In this video, I'm going to do a quick review of impulse and momentum. We're going to go over the key formulas that you need to pass your next exam. By the way, for those of you who want example problems, including a practice test on this topic, feel free to check out the links in the description section below. So let's begin.
The first formula you need to be familiar with is this. Momentum is mass times velocity. As a vector you could describe it this way.
Momentum is a vector, mass is scalar, but velocity is a vector. So let's say if you have a block that rests across a horizontal floor and let's say this is a 10 kilogram block and it's moving at a speed of 6 meters per second. The momentum of this block is going to be 10 kilograms.
That's the mass times the velocity of 6 meters per second. And so the momentum is going to be 60 with the units kilograms times meters per second. So that's how you can calculate momentum.
It's simply mass times velocity. Think of it as mass in motion. So make sure you write this down.
This is the first equation that you want to know for your test. Now, the next thing you need to know how to calculate is impulse. Now, some textbooks may use the letter J for impulse. Sometimes I've used J, sometimes I use I, I for impulse. But also, I can represent inertia too, so.
But impulse is equal to force multiplied by time. that's how you calculate impulse so imagine if you have a block and if you apply a force of a hundred Newton's for a time period of eight seconds In previous chapters, you've learned about forces. But in this chapter, with impulse, now we've incorporated time with force.
How long is the force being applied to an object? Because that's important. So here we have a force of 100 newtons being applied on this object for 8 seconds. The impulse imparted to this object is going to be F delta T. So it's 100 newtons times 8 seconds.
which is 800 newton seconds. So that's how you can calculate impulse. Impulse tells you how much force is being applied to an object for a certain amount of time period.
So an impulse of 800 can mean many things. It could be you're applying a force of 100 newtons for 8 seconds, or you're applying a large force of 1000 newtons for 0.8 seconds. So it just tells you how much force you're applying for the amount of time period combined.
Now, impulse and momentum, they're related. Impulse is equal to the change in momentum. And here's how we can derive that. So according to Newton's second law, force is equal to mass times acceleration.
And we know that the acceleration is the rate at which velocity changes with respect to time. Acceleration is the change in velocity divided by the change in time. Perhaps you remember this equation, v final is equal to v initial plus at. If you solve for acceleration, it will be v final minus v initial over time. In other words, it's the change in velocity with respect to time.
Now, if we multiply both sides by delta t, we get that f delta t is equal to m delta v. On the left, force multiplied by time gives us impulse. On the right, we know that mass times... velocity is momentum, so mass times the change in velocity is the change in momentum.
So this is related to the impulse momentum theorem. Now typically, the formula that you need to use when solving problems with this topic is this equation. This is the form of the equation that's going to be most helpful. So that is the impulse momentum theorem.
So make sure you know these three equations. Number one, momentum is mass times velocity. Number two, impulse is force multiplied by time. Number three, the impulse momentum theorem, F delta T is equal to M delta V. Now, going back to Newton's second law, particularly this form, there's a lot of other equations that we can get from this form.
So I'm going to adjust this equation. Let's say if you have a person and he has a water hose in his hand and out of it he's going to shoot out water. Let's say water is coming out of the hose at a speed of 20 meters per second. So that's how fast it's coming out. And then we need to talk about the quantity of water coming out.
There's something called the mass flow rate. So let's say water is leaving at a mass flow rate of five kilograms per second. This equation can help us to calculate the force that this fluid will exert on an object.
Let's say if you're blasting out water from the hose on a block. We can calculate the force. exerted by the water on this block so I'm going to move Delta T to this position under M so we can describe forces being mass flow rate delta M over delta T times V so instead of the change in velocity we have the change in mass in this case the velocity of water coming out of the hose is going to be relatively constant So in this problem, the mass flow rate is 5 kilograms per second. So every second, 5 kilograms of water is coming out of that hose.
Now the speed of the water is 20 meters per second. When we combine these two, we get a force of 100 newtons. That's going to be the force exerted by the water on the block. So that's another way in which you can calculate the force that a fluid exerts on an object. This could be a fluid like water, it could be a fluid like air.
