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Breaking a glass soda bottle

  1. Oct 29, 2007 #1
    http://session.masteringphysics.com/problemAsset/1011752/15/1011752A.jpg

    A glass soda bottle is emptied of soda and filled to the very top with water. A cork is carefully fitted into the top of the bottle, leaving no air between the cork and the water. View Figure The top of the bottle has a diameter of D_top = 2.00 cm and the bottom of the bottle has a diameter of D_bot = 6.50 cm. The glass breaks when it is exposed to p_max = 70.0 MPa of pressure.
    A student hits the cork sharply with her fist and the bottom of the bottle breaks. The student's fist has a mass of m = 0.480 kg and moves downward at a speed of v_i = 5.00 m/s. It collides elastically with the cork and rebounds with the same speed. The collision lasts for t = 1.20×10^−4 s. In this problem, the positive direction is upward.

    ----->What is the force that the fist exerts on the top of the bottle?
    __________________________________________________________
    I know that the acceleration a of the fist during its collision with the bottle is (0-5)/(1.20*10^-4) = -41670 m/s^2??? Somehow I think that is very wrong...
     
  2. jcsd
  3. Oct 29, 2007 #2

    learningphysics

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    The acceleration of the fist is [5m/s - (-5m/s)]/(1.20*10^-4) = 83,333m/s^2 upwards...

    So the average force the cork exerts upward on the fist is = ma = 83,333m/s^2*0.480 = 40,000N upwards.

    So the fist exerts an average downward force of 40,000N on the cork.

    What is the pressure due to this force? What is the pressure at the bottom of the bottle during the application of this force?

    The glass breaks when it is exposed to p_max = 70.0 MPa of pressure... that means that the difference of the pressures on either side has to reach 70.0MPa...
     
  4. Oct 29, 2007 #3
    Thanks I got it!
     
    Last edited: Oct 29, 2007
  5. Oct 29, 2007 #4

    learningphysics

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    Do they want the "net" force on the bottom of the bottle from inside the bottle?

    Then the pressure you need is atmospheric pressure + fist pressure + rho*g*h... multiply that by area at the bottom of the bottle...

    ie the p_0 here is atmospheric pressure + fist pressure...
     
  6. Oct 29, 2007 #5

    learningphysics

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    cool. just curious. Did you use atmospheric pressure + fist pressure + rho*g*h?
     
  7. Oct 29, 2007 #6
    No. Actually I used Pascal’s principle which states that the pressure of the fist striking the top is transmitted equally throughout the bottle. The increase in pressure at the bottom is equal to the increase in pressure at the top which means that the pressure at the top and the bottom is 127323954.4 ... and so I took that value and multiplied it with the area of bottom to get Force :)
     
  8. Oct 29, 2007 #7

    learningphysics

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    ah... cool. so they just wanted the extra force due to the fist striking.

    good job!
     
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