Conservation of Mechanical Energy Question.

In summary, the question is about a carnival game where the goal is to get a ball to land between two 20m high bumps. The ball is 5kg and there is no friction. The conversation discusses the calculations to determine the velocity needed to make the ball stay at the top of the first bump and in the middle of the two bumps. The final answer is that the velocity must be in the range of Root(392) to 28 m/s. The conversation also mentions the implications of no friction and how it affects the ball's movement.
  • #1
Azrioch
30
0
The question goes like something like this:

It is a common carnival scam to set up a game where there are two bumps each being 20mhigh. The goal is to get it to land between the two. You cannot go further than the bottom of the first hill. The bumps are one after another (the teacher did not give us a distance). He also said to ignore friction. He then wanted to know if it was possible to get the ball to land in the middle. The ball is 5kg.

I did not think the implications of their being no friction. After the test it dawned on me that with no friction it would go up and down forever never stopping and wouldn't land in the middle.

Now what I just want to know what I calculated because it made sense at the time and I still don't know. What I did was took the gravitational potential energy from the first bump which was:

Eg = (5)(9.8)(20) which was 980. Then I made Kinetic Energy equal to 980.

So, 980=(1/2)(5)(v)^2

I solved and got Root(392).

Is this the velocity needed to make it stay at the top?

After this I solved for two gravitational potential energies so I got:

1960 =(1/2)(5)(v)^2

I solved this and got 28 m/s.

So I thought this was the energy needed to propel the ball to the second bump and have it stay there. Is that correct or no? If not would you mind telling me why it is not correct?

Anyways my final answer was that the velocity must be in the open interval (Root(392), 28).

Does this make any sense?

Thanks in advance.

- Marc
 
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  • #2
Originally posted by Azrioch
I did not think the implications of their being no friction. After the test it dawned on me that with no friction it would go up and down forever never stopping and wouldn't land in the middle.

This is correct, assuming you have no friction and you start at the top of the first hill.

Originally posted by Azrioch
Is this the velocity needed to make it stay at the top?

If I understand you correctly, you start before the first hill and launch at some velocity with the intention to try and make it stop in the valley. If so then this is the velocity required to reach the top of the hill and stop there.

Originally posted by Azrioch
After this I solved for two gravitational potential energies so I got:

You don't need to solve for two grav potential energies. When the ball get's to the top of the first hill most of it's initial kinetic energy has been converted into gravitational potential energy, assuming that the initial velocity is slightly greater than Root(392). Once the ball get's to the valley bottom it has the kinetic energy that it had initially and so goes over the second hill. If however you have an initial speed slightly less than root(392) then you don't make it over the first hill. The issue here is that in the idealised world of Newtonian mechanics if you launch the ball at exactly root(392) then you can get it to the top of the first hill and have it stay there. But if it is greater then it will continue going without stopping.

Essentially problems of this type just want you to realize that if a ball goes up the kinetic energy becomes gravitational potential energy wheras on the way down (under the influence of gravity only) the grav. pot. energy turns into kinetic energy. This is why you don't multiply the grav. pot. energy by 2. Thank goodness for friction really or we either wouldn't get going or we wouldn't be able to stop.
 
  • #3
Thanks.

I overlooked the fact that upon reaching the top the potential energy would still be there and it would be enough to go over the second lump.

Thanks for the help.
 

1. What is conservation of mechanical energy?

Conservation of mechanical energy is a fundamental principle in physics that states that the total amount of mechanical energy in a closed system remains constant over time, regardless of any internal or external forces acting on the system.

2. How is mechanical energy conserved?

Mechanical energy is conserved through the exchange of potential and kinetic energy. As an object moves, its potential energy (due to its position) is converted into kinetic energy (due to its motion), and vice versa. This conversion results in the total amount of mechanical energy remaining constant.

3. What are some examples of conservation of mechanical energy in everyday life?

Some common examples of conservation of mechanical energy include a pendulum swinging back and forth, a rollercoaster going up and down hills, and a ball being thrown into the air and falling back down. In all of these cases, the potential and kinetic energy of the system are constantly exchanging to maintain a constant total energy.

4. Is conservation of mechanical energy always true?

In theory, conservation of mechanical energy is always true in a closed system. However, in real-world situations, there may be factors such as friction or air resistance that can cause some mechanical energy to be lost to other forms of energy, such as heat. This is why conservation of mechanical energy is often treated as an ideal case in physics.

5. How does conservation of mechanical energy relate to the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. Conservation of mechanical energy is a specific application of this law, focusing on the conservation of mechanical energy in a closed system. This means that although energy may change forms within the system, the total amount of energy remains constant.

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