# Video: Conservation of Linear Momentum

In this video we learn that linear momentum is conserved as a result of Newton’s Third Law of Motion, and we also learn a problem-solving framework for linear momentum scenarios.

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### Video Transcript

In this video, we’re going to learn about the conservation of linear momentum. We’ll 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 𝑝, the representation of momentum, is equal to 𝑚 times 𝑣. Knowing that, let’s imagine a scenario where we have two masses. One we’ll call mass 𝐴, and the other we’ll call mass 𝐵. These masses move towards one another and eventually collide. When this happens, Newton’s third Law of motion tells us something about this collision.

This law, which says that for every action there’s an equal and opposite reaction, tells us that the force mass 𝐴 exerts on 𝐵 in the collision is equal and opposite the force mass 𝐵 exerts on 𝐴. We know that, practically, this collision will not occur instantaneously, but rather over some span of time we can call Δ𝑡. 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 𝐴 on 𝐵, and on the right-hand side the impulse of 𝐵 on 𝐴.

If we recall the impulse–momentum 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 𝐴 multiplied by its change in speed is equal to negative the mass of object 𝐵 multiplied by its change in speed.

Say that as a next step, we add 𝑚 sub 𝐵 times Δ𝑣 sub 𝐵 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 𝐵 over this collision plus the change in momentum of object 𝐴 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, it’s conserved in every independent direction. It’s conserved in the 𝑥-direction as well as in the 𝑦 as well as the 𝑧 and so on. Let’s 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 𝑥- and the 𝑦-directions.

So, we’ll calculate the initial momentum in the 𝑥-direction as well as in the 𝑦 and the final momentum in those two directions separately too. If the firework is initially at rest, then that means it’s initial momentum in both the 𝑥- and 𝑦-directions is zero. And if we call the two pieces of the exploded firework 𝑚 one and 𝑚 two, respectively, then we can write out the final momentum in the 𝑥- and 𝑦-directions, 𝑚 one 𝑣 sub one 𝑥 plus 𝑚 two 𝑣 sub two 𝑥. And for 𝑦, 𝑚 one 𝑣 sub one 𝑦 plus 𝑚 two 𝑣 sub two 𝑦.

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 we’re given enough initial information, we’re able to solve the information we want. Let’s 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 𝑣 sub 𝑚 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 we’ll call the positive direction with the speed 𝑣 sub 𝑏 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 we’ve called 𝑣 sub 𝑚. Let’s use this conservation principle, that initial momentum is equal to final momentum, to solve for 𝑣 sub 𝑚.

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, that’s equal to the man’s mass times his velocity plus the ball’s mass times its velocity. Because linear momentum is conserved, we can equate our initial and final momentum and write that zero is equal to 𝑚 sub 𝑚𝑣 sub 𝑚 plus 𝑚 sub 𝑏𝑣 sub 𝑏.

We want to solve for the man’s velocity 𝑣 sub 𝑚. And that’s equal to negative 𝑚 sub 𝑏𝑣 sub 𝑏 over 𝑚 sub 𝑚. We know the mass of the man and the ball as well as the velocity of the ball. So, we’re ready to plug in and solve for 𝑣 sub 𝑚. When we do, we’re careful to convert the mass of the ball into units of kilograms to agree with the units of the man’s mass. When we calculate this value, we find it’s equal to negative 0.089 meters per second. That’s the velocity of the man sliding across the ice after he has thrown the ball. Now let’s 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 B’s initial speed 𝑣 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.

We’ll 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 𝑚 sub A times 𝑣 sub A. And in the eastern direction, orthogonal to the north–south axis, we have 𝑚 sub B times 𝑣 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 𝑚 sub A plus 𝑚 sub B, since the cars have stuck together, times their final speed, whatever that is, we’ll call it 𝑣 sub 𝑓, multiplied by the sine of the angle 𝛳. 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 𝑚 sub A𝑣 sub A equals 𝑚 sub A plus 𝑚 sub B times 𝑣 sub 𝑓 times the sine of 𝛳. And separately, in the eastern direction, 𝑚 sub B𝑣 sub B is equal to the sum of the two masses times their final velocity times the cosine of 𝛳. We recall that we want to solve for 𝑣 sub B, the initial speed of car B. And very importantly, we’re told in the problem statement that the masses of the two cars are equal. 𝑚 sub A is equal to 𝑚 sub B. So, if we replace every 𝑚 sub A and 𝑚 sub B we see in our two equations simply with the mass 𝑚, we see then that that mass 𝑚 cancels out from both sides of both equations.

Looking at the equation in the northern direction, we can use the fact that 𝑣 sub A and 𝛳 are given information to solve for 𝑣 sub 𝑓, the final speed of the combined car mass. 𝑣 sub 𝑓 is equal to 𝑣 sub A over two times the sine of 𝛳, or 22 meters per second divided by two times the sine of 17 degrees. If we substitute this expression for 𝑣 sub 𝑓 into 𝑣 sub 𝑓 in our equation for 𝑣 sub B and enter our value of 17 degrees for 𝛳. 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 𝑣 sub B which, to two significant figures, is 72 meters per second. That’s the initial speed of car B in this scenario.

Let’s summarize what we’ve learned about the conservation of linear momentum. We’ve seen that Newton’s third law of motion implies that a system’s 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.

We’ve 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 𝑥-direction, the 𝑦- direction, the 𝑧-direction, separately. And finally, we’ve 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.