Calculating the Force of a Fist on a Glass Bottle

[jaded18]

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?
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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...

 

[learningphysics]

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...

 

[jaded18]

Thanks I got it!

 

[learningphysics]

Thanks got it. I have another question: What is the magnitude of the force exerted on the bottom of the bottle?
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I know that the area at the top is (0.01^2*3.14) m^2 and the force as you mentioned is 40,000 that is exerted on the top. The pressure due to the force of the fist on the cork, therefore is (40000/(o.o1^2*3.14)) which = 127323954.5 pa. I also know that the area at the bottom is (0.065/2)^2*3.14 = 0.0033 m^2. I am a little lost on how to find the pressure at the bottom ... Do I just use p=p_0+density(g)(h) where p_0 is the pressure at the liquid surface so that p at bottom=101300 pa??

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...

 

[learningphysics]

Thanks I got it!

cool. just curious. Did you use atmospheric pressure + fist pressure + rho*g*h?

 

[jaded18]

cool. just curious. Did you use atmospheric pressure + fist pressure + rho*g*h?

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 :)

 

[learningphysics]

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 :)

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

good job!