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GR8677 #44
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Problem
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Mechanics }Conservation of Momentum
One could use energy, but then one would have to take into account the inertia. Momentum might be easier,
 = (p_f=MV)\Rightarrow V=\frac{m}{M}v,<br />
)
where the final momentum takes into account the fact that the final velocity of the particle is at rest (0). And, so it is (A)!
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Alternate Solutions |
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Comments |
ramparts
2009-11-05 11:30:23 | Minor trick for those of us not physics-savvy enough to realize that the angular stuff is unimportant: even if you're worried that rotational effects will come into play, in this problem ETS is extremely generous to not give any answers involving L. You know that if the rod were a point particle (L goes to 0) the answer is (A), so since none of the answers have L-dependence and can't achieve that limit, it's got to be (A). |  | segfault
2009-08-25 20:05:23 | Is it true that for this situation to occur  ? (Because the point particle comes completely to rest after the collision)
wittensdog
2009-09-28 20:03:20 |
I don't think so, not in this case.
The requirement that m = M for the first one to come to rest is when you have two point particles hitting each other head on. The fact that m = M comes from the requirement that kinetic energy and momentum both be conserved.
However, in this problem, kinetic energy is not necessarily conserved. Well, technically it is, but that's if you call the sum total of the rotation of all of the little atoms inside as part of the kinetic energy. So what happens is that the first particle has some kinetic energy, which then goes into kinetic energy of the center of mass of the stick and rotational energy of the stick. To know how much energy goes into the translation of the center of mass and how much goes into rotation, you'd need to know the moment of inertia. I guess you could look up what it would be for a uniform spatially extended stick, but I think the point remains that if you had an object with the right combination of mass and moment of inertia, the mass need not be the same as the incoming particle to bring it to rest.
Then again I could be wrong, I didn't verify this rigorously, just kind of reasoned it out in my head.
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shafatmubin
2009-10-31 19:54:10 |
If the problem is worked out in detail, it turns out that M = m indeed:rnrnM = m [ (v - 0) / (v + 0)] -> M = mrnORrnvFinalOfm = 0 = [(m - M) / (m + M) ] v -> m = M
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| | eliasds
2008-11-02 14:36:44 | I'm still confused. the equation ) =MV in no way shows the linear momentum of the particle contributing to the new angular momentum of the stick. Can someone clearly explain why angular momentum does not show up in the equation?
OrrinJelo
2009-09-24 19:01:08 |
Think of it as this: Linear momentum of the particle contributes to the linear momentum of the stick; the angular momentum of the particle contributes to the angular momentum of the stick.
How can the particle have angular momentum? Let me point out first that the particle hits the end of the stick, not at the center (where the center of mass is). If we look at the angular momentum about the center of mass of the stick, of course the particle has angular momentum! If you do the math, it will work out.
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|  | phys2718
2008-10-16 13:54:11 | I think the crucial point here is that the rod is a rigid body; thus all parts of the rod rotate about the center with the same frequency, and the linear momentum of one side of the rod due to the rotation is completely canceled by the momentum of the other side. So only the linear momentum of the center of mass needs to be taken into account. | | sabinscabin
2008-10-15 21:51:31 | I kept seeing the word "stick" in the problem; it is mentioned 4 times. So I thought that the mass "sticks" to the rod, which made me pick B.
D'oh!!!!!
511mev
2009-11-02 13:41:54 |
That's hilarious!
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| | Rune
2007-10-13 23:05:52 | I'm a bit confused because it seems to me that if the surface is frictionless and you hit a rod at one end, some of the momentum would go to rotating the rod. Why doesn't this happen?
jsdillon
2008-04-11 19:48:56 |
It does rotate. However, since the problem is asking for the velocity of the center of mass of the stick, it doesn't matter. The solution of the problem doesn't require a consideration of the conservation of angular momentum (although it is, indeed, conserved)--conservation of linear momentum is plenty.
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p3ace
2008-05-04 14:34:50 |
Only linear momentum for center-of-mass coordinates is needed.
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| | grscjo3
2006-11-03 21:01:37 | When you say that by using energy to solve this, one would have to take into account the inertia, do you mean the rotational inertia of the rod?
yosun
2007-02-22 15:11:24 |
Yes, referring to this kind of inertia: 
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