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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Apr 2007 00:13:41 IST
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I chckd in solution of Narayana. Its C
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"If you win, you shall not have to explain and if you lose, you wont be there to explain"
~ Adolph Hitler |
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9 Apr 2007 00:42:49 IST
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r u studing in narayana? i am studing in narayana rohini branch
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Beat others otherwise they will beat u |
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9 Apr 2007 14:07:02 IST
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The answer is B.Since when elastic collision takes place both K.E and momentum are conserved.But only using conservation of linear momentum u cannot prove.
ANS:B
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9 Apr 2007 14:34:55 IST
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i think the answer is B....
the first statement is correct....
when u divide the equation that u get by using conservation of momentum by the equation that u get by using conservation of energy(kinetic) u'll get the first statement.....
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9 Apr 2007 14:48:05 IST
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hey guys check on the narayna site .the ans given is (d)
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9 Apr 2007 16:43:22 IST
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then narayana is wrong bcoz its a very simple concept that in elastic collision value of coffiecient of restitution is 1 e=1 and e=V2-V1/U1-U2
as e=1 so V2-V1=U1-U2
which is statement 1 so it is right
secondly althrough linear mommentum remain conserved in every type of collision including elastic collision but that does not means relative velocity before collision will be equal to relative velocity after collision
so right answer is B bcoz both statements are correct but STATEMENT2 does not explain STATEMENT2
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9 Apr 2007 20:08:53 IST
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what will be aggregate cut off this year? can any 1 estimate ?
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10 Apr 2007 01:47:26 IST
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B is correct.....
since
velocity of seperation = e (velocity of approach)
since e=1
so statement 1 is correct
2 is also crrt
but 2 does not imply 1
some people have this doubt that e=1 only when the collision is head on and elastic but when the collision is oblique, then will e be equal to 1?????
ans: e is a material property so it only depends on the material and NOT whether collision is head-on or oblique.
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Truly Mittal
B.Tech IIT Guwahati
trulymittal@gmail.com |
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10 Apr 2007 10:36:07 IST
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Even I think it is D.
You must stress on the word relative velocity.
After a collision, the relative velocity of one w r t the other is reversed. Therefore, the relative velocities before and after the collision is not the same.
So, Statement 1 is untrue.
But Statement 2 is true for all types of collisions, whether elastic or inelastic.
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10 Apr 2007 17:37:18 IST
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hey , can sumbody tell me how Statement-2 is NOT a correct explanation for statement-1 ( in option B) ?
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10 Apr 2007 17:38:46 IST
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FIITJEE solutions
these links worked for me, i hope they work for you all tooo
PAPER 1
http://www.fiitjee.com/down/iit071.pdf
PAPER 2
http://www.fiitjee.com/down/iit072.pdf
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10 Apr 2007 17:42:57 IST
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wrong zaherva in elastic collision the velocity of aproach is always equal to velocity of seperation
This come from this two eq
Total KE initial = total KE final,..................1
linear momentum before collision =linear momentum after colision......2
in eq 1take the terms of M1 on 1 side and of M2 on other and then divide 1st eq from eq2
you will get V2-V1=U1-U2 which is wat STATEMENT 1is saying hence correct answer is B as STATEMENT 2alone can"t explain this
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10 Apr 2007 17:46:19 IST
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@catch_arnnie: Conservation of momentum takes place in ALL collisions, whether elastic or inelastic. Conservation of kinetic energy takes place only in elastic collisions, though. It is this conservation of KE that leads to the conclusion that the relative velocity of approach is equal to the relative velocity of separation.
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10 Apr 2007 20:48:45 IST
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statement 1 is wrong bcoz they are talking abt the relative speeds and not the relative velocities
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--
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10 Apr 2007 20:53:46 IST
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