Lesson Video: Collisions and Conservation of Momentum | Nagwa Lesson Video: Collisions and Conservation of Momentum | Nagwa

Lesson Video: Collisions and Conservation of Momentum Mathematics

In this video, we will learn how to apply the law of conservation of momentum to study collisions in one dimension and differentiate between elastic and inelastic collisions.

18:16

Video Transcript

In this video, we’ll learn how to apply the law of conservation of momentum to study collisions in one dimension and differentiate between elastic and inelastic collisions.

We’ll begin by recalling the relationship between the impulse produced by some force and its change of momentum. Specifically, let’s think about a body of constant mass. The impulse vector 𝐉 produced by the action of some force vector 𝐓 over an interval of time 𝑡 sub one to 𝑡 sub two is equal to the change in momentum of the body. Let’s demonstrate in particular what that looks like.

Let’s consider two particles one and two with momenta vector 𝐏 sub one 𝑖 and vector 𝐏 sub two 𝑖, respectively. During the moment at which they collide, particle one exerts a vector force 𝐅 sub one two on particle two, whilst particle two exerts a vector force of 𝐅 sub two one on particle one. These two forces result purely from the interaction between the two particles. Specifically, Newton’s third law of motion tells us that these forces are equal in magnitude but opposite in direction. So we can say that the vector 𝐅 sub one two is equal to the negative vector 𝐅 sub two one.

Due to the force that is exerted by particle two on particle one during the collision, particle one experiences a change in momentum. This change in momentum, the change in the vector 𝐏 sub one, is equal to the difference between the vector 𝐏 sub one 𝑓, that’s the final momentum, and the vector 𝐏 sub one 𝑖, that’s the initial momentum. And that of course is equal to the impulse vector 𝐉 sub one, which can in turn be considered as the definite integral between 𝑡 sub one and 𝑡 sub two. That’s the time of the collision of the vector 𝐅 sub two one with respect to 𝑡.

In a similar way, we can calculate the change in momentum of particle two. It’s equal to the difference between the vector 𝐏 sub two 𝑓, that’s the final momentum of this particle, and the vector 𝐏 sub two 𝑖, the initial momentum of this particle. But remember, we said that the vector 𝐅 sub one two is equal to the negative of the vector 𝐅 sub two one. So in turn, we can say that the impulse 𝐉 sub one is equal to the negative impulse 𝐉 sub two. This then allows us to identify that the change in momentum of the first particle is equal to the negative change in momentum of the second. We can then rearrange this formula to show a really important fact. By adding the change in momentum of the second particle to both sides, we find that the change in momentum of particle one plus the change in momentum of particle two is equal to zero.

So what does this final statement actually tell us? It says that there is no change in the total momentum of the two particles during the collision. In other words, the total momentum is constant. And this allows us to say that momentum is a conserved quantity.

Let’s define this a little more formally. The law of the conservation of momentum tells us that if we take two or more bodies in an isolated system, in other words, a system which means that there are no external forces acting on it and these bodies are acting upon each other, their total momentum remains constant. In particular, the vector sum of the momentum is equal to zero. So we know that for any collision between two particles, the total momentum is conserved if we assume there are no other interactions than the one between the colliding particles. And note that whilst we’ve provided these definitions using vector quantities, in practice, we tend to work with scalar values. And the conservation of momentum law allows us to do so with minimal concern.

So with this in mind, let’s look at solving a problem which involves finding the impulse during a collision between two objects that are moving towards one another.

Two spheres, 𝐴 and 𝐵, of equal mass were projected toward each other along a horizontal straight line at 19 centimeters per second and 29 centimeters per second, respectively. As a result of the impact, sphere 𝐵 rebounded at 10 centimeters per second. Find the velocity of sphere 𝐴 after the collision given that its initial direction is the positive direction.

We first see that we’re told that the initial direction of sphere 𝐴 is positive. So we might begin by drawing a sketch of the scenario. We have sphere 𝐴 and 𝐵 moving towards one another at 19 centimeters per second and 29 centimeters per second, respectively. Sphere 𝐴 is moving in the positive direction, which means sphere 𝐵 is moving in the negative direction. We’re also told that each sphere has an equal mass. So let’s define the mass of each of our spheres to be 𝑚 grams.

Now, in practice, we tend to work with kilograms and meters per second. But here we’re working with centimeters per second. And so it’s more usual to be consistent and use grams. We’re going to begin by calculating the initial momentum of this system. We know that momentum is conserved. So this will allow us to compare the initial momentum with the final momentum, which should then in turn allow us to calculate the final velocity of sphere 𝐴. We also know that we often calculate momentum by using vector quantities for momentum and velocity. Since scalar momentum and scalar velocity can be thought of as vectors in one direction, this means we can generalize this for scalar quantities also.

