- Apply: conservation of momentum
- Introduction to momentum
- Impulse and momentum dodgeball example
- What are momentum and impulse?
- What is conservation of momentum?
- Bouncing fruit collision example
- Momentum: Ice skater throws a ball
- 2-dimensional momentum problem
- 2-dimensional momentum problem (part 2)
- What are two dimensional collisions?
- Force vs. time graphs
In this video, David shows how to solve for the impulse and force applied during a dodgeball collision using the impulse momentum relationship. Created by David SantoPietro.
- [Instructor] This person right here is about to play dodgeball. They're just unfortunately not gonna dodge the ball. It's gonna fly in, it's gonna bounce off their head. This may or may not have happened to you. I think this probably happened to me. It's been a long time since I played dodgeball. And although, unfortunate for this person, it's a wonderful opportunity, scientifically speaking, to talk about the impulse, momentum, force, time relationship, so let's do that. Let's put some numbers on here. So, we're gonna need to know the mass of the ball. Let's say this is a .2 kilogram ball, and we're gonna need to know some other numbers. Let's say the ball comes in at a speed of about 10 meters per second. So let's say it's comin' in at 10 meters per second and let's say it leaves at a speed of five meters per second. So it's probably gonna recoil with a little less speed than it came in with. It comes in with 10, leaves with five, and let's say that time, time right here, the time period that it's actually in contact with the person's face, let's say the time, when the ball is getting kind of compressed and then recoils and expands again, let's say the time that it's actually in contact is about .02 seconds, or about 20 milliseconds. So, knowing this information, we can ask all kinds of questions. One of them is: What was the impulse on the ball from the person? Now, the definition of impulse, we use the letter J for impulse, that always seemed a little weird to me. There's no J in impulse. I end up calling it jimpulse, just so I can remember that it's impulse, and there's a J for it. So the jimpulse, or the impulse, is defined to be the force acting on the object, multiplied by the time duration during which that force is acting. In other words, the impulse, from a force, is equal to that force, multiplied by how long that force was acting on the object. So, if we knew the force on this ball, we could use this formula to get the impulse, but we don't. I don't know the force that this person's face is exerting on the ball, so I can't use this formula to solve for the impulse. But, there's an alternate formula for impulse. If you're talking about the net impulse, in other words, the impulse from all forces on an object, like this ball, that should just equal the change in momentum of that object, like the change in momentum of this ball. So if we can figure out the change in momentum of this ball, we can figure out the net impulse on this ball. And since it's the net impulse, and this formula appears also true, this is equivalent, which is saying that it's the net force, multiplied by the time duration, during which that net force is acting. This is hard for people to remember, sometimes my students like to remember it as Jape Fat. So, if you look at this, it looks like J-A-P, this kinda looks like an E, F-A-T. So if you need a way, a pneumonic device, to remember this, Jape Fat is a way to remember how impulse, change in momentum, force, and time, are all related. So let's do it. We can't use force because we don't know it yet, but I can figure out the change in momentum 'cause I know the velocities. So, we know that the change in momentum is gonna be P final, the final momentum, minus the initial momentum. What's my final momentum? My final momentum is M times V, so it's gonna be mass times V final, minus mass times V initial, and my mass is .2, so I've got a mass of 0.2 kilograms. My final velocity is five, because the ball recoiled to the right with positive five. Positive five 'cause it's moving to the right. I'm gonna assume rightward is positive. Then minus, the mass is .2 again, so 0.2 kilograms. My initial velocity is not 10. This is 10 meters per second to the left, and momentum is a vector, it has direction, so you have to be careful with negative signs here. This is the most common mistake. People just plug in positive 10, then get the wrong answer. But this ball changed directions, so the two velocities here have to have two different sides, so this has to be a negative 10 meters per second, if I'm assuming rightward is positive. This leftward velocity, and this leftward initial velocity, has to be negative 10. And, if you didn't plug that in, you'd get a different answer, so you gotta be careful. So, what do I get here if I multiply this all out? I'm gonna get zero, no, sorry, I'm gonna get one kilogram meters per second, minus a negative two kilogram meters per second, and that's gonna give me positive three kilogram meters per second is the impulse, and that should make sense. The impulse was positive. The direction of the impulse, which is a vector, is the same direction as the direction of the force. So, which way did our face exert a force on the ball? Our face exerted a force on the ball to the right. That's why the impulse on the ball is to the right. The impulse on this person's face is to the left, but the impulse on the ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and bounce back to the right. That's