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Ping-Pong balls in a room?

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Estimate the number of Ping - Pong balls that would fit into a typical-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. (Assume 25% of the space cannot be filled due to spherical packing.)

    (a) Find the volume of the room (in m^3)

    (b) Find the volume of a ball (in cm^3)

    (c) Find the number of balls (order of magnitude only) (balls)

    2. Relevant equations
    75% of 1000m^3 = 750m^3
    750m^3(1 000 000 cm^3/1m^3)

    3. The attempt at a solution
    (a)I said that an average room had 1000m^3
    It said "Your response differs from the correct answer by orders of magnitude."

    (b) I said the volume of a ball had 100cm^3
    It said "Your response differs from the correct answer by more than 100%."

    (c) I said the number of balls equaled 7.5e6
    It said "Your response differs from the correct answer by more than 100%."

    Any help would be greatly appreciated!
  2. jcsd
  3. Feb 17, 2010 #2
  4. Feb 18, 2010 #3
    a) Do you really think a typical room is 1000m^3? that would be 20 x 20 x 2.5 m^3
    b) google for "size of pingpong ball". Use formula for volume of sphere.
  5. Feb 18, 2010 #4
    volume of pingpong ball = 4/3[tex]\pi[/tex]r3

    Volume of room = Length*Width*Height

    let's say that a standard room has a length of 7m width of 4m and a height of 2.5m. and the diameter of a pingpong ball is 4cm or 0.04m.

    Volume of room = 7*4*2.5 = 70m^2
    Volume of room @ 75% = 70*0.75 = 52.5m^3

    volume of a pingpong ball = 4/3([tex]\pi[/tex]0.02^3)
    = 0.0000335m^3

    Fitting pingpong balls into room = volume of room / volume of pingpong ball.
    = 52.5/0.0000335
    = 1,567,164 ping pong balls.
  6. Feb 18, 2010 #5

    All of this was very helpful. I did the math on my own and my answer matched yours. I went to enter it in and this is what I got.

    Attempt 1:
    (c) Find the number of balls (order of magnitude only)
    1.56e6 Balls
    Your response differs from the correct answer by 10% to 100%.

    Attempt 2: (I tried rounding)
    (c) Find the number of balls (order of magnitude only)
    1.57e6 Balls
    Your response differs from the correct answer by 10% to 100%.

    I only have one more shot at the answer and I know I am correct with the math. What am I doing wrong?
  7. Feb 18, 2010 #6
    Do they not state the size of the room? and specific size of the ball? that's probably why the answers are wrong, because the values we're using are just random.
  8. Feb 18, 2010 #7


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    Gold Member

    Sirsh, this is in violation of https://www.physicsforums.com/showthread.php?t=5374".
    Please avoid this in the future.
    Last edited by a moderator: Apr 24, 2017
  9. Feb 18, 2010 #8
    I'll be sure to not let it happen again.
  10. Feb 18, 2010 #9


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    Homework Helper
    Gold Member

    I haven't gone through the math myself. But one thing I've noticed that might help is that the program stipulated in the question (c) "order of magnitude only". However you have mentioned that you were entering things like "1.57e6". That's far more precise than just the order of magnitude.
  11. Feb 18, 2010 #10


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    Homework Helper

    That's an epic programming fail. The question asks for an order of magnitude. 10-100% is clearly within an order of magnitude.
  12. Aug 28, 2012 #11
    I think the issue here is answer format. A correct answer would be written as " ~ 10x #/balls in a room of x*y*z size ", where x clearly denotes the order of magnitude on base-10. Still seems like a sensitive answer to write in online submission though. Note that x #/ping pong balls is clearly separate from volume.
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