So, we know that we can define the initial momentum of sphere 𝐴 as 𝑝 sub 𝐴, 𝑖. And it’s its mass times velocity. That’s 𝑚 times 19 or 19𝑚. In a similar way, we can calculate the initial momentum for sphere 𝐵. 𝑝 sub 𝐵, 𝑖, that’s the initial momentum, is mass times velocity. So that’s 𝑚 times negative 29 or negative 29𝑚. Then, we can calculate the total momentum in the system — let’s call that 𝑝 sub net 𝑖, the initial net momentum — by finding the sum of the initial momentum of 𝐴 and 𝐵. That’s 19𝑚 plus negative 29𝑚, which is negative 10𝑚.

Now that we’ve identified what’s happening before the collision, let’s think about what’s happening immediately after. We’re told that the spheres rebound. In other words, they move immediately after the collision in opposite directions. We’re told sphere 𝐵 rebounds at 10 centimeters per second, so it travels in the opposite direction at this velocity. We’re trying to find the velocity of sphere 𝐴. So let’s define that to be 𝑣 centimeters per second. And we’re assuming it’s moving in the negative direction.

Then, we calculate the net momentum by considering the final momentum of each sphere. The final momentum of sphere 𝐴 is going to be negative 𝑚𝑣, whilst the final momentum of sphere 𝐵, let’s call that 𝑝 sub 𝐵, 𝑓, is mass times velocity, that’s 10𝑚. Then, we find their sum to find the final net momentum. That’s negative 𝑚𝑣 plus 10𝑚.

Now, according to the principle of conservation of momentum, the net momentum before the collision must be equal to the net momentum after the collision. In other words, negative 10𝑚 must be equal to negative 𝑚𝑣 plus 10𝑚. Let’s clear a little bit of space and solve this equation. To do so, we might first begin by noticing that each expression in this equation contains a factor of 𝑚. We also defined 𝑚 to be the mass of each object, so it cannot be equal to zero, meaning that we can divide our entire equation by 𝑚. When we do, we get negative 10 equals negative 𝑣 plus 10. Subtracting 10 from both sides, and our equation becomes negative 20 equals negative 𝑣.

And so the velocity of the sphere after the collision is 20 centimeters per second. But of course, we define this motion to be to the left. Since velocity is directional and we define the direction to the right to be positive, we have to say that the final velocity of sphere 𝐴 is negative 20 centimeters per second.

Now, at this stage, it’s worth noting that we could have alternatively modeled the motion after the collision slightly differently. In other words, we could have defined the arrow 𝑣 centimeters per second to be moving to the right. If we’ve done that at this stage, we would have ended up with 𝑣 equals negative 20 centimeters per second, thereby still indicating that the object is moving to the left. Either way, the velocity is negative 20 centimeters per second.

We’ll now demonstrate how this process can determine the impulse of the force of the collision of bodies moving in opposite directions as well as the velocity of a body after the collision.

Two spheres of masses 200 grams and 350 grams were moving toward each other along the same horizontal straight line. The first was moving at 14 meters per second and the second at three meters per second. The two spheres collided. As a result, the first sphere rebounded at seven meters per second in the opposite direction. Given that the positive direction is the direction of motion of the first sphere before the impact, determine the impulse 𝐼 the second sphere exerted on the first one and the speed 𝑣 of the second sphere after impact.

Now before we go any further, we might observe that with our speeds, we’re working in meters per second, whilst with the masses, we’re working in grams. So we’re going to convert their masses to SI base units of mass, the kilogram. To do so, we divide each by 1000. We get 0.2 kilograms and 0.35 kilograms, respectively.

So how do we calculate the impulse 𝐼 that the second sphere exerted on the first? Well, first, we know that the magnitudes of the impulses on the spheres are equal. We also have information about the velocity of the first sphere before and after the collision. This means we can calculate the change in velocity of the first sphere due to the collision. If we call that the change in 𝑣 sub 𝑎, where 𝑎 is the name of the first sphere, then we can say that this is equal to its final velocity minus its initial velocity.

Now, since it was initially moving in the positive direction, it will rebound and move in the negative direction. This means if it’s for traveling at a speed of seven meters per second, its velocity is negative seven. So the change in velocity is negative seven minus 14. And that’s negative 21 meters per second. Then, we can calculate the impulse on the first sphere by finding its change in momentum. Momentum is of course mass times velocity. So we could calculate the momentum before and after and subtract them. Or alternatively, we can multiply the momentum by the change in velocity. For our first sphere, we can then say that its mass is 0.2 and its change in velocity is negative 21. So the impulse on this sphere is 0.2 times negative 21, and that’s negative 4.2. And the units we use are newton seconds.

