Finding the horizontal distance that the cork flew

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Homework Statement


A 14cm diameter champagne bottle rests on its side on top of a frictionless table. Suddenly, the cork pops and the bottle slides backward for a distance of 22 cm in .42 s. If the mass of the bottle is 500 times the mass of the cork, find the distance from the original position the cork will land on the table.

14 cm diameter
bottle final distance is 22 cm
time is .42 seconds
mass of bottle is 500 times mass of cork


Homework Equations


p = mv
Sf = Si + Vi*t + (1/2)(a)(t^2)
Vf^2 = Vi^2 + 2a*S
Vf = Vi + a*t


The Attempt at a Solution


I got a little lost.
I figured:
-22 cm = 0 +Vi*t
Vi = -52.381 cm/s, Vi for champagne bottle.

I'm having problems getting one more variable (the Vi for the cork). I'm not sure how to derive it.
Once I get the Vi for the cork, I should be able to solve the problem.
 
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Hi jheld,

jheld said:

Homework Statement


A 14cm diameter champagne bottle rests on its side on top of a frictionless table. Suddenly, the cork pops and the bottle slides backward for a distance of 22 cm in .42 s. If the mass of the bottle is 500 times the mass of the cork, find the distance from the original position the cork will land on the table.

14 cm diameter
bottle final distance is 22 cm
time is .42 seconds
mass of bottle is 500 times mass of cork


Homework Equations


p = mv
Sf = Si + Vi*t + (1/2)(a)(t^2)
Vf^2 = Vi^2 + 2a*S
Vf = Vi + a*t


The Attempt at a Solution


I got a little lost.
I figured:
-22 cm = 0 +Vi*t
Vi = -52.381 cm/s, Vi for champagne bottle.

I'm having problems getting one more variable (the Vi for the cork). I'm not sure how to derive it.
Once I get the Vi for the cork, I should be able to solve the problem.

What is conserved in the process of the cork shooting out of the bottle? If you write a conservation equation for that quantity you should be able to find the velocity of the cork right after it leaves the bottle.