Recent content by Morhas

I think I understand this, but what does it mean physically speaking in terms of mass and the velocity?

Homework Statement http://img29.imageshack.us/img29/4863/12544262.jpg Homework Equations The Attempt at a Solution I don't know if I did this right. If I did, I can't find what meaning the inverse of this functions has :S. Any help is appreciated...

To start, draw a free body diagram of both situations to see all the forces acting on the objects and how they're acting. Then come up with a Fnet statement, by looking at all the forces.

Thats why I wan't to see an attempt at a solution first.

Fg= Fc mg=mv^2/r g=v^2/r v=\sqrt{gr} Eg(initial)= Eg(at top of ramp) + Ek mgh1 = 2mgr + 1/2 mv^2 gh1=2gr + 1/2 v^2 gh1= 2gr +1/2 rg (subbed in earlier value of v) h1=2r +1/2r h1= 2.5r h1= 2.5(0.15) h1=37.5cm Maybe? The ball will have gravitational potential energy AND kinetic energy, in the...

Yeah, either way it doesn't make sense. The height has to be at least 0.3m. Hmmmm I like this. I think I have it 1 sec.

OK, so at the top of the loop Fg <= Fc for it to not fall. So if we say... Fg= Fc mg=mv^2/r g=v^2/r v=\sqrt{gr} Once we know the velocity it needs to reach at the top of the loop... Ek(at top of loop)=Eg(at top of ramp) 1/2mv^2 = mgh v^2=2gh \sqrt{gr}^2=2gh gr=2gh r=2h h=r/2 h=0.3/2 h=15cm...

It looks like since you arn't give any actual values in the question, you can only answer the question in terms of the variables given. Try using the formula for completely inelastic collisions. M1v1+m2v2 = (m1 + m2) V' OR (mass 1)(initial velocity of mass 1) + (mass 2)(initial velocity of...

All you've done is told us a story about a hockey puck. What is the question? (Unless I'm blind) :smile:

Ahhh I see. So What I've done is drawn the two vectors as two sides in a triangle, with the third side being the initial momentum vector. Then I solved with sin law. Thanks for the post!

Bump. Can anyone explain this? My friend and I have a test tommorow and we still don't see how this is possible. Thanks.

Homework Statement Two hockey puck of equal mass undergo a collision on a hockey rink. One puck is initiall at rest while the other is moving with a speed of 5.4m/s. After the collision, the velocities of the pucks make angles of 33 and 46 (degrees) relative to the initial velocity of the...

So if I lift a brick a foot off the ground diagonally over a 1000m distance, it would take the same amount of work to lift it straight up? I have done the same work vertically, but I have also done fd horizontally. (Sorry that was poorly worded)

I'm afraid I don't understand that argument.The horizontal velocity will remain constant, but I don't see that being relevant since a force was still required to start it. While the additional work done didn't achieve any gravitational potential energy, additional work was done.

The question didn't even specify it was a frictionless system, so there would be air resistance as well. @Lowly Pion: I'm not sure I understand. Although work is still being done and converted to gravitational potential energy, wouldn't an additional force have to be applied to start...