Potential Energy, Conservation of Energy problem


by pezzang
Tags: conservation, energy, potential
pezzang
pezzang is offline
#1
Nov7-03, 07:58 PM
P: 29
Hi, I need an explanation for the following question. I also wrote down my answer. TELL me what you think and give me clues to solve this question.

There is a picture of two tracks(lines) that are horizontal and parallel to each other. In the middle of tracks, both have "bumps," Track A has concave down bump, Track B has a concave down bump.

(a) Two smooth tracks of equal length have "bump" - A up, and B down. Both "bumps" have the same curvature. IF two balls start simultaneously with the same initial speed, will they complete their journey at the same time or not? If not, which will arrive first? Explain your answer clearly.
-> My guess is that both tracks will complete their journey at different time. But, I am not exactly sure why and how they will complete at different time. Please teach me.

(b) If the initial speed of the balls is 2m/s and teh speed of the ball at teh bottom of the curve on track B is 3m/s, will the speed of the ball at the top of the curve on Track A be greater than, less than, or equal to 1m/s? Explain your answer clearly.
-> The speed will be exactly 1m/s because w = mgh can be applied. Because the mass, height and g(9.9m/s^2) are the same in both cases, the answer will be 1m/s.

Am I right??? Or am I totally ignorant?
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Ambitwistor
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#2
Nov7-03, 08:23 PM
P: 837
Originally posted by pezzang
(a) Two smooth tracks of equal length have "bump" - A up, and B down. Both "bumps" have the same curvature. IF two balls start simultaneously with the same initial speed, will they complete their journey at the same time or not? If not, which will arrive first?
Use conservation of energy to determine what happens to the speed of a ball when it increases or decreases in height.


(b) If the initial speed of the balls is 2m/s and teh speed of the ball at teh bottom of the curve on track B is 3m/s, will the speed of the ball at the top of the curve on Track A be greater than, less than, or equal to 1m/s? Explain your answer clearly.

-> The speed will be exactly 1m/s because w = mgh can be applied. Because the mass, height and g(9.9m/s^2) are the same in both cases, the answer will be 1m/s.
Well, in both cases, the work done has the same magnitude (but opposite sign). However, the work done is the change in kinetic energy, which is proportional to the change in the square of the velocity. So you can't say that there is the same magnitude of change in the velocity; there is really the same magnitude of change of the square of the velocity.

I may be misunderstanding the problem, however. I interpreted it like this: the balls are initially at zero height with speed 2 m/s. The ball on track B ends up at height -h with speed 3 m/s, and the ball on track A ends up at height +h with unknown speed.

If so, the problem seems to be a trick question: given the information stated, it is impossible for the ball to reach the top of track A; it doesn't have enough kinetic energy to start with.
HallsofIvy
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#3
Nov8-03, 08:13 AM
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(a) Two smooth tracks of equal length have "bump" - A up, and B down. Both "bumps" have the same curvature. IF two balls start simultaneously with the same initial speed, will they complete their journey at the same time or not? If not, which will arrive first? Explain your answer clearly.
-> My guess is that both tracks will complete their journey at different time. But, I am not exactly sure why and how they will complete at different time. Please teach me.
The ball moving along A loses kinetic energy (and speed) as it goes up the bump but gains it back as it comes down. The ball on B gains first and then loses but the total time to go up and then down is exactly the same for both A and B. The two balls will arrive at the end at exactly the same time with exactly the same speed.

(b) If the initial speed of the balls is 2m/s and teh speed of the ball at teh bottom of the curve on track B is 3m/s, will the speed of the ball at the top of the curve on Track A be greater than, less than, or equal to 1m/s? Explain your answer clearly.
-> The speed will be exactly 1m/s because w = mgh can be applied. Because the mass, height and g(9.9m/s^2) are the same in both cases, the answer will be 1m/s.
The ball on track B has gained enough kinetic energy to gain 1 m/s in speed by losing potential energy. The ball on track A goes up the same distance B goes down and so gains the potential energy B lost, loses the kinetic energy B gained. Since the two balls have the same mass, A loses the same speed that B gained. Your answer is exactly correct.

Doc Al
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#4
Nov8-03, 11:19 AM
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Potential Energy, Conservation of Energy problem


Originally posted by HallsofIvy
The ball moving along A loses kinetic energy (and speed) as it goes up the bump but gains it back as it comes down. The ball on B gains first and then loses but the total time to go up and then down is exactly the same for both A and B. The two balls will arrive at the end at exactly the same time with exactly the same speed.
Assuming that ball A has enough energy to make it over the bump, both balls arrive at the end with the same speed, but not at the same time. Think of it this way: when ball A goes over the bump, it moves slower than it's original speed; when ball B goes over the dip, it moves faster. Ball B wins the race.


The ball on track B has gained enough kinetic energy to gain 1 m/s in speed by losing potential energy. The ball on track A goes up the same distance B goes down and so gains the potential energy B lost, loses the kinetic energy B gained. Since the two balls have the same mass, A loses the same speed that B gained. Your answer is exactly correct.
Ball A loses the same amount of KE (not speed) as ball B gained. The original KE of each ball is 1/2 m v^2 = 1/2 m (4). Ball B speeds up to 3 m/s, KE = 1/2 m (9); an increase of 1/2 m (5). Ball A would need at least 1/2 m (5) worth of KE to just make it over the bump. It won't make it. (As Ambitwistor already pointed out.)


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