hi welcome back to ap physics 1 review this is brian brown coming to you again from beautiful central new jersey i know miss gv did a great job with energy and work in the last unit and in this unit i'm going to talk about impulse and momentum what are the key takes from momentum and impulse well impulse has to act on an object in order to change its momentum an impulse is related to the force and the time that that force acts on an object and the total momentum of a closed system is a conserved quantity i know that msgv got into that concept of conserved quantities last unit turns out that momentum is is also important for that reason all right so we are going to look at momentum and impulse by defining the key terms looking at the equations discussing the relevant graphs and understanding when we can apply these ideas and when we can't we'll take a look at both multiple choice and free response questions as well so let's get started first we'll start with momentum it's a quantity that measures mass in motion it's given the symbol p and the equation for momentum is a nice and easy one it's p equals mv it's on the equation sheet just in case you forget it you see that it has the arrow over p which means it's a vector and it goes in the same direction as the velocity vector quick example if a cart of five kilograms goes is traveling at 10 meters per second and a larger object is 10 kilograms traveling at 5 meters per second how do those momentums compare well we see that they have the same magnitude but remembering that they are the impulse sorry that momentum is a vector quantity we know that the momentums are opposite each other because one of the momentums is pointing to the right and the other is pointing to the left we've seen that in order to change the velocity a net force has to act on the object whether it's speeding up slowing down or changing directions if i look at newton's second law from our old perspective right the acceleration is the net force divided by the mass and just rearrange a couple things to look at make it look like momentum we can make a similar statement in order to change the momentum a net force must act for a certain time and we know that acceleration is really the change in velocity per time and the right side stays the same net force over mass but rearranging this equation i see that i can get a mass times velocity or change in velocity on one side which is which is the change in momentum and that's equal to the net force times the time newton's second law was really written in terms of momentum when newton first introduced it so this equation is on the ap physics 1 equation sheet again it's really another form of newton's second law but it only holds true when the mass remains constant the only change on the left hand side is that the velocity is changing the mass has to remain constant and m delta v since it's mv final minus mv initial is really the change in momentum of an object the term on the right f net times delta t is the impulse that acts on the object and it's also a vector quantity which has the same direction as the net force another relationship is that the change in momentum and the impulse also have to have the same direction and this leads us to the impulse momentum theorem which in word says the change in momentum of an object has to equal the impulse that acts on it when can we apply this well whenever there's a change in momentum of a single object that results from a single large force ax that acts on that object typically that one force is so much larger than the other forces that it effectively becomes the net force on the object newton's third law can also be related to impulse and momentum instead of saying forces are action reaction pairs we can say that impulses are part of action reaction pairs so the impulse exerted by object a on object b is equal and opposite to the impulse exerted by object b on object a so there's lots of examples that come to mind one is when a tennis ball hits is being hit by a tennis racket right and if you're a fan of any other sport just fill in your your sport of choice here another situation where we look at an action reaction pair is a bumper car bouncing off a wall or bouncing off another car we know that those are action reaction pairs if a force isn't constant then this equation changes slightly to become the average force times the time not the net force this one's not on the equation sheet even though some items may ask about the average force that's part of an impulse another key term key idea in this unit is the conservation of momentum any conservation law in physics is a big idea the conservation momentum again says that if you're in a closed system the total momentum of that system will remain constant right the closed system you get to define the systems so when you define it define it such that the system does not exchange any momentum with its surroundings typical closed systems we consider are collisions between carts or carts that might be at rest and there's a spring between them and the spring is released in general we apply conservation momentum to two categories right collisions between objects or explosions whether they're mechanical or chemical it could be a firecracker that explodes um just like a cart with a spring between it or maybe even there's magnets repelling the two carts when we look at the conservation momentum it's important to look at two different points in time and we're equating the momentum so we can say that the total momentum of system has before an event which the event could be the collision has to equal the total momentum of the system after the collision in a two-object system we can write this all out in one big equation right the mass times the velocity of each object before the collision has to equal the