momentum problem#2

[bluesky]

A(50 kg) and B(60 kg) are sitting at the two extremes of a 4m long boat,standing still in water.They came at the middle.Neglecting water friction,find the displacement of the boat.

I did it in the following way.Check if I am right.

I first found the initial position of CM=(32/15)m [A at 0,B at 4m,Boat CM at 2m]
I then found the final position of CM=2m
So,displacement of the CM=[(32/15)-2](-i) m assuming x axis pointing right.

This gives displacement~13.3 m which is close to given answer 13cm.
But I am not convinced.Since,there is no external force,the CM should be static...
Please help.

[ronthepon]

The center of mass will remain at the same point.

For that, the entire boat will move in the water, when the masses placed on it are shifted.

Notice what exactly the qustion is about. It's about how the boat moves. Understand that for the center of mass to remain at the same point, the boat must move.

And that's all there to the question.

BTW, is the answer in cm or m? Beacuse your answer of 13 cm is nowhere close to 13m

[bluesky]

I am sorry I forgot to provide the mass of the boat.It is 40 kg

X_Ab=X_Aw-X_bw
X_Bb=X_Bw-X_bw
Now use the definithion of CM,and use that delta R_CM=o
In RHS,replace X_Aw and X_Bw by (X_Ab+X_bw) and (X_Bb+X_bw)

Another method:
Let me choose the origin of co-ordinates as the geometrical midpoint of the boat.The boat moves relative to water and the origin of co-ordinates will change relative to a frame attached to water.Whereas the position of CM will be stationary relative to water.That is,the boat will move,but, CM will not move relative to water.It is also clear from the statement of the problem.

So they key to this problem is to find the position of the CM relative to the center of the boat, both before and after A and B move. The difference in the positions of CM relative to the center of the boat must be the distance the boat moved in the water to keep the CM at the same location in the water.
So... let's do "after" first. When they are both at the center of the boat, the CM is in the middle, relative to the middle, its position is zero!

Ok, now lets find the cm when they are at opposite ends of the boat. We need to consider the masses and positions of all three objects: A, B and M (the mass of the boat) relative to the center of the boat.

calc the pos of cm with:

x_cm = ( 1 / total mass) ( m1x1 + m2x2 + m3x3 )

person A m = 50, x = -2 meters
person B m = 60, x = 2 meterrs
boat 40 kg,x=0

xcm = ( 1 / 50 + 60 + 40) ( 50*-2 + 60*2 + 40*0 ) =

= ( 1 / 150 ) ( 20 ) = 0.133 meters

So,here it is the way to do it.The approach really does not make any problem because,we are interested in the difference.We have used a frame relative to which the CM changes,but,relative to water CM does not change at all.