Question

Six Identical particles each of mass 'm' are arranged at the corners of a hexagon each side of length L. If mass of one particle was removed then shift in the centre of mass is?

Answer 1

Since, the hexagon is regular and the masses are equal, therefore by symmetry center of mass lies at the center of the hexagon.
Next, one of the mass is doubled, it is same as adding one more particle of mass 'm' at the one of the corner without disturbing the original arrangement.
Initially we can assume the entire mass(6m) to be lying at the center because it is a place where center mass of the system was present.
Now we have added an additional mass 'm' at one of the corner so the entire question turns into : find the center of mass of a system with 6m mass placed at the center of a regular hexagon and m mass is placed at one of its corner.
The separation between the center and the corner equals to :
L/2÷(sin30∘) = L

The position of COM from

center = mL/7m

=L/7

This also equals to the shift in center of mass because initially COM was present at the center .
So, shift =L/7

alt sol :

$6m_1{x}_1=m_1{x}_2^2$

$x_2=6x_1$

$x_1+6x_1=L$

$x_1=l/7$