Well I did the velocity time graph, there is a hump in the middle of the graph for the ditch in it. So the displacement under that hump is more than the displacement under the normal straight triangle+rectangle. Therefore I conclude that it has covered more distance in the same amount of time...
Goes up, and then the acceleration becomes the same as the upper ball, ie a=gsinθ, but with a different initial velocity, therefore yielding a higher final velocity as well. But what the problem is that does it lose that initial impulse of high velocity when it comes up the ramp again? Does it...
I made a VT graph but doesn't really say anything. The first without the ditch, which comes out to a straight line due to the accelration being constant, ie. a=gsinθ. The other has a bump in it for the extra acceleration caused due to the extra slope.,then the bump is eliminated (thinking that...
Well I still cannont figure it out. It's rather more confusing now... Simple question which one arrives there first? Or do they arrive there at the same time?
Yeah, I suppose the figure was misleading. The tracks can be thought of as being parallel and equal in length, except one of them has that dip factor. The height is the same as well. And if one comes to think of it the net acceleration will be the same as well, as a = gsinθ.
I was thinking of something around the lines of energy conservation.
1/2mv^2 = mgh
Since mgh is the same for both the balls, v, has to be the same as well. If they are traveling at the same velocity and covering the same distance, then they must arrive at the end at the same time. Am I...
Hi! This is my first post so hello to everyone. :) My name is john, and no I don't like big rear-ends.
Anywho back to the burning question that I have. I think this most likely has been beaten to death before, but none the less here it goes:
This is the classic high road low road question...