As long as you know the mass flow rate and the speed at which that fluid is moving, you can calculate the force. I should really highlight this equation. So this is the fourth equation you want to know. Force is equal to mass flow rate multiplied by the velocity. Now for those of you who are taking calculus with physics, the next formula is going to be more applicable to you.
So going back to this equation, instead of trying to get the mass flow rate from this equation, We're going to focus on this part. We know that mass times velocity is momentum. So mass times the change in velocity is the change in momentum. So this is another way to describe force. Force can also be described as the rate at which momentum is changed.
changing. In other words, force is the derivative of the momentum function with respect to time. So if you know the function for momentum, you can get the function for force with respect to time. So f of t, this is force as a function of time, is the derivative of the momentum function. So these are some other ways in which you can calculate the force acting on an object if you know the momentum function.
So I would write these equations, number 5 and number 6. You can add those to your notes. Now let's talk about the conservation of momentum. So let's say we have a horizontal frictionless floor and we have block one.
Let's say block one is moving at a speed of two meters per second and it strikes block two. Afterward, the two blocks, they stick together. And they're moving at the same speed. And you want to find out what this final speed is. Here we have what is known as an inelastic collision.
Whenever you have two objects colliding, if they stick together, it's a collision, but it's inelastic because you're going to have loss of kinetic energy. collisions both momentum and energy is conserved kinetic energy but for inelastic collisions only momentum is conserved so this is an example of an inelastic collision anytime they stick together and they don't bounce off you're dealing with an inelastic collision momentum is conserved but kinetic energy is not conserved for an inelastic collision So before the collision, we have the momentum of object 1, m1v1, plus the momentum of object 2. After the collision, we're going to have the momentum of both objects as well. So this is the conservation of momentum formula. You want to make sure you write this down. The basic idea behind this equation is that the total momentum of the system before the collision is equal to the total momentum of the system after the collision, assuming no external forces are acting on the system.
Because if there are external forces, they will change the total momentum of the system. So this only holds true if there are no external forces acting on the system. So make sure you add this equation. This is number 7. Now for this particular problem, because they stick together, V1 final and V2 final, they're the same.
They just equal Vf. So for this particular situation, you could shorten this equation to this form. It's going to be m1 plus m2 times the final speed because they stick together. So they move at the same final speed. So for any conservation of momentum problem, you could use equation 7. But if they stick together at the end, the two objects, after they collide, you could use equation 8 to solve it.
Now let's say if we have a ball moving at a speed of 6 meters per second, and then strikes ball 2, which was initially moving at 2 meters per second, and then they bounce off each other. Ball 1 goes back in this direction, let's say at 3 meters per second, and you're trying to find out the speed at which ball 2 moves to the right. In other words, you're trying to find V2 final.
A lot of times, not always, if they bounce off each other, it could be an elastic collision. But you want to make sure the problems specify that. So if it's a perfectly elastic collision, that means that kinetic energy is conserved, as well as momentum. So for this problem, you would need to use equation 7 to solve it.
So you have the conservation of momentum equation. Now, you're also, let's say if you don't know this speed. Because if you know V1 final, you can probably just use this equation to get the answer.
But if you don't know V1 final, you need another equation to solve this. So this is when you need to use conservation of energy. That is, the total initial energy is equal to the total final energy. So this would be KE initial or KE1 plus KE2 equals KE1 final plus KE2 final. Now, this form of the equation, there's going to be a lot of work, a lot of algebra if you use it.
But if you want to get the answer faster, you could use the simplified equation that comes from this equation. And that equation is this, which we'll describe as number one. 10. v1 plus v1 final is equal to v2 plus v2 final. So if you have an elastic collision where you're missing the final speeds of both objects, you need to use equation 7 and equation 8 to solve it.
So you need to write a system of equations to figure that problem out. If you're given v1 final but not v2, you could just use equation 7. But if you're missing two variables, you have to use two equations to solve two variables. So you'll need equation 7 and 10. I do have example problems on this on my other videos on YouTube, so feel free to take a look at that when you get a chance.
And you might see this also on my practice test too. So that's basically it for this video. Those are the main equations you need for this chapter on impulse and momentum, as well as elastic collisions and inelastic collisions. Thanks for watching.