Physics Help Forum Two particles

 Jul 14th 2013, 05:13 AM #1 Junior Member   Join Date: Jul 2013 Posts: 7 Two particles Hi my friend....i need solve it...very very need it.... thnak youfor help. __________________________ Two particles of masses M and m needs to move on the horizontal circle. (At t = 0 the heavier particle velocity u0 and v0 another call and assume u0> v0) Resilience Factor e got to get this deal. A) After n collisions between particles quickly got dig it. B) as n tends to infinity, the two sequences from a dig it up. (Terminal velocity of particles) C) locate and limit energy consumption and the total sequence of the n th n tends to infinity, find the energy dissipation. _________ thank you
Jul 14th 2013, 11:55 AM   #2

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 Originally Posted by hakimghaziali2 Hi my friend....i need solve it...very very need it.... thnak youfor help. __________________________ Two particles of masses M and m needs to move on the horizontal circle. (At t = 0 the heavier particle velocity u0 and v0 another call and assume u0> v0) Resilience Factor e got to get this deal. A) After n collisions between particles quickly got dig it. B) as n tends to infinity, the two sequences from a dig it up. (Terminal velocity of particles) C) locate and limit energy consumption and the total sequence of the n th n tends to infinity, find the energy dissipation. _________ thank you
I'm sorry, what is v0? What do you mean by "dig it?"

-Dan
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Jul 14th 2013, 04:03 PM   #3
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 Originally Posted by topsquark I'm sorry, what is v0? What do you mean by "dig it?" -Dan
v0 is lighter particle velocity
and

A)After N collisions
What is the speed of the two particles?

 Jul 15th 2013, 02:43 AM #4 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,814 I am going to assume that your "resilience factor" (e) is actually the "coefficient of restitution." Please correct me if I'm wrong. Calling the speed of M after the first impact u and the speed of m after the impact v we have (note the difference between your notation and the notation in the Wikipedia link!) We also know that momentum is conserved in the collision, so Two notes: 1) The u's and v's in the equations are vectors in 1 - D. That is to say pick an arbitrary direction (of either u_0 or v_0) to be positive. 2) You want to solve for v and u in terms of u_0 and v_0 so two equations are enough for this. See what you can do with this and if you get stuck just let us know. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
Jul 15th 2013, 01:05 PM   #5
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 Originally Posted by topsquark I am going to assume that your "resilience factor" (e) is actually the "coefficient of restitution." Please correct me if I'm wrong. Calling the speed of M after the first impact u and the speed of m after the impact v we have (note the difference between your notation and the notation in the Wikipedia link!) We also know that momentum is conserved in the collision, so Two notes: 1) The u's and v's in the equations are vectors in 1 - D. That is to say pick an arbitrary direction (of either u_0 or v_0) to be positive. 2) You want to solve for v and u in terms of u_0 and v_0 so two equations are enough for this. See what you can do with this and if you get stuck just let us know. -Dan
hi my friend ...thank you for help
i need solve it....i cant understandYplease solve it for me ... tnx

Last edited by hakimghaziali2; Jul 15th 2013 at 01:11 PM.

Jul 15th 2013, 02:49 PM   #6

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 Originally Posted by hakimghaziali2 hi my friend ...thank you for help i need solve it....i cant understandYplease solve it for me ... tnx
I'm not going to solve it for you. I will help, but you have to contribute as well otherwise you will learn nothing from this. Bad idea when exam time comes around.

The battle plan is to do the first collision and see what happens with the speeds, etc. Then we can take a look at what we need to do next.

First, let's look at the table from the top. I'm going to pick a positive coordinate direction to the right. I'm going to start the problem from the bottom of the circle, and I'm going to put M on the left with speed v_0 and m on the right with speed u_0. So at the top of the circle we've got M going to the right at speed v_0 in the positive direction and m going to the left at speed u_0 in the negative direction.

The momentum equation gives:

The coefficient of restitution equation gives:

Now, we have two equations in two unknowns. Let me know how you are doing at this point and let me know how your equation solving gets you. Give it a good try...I will help out on any mistakes.

-Dan
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Jul 15th 2013, 03:11 PM   #7
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 Originally Posted by topsquark I'm not going to solve it for you. I will help, but you have to contribute as well otherwise you will learn nothing from this. Bad idea when exam time comes around. The battle plan is to do the first collision and see what happens with the speeds, etc. Then we can take a look at what we need to do next. First, let's look at the table from the top. I'm going to pick a positive coordinate direction to the right. I'm going to start the problem from the bottom of the circle, and I'm going to put M on the left with speed v_0 and m on the right with speed u_0. So at the top of the circle we've got M going to the right at speed v_0 in the positive direction and m going to the left at speed u_0 in the negative direction. The momentum equation gives: The coefficient of restitution equation gives: Now, we have two equations in two unknowns. Let me know how you are doing at this point and let me know how your equation solving gets you. Give it a good try...I will help out on any mistakes. -Dan
i dont know how can find the "e"
and how can find the answer "A" and "B" and "C"
if you solve it i can understand it.

Jul 15th 2013, 03:31 PM   #8

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 Originally Posted by hakimghaziali2 i dont know how can find the "e" and how can find the answer "A" and "B" and "C" if you solve it i can understand it.
I should have mentioned that the problem will be in terms of e as well. So the restitution equation will end up in terms of e, u_0, and v_0.

If you don't want to do the work, that's up to you. I will not give you any more information until you provide some effort. I don't care if you get it wrong, just put something into it.

This is a help site. I'm not going to do your homework for you.

-Dan
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Jul 15th 2013, 03:53 PM   #9
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 Originally Posted by topsquark I should have mentioned that the problem will be in terms of e as well. So the restitution equation will end up in terms of e, u_0, and v_0. If you don't want to do the work, that's up to you. I will not give you any more information until you provide some effort. I don't care if you get it wrong, just put something into it. This is a help site. I'm not going to do your homework for you. -Dan
ok
i dont know how can find the "e".. just tell me that

Jul 15th 2013, 06:46 PM   #10

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 Originally Posted by hakimghaziali2 ok i dont know how can find the "e".. just tell me that
You can't. Your solutions will be in terms of e, u_0, and v_0. To make this clear, in case you didn't understand the question, we are not going to have a nice numerical solution to this problem. It may or may not depend on u_0 or v_0 (I haven't fully solved this yet myself) but it will certainly depend on e.

Hint: Solve the e equation for u. Then substitute that "value" of u into the momentum conservation equation.

-Dan
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