M2 Collisions Help required Please... Watch

bruceleej
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#21
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#21
(Original post by sohanshah)
so the 3 i get are:

u1-v1=e(V-u)

V+u=v1+u1

I=mu1-mu

is this right???
Yes thats right, now time for some algebra. Eliminate v1 and u1
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Kyalimers
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#22
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#22
lol tht is difficult
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Kyalimers
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#23
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#23
lol are you sure these equations are correct because I get something humongous:

2I=[(e(V-u)+V+u)m]/2mu
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bruceleej
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#24
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#24
Lol, easiest way is to sub for V1 using the first one into the second.
You should get
Spoiler:
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 (v+u)=[2u_1 - e(v-u)]
.
Then sub for u1 using the momentum equation into the equation you get from doing the above.
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bruceleej
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#25
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#25
(Original post by sohanshah)
lol are you sure these equations are correct because I get something humongous:

2I=[(e(V-u)+V+u)m]/2mu
You must have made a slight mistake somewhere.
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Kyalimers
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#26
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#26
I'm a -u out in the brackets - (so close yet so far)

I have

I= 0.5m(V+u+e(V-u)-u)

It should be....

I= 0.5m(V+u+e(V-u)-2u)
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Kyalimers
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#27
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#27
got it - thankyou so much bruceleej

You are a legend - I will rep you as much as poss over next week lol - please remind me to as well - you deserve it...
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bruceleej
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#28
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#28
Its cool
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bruceleej
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#29
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#29
Now for question 8, Im still trying to eliminate the v1's and the v2's..lol...So I will get back to you on that one cant see a nice way to do it yet{but I think using the difference of 2 squares somewhere in the Loss in KE equation might help to split the terms}.Hmm lets see.
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Kyalimers
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#30
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#30
lol im waiting........ lol joking
Leave it if u cnt do it - uve done enough lol

I get upto

2E=m1u1^2+m2u2^2-m1v1^2-m2v2^2

and

v2-v1=e(u1-u2)

BTW YOU ARE VERY VERY VERY CLEVER... u at uni doing maths???? (take it as a compliment please-no offence intended)
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bruceleej
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#31
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Why, thank you. Ohh and No Im not doing maths at Uni. Im on a gap year gonna do Electronic & Electrical Eng next academic yr.
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Kyalimers
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#32
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#32
Wow cool. I wish one day I can do maths like you can

Unfortunately I'll be dead by then or I'll be studying all day every day for a hundred years lol
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Kyalimers
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#33
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#33
i found the third eq. i think: u1m1+u2m2=v1m1+v2m2
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bruceleej
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#34
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#34
ohh yea thats right.
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Kyalimers
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#35
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#35
algebra again lol
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Kyalimers
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#36
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#36
I reckon I'll challenge my teacher to simultaneously solve these 3 lol

Thanks so much again

Without you today i would have died
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bruceleej
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#37
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#37
(Original post by sohanshah)
I reckon I'll challenge my teacher to simultaneously solve these 3 lol

Thanks so much again

Without you today i would have died
Lol, that should be quite interesting..Thats if he hadnt done so in the past and has a little note somewhere with the solutions.

But anyway, I managed to complete 8;
Spoiler:
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Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 646x697
e=\frac{v_2 - v_1}{u_1 - u_2} \Rightarrow e(u_1 - u_2)=v_2 - v_1 \, \, \, \, (\mathrm{I}) \\



\mathrm{cons. of mom.} m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\Rightarrow (u_1 - v_1) = \frac{m_2}{m_1}(v_2 - u_2) \, \, \, \,(\mathrm{II})\\



\mathrm{For Loss in KE}, E = 0.5m_1({u_1}^2 - {v_1}^2) + 0.5m_2({u_2}^2 - {v_2}^2) \,\, \, \, (\mathrm{III})\\





\mathrm{Sub (\mathrm{II}) in (\mathrm{III})}:

2E=m_1(u_1 + v_1)\frac{m_2}{m_1}(v_2 - u_2) + m_2(u_2 - v_2)(u_2 + v_2)\\

2E = m_2[(u_1 + v_1)(v_2 - u_2) + (u_2 - v_2)(u_2 + v_2)]\\

2E= m_2(v_2 - u_2)[(u_1 + v_1) - (u_2 + v_2)]\\

\mathrm{Sub in for v_2 using (I)} \Rightarrow \, 2E=m_2(v_2 - u_2)[u_1 - u_2 - e(u_1 - u_2)] \,\,\,\,\,\,\mathrm{(IV)} \\



\mathrm{Now From (I) and (II)}; (u_1 + e(u_1 - u_2) - v_2) = \frac{m_2}{m_1}(v_2 - u_2)\\

\Rightarrow \,\, m_1[u_1 + e(u_1 - u_2)] - m_1 v_2 = m_2 v_2 - m_2 u_2\\

m_1 [u_1 + e(u_1 - u_2)] + m_2 u_2 = (m_1 + m_2)v_2 \,\,\,\ \mathrm{(V)}





\mathrm{Sub (V) in (IV)}; 2E = m_2(\frac{m_1 u_1 + m_1 e (u_1 - u_2) + m_2 u_2}{(m_1 + m_2)} - u_2)[u_1 - u_2 - e(u_1 - u_2)] \\



\Rightarrow 2E=m_2(u_1 - u_2)(1-e)[\frac{m_1 u_1 + m_1 e (u_1 - u_2) + m_2 u_2 -u_2(m_1 + m_2)}{(m_1 + m_2)}\\



2E= m_2(u_1 - u_2)(1-e)[\frac{m_1 u_1 + m_1 e (u_1 - u_2) - m_1 u_2}{(m_1 + m_2)}\\

2(m_1 + m_2)E= m_1 m_2 (u_1 - u_2)(1-e)[(u_1 - u_2)(1+e)]\\

\therefore \, 2(m_1 + m_2)E= m_1 m_2 (u_1 - u_2)^2(1-e^2) \mathrm{Q.E.D}
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Kyalimers
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#38
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#38
wow bruceleej lol Thats quite something. U r amazing lol - I must have kept u busy doin that q lol

Thnx
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Kyalimers
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#39
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#39
I understand hardly any of it lol

I lose you nearly immediately after you start subbing the 3 equations lol

But maybe I can maintain some pride after having seen my teacher do this inside 7 minutes flat. he showed us not the working lol as he thinks this is a capable challenge for a 15 yr old..... hahah lol thankyou so much
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