Recent content by rie_06

in solving .280(v1) = .280(v1)' + m2(0.5v1) to get m2 by itself - i divide both sides by 0.5v1 .14 = .14 + m2 m2 = 0 doesn't seem right? what am i doing wrong?

i meant you are correct on the 0.5v1' Thank you

you are correct - let me work it some more and see what i come up with - thank you very much!

ok - let me try it without cancelling the mass m1v1 = m1v1' + m2v2' .280(v1) = .280(v1') + m2(.5v1) is that how i should proceed?

thanks for suggesting using v1' and v2' I did use them in my equation but forgot to mention them in the post so m1v1 = m1v1' + m2v2' i then get v1 = v1' + .5v1 then .5 = v1' does that make sense? should i continue with the next equation from here - or do i need to find the mass of the 2nd...

i agree - that's where my confusion is! The question in the book does not state a velocity for the first ball.

280kg ball has elastic collision with 2nd ball which is initially at rest. Then 2nd ball moves with .5 velocity of 1st ball - what is mass of the 2nd ball? what percentage of KE gets transferred to 2nd ball Homework Equations I believe i use m1v1 + m2v2 = m1v1 +m2v2 (before & after)...