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November 29, 2014

November 29, 2014

Posted by **collisions** on Sunday, July 3, 2011 at 5:30pm.

- physics -
**Damon**, Sunday, July 3, 2011 at 6:32pmNot having your diagram, I do not know if the first mass is at 50 degrees from the y axis in quadrant 1 or quadrant 2

Therefore I can only tell you what to do

X momentum before = .15 * .9 cos40 (or -cos 40) + .26 *.540

=

X momentum after = .15*.64 cos 11 +or -.26*V2 cos phi

Y momentum before = .15*.9 cos 50 + 0

=

Y momentum after = .15*.9 sin 11 - .26*V2 sin phi

and

(1/2).15 (.9)^2 + (1/2).26(.54)^2

=

(1/2).15(.64)^2 + (1/2).26(V2)^2

- physics -
**collisions**, Sunday, July 3, 2011 at 7:08pmits in quadrant 2, going in negative y direction. i already solved the answer though thanks!

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