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An 8.00 -kg point mass and a 15.0 -kg point mass are held in place 50.0 $\mathrm{cm}$ apart. A particle of mass $m$ is released from a point between the two masses 20.0 $\mathrm{cm}$ from the 8.00 -kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

2.2 $\times$ 10$^{-9}$ m/s$^2$ towards the 8 kg mass

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Hope College

University of Sheffield

University of Winnipeg

Lectures

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Okay, so in problem 25 there are two masses, so I will draw them here. Different colors. There is this eight kilogram mass which miracle and one. And then there is a 15 kilograms mess. Joeckel Mt. And we are told that those two masses are 50 centimetres apart or zero point five meters. And then we placed a particle of mass stem, Um, 20 centimeters from and one 0.2 meters. Always best to use meters because all the Constance have meters in them. So if you use meters all the time, you know you'll never get a room. This particle, we don't know the mass. We just know it's Masten. But the good news is we're not gonna need that thought. You have this mass here? Um um let's do a free brake diagram on top of that for domestic end. There is a strike in blue still, because this is force from M two, and it also feels a force from anyone. Okay, so for mass m, this son of forces, um, is equal to before doing this. That's drawn acts only one dimension, Simeon drama X axis that so f do is positive. According to this access on some ring. If to here. Let's, um let's rename distances. So this distance will be yuan and this distance will be be too. And then So So that was left to hear. And then if one it's also applied on mass m. But this one will be negative because its opposite directions x so g m and one over. Oh, Rhett, do you want here? That was I told you to over Do you want square? And from Britain's Lo, we know that the sum of the forces equals two mass times acceleration. And this acceleration is actually what we want to find in the problem. So acceleration, I'm going to start from here. And then well, what happens is that you can cancel out those masses. So that's why we don't actually need the value of the mass in question. The mass in the mill. Okay, so I'm gonna put the big G out of a parent. Is this and I got into over you two squared minus and one over the one squared. Yeah, that's correct. So we have all of this. M two is 15 k and one is eight kilograms. He too is pretty easy to find. It's gonna be minus. It's going to be 0.5 minus you too. So it's your country, Um and we know that big G, the gravitational constant. It's 6.64 times 10 to the minus 11 and you didn't hear square squared. Okay, so em to I said, it's 15 kilograms. Two is your country years and one is eight kilograms and the one is 0.2 meters kit and the result that this is minus 2.2 plans tend to be minus nine meters per square second. So the questions specifically asks for the magnitude and direction. So when you want the magnitude yeah, to the absolute value. So I don't know, I work ex, she wrote. A magnitude of the exploration is the absolute value of it. So here, absolute value of that means we're removing the minus sign 2.2 virus 10 to the minus nine meters per second squared and then the direction Well and this diagram, it could be left it right. But that depends on where we placed a mass. Maybe someone else would have pleased them two on the left, so instead we will say. Well, since there's a minus sign, uh, we know it's going opposite of the X axis. Um, so the words m one And that is the answer, given the manual, too. So we could start by saying towards ex negative. But that is highly dependent. How we placed her access. And so something that is not dependent that that is the words. And one which was eight kilograms. Yep. So that's it for his problem, then.

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