Recent content by Sjoyes

Right! Missed that! Thanks again for all your help.

OK, now that I have the angle of v'2, I think I know how to figure out v'1 and v'2; Using conservation of momentum... In the x direction m1v1 + m2v2 = m1v'1 + m2v'2 (2)(3) + (2)(0) = (2)v'1cos(30) + (2)v'2cos(60) 3 = v'1cos(30) + v'2cos(60) In the y direction m1v1 + m2v2 = m1v'1 + m2v'2...

Well I was thinking that the 2ABcos(angle) term had to be 0 because the rest of the equation (C^2 = A^2 + B^2) is analogous to the conservation of NRG eqn ( (1/2)mv1^2 + (1/2)mv2^2 = (1/2)mv'1^2 + (1/2) mv'2^2 ) and because of that, 2ABcos(angle) had to be zero. But the only for sure way that...

Yes, sorry, just a typo. So since the 2(mv'1)(mv'2) cos (30deg + alpha) is 0... then that just leaves me with mv1^2 = mv'1^2 + mv'2^2... but then I still have two unknowns. So I would need another equation to substitute into that to figure out one of the unknowns. Now I am confused again...

OK, so since sin^2(angle) + cos^2(angle) =1 is in both A and B terms, then everything would simplify to: mv1^2 = mv'1^2 + mv'2^2 - 2(mv'1)(mv'2) cos (30deg +theta)

That would be Pythagorean theorem... c^2 = a^2 + b^2 so the first term (on the right) would become |sgrt((mv'1cos(30deg))^2 + (mv'1sin(30deg))^2)|^2... and so on... also in the second half of your equation (the -2AB...), is the theta referring to the 30 deg? or the 30 deg +alpha?

OK, I think I understand a little better now? Just like in kinematics, adding vectors A and B tip to tail, will be equal to vector C, and like you said, C will be equal to the vector of the first ball because of conservation of momentum. So in the x direction (v1i)^2 = (v1fcos30)^2 +...

I am so very lost. I don't think I understand where you are trying to hint me towards using the law of cosines. Is it for the whole system, or just one isolated vector? i.e. C = v1 or am I just using it to find v1' and A and B will be the x and y component of v1'? Also is this referring to...

So are you saying that: v1^2 = v1'^2 + v2'^2 - 2(v1')(v2')cos(theta) ?? then would v1' and v2' just be the rearrangement of the x and y components of the momentum eqns in the Pythagorean theorem? i.e. v1' = sqrt(((v1 - v2'cos(theta))/cos30)^2 + ((v2'sin(theta))/sin30)^2) and v2' =...

Homework Statement A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown. If the first ball moves away with angle 30 ° to the original path, determine a. the speed of the first ball after the collision. b. the speed and direction of the second...