A question about momentum, thank you in advance. The concept may be easy but perhaps I missed out on something, in attached file, thanks in advance, The answer is A.
Do you know what a) momentum is? b) Principle of conservation of momentum is? c) Elastic and non elastic collisions are? if yes .. answer briefly then ll help u :) if no.. better read up on them before u make an attempt to solve this question
Yes I do know all of these (: Thankssss I do appreciate that
@Mashy
can u tell me briefly each of them? :P
I know that the principle states that M1u1 + M2+u2 = M1v1 + M2v2 U= velocity before collision and V= velocity after collision In inelastic collisions the Kinetic energy is not conserved, only momentum is In Elastic collisions KE AND momentum are both conserved :33 Lmao yeah
ohww. ok.. cool :P so now.. so lets apply this shall we. can u tell me what is the total momentum of the two spheres before collision? .. lets take right as positive!
Okay ^-^
2mu + mu = The total momentum before collision (: Am I right?
nope.. right is positive.. so the sphere one has +2mu and spehere two (which is coming left ) has -mu.. so whats the total?
Awwwwww !!!
I guess that's the only thing I ALWAYS missed out xD haha Thank you, the total is , 2mu-mu ::))
so , 2mu-mu = 1mu....?
yes.. correct.. so we know that the final moentum must also be +mu next.. what is the initial k.e? the final k.e should also remain the same.. do this check to all the 4 cases..
So the KE is conserved too>? It is an elastic collision then , right?
yea.. thats mentioned in the problem :P
Oh god. Hahahaa xD YES IT IS. Lmao
0.5 X 2m X u^2 + 0.5 X m X u^2
ll brb.. try continuing.. dinner time.. and yes.. that is the initial total k.e.. just add them
All right enjoy your meal and TYT (: I will work on it and I will leave the answer here :) Thanks a lot.,
enjoy ur dinner ;)
I can't add them properly god, But 3mu^4..... Does that make any sense ._. , Lol, I tried :33
And half + half = 1 xD
\[2\frac{m u^2}{2} + \frac{m u^2}{2} = 3 \frac{m u^2}{2}\]
Oh okay so that's how it is
I get what you did up there, but the next step is to find the after the collision values right?(:
the total k.e. after the collision must also remain the same.. right? so now you can check for each case, what is the momentum and what is the final k.e the one in which the momentum = initial value, and the k.e. = initial value is the answer
Yes the total KR after the collision will also be 3 mu^2/2 And I get you but I am sorry what do you mean by initial value? As in before the collision?
Ke*
yes.. initial value.. as in the momentum and ke we JUST calculated.. REDO The calculations for all the cases AFTER THE collisions
Awwww Okay I will do so , thanks (:
np , ur wlc :D
:)
Hold up ,.. But I already got the answers for KE and momentum I mean after the collision the momentum is (mu) and the KE is ( 3 mu^2/2 ) so thats after the collision but how do I put them together in order to get answer A..??
@Mashy
you re do the calculation . like for example in case d the total momentum = 3m (u/3) = mu .. so momentum is conserved total k.e = 3m (u^2/9) = mu^2/3.. (which is not as much as before which was 2mu^2/3) thus k.e is not conserved.. hence d cannot be the option similarly try with others
Awwwwww now I get it all right no more questions thanks A LOT. :)))))) You meant compare with the options (: Have a nice day!
yea yea :D.. besides its already 11 30 pm.. so ll have a good night.. good day to you :P
Aww good night it is 9:52 PM where I am haha (: Sleep tight!
:o its 8:52 here , where are u from @MonsterIsEnergy
@ikram002p I am from Iraq (: But I do not live there so yeah (:
oh :o where so u live ?
Dxb
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