Just to begin, uhh. I'm 75% sure my work makes logical sense [I hope], and that my problem lies in a computational error. But I really don't know why my answers aren't working out. I've now gone through the problem in 3 different ways, and I keep getting the same answer, but it's apparently still incorrect. If it is an arithmetic error, sorry for making you read through all of this to fix such a stupid mistake >.> 1. The problem statement, all variables and given/known data A 12.7g object moving to the right at 26.5cm/s overtakes and collides elastically with a 10.5g object moving in the same direction at 10.7cm/s. Find the velocity of the faster object after the collision. Answer in units of cm/s. Find the velocity of the slower object after the collision. Answer in units of cm/s. 2. Relevant equations m1v1 + m2v2 = m1v3 + m2v4 .5m1v1^2 + .5m2v2^2 = .5m1v3^2 + .5m2v4^2 And, sorry in advanced for the ugly way that equation appears. And for the work that is likely to look just as messy. 3. The attempt at a solution m1 = 12.7g ; v1 = 26.5cm/s m2 = 10.5g ; v2 = 10.7cm/s I plugged those two into the first equation and got: 448.9 = 12.7v3 + 10.5v4 Rearranged for v3: v3 = 35.35 - 0.827v4 I then plugged in the m1, v1, m2, and v2 into the second equation and got: 5060.36 = 6.35v3^2 + 5.25v4^2 Plugged in the v3 from the previous equation: 5060.36 = 6.35 ( 35.35 - 0.827x )^2 + 5.25 x^2 AND HEREEEE I FIND MY PROBLEMO. My TI-89 keeps giving me two answers: v3 = 28 [in which case v4 = 8.89] or v3 = 10.7 [and thus v4 = 29.8] Problemo Uno: Why are there two answers? ? And both seem to work when plugged back into the original equations? This point makes me most uncomfy and confused. Problemo Dos: If it's an elastic collision, shouldn't one of the objects be moving in a negative direction now? Perhaps my thought process is wrong here though, but idk. ///////////////////// Thanks in advanced to whoever offers help yay. And I know I'm a noob leeching off these forums, so just so I don't feel like I'm nomming up the answer and running away, I'll stay around to help someone else if I can xD.