1. Nov 10, 2006

### MCHammer

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 in which two balls are relased from a similar incline (same height/ angle to the horizontal and the same length). Using the Energy principles discuss which ball will reach the end first.

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My two cents: As the inclines in the second road are equal, and the downward force given by the first small incline will be taken away by the second upward incline, therefore it will have no difference and they will reach the bottom at the same time. Am I right? Please go easy on me I am a noob. :D

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2. Nov 10, 2006

### Staff: Mentor

What determines the speed of the ball? (Think energy, just like it suggests.) Which ball is moving faster most of the time?

3. Nov 10, 2006

### MCHammer

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 right? :)

4. Nov 11, 2006

### Staff: Mentor

Oh, really? Certainly mg is the same, but is h the same for both balls?

5. Nov 11, 2006

### PhanthomJay

I believe the figure is misleading. From the problem statement, it would appear that they both start at the same height h and end at the same height 0, but that the second ball dips and deviates to the low road then rises back to the high road before reaching bottom. In which case the problem is just as interesting. They would end up with the same speed at the bottom, as per the stated conservation of energy principle, but which ball wins the race, if either?
( Hammer, I enjoyed your music many years ago. Still rapping?)

Last edited: Nov 11, 2006
6. Nov 11, 2006

### Staff: Mentor

You're correct that the problem does state that both balls start and end at the same height, so the figure should match.

But the only thing that matters is change in height, not absolute height. Balls rolling down parallel tracks, but one higher than the other, will have the same speed at all points, so you get the same answer.

7. Nov 11, 2006

### PhanthomJay

Yes, but I'm thinking along the lines of say a ball on an incline starting at height h, versus a ball on a say an up an down roller coaster, also starting at height h,where they don't have the same speed at all horizontal points, but they end up at the same speed, but one of them wins, and it is not all that clear to me for a specific case which one wins.

8. Nov 11, 2006

### MCHammer

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.

9. Nov 11, 2006

### PhanthomJay

No-ooo.For when that ball on the second track takes a dip, theta increases, thus it's acceleration, and hence its speed, increases. The more in line the motion is with the vertical acceleration of gravity, the greater its speed. So even though the path taken by the second ball is longer, that does not necessarily imply it will take longer. Will it?? Won't it? Or is it a tie ballgame??

10. Nov 11, 2006

### QuantumCrash

Hmmm. On one hand one ball has to travel further. But on the other hand the same ball will travel faster. But does it really mean that they will reach the end at the same time? The same speed at the end does not automatically mean the same time does it?

11. Nov 11, 2006

### PhanthomJay

That is right, it does not. Like if you had a coaster with a lot of dips and rises, dips and rises, resulting in fast speeds then slower, fast then slower, etc., it would still arrive at the bottom with the same speed as one on an incline that started at the same height. But it would take a lot longer to get there. On the contrary, if you has a coaster with a parabolic curve, no rises, of the same height as the incline, it would take faster to get there, even though it had to travel a longer path, arriving at the same speed. But now back to Hammer's question, well???? (If you're looking for an answer from me, I finally figured it out based on Doc Al's hint on the very first response. Thanks, Doc.).

Last edited: Nov 11, 2006
12. Nov 11, 2006

### MCHammer

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?

13. Nov 11, 2006

### Staff: Mentor

Yes, that's the question all right. What do you think? What can you conclude about the relative speeds of the balls? (They both start out with the same speed, but do they keep the same speed?)

14. Nov 11, 2006

### MCHammer

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 it will take the same amount of acceleration to get off that bump). Which tells me nothing. Maybe it will help you guys.

(It is a general graph. Don't kill me)

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15. Nov 12, 2006

### Staff: Mentor

When the ball falls into the ditch does it speed up or slow down?

16. Nov 12, 2006

### MCHammer

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 travel at the same velocity as the upper ball?

Well It seems like we are answering questions by other questions... lol. Well I am getting desperate.

Last edited: Nov 12, 2006
17. Nov 12, 2006

### MCHammer

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. Thus it reaches the end first (second ball).

18. Nov 12, 2006

### PhanthomJay

You are on the right track. Maybe this example will help: Consider two cars going from A to B on a level road. The first one starts at A at 5mph,and moves at constant speed, and ends up at B traveling at 5mph. The second one starts at 5mph, then guns it to 60, then slows to 5 mph as it reaches point B. So both cars have the same start speed and the same end speed. But you know darn well that the 2nd car gets there first. So back to Doc Al's hint way back when: Which ball is moving faster most of the time? Or, which one has the greater average speed?

19. Nov 12, 2006

### MCHammer

Does anyone have a formal answer or I would have to beat around the bush for another day? (he he bush...)

20. Nov 12, 2006

### BobG

You know the kinetic energy depends on the ball's location.

$$ME=1/2mv^2_f - 1/2mv^2_i + mgh_f - mgh_i$$

Both balls had the same height and velocity when the second ball entered the dip. Both balls had the same height and velocity when the second ball exited the dip.

You only have to worry about what's happening while the ball is in the dip.

You also know that:
$$v_f = v_i +at$$

Since velocity was the same entering the dip and exiting the dip, the at for both balls has to be equal.

The first ball has a constant acceleration. The second ball has a higher acceleration upon entering the dip, but the acceleration decreases as the ball exits the dip. It will be similar to letting one ball roll down a ramp while a second ball swings on a pendulum from the start of the ramp to the end of the ramp.

Last edited: Nov 12, 2006