Finding Distance Moved by Swapping Positions of Two Masses on Boat

[dibilo]

2 man, mass m1 and m2 stands on a boat mass M at L distance apart on static water. they slowly swop their places. find the distance the boat move due to this.

------------------------------------------------------------
i know that the centre of mass remains unchange, and at start,
Xcm=m1x1+m2x2+Mxb/m1+m2+M

then after changing place, i let y be the distance moved, Xcm becomes
Xcm'=m1(x2+y)+m2(x1+y)+M(xb+y)/m1+m2+M

do i simply divide the 2 equations? then i'll end up with lotsa varibles...

then i am pretty much stuck here.. pls offer ur advice, hints, help ... thanks in advance :)

 

[HallsofIvy]

First, state clearly what it is you are trying to determine What do you mean by "distance the boat moves"? What coordinate system are you using?

The center of mass of the entire system (boat and men) remains unchanged but the center of mass of the boat alone WILL move- that's the motion you want to find. I would suggest taking some fixed point (that will not move as the boat moves) in front of the bow of the boat as x= 0 and measure everything from that.

I do notice that there is no L in your equations. It seems to me that the distance between the two men is of importance.

The center of mass of the system, initially, is
X_cm= (m₁x₁+m₂x₂+Mx_b)/(m₁+m₂+M) as you say, but, taking x₁ to be the position of the man in the bow, the position of the other man, x₂= x₁+ L so this is the same as X_cm= (m₁x₁+m₂(x₁+L)+Mx_b)/(m₁+m₂+M)

After the men exchange places we have
X_cm= (m₂x₁+m₁(x₁+L)+Mx_b2)/(m₁+m₂+M) - the only difference is that m₁ and m₂ have swapped places and x_b2 is the new position of the center of mass of the boat alone. No need to divide anything- since X_cm is the same in both equations, set the right sides equal:

(m₁x₁+m₂(x₁+L)+Mx_b)/(m₁+m₂+M) = (m₂x₁+m₁(x₁+L)+Mx_b2)/(m₁+m₂+M) and solve for x_b- x_b2.

There are not "lotsa variables"- all except x_b and x_b2 are given constants. x_b- x_b2 is the distance the boat moves.

 

[M98Ranger]

To find the distance the boat moves, we can use the conservation of momentum principle. This states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the system includes the boat and the two masses on it.

Initially, the total momentum of the system is given by:

P = m1v1 + m2v2 + Mv

where m1 and m2 are the masses of the two men, v1 and v2 are their initial velocities, and M is the mass of the boat.

After swapping places, the total momentum of the system becomes:

P' = m1v2 + m2v1 + Mv'

where v' is the final velocity of the boat after the men have swapped places.

Since the total momentum of the system remains constant, we can equate the two equations:

m1v1 + m2v2 + Mv = m1v2 + m2v1 + Mv'

Rearranging this equation, we get:

m1(v1 - v2) = M(v' - v)

Since the men are swapping places slowly, their velocities can be considered approximately equal. Therefore, v1 ≈ v2.

Substituting this into the equation, we get:

m1(v1 - v2) ≈ m1(0) = 0

Solving for v', we get:

v' = v

This means that the final velocity of the boat is equal to the initial velocity. We can use this information to find the distance the boat moves by using the formula:

s = ut + 1/2at^2

where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

Since the boat is moving at a constant velocity, the acceleration is 0. Therefore, the formula simplifies to:

s = ut

Substituting in the values, we get:

s = v*t

Since we want to find the distance the boat moves, we can rearrange the equation to solve for t:

t = s/v

Substituting this into our original equation, we get:

Xcm' = m1(x2 + s/v) + m2(x1 + s/v) + M(xb + s/v)/m1 + m2 + M

We can simplify this equation by dividing both sides by m1 + m2 + M: