A 14 cm 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 22.0 cm in 0.61 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.

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The total momentum is the same before the bottle is opened and after the cork pops up. Initial the momentum is zero so that this means

` 0 = M*v_1 +mv_2`  or equivalent `v_2 =-(M/m)*v_1`

For the force that acts on the bottle we can write the theorem of momentum variation:

`F = (M*v_1)/t `` ` or rearranging the terms `F/M =v_1/t` which means

`a_1 =v_1/t`

The motion of the bottle is uniform decelerated so that one can write

`d = v_1*t -(a*t^2)/2 =v_1*t -(v_1/t)*t^2/2 =v_1*t-(v_1*t)/2=(v_1*t)/2`

Thus the initial speed of the bottle is

`v_1 =(2*d)/t = (2*0.22)/0.61 =0.72 m/s`

This gives for the speed of the cork the value

`v_2 =(M/m)*v_1 =500*0.72 =360.66 m/s`

Now, the cork will follow a parabolic falling trajectory. On the vertical axis there is free fall (uniform accelerated motion). Since the diameter of the bottle is 14 cm the falling time is

`(D/2) =(g*t_2^2)/2`

`t_2 =sqrt (D/g)=sqrt(0.14/9.81) =0.1195 s`

On the horizontal axis there is uniform motion with initial speed ` ` `v_2`

The total horizontal distance that the cork will travel is

`s = v_2*t_2 =360.66*0.1195 =43.02 m =43 m`

The cork will land on the table at 43 m from its original position.

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