Recent content by gutupio

Thank you so much, I've been wracking my brain over this problem for 5 days. It feels so good to have an answer.

If i do that, then the horizontal component of the moment becomes m1v1cos(180) + m2v2cos(180) = 0, or -m2v2=m1v1. So they have the same magnitude, just opposite directions. I solved that equation for v1=(-m2v2)/m1. Then, with conservation of K.E. I did Kball,i + Kwedge,i = Kball,f + Kwedge,f...

This force is the vertical component of the force of the ball striking the wedge right? So it'd be mballgsin(45). It's also the same as the horizontal force, shich would cause the movement, at a Force of 20.81 N. If you divide by the mass of the wedge, you get 4.162m/s^2, which would be the...

To be more clear, if i do: m2<v2 cos(180), v2sin(180)> + m1<v1cos(45), v1sin(45)> = m1<0, 4.50> since cos(180) is -1, and sin(180) is 0, this leads too: m2<-v2 ,0> + m1<v1cos(45), v1sin(45)> = m1<0, 4.50> This leads to two equations: -v2m2 + m1v1cos(45) = 0 m1v1sin(45) = 4.50 These numbers...

Homework Statement A ball is dropped straight down onto a wedge that is sitting on top of a frictionless surface. The wedge is at an angle of 45degreees and has a mass of 5.00kg. The ball has a mass of 3.00kg. If the ball collides with the wedge at 4.50m/s, and the collision is instantaneous...