# Ball falling on to parabolic track

1. Oct 25, 2008

### jemstone

1. A 50 g ball is released from rest 1.0 m above the bottom of the track. It rolls down a straight 30 degree segment, then back up a parabolic segment whose shape is given by y = 1/4x^2, where x and y are in m.

There is a picture shown too, but I can't figure out how to get it in the post. I figure that enough information is given to understand the problem.

2. KE = 1/2 mv^2
PE = mgy
KEf + Ugf = Ki + Ugi

3. I really don't know what I'm doing, but we know that:
m=.50 kg
yi = 1
xf = 1 / (tan30) = 1.73
I think the question is asking me to find the final y which I could find from the equation of the parabola, but then I need x. I'm just really lost.

2. Oct 25, 2008

### JoAuSc

Re: Energy

You weren't too clear on what the question states, so I'll just mention that the potential energy is the same for any x, if we have the same y. Also, when a ball rolls to the top of a hill or something, its kinetic energy must be zero because, for a brief moment, it isn't moving.

3. Oct 25, 2008

### jemstone

Re: Energy

I think I'm having a hard time approaching this problem. I know that I'm looking for the final y value.

I also am thinking that since this has an inclined plane, the gravity isn't going to be g, but gsin30 (but it's possible that I am making this more complicated than it is).

So maybe we use:
(1/2)mvf2 + mgsin30yf = (1/2)mvi2 + mgsin30yi
0 + (.05 kg)(9.8)(sin30)yf = 0 + (.05 kg)(9.8)(sin30)(1m)
.245yf = .245
yf = 1

So that means that I'm not using the parabolic equation and the final height is the same as the initial height? Does this make sense?

4. Oct 26, 2008

### jemstone

Re: Energy

Ok I think I figured it out. I made it harder than it should be. Regardless of the shape of the trajectory, the object will go to the same height, 1m.