Recent content by areverseay

Yes, I think we had different views of which orientation everything was. But, I've figured out the correct answers for all of the parts, so it's fine. Thanks for both of your help, explicitly writing the equations for conservation of momentum did indeed help a lot!

Both of these replies are very helpful, but I'm still stuck on the horizontal component. So, the vertical component of the momentum of B is the same as the original momentum of A, since there is no vertical component in the new momentum of A, as cos90 = 0 / they are perpendicular. But how do you...

Thanks for this, I've made this post just the first question. In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't...

vA = 3u/4 and vB = u/4, and 1/8 KE is lost. I can't get to these answers however: for the first part, I got to u = vA + 3vB using conservation of momentum, and the fact that particle B is at an angle, hence I would think its momentum should be 10mvBsin(arcsin(3/5)). Doing the same for A with...