mass times the velocity of each object after the collision again remember each of these are vectors so direction counts and if it's a two-dimensional problem we're gonna need to apply the conservation momentum in both the x and the y direction separately one of the most common graphs that we come across in this unit is a forced time graph this gives us a lot of information again it's really the net force versus time and what's the area represent here well the area represents the impulse if i have a constant force you can see that the area of the rectangle would just be the force times the time over that interval if i don't have a constant force maybe this will form a triangle as part of the graph that still we can apply the same idea that the area is the impulse in these two cases let's say the situation the purple situation that lasts for two seconds and the green situation where the the time lasts for eight seconds the impulse is the same in each case this is a really nice visual these are these situations have the same impulse on the object and we use this idea if the time is large then you can have a smaller force acting an object lots of safety devices have this feature in mind right airbags seat belts all of the and whenever you're you're colliding with something that has some give to it you're increasing the duration of the force so you can decrease the decrease the amount of force to have the same impulse on the object so let's apply these to these ideas to some different situations okay let's go back to our ping pong ball example and drop it onto two different surfaces in one case i drop it on to the hard desk and we see that the ping-pong ball bounces in the second situation the ping-pong ball is dropped onto a soft cloth there's very little bounce to that so what can you claim about the magnitude of the impulse of the ping pong the impulse on the ping-pong ball in each of these trials are they the same impulse well they're dropped from the same height they hit something maybe it's the same impulse or maybe one is greater than the other or perhaps you need more information to figure out the impulse a lot of people get confused on this type of question because there's no information about the force or the time duration of that force when the ping-pong ball is in contact with either the table or the cloth and so they just go right away to option d because it's there and you don't know the force or the time that's a mistake because the impulse is always equal to the change in momentum so when you see the word impulse think change in momentum if you don't know the force in time because it's the definition behind the impulse momentum theorem so now we have to phrase the question a little differently but i think we can answer it which of the ping pong which of the situations does the ping ball have a larger change in momentum even this part's a little tricky because we have to go back to the situation and define our reference frame if i call the upwards direction positive the velocity that the ball has just before it hits the table top is downwards and let's call it negative v the ping-pong ball bounces almost to the same height that it started so after it hits the table it returns and it's at that moment it leaves the table it's going upwards with positive v almost as much as with almost as much speed as the negative v in the second case the ball has the same speed going downwards initially just before it hits the cloth but then stops so its final velocity is zero which of these is the bigger change in momentum again you have to go back to the direction in the first case it's a large change in momentum because it's positive v minus a negative v for the change in velocity in the second case your change in velocity is just v it's zero minus negative v so you see that in the first situation the the change in velocity is double therefore the change in momentum is double since the masses are the same so the answer is trial one the bottom line is when objects bounce off of each other when an object bounces off the surface it has a greater change in momentum than if it just stops when it hits that surface let's take a look at a free response question this one has many different parts to it so we have a smaller car smaller toy car going to the left with a velocity v that's going to collide with a larger toy truck that's going three b that has a mass of 3m okay it's given that after the collision the car bounces so that its final speed is v over two so it's not going as fast as it initially was and it's going to the right so we're going to use this representation to explain conservation of momentum these are velocity mass graphs so the velocity is not changing as a function of mass that's not what this representation is trying to show it's just showing how much momentum is there in the for each of these objects at these different points they're more like bar graphs but they're looking at the area so it's a little different and so i want to walk through this problem a little bit in order to get this we actually have to do some calculations so we can figure out all the different uh options so i'm going to use the conservation momentum to determine the final velocity of the toy truck just after the collision again we kind of have to define our positive direction so i'll just stick with the standard to the right is positive before the collision with the toy truck isn't moving but the toy car has a velocity of negative v since it's moving to the left so the initial momentum of the system is negative mv after the collision we don't have all the information we need we know that the car bounces to the right at v over two so its momentum is m times v over two after the collision it's the tortrux moment uh velocity that we don't know so we're gonna have