Professor Anderson

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>> Hello, class. Professor Anderson here. Let's take a look at one of your homework problems. This is an impulse problem. And the idea is this. We have an object, say a tennis ball, that is going to hit the wall and bounce back off the wall. And while it's in contact with the wall, we can map out the force as a function of time. And we want to figure out something about that force. So let's say the force does something like this. It ramps up as the ball is getting compressed against the wall. It's fairly constant for some amount of time while the ball is compressed. And then it springs back off and goes back down to zero. So if we think about this problem, what we know is the area under the curve is important. And in fact, the area under the curve tells us the impulse that is delivered to the ball. Impulse we write with a J. That's equal to the area under the curve. But impulse J is simply change in momentum, delta P. And in this case, we know what? If it goes in with Vi, comes out with Vf, then delta P is mVf minus mVi. And in the case of something like a super ball, let's say that it goes in with speed V and comes out with speed V. Alright. If I made both of those V's, this whole thing would go to zero. We know that can't be right. But what we do know is that V initial is to the left. And we're going to say that's a negative. And so look what happens. We end up with a negative negative. We end up with two mV. OK. So if we know those numbers, m and V, then we can relate it back to the force because the force is related to this area under the curve. So let's see how that works. And let's try it with some real numbers. Alright, so let's plot this again. Here is F. And we're going to say that it does this sort of shape. And this is what we're calling F max. And let's say we want to figure out what that F max is equal to. Alright. This is two. This is four. This is six. This is zero seconds. Now if I want to calculate the area under this whole thing, one thing that I notice is this first triangle is exactly the same as the last triangle. And so if I take this last triangle and move it over here, I can fill in this gap right here. OK. So the area is just F max times delta T, where now delta T is going to be zero to four. Alright? So we'll write this as F max times four seconds. And now it looks like we have all the information we need. OK? Two mV is equal to F max times four. OK. So we just calculated that the area under the curve was equal to F max times four. If you look at the units here, this is milliseconds. So this should be four times ten to the minus three seconds. And that's equal to this twice times the mass times the speed V. And so now we can solve this for F max. We get two times the mass, which we said was 0.06. V was 32. And the bottom becomes four times ten to the minus three. And all this is SI units now. And if you plug those into your calculator, you should get 960 newtons. Alright. Try that with your numbers. Hopefully it's clear. And if not, come see me in my office. Cheers.

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>> Hello, class. Professor Anderson here. Let's take a look at one of your homework problems. This is an impulse problem. And the idea is this. We have an object, say a tennis ball, that is going to hit the wall and bounce back off the wall. And while it's in contact with the wall, we can map out the force as a function of time. And we want to figure out something about that force. So let's say the force does something like this. It ramps up as the ball is getting compressed against the wall. It's fairly constant for some amount of time while the ball is compressed. And then it springs back off and goes back down to zero. So if we think about this problem, what we know is the area under the curve is important. And in fact, the area under the curve tells us the impulse that is delivered to the ball. Impulse we write with a J. That's equal to the area under the curve. But impulse J is simply change in momentum, delta P. And in this case, we know what? If it goes in with Vi, comes out with Vf, then delta P is mVf minus mVi. And in the case of something like a super ball, let's say that it goes in with speed V and comes out with speed V. Alright. If I made both of those V's, this whole thing would go to zero. We know that can't be right. But what we do know is that V initial is to the left. And we're going to say that's a negative. And so look what happens. We end up with a negative negative. We end up with two mV. OK. So if we know those numbers, m and V, then we can relate it back to the force because the force is related to this area under the curve. So let's see how that works. And let's try it with some real numbers. Alright, so let's plot this again. Here is F. And we're going to say that it does this sort of shape. And this is what we're calling F max. And let's say we want to figure out what that F max is equal to. Alright. This is two. This is four. This is six. This is zero seconds. Now if I want to calculate the area under this whole thing, one thing that I notice is this first triangle is exactly the same as the last triangle. And so if I take this last triangle and move it over here, I can fill in this gap right here. OK. So the area is just F max times delta T, where now delta T is going to be zero to four. Alright? So we'll write this as F max times four seconds. And now it looks like we have all the information we need. OK? Two mV is equal to F max times four. OK. So we just calculated that the area under the curve was equal to F max times four. If you look at the units here, this is milliseconds. So this should be four times ten to the minus three seconds. And that's equal to this twice times the mass times the speed V. And so now we can solve this for F max. We get two times the mass, which we said was 0.06. V was 32. And the bottom becomes four times ten to the minus three. And all this is SI units now. And if you plug those into your calculator, you should get 960 newtons. Alright. Try that with your numbers. Hopefully it's clear. And if not, come see me in my office. Cheers.