Elastic momentum problem

1. http://i1269.photobucket.com/albums/jj597/bubakazouba/fd9bfcc2.png [Broken]
2. m1u1+m2u2=m1v1+m2v2
3. I just don't get i know that in an elastic collision velocity of approach=velocity of seperation it can be C or D why is the answer D?

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Elastically = kinetic energy is conserved.

CAF123
Gold Member
1. http://i1269.photobucket.com/albums/jj597/bubakazouba/fd9bfcc2.png [Broken]
2. m1u1+m2u2=m1v1+m2v2
3. I just don't get i know that in an elastic collision velocity of approach=velocity of seperation it can be C or D why is the answer D?
An elastic collision is one in which kinetic energy is conserved. For c) we have the eqn: $$\frac{1}{2}mv^2 = \frac{1}{2}m(\frac{1}{4}v^2) + \frac{1}{2}m(\frac{1}{4}v^2),$$ and this gives, cancelling out m since mx and my are the same;$$\frac{1}{2}v^2 = \frac{1}{8}v^2 + \frac{1}{8}v^2$$ which doesn't make sense.

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yeah i think i got it
what just confuses me is that the law of conservation of KE:
$\frac{1}{2}$m$_{1}$u$_{1}$$^{2}$+$\frac{1}{2}$m$_{2}$u$_{2}$$^{2}$=$\frac{1}{2}$m$_{1}$v$_{1}$$^{2}$+$\frac{1}{2}$m$_{2}$v$_{2}$$^{2}$
is used to derive the Relative velocity Rule:
(u$_{1}$-v$_{2}$)=-(v$_{1}$-v$_{2}$)

so why does the relative velcotiy rule apply to answer C while the law of conservation of KE doesn't?