Next, we know that an impulse of equal magnitude, but opposite in sign, acts on the second sphere. In other words, the impulse on the second sphere must be 4.2 newton seconds. But of course, we can link that to its change in momentum. This means it’s equal to its mass times its change in velocity. Now, since it was initially moving in the negative direction at a speed of three meters per second, its change in velocity is 𝑣 minus negative three. So its change in momentum is 0.35 times 𝑣 minus negative three. This right-hand side is equivalent to 0.35 times 𝑣 plus three. And we calculated the impulse on the second sphere was 4.2. So we can say 4.2 equals 0.35 times 𝑣 plus three.

Next, we divide through by 0.35. Now, 4.2 divided by 0.35 is simply 12. Then, we subtract three from both sides. 12 minus three is nine. So we’ve calculated that 𝑣 is equal to nine meters per second. And so we’ve answered the question. The impulse that the second sphere exerted on the first, which we defined earlier to be 𝐽 sub 𝑎, is in fact 𝐼 equals negative 4.2 newton seconds. Similarly, the speed 𝑣 of the second sphere is the magnitude of nine meters per second, which is simply nine meters per second.

In the two examples we’ve seen so far, we demonstrated how to use conservation of momentum to calculate various quantities after a collision. It’s worth noting at this stage though that whilst momentum is conserved, this doesn’t necessarily mean kinetic energy is conserved. In fact, there are three scenarios. The first is said to be an elastic collision. This is a collision during which the kinetic energy is conserved. In an inelastic collision, part of the kinetic energy is dissipated in friction between the particles. Then finally, we have something called a perfectly inelastic collision. Now, in this collision, kinetic energy is dissipated in friction as in the previous definition. But the particles are also bonded together, and this results in them having the same velocity after collision. This might sometimes be called coalescence.

In our final example, we’ll look at the motion of two particles after an inelastic collision.

Two spheres are moving along a straight line. One has a mass of 𝑚 and is moving at speed 𝑣, whereas the other one has a mass of 10 g and is moving at 36 centimeters per second. If the two spheres were moving in the same direction when they collided, they would coalesce into one body and move at 30 centimeters per second in the same direction. However, if they were moving in opposite directions, they would coalesce into one body, which would move at six centimeters per second in the direction the first sphere had been traveling. Find 𝑚 and 𝑣.

With the information we’ve been given, we’re going to set up a pair of simultaneous equations in 𝑚 and 𝑣 using conservation of momentum. In particular, we’re going to use the formula momentum is equal to mass times velocity.

Let’s begin by identifying the direction to the right to be positive. Then, we can say the momentum of the first sphere is 𝑚 times 𝑣. The momentum of the second sphere is 10 times 36. That’s 360, giving us a total or net momentum before the collision of 𝑚𝑣 plus 360. By the conservation of momentum, we know this must be equal to the net momentum after the collision. The new mass of the body is 10 plus 𝑚, whilst its velocity is 30. Since it’s in the same direction, it’s still positive, meaning the momentum after the collision is 30 times 10 plus 𝑚. By distributing parentheses and rearranging, we can form an equation, 60 equals 30𝑚 minus 𝑚𝑣.

Now that we have this equation, let’s complete the process for the second scenario. In this scenario, these spheres are moving in opposite directions. This means the velocity of the second sphere must be negative 36. So the net momentum is 𝑚𝑣 minus 360 before the collision. After the collision, they coalesce, giving us a total mass of 10 plus 𝑚. And they move at six centimeters per second, still in the positive direction. This time, if we distribute the parentheses and rearrange, we get the equation negative 420 equals six 𝑚 minus 𝑚𝑣.

And now we might observe that we can quite quickly calculate the value of 𝑚 here. We have a system of equations in which we have the same term negative 𝑚𝑣. Let’s define these equations to be one and two, respectively. Then, if we subtract one equation from the other, and it doesn’t matter which way round we do it, that negative 𝑚𝑣 term will disappear. 60 minus negative 420 is 480. Then, 30𝑚 minus six 𝑚 is 24𝑚, and negative 𝑚𝑣 minus negative 𝑚𝑣 is zero. If we then divide through by 24, we find that 𝑚 is equal to 20 or 20 grams.

Then, we can find the value of 𝑣 by substituting it into either of our earlier equations. When we do, we get 60 equals 30 times 20 minus 20𝑣, which gives us negative 540 equals negative 20𝑣. Finally, we divide through by negative 20, and we find 𝑣 is equal to 27 centimeters per second. 𝑚 is 20 grams and 𝑣 is 27 centimeters per second.

We’re now going to recap the key points from this lesson. In this lesson, we saw that the momentum of a body vector 𝐩 is given by 𝑚 times the velocity vector 𝐯 but that we generally work with the scalar form 𝑝 equals 𝑚𝑣. We also saw that if we’re working in a closed system, the total or net momentum is conserved. In other words, the total momentum before the collision is equal to the total momentum after. We saw that impulse is equal to the change in momentum. And we learned what it means for a collision to be elastic, inelastic, and perfectly inelastic.

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