bullet

Cork of a Champagne Bottle Pops out, Rises Vertically, and fall to the Floor

A champagne bottle is held upright 1.2 m above the floor as the wire around its cork is removed. The cork then pops out, rises vertically, and falls to the floor 1.4 s later.
What was the cork's initial velocity?

 

a)

3.2 m/s

 

b)

4.9 m/s

 

c)

7.3 m/s

N

d)

6.0 m/s

 

e)

None of the above.

What height above the bottle did the cork reach?

 

a)

2.12 m

 

b)

1.43 m

 

c)

2.21 m

N

d)

1.84 m

 

e)

None of the above.

Solution:

Solution:

For solving this problem, use can be made of the time needed by the cork to reach the floor, t = 1.4 s, which we know is 1.2 m below the launching point of the cork. This quantity correspond to the cork displacement of y = -1.2 mhe initial velocity of the cork can be calculated using the formula g is the acceleration of gravity. Thus, solving for the initial velocity, v0y, in the previous relation,  . And the initial velocity becomes .

 

Knowing the initial velocity of the cork, the maximum height can be calculated from the formula, , (F_F 5), since at the velocity of the cork is zero, . The maximum height can be calculated, . Substituting the numerical values .

its final velocity?

 

a)

- 6.5 m/s

N

b)

- 7.7 m/s

 

c)

- 8.5 m/s

 

d)

- 5.4 m/s

 

e)

None of the above.

Solution:

At the floor level, the final velocity of the cork is obtained directly from the formula that evaluated at the known numerical values results in  .

The numerical calculation with MathCad is

 

 Up Soccer Kick up Falling with Air Resistance Ball Up Edge of Building Cork of Champagne Water Balloon

bullet
by Luis F. Sez, Ph. D.    Comments and Suggestions: LSaez@dallaswinwin.com