Conservation of Mechanical Energy Question.

[Azrioch]

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

 

[Azrioch]

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.