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Frictionless Environment

  1. Oct 7, 2003 #1
    [SOLVED] Frictionless Environment

    K...heres a question for you people out there....I am totally lost and cant seem to figure this out. Ive been trying for about 2 weeks now and still can't figure it out...maybe someone in here can help me

    You are standing on the edge of a frozen pond where friction is negligible. In the centre of the ice is a red circle 1.0m in diameter. A prize of a mega dollar will be offered if you can apply all three of newtons laws of motion to get to the red circle and stop there. Describe what you would do to win the prize?

    Well thats it hopefully someone can help me
  2. jcsd
  3. Oct 7, 2003 #2
    It's really not so hard. Newton's three laws are all very similar. I think I can figure out what to do if you use the laws in the form:
    1) dp/dt=0 if Fnet=0
    2) F=dp/dt
    3) dp1/dt=-dp2/dt or dP/dt=constant

    I would carry two weights of equal mass. I will throw them either before me or behind me to change my speed. I throw one weight behind me to get going and I throw another weight in front of me (with the exact same speed) to stop. Does that make sense to anyone?

    Do I get the megaprize?
  4. Oct 8, 2003 #3
    now that would be impressive to see.


    Are you actually allowed to carry stuff? If not, I'd recommend filling up on a few pints of water before the challenge.

    Only feasible for males though, really...
  5. Oct 8, 2003 #4


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    Good try, Stephen, but you didn't do the math!

    Take your mass to be M and the mass of the two weights to be m each. When you throw one weight with speed v, it will have momentum mv. To conserve momentum, you and the remaining weight must have momentum -(M+m)v so your speed is m/(M+m) v.

    When you reach the center circle, you throw the remaining weight in the opposite direction to which you threw the first with the same speed. Since your original momentum is -(M+m)v, if we take your final speed to be u, conservation of momentum gives Mu- mv= -(M+m)v
    or Mu= -Mv-mv-mv= -(M+2m)v and your final speed is u= ((M+2m)/M)v, not 0. You have to take into account that your mass when you throw away the first weight is greater than when you throw away the second.
  6. Oct 8, 2003 #5
    Which wouldn't be a problem for my solution, since you would be a little lower on pressure after the initial burst.

    Seriously though, are you allowed to use outside influences like weights?
  7. Oct 8, 2003 #6
    well according to my physics teacher he has said that u only yourself and your clothes....i was given a clue today about it...he said neglect air resistance, use your clothes somehow and "be gentle" thats about all i know right now
  8. Oct 8, 2003 #7
    I had a few theories about it but then the more I though about it the more I saw that they were incorrect or improbable....my first theory was similar to yours...except I would have my clothes in 2 piles(brr..cold) ....I would lay them on the ice and run and jump on to them. once I reached the centre I would throw the 1st pile forward at the same speed that I was traveling...but this would be very difficult to do. For one both those pile of clothes would have to weigh the same as I and 2 I would most likely just slow down a little.

    This one question has been hanting me for a while....I even started my own personal organization/joke about it for all who are getting frustrated with it...lol....the FOQ -13 ( The Frustrated Organzition of Question 13) kinda stupid but whatever...

    Thanks for the help so far and keep giving your ideas :smile:
  9. Oct 8, 2003 #8
    Ok, ok, I was lazy. Instead, I'll hire someone to throw a weight to me when I reach the cirlce and I'll only carry one weight to begin with. And to avoid seeming lazy again,
    (M+m)V=k=0 since V=0
    When I throw the weight,
    -mv + M(V2)= 0
    When I catch the weight,
    -mv + M(V2) = 0 since M(V2)=mv
    I've stopped (I think)

    Or since we only have our clothes, I'll use a button, rather than a weight.
    Last edited: Oct 8, 2003
  10. Oct 9, 2003 #9
    It seems to me that the proviso about using all three laws is a red-herring since you really couldn't prevent them from being involved if you wanted to.

    I have several ideas but a choice is dependent on some idea of how far it is from the shore to the circle. Is this given? Likewise Is this a race situation?

    I'm also not sure what the important elements are. Is this really meant to be about doing this on a frozen pond or would an exceptionally smooth giant marble slab with a thin layer of oil be better? (In oher words, is it the negligible friction he wants you to overcome? Or is this more on the lines of "Tactics for Traveling on Ice for Rescue Teams"?

  11. Oct 9, 2003 #10
    the question is meant to be that how would you deal with a non friction environment to reach the centre an an area...the size of the pond can differ unless your talking about just simply stepping into the centre.....you must neglect things like air resistance and base your idea solely on what you have...that being you and your clothes...dont think as the situation as ice but more just a frictionless surface you must reach the centre of...you begin on the outside where friction is involved. You are also correct in associating that no matter what u do u are going to use newtons laws of motion so in a direct term....How do you reach the centre and remain there....so in saying you must overcome this force...or rather lack of force you are correct

    Thanks with the help so far
    Last edited by a moderator: Oct 9, 2003
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