### Video Transcript

In this video, weโre going to learn about the conservation of linear momentum. Weโll learn why it is that linear momentum is conserved. And how we can use this fact to help us solve problems.

To start out, imagine that you are the manager of a fireworks production company. Before every show, once the fireworks launchers are put in place, one of your roles is to mark off a perimeter around these launchers. This perimeter forms the boundary of the closest people are allowed to get to the firework show itself. To figure out where that perimeter should be, you want to know the furthest possible point from the launchers that a piece of the exploded fireworks might travel. To get an idea for that maximum possible distance, it will be helpful to know about the conservation of linear momentum.

We know that the linear momentum of an object is equal to its mass times its velocity. Written as an equation, lower case ๐, the representation of momentum, is equal to ๐ times ๐ฃ. Knowing that, letโs imagine a scenario where we have two masses. One weโll call mass ๐ด, and the other weโll call mass ๐ต. These masses move towards one another and eventually collide. When this happens, Newtonโs third Law of motion tells us something about this collision.

This law, which says that for every action thereโs an equal and opposite reaction, tells us that the force mass ๐ด exerts on ๐ต in the collision is equal and opposite the force mass ๐ต exerts on ๐ด. We know that, practically, this collision will not occur instantaneously, but rather over some span of time we can call ฮ๐ก. If we multiply both sides of this equation by that time interval, then we see that each side now represents an impulse, on the left-hand side the impulse of ๐ด on ๐ต, and on the right-hand side the impulse of ๐ต on ๐ด.

If we recall the impulseโmomentum theorem, which says simply that the impulse an object experiences is equal to its change in momentum, then we can rewrite our expression for impulse to involve the masses and change in speeds of each of our objects. In particular, the mass of object ๐ด multiplied by its change in speed is equal to negative the mass of object ๐ต multiplied by its change in speed.

Say that as a next step, we add ๐ sub ๐ต times ฮ๐ฃ sub ๐ต to both sides of this equation, cancelling that term out on the right-hand side. We now have an equation that tells us that the change in momentum of object ๐ต over this collision plus the change in momentum of object ๐ด is equal to zero. That is, overall, there is no momentum change over this interaction.

Another way to express this is to say that momentum is conserved. That is the momentum at the beginning of an interaction is the same as the overall momentum after the interaction. This is what the conservation of linear momentum means. It means if we count up all the momentum in a scenario before an interaction and then count up all the momentum in the scenario afterwards, those two amounts will be the same.

Something important to recall about the conservation of linear momentum is that since momentum is a vector, itโs conserved in every independent direction. Itโs conserved in the ๐ฅ-direction as well as in the ๐ฆ as well as the ๐ง and so on. Letโs consider for a moment how to practically use this principle of the conservation of linear momentum.

Imagine that we had a firework which when launched up in the air explodes into two different pieces. Say we want to use the fact that linear momentum is conserved to solve for the final speed of one of those two exploded pieces. If we set up a coordinate system, an origin, we can begin by calculating the initial as well as the final momentum involved. Since momentum is a vector, we can apply this conservation principle in both the ๐ฅ- and the ๐ฆ-directions.

So, weโll calculate the initial momentum in the ๐ฅ-direction as well as in the ๐ฆ and the final momentum in those two directions separately too. If the firework is initially at rest, then that means itโs initial momentum in both the ๐ฅ- and ๐ฆ-directions is zero. And if we call the two pieces of the exploded firework ๐ one and ๐ two, respectively, then we can write out the final momentum in the ๐ฅ- and ๐ฆ-directions, ๐ one ๐ฃ sub one ๐ฅ plus ๐ two ๐ฃ sub two ๐ฅ. And for ๐ฆ, ๐ one ๐ฃ sub one ๐ฆ plus ๐ two ๐ฃ sub two ๐ฆ.

Once all these terms are written out, we use the fact that linear momentum is conserved to set the initial and final momenta in each direction equal to one another. Then, if weโre given enough initial information, weโre able to solve the information we want. Letโs get some practice with the conservation of linear momentum through a couple of examples.

A man of mass 62 kilograms stands at rest on an icy surface that has negligible friction. He throws a ball of mass 670 grams, giving the ball a horizontal velocity of 8.2 meters per second. What horizontal velocity does the man have after throwing the ball?

We can label this horizontal velocity of the man ๐ฃ sub ๐ and start off by drawing a sketch of the situation. In this situation, initially, we have a man holding a ball and standing at rest on a frictionless surface. The man then throws the ball, moving in what weโll call the positive direction with the speed ๐ฃ sub ๐ of 8.2 meters per second. Because momentum is conserved, we know that as a result of throwing the ball, the man himself will move to the left with some velocity weโve called ๐ฃ sub ๐. Letโs use this conservation principle, that initial momentum is equal to final momentum, to solve for ๐ฃ sub ๐.

