Find distance of Champagne cork

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Homework Help Overview

The problem involves a champagne bottle resting on a frictionless table, where the cork pops and the bottle slides backward. The objective is to determine the distance the cork will land from its original position, given the mass relationship between the bottle and the cork.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and initial calculations of velocity and distance. There is an exploration of the relationship between the masses of the cork and bottle, and the implications of these values on the final calculations.

Discussion Status

Some participants have provided calculations based on momentum conservation, while others have acknowledged the need for further assistance in progressing through the problem. There is an ongoing exploration of the equations involved and the assumptions made regarding the system.

Contextual Notes

Participants are working under the constraints of the problem statement, including the mass ratio of the bottle to the cork and the initial conditions provided. There is also a noted lack of consensus on the next steps to take in the calculations.

Ienjoytrig...
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Homework Statement


A champagne bottle (diameter = 0.014 m) rests on its side on top of a frictionless table. The cork pops and the bottle slides backwards a distance of 0.3 meters in 0.72 seconds. If the bottle’s mass is 500 times greater than the cork’s mass, find the distance from the original position the cork will land on the table.

M= Bottle mass
m= cork mass

Homework Equations


vbottle=d/t
pbottle=Mv

The Attempt at a Solution


vbottle= (.3m)/(.72s)= .42m/s
pbottle= (.5kg)(.42 m/s) = .21 kg m/s

Obviously, I substituted M for .5kg making m= .001kg.

This is where I got stuck and I just need someone to help me with my next equation. From there I should be able to get through this.
 
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use conservation of momentum
 
Thanks, cupid.

Using CoM I got:

0kg m/s + 0kg m/s = (.21kg m/s) + pc, f
pc, f = -.21kg m/s

p=mv
-.21kg m/s = (.001kg)(v)
v = -210m/s

v = vi + at
-210m/s = 0m/s + (-9.8m/s2)(t)
21.43s = t

x = xi + vit + 1/2 a t2
x = 0m + 0m/s(21.43s) + 1/2 (9.8m/s2)(21.43s)
x = 2550.3m
 
Last edited:
Nice ;)
 

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