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Is this problem possible?

  1. Nov 30, 2003 #1
    two people toss a raw egg in the egg toss. the force required to break the shell is 5N. the mass of the egg is 50g. estimate the maximum separation distance for the egg throwers. make whatever reasonable assumptions you need to about the launch angle and hand movement.

    isnt an initial velocity needed for this problem?
     
  2. jcsd
  3. Nov 30, 2003 #2
    Estimate!

    :smile:
     
  4. Nov 30, 2003 #3

    Doc Al

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    That's what you've got to figure out. How fast can you toss that egg without breaking it?
     
  5. Nov 30, 2003 #4
    Suppose the tossers are very, very, very strong. Then what?
     
  6. Nov 30, 2003 #5

    Doc Al

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    Hint: How strong do they have to be to exert 5 N?
     
  7. Nov 30, 2003 #6
    okay, so i figured the acceleration for the egg to break to be 100 m/s^2.
    im estimating the max distance to be 100 m.
    im stuck. how do i figure in the 100 m/s^2? the acceleration in both directions is gonna be constant. 0 in the x direction and 9.8 in the y direction.
    so if the tossers are very strong, they would throw the egg very hard. leaving a high initial velocity. but that doesnt help with how to get the acceleration so high.
     
  8. Nov 30, 2003 #7

    Bystander

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    How long is your arm?
     
  9. Nov 30, 2003 #8
    oooohhh, use centripetal acceleration to get initial velocity?
     
  10. Nov 30, 2003 #9

    Doc Al

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    The max distance is what you're supposed to be figuring out. Don't guess at that answer!

    You figured out the max acceleration the egg can withstand. Now use Bystander's hint!
     
  11. Nov 30, 2003 #10
    using centripetal acc. i found the velocity to be 31.6 m/s with a radius(arm length) of .5 m. im assuming the launch angle to be 60 deg from the horizontal.
    i have a feeling im way off though.
    am i getting close?
     
  12. Nov 30, 2003 #11
    Trout or Carp

    You sound like you are from a town named after a fish.
     
  13. Nov 30, 2003 #12
    t

    Tom, are you still around. I looked at it like Work = mV^2 where V is the horizontal velocity. I don't know if I am right.
     
  14. Nov 30, 2003 #13
    uhhh how do you know my name?
     
  15. Nov 30, 2003 #14
    ?

    Because you told me about this place at church today.
     
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