to find that but we know its mass is 3m so the variable we're solving for is this v2f the velocity of the toy truck after the collision i have my algebraic statement already set up based on the conservation momentum i can sum the initial momentums to get negative mv and i'm summing the final momentums and my only variable is that i'm going to solve for is v2f so it's a symbolic situation that we're we're solving in this case and v2f will equal negative v over 2. the truck makes sense will bounce to the left after this collision the truck is certainly going to go to the left if it was stationary at a speed of v over two whatever the initial speed was over two interesting that the truck goes back at the same speed that the car goes to the right after the collision but that's how this math works out so now i can return to the graph again let's just separate out the the objects the final velocities we have and i just listed them here and i've got to decide on my scale i know that the larger mass is 3m so i'm going to make every two blocks worth m that way six blocks will equal 3m and for the velocity scale i'm just going to make it so every six blocks equals v to give me a nice scale here well the easiest one to deal with is the toy truck before it's mass its 3m mass which is 6 blocks long have no velocity so there's no momentum right there's no height to that graph now let's go over to the tory car's initial momentum its mass is m so i know that's two blocks wide and its velocity is v but its velocity is negative v so i drew that down six units and so again it's it's like a bar chart but i'm looking at the area it's an area chart so its momentum works out to be negative mv in this case let's go to the car's momentum the final momentum after the collision so again its mass is 2 but now its velocity is v over 2 which makes the height only 3 units tall now that i have that information i could actually just go in and graph the truck's final momentum i know that it needs to be the areas before and after need to equal each other so i need some negative area here and the negative area comes about from the 6 units wide of the toy truck and its velocity was negative v over two so now i look at the areas before and the total area after they equal each other another question that could be asked in any collision is to justify whether the collision is elastic and show how you know that this gets back to the definition of elastic collisions elastic collisions have are defined uh to be collisions where the kinetic energy is conserved not just energy we know that the conservation of energy is a big idea that always works but an elastic collision tells us that the kinetic energy before the collision has to equal the total kinetic energy after the collision and momentum is always conserved but this is the definition of an elastic collision so i need to look at it in terms of the kinetic energy so before the collision how much kinetic energy is there well the tor truck's not moving it has no kinetic energy and the kinetic energy of the toy car is just a half mv squared i don't have to worry about the negative i mean first of all when i square a negative it becomes positive and secondly energy is a scalar quantity you can't have negative kinetic energy after the collision it's a little bit more complicated because both objects are moving so i need to figure out the kinetic energy of each object but i know the velocities and masses of each object so if you walk through the algebra you can see that the car has a kinetic energy of 1 8 mb squared and the truck has a kinetic energy of 3 8 mb squared that sums to be a half mv squared which is the same as the kinetic energy before the collision so this collision is elastic just because they bounce off each other isn't enough to justify that the collision is elastic you can still lose kinetic energy in a collision that bounces um a perfectly inelastic collision is a collision where the objects stick together so just a couple more terms that come about with collisions okay so we use the conservation of momentum to figure out the final velocity of the object but we could also take a look at this item and look at the impulses on each object so the first part of the question asks for the impulse on the car again in order to find the impulse of the car we only are concerned with the the force and time the impulse on the car not on the truck in this case so i can look at we already have these graphs for the car's initial and final momentum so i know that the impulse has to be the chain the same as the change in the momentum of the car since i had six blocks worth of momentum and a negative 12 blocks worth of momentum initially that change is is 18 blocks in this case of momentum and the impulse therefore has to be a positive 18 which it relates to three halves mv because remember the the scales that we used kind of dictate how large that's going to be but visually it makes sense if i was asked about the impulse on the truck then i would look at the i could do that two ways i could look at the truck's change in momentum or i can look at it in terms of the action reaction pair with the car and so i know that the truck has an impulse on the car of three halves mv so the car has an impulse on the truck of negative three halves mv which is a nice way of solving the problem and still leads to the solution that the velocity of the truck is negative a half feet let's take another look at conservation of momentum let's say carlos is holding a bowling ball on a skateboard and is going to toss the bowling ball to angela who's going to catch the bowling ball i don't recommend you try this one at home but we could work this one out nonetheless the question says which move which student moves faster after the situation is done so after carlos