Expert: Daniel Mazur - 10/24/2008

Question
Hey Mr. Mazur,
 I have been trying this question over and over, finally I give up..if you would please look over this problem and if you don't mind explain how to do it , I'd really appreciate it . Thanks you Sir.

Q:Two people are standing on a 2.3-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. One person throws a 6.1-kg ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is 130 kg. Because of the throw, this 130-kg mass recoils. How far does it move before coming to rest again?   ..........m  

Answer
Hi Patrick,
the key to the solution lies in the fact that there is no friction with the ground and, therefore, that the *center of mass* must be located at the same place with respect to the ground all the way through the action. Look for the distance of the center of mass before and after the throw and that will be the answer.
It helps our solution, if the platform with people (everything except the ball) is mirror-symmetrical. In the side view (so that the platform motion will happen from left to right), place the "zero" coordinate in the middle of the platform. With the ball held at the right end (coordinate +1.15m), calculate the position of the center of mass (it will be a few centimeters to the right from zero) with respect to the center of the platform. Then after the ball has moved to the left end, the situation is exactly symmetrical with the beginning (from the point of view of the platform), just the position of the mass center will be to the left from platform center. The mathematics, naturally, works equally well for any choice of system of coordinates:
From the conservation of mass center position:
(Eq.1)  (M*r1_pre+m*r2_pre)/(M+m)=(M*r1_past+m*r2_past)/(M+m)
Where M=130kg, m=6.1kg, r1_pre=0.00m (position of the mass center of the platform WITHOUT the ball, before the throw), r1_past=? (that is the result we seek), L=2.3m, r2_pre=L/2=1.15m (position of the ball before the throw), and
(Eq.2)  r2_past=r2_pre-L+r1_past (position of the ball after the throw).
Now eliminate the (M+m) in Eq.1, substitute for r2_past from Eq.2, and r1_pre=0. After some manipulation we get
(Result)  r1_past = L*m/(M+m) = 10.3cm
And that is your result. You are welcome, come again if needed.
Cheers,
Daniel