 Okay so I'm going to go over the test. This is the Tuesday class. If you're the Thursday class, you can look if you want. So the first one had this ball going up a ramp and I'm going to call that the X direction. And I gave the acceleration so it's going to go up and it's going to slow down. And there's two balls. So one of them starts with the higher initial speed and then a little bit later one with the lower initial speed. The first thing was to make a sketch and I've already put the answer over here just to save time. Okay so the first one is just going to look like this. It's just a parabola. This is the time and this is X. You did something just like that before. Now the second one starts a little bit later and what's going to be different about it is that it's going to have this initial slope isn't going to be as great because it's going to be slower. So you don't really know where it's going to cross on a sketch but you do know it's going to cross so I'm just going to do like this. And so that's the point where they meet. Okay that's the exaggerated view but you get the idea. So that's your sketch. It doesn't matter about the numbers. It doesn't matter where you put these as long as they cross. So what's next? The next part said when do they meet and where do they meet? So here's what I have. This is the first ball. Okay I was doing that too. Let's see. Okay that's fine. I was trying to block the reflection but oh well. It's going to fall. Okay here's the first position. I said it started at X equals zero at time T equals zero and it started with this velocity point three minus one half the acceleration. The minus really goes with the acceleration but it doesn't matter. T squared. So that's just your basic kinematic equation for ball one. So I have to rewrite that. All I did was multiply one half times point four. For ball two the difference is it started half a second later so instead of T I'd have a different time. But if I just tracked off half a second from that time I get the original time. That's how it starts half a second later. I agree that was a little trickier than I wanted it to be. But still. And the same thing for the T over here. So it has the same acceleration but it has T minus half point five squared for that time. So now you just do exactly what you did before. You just set those two X's equal to each other and find the time. So here I just, all I did was I multiplied this out to get that. And this I have to multiply T minus point five squared. I get that and then I multiply by the negative point two and I get this. Then I set the two equal to each other. That's TX1, that's X2. The T squareds cancel because they have the same acceleration. And so I get something a little bit simpler. I solve for T I get point eight six seconds. And then where do they meet? I can either plug into the X1 equation or the X2. But either way, I plug in here's X1. I plug in point eight six seconds. I get point one one meters. That's it. So I was pretty generous on the credit for this one since I knew it was a little bit more difficult than I wanted it to be. You know, if you're setting up equations like this and it's clear that you're knowing you need to solve that and you have a sketch. I think I gave you full credit or at least close. A lot of people put the sketches, they put constant straight lines. I mean that's clearly not straight lines. They're not going to meet in that case. You're just making up stuff. Don't make up stuff. Okay, the next question was a little more straightforward. And it was straight from the lab. I gave you mass and acceleration data. So I gave you, for this case, here's a car with weight over the end and then you find the acceleration. If you plot force versus mass, the total, what did I plot? M2. No, I'm sorry. You're plotting force versus acceleration where this force is M2 times g. So it's a mass and then that times g is the force accelerating the whole thing and then I plot it versus acceleration. You do get a straight line. You fit that straight line and you should get a slope around 0.6 newtons per meters per second squared because the force of the newtons, acceleration is in that. And one newton is a kilogram meter per second squared. Kilogram meters per second squared. So this is just the same as 0.65 kilograms. And since f equals ma, that's the total mass, if I plot f versus a, the mass would be the slope. So the total mass is 0.65 kilograms. That's all there was to that one. You can see what I'm writing here. I think you can. I'm going to assume that you can. Okay. Number three is a ball shot off of a table. I'm not going to go over the whole solution. Here's v0 theta and here's y0 and you want to find out where it lands x. So you have two equations. You have the x equation, which says, let's call this x equals 0, y equals 0 right there. So x final is x0, which is 0, times the x velocity v0 cosine theta t. That's it. So I know theta. I know v0. I don't know t. But once I do, I can plug it in right here to get it. For the y equation, I have y equals, the final y equals y0 plus the y velocity, initial y velocity v0 sine theta t minus 1 half gt squared. Now in this case, I know it ends up at the ground. So this is 0. So now I know that number. I know that number. I know that number. So I just have the quadratic, I can just use a quadratic equation to solve for time. And then I can get that time and plug it into here. And when you do that, you get x is about 1.9 meters. That's it.