why this impulse has a positive direction to it. Now, if you've been paying attention, you might be like, wait a minute, hold on. What we really did was we found the change in momentum of the ball, and when we do that, what we're finding is the net impulse on the ball. In other words, the impulse from all forces on the ball. But what this question was asking for was the impulse from a single force. The impulse from just the person's face. Now, aren't there other forces on this ball? Isn't there a force of gravity? And if there is, doesn't that mean what we really found here wasn't the impulse from just our face, but the impulse from the person's face and the force of gravity during this time period? And the answer is no, not really, for a few reasons. Most important reason being that, what I gave you up here was the initial horizontal velocity. This 10 meters per second was in the X direction, and this five meters per second, I'm assuming is also in the X direction. 'Cause if I'm taking the initial velocity in the X, and the final velocity in the X, and I take the difference in momentum, what I really found was the change in momentum in the X direction. When I do that, I'm finding the net impulse in the X direction, and there was only one X directed force during this time and that was our face on the ball, pushing it to the right. There was a force of gravity. That force of gravity was downward. But what that force of gravity does, it doesn't add or subtract any impulse in the X direction. It tries to add impulse in the downward direction, in the Y direction, so it tries to add vertical component of velocity downward, and so we're not even considering that over here. We're just gonna consider that we're lookin' at the horizontal components of velocity. How much velocity does it add vertically, gravity? Typically, not much during the situation, because the time period during which this collision acted is very small and the weight of this ball, compared to the force that our face is acting on the ball with, the weight is typically much smaller than this collision force. So that's why, in these collision problems, we typically ignore the force of gravity. So, we don't have to worry about that here. That's not actually posing much of a problem. We did find the net impulse in the X direction since our face was the only X directed force, this had to be the impulse our face exerted on the ball. Now, let's solve one more problem. Let's say we wanted to know: What was the average force on this person's face from the ball? Well, we know the net impulse on the ball, that means we can figure out the net force on the ball, because I can use this relationship now. Since I know that the net impulse on the ball in the X direction should just equal the net force on the ball in the X direction, multiplied by the time interval during which the force was applied, I can say that the net impulse on the ball was three kilogram meters per second, and that should equal the net force on the ball in the X direction, which was supplied by, unfortunately, this person's face, multiplied by the time interval, which is 0.02, 20 milliseconds. So, now I can solve. The force, the net force on the ball, during this time interval in the X direction, was three, divided by .02. If I take three kilogram meters per second and I divide by .02 seconds, I'm gonna get 150 Newtons was the net force on the ball. We got a positive number, and that makes sense, because this person's face exerted a positive force on this ball, 'cause the force was exerted to the right. So, these were positive, and the impulse from the face on the ball should be going the same direction as the force from the face on the ball. So, this is the force on the ball by the person's face, but notice this question is asking: What was the average force on the person's face from the ball? Not on the ball by the face. You might think, oh no, we gotta start all over, we solved for the wrong question, but we're in luck. Newton's third law says that the force on the face from the ball should be equal and opposite. So, this force on the face from the ball has got to be equal and opposite to the force on the ball from the face, so that's what we found here, the force on the ball from the face, that means the force on the face from the ball is gonna have the same size. It's gonna be 150 Newtons. It's just gonna be directed in the leftward direction, that means it's gonna be a negative force, so technically, you could say, this would be negative 150 Newtons on the face from the ball. So, to recap, the impulse, from an individual force, is defined to be that force, multiplied by the time interval during which that force is applied. And if you're talkin' about the net force in a given direction, multiplied by the time interval, you'd be finding the net impulse in that direction, and this also happens to equal the change in momentum in that direction. So, in other words, if there is a net impulse in a given direction, there's gotta be a change in momentum in that direction by the same amount. And one convenient way to remember how are these related, is you could use the pneumonic device, Jape Fat. I have no idea what Jape Fat means, but it helped me remember that the net impulse equals the change in momentum, and that also equals the net force, multiplied by the time interval, during which that force was applied. And finally, remember that during these collisions, there's always an equal and opposite force exerted on the two objects participating in the collision.