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Homework Help: Year 12: Cambridge Physics Problems (2D kinetic theory)

  1. Jun 15, 2012 #1
    A square frame of side 10cm rests on a table. Confined within the frame are 500 ants, each of mass 0.001g. The ants rush about randomly with constant speed 0.02m/s, colliding with each other and with the walls. Assuming the collisions are perfectly elastic, calculate for force on each side of the square due to the ants' movement.

    Assuming a random motion factor ( I don't really know what's the proper name for it) of 1/2 instead of 1/3 as we are dealing with a 2-dimensional motion, and applying the usual kinetic theory formula, answer is 10-6N.

    What I don't entirely understand is this:
    Are we trying to postulate that the net vector force acting in the region is zero? If that's the case then why the need to have the ants moving "parallel to the sides of the frame"? The resultant force will still be zero if they move at 45° to the planes AND perpendicular to each other.

    Can anyone please provide a more elaborate explanation on the bold statement? Thank you!
  2. jcsd
  3. Jun 15, 2012 #2
    Lets called the directions north,south east and west.
    What it is saying is that at any instant approximately 50% of the ants are moving in the north-south direction and the other 50% in the east-west direction.
  4. Jun 15, 2012 #3
    Thanks! Could you please help me out with a proper explanation for my quote above? I'm not really sure whether my explanation is correct, or how could I improve it.
  5. Jun 15, 2012 #4
    Yes, your answer is correct. 3 is the term that comes in to describe how many possible dimensions in which the ant(or molecule, in terms of kinetic theory) could have gone in case of a three dimensional cube. But since you need the force only on one side, you divide it by 3. The same logic is applicable to a two dimensional square.

    This is basically the assumption that at any given time half of the ants are moving east-west, and the other half are moving north-south.
  6. Jun 15, 2012 #5
    I am confused, don't we NEED the time of collision of the "ant" and the side of the square block to calculate the force ?

    we can calculate mv-mu=impulse
    But to calculate the force we NEED the time interval during which the collision took place.
    That wil give us the force exerted by ONE ant on one side of the container, then we multiply it by , say 250, (using the above argument) to find the force on ONE side.
  7. Jun 15, 2012 #6
    Yes, you do need the time. But, if you carefully observe the problem, you will see it hidden in the given information :wink:
  8. Jun 15, 2012 #7
    You're right, but in this case you obtain the force exerted on the sides by deriving the formula for Nm<c>^2/3 (which in this case the denominator is 2, not 3).
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