Our next concept is impulse. Impulse is simply when a force is applied, for a time, you'll see what I mean by that, for a time, usually just a specific period of time, to change momentum. Change, I'll just put p, all right? So to see it, we can go back to Newton's Second Law, the real Newton's Second Law, F = dp dt, and then we could apply the fundamental theorem of calculus, or we could bring the dt over here and integrate both sides, whatever you want to do. You know that this is really p2- p1. If we look at the momentum change from 2 times that we call dt, it looks like that, and therefore, you'd have F times dt. So if you integrated that side, it would look like the integral of f dt from t1 to t2. So to find the change of momentum from time 1 to time 2 is the integral of F times dt, all right? So what we could also call this, this is just delta p, right? So the impulse I is equal to delta p, and it's equal to the integral from t1 to t2 F dt. So this is the impulse, and it is the change in momentum. That's all it is, it is a literally delta p. It has units of kilograms meter per second. It is a vector. Okay, so it's just a fancy way of saying the change of momentum. You might see it written without the integral. So you need the integral if, in this case, of course, if the force varies with time, if the force varies with time, and if you're good at your calculus, you know what this looks like if the force does not vary in time, then it's just F times delta t. So you can also say delta p equals F times delta t, for if you have a constant force applied for some time. Let's think about reality though. So what if I take this ball, right, and I drop the ball and it bounces back up, or I throw it down and it bounces back up. As it falls, it clearly has a momentum down, mv is pointing down, the p vector is down, and it's getting bigger because the gravitational force is increasing the p vector. And then it hits the surface, and suddenly, it has momentum up, because it received a brief impulsive force from the table, right, the table pushed it up. So let's see, what do we think that really looks like? So impulsive forces are usually complicated. If you think about it, the rubbery ball is going to act like a spring. So if we were to plot the real force versus time, force versus t, it would kind of be nothing, then it starts to touch. We need a big force, and then the force goes away as it pulls away, something like that, and the impulse would be the area under that curve, area of the Ft curve. But often, when we do problems with impulse and we talk about impulse, we often define just the average force, right? So we could do it this way, or we could say, well, it was kind of like this, all right? Where this is F average, and you pick that average so that the area under this curve, F average times delta t is the same as the area under the true impulsive curve. So the problems, you'll often see a problem written to say, it felt an average force of 20 newtons for 5 milliseconds. And then you know, for the impulse, you just multiply 20 newtons times 5 milliseconds, and you don't worry about what the force really looked like in real time. Also in these kinds of problems, one thing to watch for is to keep in mind the vector nature of the momentum and the impulse and everything. So for instance, if you were trying to calculate the impulse of the bouncing ball, right, the impulse that cause the ball to bounce, you would have here, if we call this the positive z direction, and it's falling at v initial, then at first, your momentum vector is down, right? So p initial is minus m, I'll write the magnitude, the speed, right, I had, or ijk had, right? It's down, and then it says it bounces back up at v final. So keep mind, it's going the opposite direction. So p final is in this case positive, because it's going up, and the speed final in the k hat. So keep in mind that the delta p that's equal to your I, it's p final minus p initial, but vectors, all right? So final m v final I hat minus, then minus m v initial I hat. So the point is, you don't just subtract the two speeds, right, they're opposite directions. You actually add the speeds to get the change in speed because we're subtracting what is actually in the opposite direction, negative m, times that speed so it ends up becoming the sum of the two. That's just a little thing, intuitively you get it. Sometimes when you're doing problems you kind of forget that it's actually a big change, and then in the end, this change, the direction of that change vector is up, right, final up minus initial down. The total vector is up, the impulse is up, which makes sense because the force is up, the table pushes up on the ball.
