ME274 - Basic Mechanics II: Homework Problem 3/205

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Oct 16, 2008

Homework Problem 3/205

Any questions?

6 comments:

Upasana said...: Hi,

I was getting a little confused about the role played by the weight of the particle in this problem. When I apply the impulse momentum equation in the y direction would I include the weight as a force acting on the particle even though it is constant with change in time?

Thanks
-Upasana; October 19, 2008 at 6:31 PM
BoilerBrian said...: This problem is actually a little tricky. Since it's in a horizontal plane the weight plays no part in the forces. It took my roommate and I a few minutes to figure that one out.; October 19, 2008 at 8:31 PM
J.T. Kinsey said...: I was having the same problem... whether to use the weight of the particle in the impulse momentum equation. So thank you for answering that. Even when I do not use the weight, I get a a velocity almost twice the right answer.

Maybe my math is wrong. I used mv2 = mv1 + int(R)dt. With R as the F given.

If I'm missing anything please let me know.

-Ted; October 19, 2008 at 11:06 PM
BoilerBrian said...: That looks right to me. That equation should give you the y-component of the velocity and then you use the x-component of the initial state (because it's constant) to find the speed that it asks for in the problem. Make sure you solve for v2 not just mv2.; October 20, 2008 at 8:27 AM
FrankTheTank said...: You need the weight in the equation to get the right answer. Since this is a momentum equation, if you leave the weight out, I'm guessing you came up with an answer of 16.8 m/s for v2. Dividing through by your 2.4 kg will give you the correct v2.; October 20, 2008 at 10:31 AM
rpmccaul said...: I was confused for awhile too, but it was just little mistakes, be sure you are using vector form and solving for magnitude at the end for the velocity, then use the vector form to find the angle. pretty simple in hindsight.; October 20, 2008 at 10:46 AM

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