Distance change

Question

A 14 cm diameter champagne bottle rests on its side on top of along frictionless table. Suddenly, the cork pops and the bottle slides backward moving a distance of 22.0cm in 0.44s. If the mass of the bottle is 500 times the mass of the cork, how far from the original position will the cork land on the table? Assume that the width of the cork is negligible.

Answer 1

This problem can be broken down into two parts:

 

First part:  The explosive ejection of the cork.    The principal of conservation of momentum applies to this event.  Before the ejection, the system momentum is zero, so it must also be zero just afterwards.   Just afterwards the velocity of the bottle is  .22 m / .44 s  = - 0.5 m/s.  The minus sign is there because the velocity is to the left (backwards).   If the mass of the bottle is M,  its momentum, p_{B ,}  is p_B =  - 0.5 M.     To conserve momentum the momentum of the cork just after, p_C ,  must be + 0.5 M.   So  p_C = v_C m = + 0.5 M.    Here v_C is the velocity of the cork just after, and m is the mass of the cork.  Working this equation and using  M/m = 500,  it is found that v_C = 250 m/s.  This velocity is horizontal to the right.

 

Second Part:   Flight of the cork.    Standard kinematics can be used in the analysis.   The time of flight of the cork is the time it takes an object to fall 0.07 m straight down starting with zero vertical velocity.  The standard kinematic formula for this is   t = sqrt[ 2 (0.07) / g]  where g = 9.81.   This works out to be 0.12 s.   The horizontal distance covered by the cork in this time is  250 m/s x 0.12 s = 30  m.