throws it and after angela catches it and explain it qualitatively right we'd not we're not supposed to justify it in this case with a mathematical solution are they going the same speeds in opposite directions i think there's more going on here than just that right this if we define the system to be angela carlos and the bowling ball we can use the conservation momentum to justify this initially before the ball is thrown there is no momentum in the system if the total momentum is zero there's nothing moving that means that after the events there still has to be zero total momentum in my system carlos is moving backwards with a certain velocity angela and the bowling ball are moving forwards with a certain velocity since carlos and angela have the same mass but angela is holding the bowling ball there's more mass going to the right than to the left which means the velocity of the mass to the right is less than the velocity of the mass of the left so carlos is going faster it's a nice application of the conservation momentum with a with you know a careful consideration of the system okay there's more analysis that we can do with this one we can take a look at the an algebraic way of figuring out the final velocity so again whenever we want to solve a vector problem algebraically we need to define our positive direction so let's say the right is positive but we're going to do this one not as one big system because there's really two events carlos is throwing the ball so before carlos throws the ball there is no total momentum in the system after carlos throws the ball carlos goes backwards with a certain velocity and the ball goes forwards with a velocity v so we need to and it's given that the the ball moves forwards with a velocity v so how fast is carlos moving well we can use conservation momentum zero our initial total momentum has to equal the sum of the momentums afterwards so we see that the velocity of carlos afterwards is negative little m the mass of the bowling ball times its velocity divided by carlos's mass so that's our expression in terms of the variables we were asked to again it's important to in this case to keep that as negative to show that the direction of the velocity is to the left for the second part of the problem we want to figure out the angela's and the bowling ball's velocity after she catches the ball so this system i'm going to switch it up a little bit and i'm just going to say the system is angela and the bowling ball so initially there is momentum in this system the ball is moving with a velocity v angela's not afterwards you should recognize that this is a perfectly inelastic collision because angela catches the ball right they stick together so the total momentum before they stick together is mv the total momentum after is the combined mass of angela and the bowling ball times the velocity of angel in the bowling ball afterwards so now when i solve for the final velocity of angela and the ball it's the bowling ball's mass times velocity over the combined mass now does that how does that compare to the answer that we gave initially in terms of their final velocities well that actually looks good because if i divide the same quantity mv divided by a larger mass in the second case i know that i'm going to have a smaller velocity so the math supports the answer that we gave in part a the second part of this problem talks about the mechanical energy so um we're looking at the mechanical energy before the ball is thrown while the ball is traveling in the air between and at the end after angela catches the ball we want to take a look at the total mechanical energy at those three times um there is no potential energy change so i don't have to consider any elastic or gravitational potential energies so i'm just looking at kinetic energy in terms of defining the mechanical energy we know there's the least amount of kinetic energy in the beginning because there's no velocity to any of the objects so there's no kinetic energy in the collision we already decided that the collision between the bowling ball and angela is perfectly inelastic that means that kinetic energy is not conserved in that part so there's more mechanical energy before angela catches the bowling ball than after so the ranking would be e2 is greater than e3 is greater than e1 okay so what should we take away from the impulse momentum unit first you notice in in most the problems we sketched the motion diagram of the velocity before and after the collision or explosion this is going to help visualize momentum at the critical times remember you're only looking just before the event or just after going back to the ping-pong ball it didn't matter that they started at rest it was the velocity that the ping-pong ball had just before it struck the table and just after impulse on and change in momentum of are interchangeable terms it's a way of kind of hiding something in a physics question to talk about one rather than the other when it makes more sense to change those terms please make that change the impulse momentum theorem applies to a single object or system with an external force causing a change in momentum that's when you can use the impulse momentum theorem if you're really focusing on a single object or system with the external force the conservation momentum applies to a closed system right when you want to capture all the objects involved in the event the last part the last note is that remember that impulse momentum are both vectors so pay attention to directions and include signs positive and negative when you need to we're going to do great miss g miss gv will be with you for the rest of the way and i hope you feel confident that the ideas of ap physics 1 are are clear to you and i'm sure you're going to do fantastic on the exam good luck you