As we consider the initial condition of our system, since the man and the ball are both at rest, that means their speed is zero and, therefore, the overall initial momentum of our system is also zero. When we consider the final momentum of our system, after the ball has been thrown, thatโs equal to the manโs mass times his velocity plus the ballโs mass times its velocity. Because linear momentum is conserved, we can equate our initial and final momentum and write that zero is equal to ๐ sub ๐๐ฃ sub ๐ plus ๐ sub ๐๐ฃ sub ๐.

We want to solve for the manโs velocity ๐ฃ sub ๐. And thatโs equal to negative ๐ sub ๐๐ฃ sub ๐ over ๐ sub ๐. We know the mass of the man and the ball as well as the velocity of the ball. So, weโre ready to plug in and solve for ๐ฃ sub ๐. When we do, weโre careful to convert the mass of the ball into units of kilograms to agree with the units of the manโs mass. When we calculate this value, we find itโs equal to negative 0.089 meters per second. Thatโs the velocity of the man sliding across the ice after he has thrown the ball. Now letโs consider an example involving two masses which initially are in motion.

Two cars of equal masses approach a frictionless perpendicular intersection. Car A travels north at 22 meters per second and car B travels east. The cars collide and stick together, traveling at 17 degrees north of east. What was the initial speed of car B?

We can label car Bโs initial speed ๐ฃ sub B and begin our solution by drawing a sketch of the situation. In this scenario, initially, we have two cars approaching one another at an intersection. Car A is traveling north at a speed given as 22 meters per second, and car B is traveling to the east at an unknown speed. At the center of the intersection, the cars collide and, finally, they move together as one mass in a direction given as 17 degrees north of east. Based on this information, we want to solve for the initial speed of car B.

Weโll use the principle that the total initial momentum in a system is equal to the total final momentum in that system. In our scenario, initially and finally, we have momenta pointing in two different directions. In the northern direction, we have ๐ sub A times ๐ฃ sub A. And in the eastern direction, orthogonal to the northโsouth axis, we have ๐ sub B times ๐ฃ sub B.

If we then consider the final momentum in our system, again both in the north and eastern directions separately, to the north we have our combined mass ๐ sub A plus ๐ sub B, since the cars have stuck together, times their final speed, whatever that is, weโll call it ๐ฃ sub ๐, multiplied by the sine of the angle ๐ณ. And to the east, we have the same expression except now we use the cosine of that angle to point in the direction east of north.

Applying our conservation principle, we can write in the northern direction that ๐ sub A๐ฃ sub A equals ๐ sub A plus ๐ sub B times ๐ฃ sub ๐ times the sine of ๐ณ. And separately, in the eastern direction, ๐ sub B๐ฃ sub B is equal to the sum of the two masses times their final velocity times the cosine of ๐ณ. We recall that we want to solve for ๐ฃ sub B, the initial speed of car B. And very importantly, weโre told in the problem statement that the masses of the two cars are equal. ๐ sub A is equal to ๐ sub B. So, if we replace every ๐ sub A and ๐ sub B we see in our two equations simply with the mass ๐, we see then that that mass ๐ cancels out from both sides of both equations.

Looking at the equation in the northern direction, we can use the fact that ๐ฃ sub A and ๐ณ are given information to solve for ๐ฃ sub ๐, the final speed of the combined car mass. ๐ฃ sub ๐ is equal to ๐ฃ sub A over two times the sine of ๐ณ, or 22 meters per second divided by two times the sine of 17 degrees. If we substitute this expression for ๐ฃ sub ๐ into ๐ฃ sub ๐ in our equation for ๐ฃ sub B and enter our value of 17 degrees for ๐ณ. Then we see our expression simplifies a bit and that factors of two cancel and the cosine of one angle divided by the sine of the same angle is equal to the cotangent of that angle.

So, entering these values on our calculator, we find a result for ๐ฃ sub B which, to two significant figures, is 72 meters per second. Thatโs the initial speed of car B in this scenario.

Letโs summarize what weโve learned about the conservation of linear momentum. Weโve seen that Newtonโs third law of motion implies that a systemโs overall change in momentum is zero. That is the initial momentum in the system is equal to the final momentum in that system. This is what it means that linear momentum is conserved.

Weโve also seen that since momentum is a vector quantity, it is conserved in each orthogonal direction, or each perpendicular direction, of a scenario space, for example the ๐ฅ-direction, the ๐ฆ- direction, the ๐ง-direction, separately. And finally, weโve seen that a problem-solving strategy for linear momentum involves considering the initial and final states of a scenario as well as each component direction.