Physics question mechanics... See attached figure?

Question

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A loop of light inextensible string passes over smooth small pulleys A and B. Two masses m and M are attached to the string at the respective positions. Then the condition between m, M and n so that m and M will cross each other will be?
The masses are point masses and the pulley has negligible radius and are at the same elevation

This is how I did it
Method I]
The final position is dictated by the total length of the string (L(1 + n)), the distance between pulley and m(or M) will be (L(1 + n) / 2).
We find the distance traveled by m by Pythagoras theorem as L√[(n + 3)(n - 1)] / 2 = x
Total distance between the blocks in the initial position is L√[(n - 1)(n + 1)]
We also find the distance travelled by M in the same time as L√[(n - 1)(n + 1)] - x
velocity at that time of the blocks will be non zero
Applying Work energy theorem,
mgx - Mg(L√[(n - 1)(n + 1)] - x) > 0
=> mgx > Mg(L√[(n - 1)(n + 1)] - x)
=> m / M > (L√[(n - 1)(n + 1)] - x) / x
=> m / M > (L√[(n - 1)(n + 1)] - L√[(n + 3)(n - 1)] / 2) / L√[(n + 3)(n - 1)] / 2
=> m / M > (2√[(n + 1)  / (n + 3)]) - 1
Which is the correct answer

Method II]
We note that the masses meet at the centre of mass of the system
mx - M(L√[(n - 1)(n + 1)] - x) = 0 [If we define origin as the final location in method I] 
=> m/M = (L√[(n - 1)(n + 1)] - x) / x
=> m / M = (2√[(n + 1)  / (n + 3)]) - 1 [For the bodies to meet]
=> m / M > (2√[(n + 1)  / (n + 3)]) - 1 [For the bodies to continue their motion]
Note that this is the same answer as in method I

Now I see there is a flaw in the second method that the centre of mass keeps changing its position which is governed by external forces gravity and reaction force on the pulleys.
In solving the problem I have assumed that the centre of mass of the bodies does not change.[That is why m1x1 + m2x2 =  0 (final position of the bodies is zero)]
If the position of Centre of mass is not same as the initial position then I should have got an different answer. So this means that the centre of mass doesn't change its position inspite of external forces.

Can someone explain why?? Or maybe the centre of mass goes down and comes up.

Also new methods are welcome

? · Accepted Answer

Conceptually your two methods are the same. Due to conservation of energy,

K + U  = U_0.

Here, K is the kinetic energy of the system, U is its potential energy, and the subscript "0" refers to the original position. Masses will meet if in the meeting point K is non-negative, i.e.

U_0 - U >= 0.

In the first method, you calculate the change in the potential energy U_0-U explicitly. Alternatively (the second method) one can write U  as

U = (m+M) g X,

where X  is the coordinate of the mass center. Hence, the condition U_0 - U >=0 is the same as X_0 - X >=0.

The masses just meet and do not continue motion if X=X_0. Note that the condition X=X_0 refers only to the original and the final positions, and it does not mean that the mass center does not change. In the beginning, the mass center goes down and the system starts moving. Then the mass center moves up to its original position and the motion stops.

*****************
Soham Shanbhag, if masses m1 and m2 have coordinates x1 and x2, the coordinate of the mass center is

X = (m1*x1 + m2*x2)/(m1+m2).

Let's measure positions from the original level of mass m1 = m. The positive direction is upwards. If the distances between m1=m and m2=M and the pulley are s1 and s2 respectively, then the coordinates of m and M are

x1 = - √(s1^2 - L^2),
x2 = - √(s2^2 - L^2).

Taking into account that s1 + s2 = (n+1)L, you can express the coordinate of the mass center explicitly via displacement of one of the masses. For a better understanding of what is going on, we can diferentiate x1 and x2 with respect to s1,

d x1/ d s1 = - s1/x1 = - 1/sin α1,
d x2/ d s1 = d x2/ d s2 * (d s2/ d s1) = -d x2/ d s2 = s2/x2 = 1/sin α2.

Here, α1 and α2 are the angles between the lines connecting m1 and m2 with the pulleys and the horizontal line. The derivative dX/ds1 is proportional to

dX/ds1 propto -m/sin α1 + M/sin α2.

Originally α1 = 0 and the above derivative is negative. While m goes down, s1 increases and X decreases (center of mass goes down). When m continues going down, α1 increases and α2 decreases. Eventually the point is reached where m/sin α1 = M/sin α2. After this point, X starts increasing (center of mass goes up). m goes down until is stops. At this point M also stops, so that the kinetic energy of the system is zero. The potential energy of the system should be equal to its starting potential energy, which implies that the mass center returns to its original position.

? · Answer

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Physics question mechanics... See attached figure?

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