â Its 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 v1

Â final and break it up by using cos of theta.

Â Now that gives me the horizontal component to 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, so I'm going to leave this positive.

Â And if it's supposed to come out negative and maybe go left afterward,

Â it will let us know and actually come out negative at the end of our problem.

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

Â Since this is all the instruction solving the v there will give me the x component

Â of my speed.

Â Giving myself some more room so

Â we can sub in values I know each marble had a mass of 0.02.

Â The initial one was traveling with a of 2.

Â I've already set up all my positive negative sides.

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

Â We're 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.

Â That final speed of that first marble, 0.5 meters per second, and

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

Â cosine of 38, plus 0.02, and I'm solving for the final x component.

Â Of that second marble.

Â Solving, I get v2fx to be

Â 1.6 meters per second.

Â That's just the x component.

Â We're not done yet.

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

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

Â In the distance formula, or Pythagorean theorem,

Â however you'd like to think about it to solve for the total.

Â So, setting it up then, some of them went to initial in the y direction.

Â Always starting off with your fundamental principle.

Â I notice, looking up at my diagram, again.

Â Nothing is moving in the Y direction at the beginning of this problem.

Â There is no momentum in the Y direction.

Â 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 the momentum's over here on the right hand side,

Â needs to also be zero.

Â Coming down then, no momentum in the Y direction to begin with at the beginning

Â of the problem my M2, I don't know what direction it is, so

Â 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 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 components.

Â And looking up at my diagram again red vertical component is going to be 25 the B

Â one final times the sign of that 38 to give you 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 zero, the mass again for both, 0.02.

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

Â 0.02.

Â The final speed of the first marble, 0.5 meters per second.

Â Sine of 38.

Â Solving for that, in the right

Â direction with .31 meter per second.

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

Â how fast the marble was moving in the y direction.

Â Move to the right and 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 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'm going to use my distance formula.

Â V2f = square root V2f in the x direction squared + V2f in the y direction squared.

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

Â V2f Subbing in my numbers,

Â 1.6 squared, .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 this by using trig, using the sides of this triangle.

Â That means that since I already know the horizontal and vertical components,

Â why don't we use tangent.

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

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

Â Solving, then, I get 10.97 degrees and I have to look

Â at my picture at this moment and notice that it went east and it went north.

Â It went to the right and up.

Â So my angle then is that many degrees north of east.

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

Â So, now that we found the final speed of this marble and it's 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.

Â Elastic collisions, where kinetic energy is conserved,

Â 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's not conserved.

Â But then there are just regular inelastic collisions, and

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 initial of the problem and

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

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

Â kinetic energy was conserved, if they are not equal, and

Â we've lost energy well that's what should happen in most realistic problems.

Â That's going to be inelastic, they didn't stick together, so

Â it's not perfectly inelastic, 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

Â in 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 1/2(.02), it had a speed of being two meters per second,

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

Â And I calculate that initially the system had .04 Joules of kinetic energy.

Â After the collision there are two objects moving though.

Â You have the first object with it's final kinetic energy, and

Â you have the second with it's final kinetic energy.

Â 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 of those numbers now that we've done problems using momentum and

Â found the speed.

Â So, let's do that.

Â K final will equal one half, .02, now notice that term shows up in both,

Â they have the same mass, 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, I've pulled out the 1/2 and the m.

Â Solving, I get a final kinetic energy 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 had .04 joules,

Â now we have something more like .029, about 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 inelastic collision.

Â Remember, they don't stick together.

Â It could be elastic.

Â It could be inelastic.

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

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