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Conservation of momentum

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?
Reply 1
Avatar for soup
soup
11
Do you have an answer to this question? If you do i would like to see please
Reply 2
Avatar for Sasukekun
Sasukekun
OP
7
Original post by soup
Do you have an answer to this question? If you do i would like to see please


Nope, sorry :|
Reply 3
Avatar for soup
soup
11
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
Reply 4
Avatar for Sasukekun
Sasukekun
OP
7
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.
Reply 5
Avatar for soup
soup
11
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
Reply 6
Avatar for Sasukekun
Sasukekun
OP
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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