Ball falling on to parabolic track

AI Thread Summary
A 50 g ball is released from a height of 1.0 m and rolls down a 30-degree incline before moving up a parabolic track described by y = 1/4x^2. The discussion revolves around the conservation of energy principles, specifically kinetic and potential energy equations. The user expresses confusion about determining the final height using the parabolic equation, while also considering the effects of gravity on an inclined plane. Ultimately, they conclude that the final height remains 1 m, regardless of the track's shape, due to energy conservation. The problem highlights the relationship between potential energy and height in different trajectories.
jemstone
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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.
 
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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.
 


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?
 


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.
 
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