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If I am on one end of a still boat and drop a ball in the water, the center of mass will change (since the ball is gone), but the boat will not experience any acceleration. How does the center of mass move with net force being 0 in the horizontal direction? $\sum F_x=0$ and $v_x=0$, yet the system moved.

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  • $\begingroup$ x-direction would be horizontal, sorry if it was not clear $\endgroup$ Commented Oct 27, 2021 at 20:14
  • $\begingroup$ I'm not sure why this question is confusing to you. $\sum F_x = 0$ and $V_{ix} = V_{x} = 0$, but you're referring to the change of centre of mass in the vertical direction. Why should this be unexpected? $\endgroup$
    – gmz
    Commented Oct 27, 2021 at 20:18
  • $\begingroup$ I was referring to the change of mass in the horizontal direction. Because when the mass leaves the boat, the center of mass of everything on the boat shifts away from u and towards the middle of the boat (assuming i am on one end of the boat) How can change in center of mass happens when Fnet=0 $\endgroup$ Commented Oct 27, 2021 at 20:32
  • $\begingroup$ This is to be expected. Imagine a symmetrical dumbbell; if you detach one end, the center-of-mass of the new dumbbell will change, even with no net forces. Once you consider a different system, there is no reason for things to stay constant, which should be clear from the definition of the center-of-mass. $\endgroup$
    – gmz
    Commented Oct 27, 2021 at 20:43
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    $\begingroup$ I see that I misinterpreted the question. $\endgroup$
    – march
    Commented Oct 27, 2021 at 22:36

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For the horizontal component of the center-of-mass

The horizontal component of the center-of-mass changes trivially because you have removed the ball from the system. The $x$-component center of mass doesn't change if you keep the ball in the system, because it only moves vertically. Hence, there is no inconsistency here.

For the vertical component of the center-of-mass

At the moment that the ball is released, the gravitational force downward on the boat system decreases, leaving a net force upward because the buoyant force doesn't change. As a consequence, the boat will accelerate upward until the buoyant force cancels the gravitational force. There will be some (likely overdamped) oscillations around the equilibrium point until the boat settles into the new equilibrium, slightly higher in the water than it used to be.

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