# Elastic momentum problem

bubakazouba
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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## Answers and Replies

voko
Elastically = kinetic energy is conserved.

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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bubakazouba
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?
and Thanks for reply :)

Gold Member
The collision is elastic as said in the question. This means the kinetic energy of the two body system before the collision is equal to the kinetic energy of the system after the collision. So by a method similar to that in my previous post, you can eliminate a) as well.
Now from the equations that you can derive (by solving simultaneously the two eqns given by conservation of momentum and kinetic energy conservation) you see that the velocity of body x and body y are exchanged, ie d) is correct. So essentially the key word is 'elastic' here.

bubakazouba
Yes I finally understood it,thank you very much CAF123 :)