Walking in a Boat (Center of Mass?)

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SUMMARY

The problem involves a 64 kg boat and a 32 kg child, where the child walks from the near end to the far end of the boat. The initial distance from the pier is 8.5 m, and the child’s movement affects the boat's position due to the conservation of the center of mass. When the child reaches the far end of the boat, the new distance from the pier is calculated to be 16.5 m. The absence of friction ensures that the center of mass remains constant, allowing for straightforward calculations based on the initial and final positions of the child and the boat.

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Homework Statement



A(n) 64 kg boat that is 8 m in length is initially 8.5 m from the pier. A 32 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle.

Assume: There is no friction between boat and water.

How far is the child from the pier when she reaches the far end of the boat?

Homework Equations



I am not quite sure if this is a Center of Mass question, but if it is the equation is:

X = m1*x1 + m2*x2/m1 + m2
m = mass
x = distance

The Attempt at a Solution



I really don't know where to begin with this problem. I just need a little push or hint to catalyst the problem solving process.

I understand that when the child is at the far end of the boat he will be 16.5 m away from the pier, but I'm not sure what to do next.
 
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Figure that since there is no friction, there is no external force, so the center of mass of both won't change.

Now figure where the center of mass is originally with respect to the center of mass of just the boat. With the child at the other end, how much must the center of mass of the boat shift to account for the change in the position of the child. That distance is how much nearer the pier the boat will move.
 

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