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# Conservation of momentum Watch

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1. Always get stuck on questions like this;

"A proton makes a low-energy head-on collision with an unknown particle and rebounds straight back along its path with 4/9 of its initial kinetic energy. Assuming that the collision is elastic and that the unknown particle is originally at rest, calculate the mass of the unknown particle."

I realise momentum and kinetic energy are conserved in this situation, and I've arranged and re-arranged the several conservation equations but can't seem to get the answer. It's obviously about re-arranging it the correct way but I simply can't...can anyone help?
2. Do you have an answer to this question? If you do i would like to see please
3. (Original post by soup)
Do you have an answer to this question? If you do i would like to see please
Nope, sorry :|
4. (Original post by Sasukekun)
Nope, sorry :|
Oh that's quite annoying but i'll tell you what i did (even though it's probably wrong)

So we know that the final KE of the proton if 4/9 of the initial KE so we can say
4/9(1/2mv^2) = 1/2mu^2
Divide both sides by 1/2m to get
4/9 = u^2 / v^2
Rearrange to get
u = 3/2 v
So by conservation of momentum we know
m1u1 = m1v1 + m2v2 but u1 = 3/2v1 so
3/2 m1v1 = m1v1 + m2v2
1/2 m1v1 = m2v2
m1v1 = 2 m2v2
If we put that into m1u1 = m1v1 + m2v2 we get
m1u1 = 3m2v2
Now we need to find a relationship that involves u1 and v2 so go back to our KE relationships and we find that
5/9(1/2 m1u1^2) = 1/2 m2v2^2
Rearrange to get
u1 ^2 = (9m2v2^2) / (5m1)
Now we square m1u1 = 3m2v2 to get
m1 ^2 u1^2 = 9m2^2 v2^2
Now put in u1 ^2 = (9m2v2^2) / (5m1)
And a bit more rearranging we see
9/45 m1 = m2
m2 being mass of the unkown and m1 mass of proton

Hope this explains it well but it probably wrong
5. (Original post by soup)
Oh that's quite annoying but i'll tell you what i did (even though it's probably wrong)

So we know that the final KE of the proton if 4/9 of the initial KE so we can say
4/9(1/2mv^2) = 1/2mu^2
Divide both sides by 1/2m to get
4/9 = u^2 / v^2
Rearrange to get
u = 3/2 v
So by conservation of momentum we know
m1u1 = m1v1 + m2v2 but u1 = 3/2v1 so
3/2 m1v1 = m1v1 + m2v2
1/2 m1v1 = m2v2
m1v1 = 2 m2v2
If we put that into m1u1 = m1v1 + m2v2 we get
m1u1 = 3m2v2
Now we need to find a relationship that involves u1 and v2 so go back to our KE relationships and we find that
5/9(1/2 m1u1^2) = 1/2 m2v2^2
Rearrange to get
u1 ^2 = (9m2v2^2) / (5m1)
Now we square m1u1 = 3m2v2 to get
m1 ^2 u1^2 = 9m2^2 v2^2
Now put in u1 ^2 = (9m2v2^2) / (5m1)
And a bit more rearranging we see
9/45 m1 = m2
m2 being mass of the unkown and m1 mass of proton

Hope this explains it well but it probably wrong
Hey thanks a lot this seems excellent, although how do you do the bolded bit? I can't seem to see it myself... xD Sorry.
6. Have you substituted u1^2 into m1 ^2 u1^2 = 9m2^2 v2^2 ???
If you have the m1^2 divided by m1 gives m1 and then you just divide by m^2
7. (Original post by soup)
Have you substituted u1^2 into m1 ^2 u1^2 = 9m2^2 v2^2 ???
If you have the m1^2 divided by m1 gives m1 and then you just divide by m^2
I see that now ^^, Thanks very much for your help.

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Updated: November 28, 2010
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