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Conservation of Mechanical Energy Question.

  1. May 5, 2003 #1
    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
  2. jcsd
  3. May 5, 2003 #2
    This is correct, assuming you have no friction and you start at the top of the first hill.

    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.

    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 realise 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.
  4. May 7, 2003 #3

    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.
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