Question

Four identical particles each of mass $1Kg$ are arranged at the corners of a square of side length $2\sqrt 2 m$. If one of the particles is removed the shift in the centre of mass is
(A) $\dfrac{8}{3}m$
(B) $\dfrac{4}{3}m$
(C) $\dfrac{2}{3}m$
(D) $2m$

Answer 1

Hint Four particles and their masses are given. It is given that the four particles are arranged in a square. The length of the side of the square is given. We have to find the shift in the centre of mass, if one of the particles is removed. To find that we have to find the position of centre of mass of the four particles and then position of centre of mass of the any three particles after that we have to find the shift in the centre of mass.

Complete step by step answer
Let us consider two particles joining a line, the position of centre of mass of line joining the two particles defined by x axis is given by
$X = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$
X is the position of centre of mass
${m_1},{m_2}$ is the mass of the particles
${x_1},{x_2}$ is the distance of the particle from the origin
Let us consider n particles along a straight line taken in the x- axis, the position of the centre of the mass of the n system of particles is given by
$X = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + .......... + {m_n}{x_n}}}{{{m_1} + {m_2} + ....... + {m_n}}}$
X is the position of centre of mass
${m_1},{m_2},{m_n}$ is the mass of the particles
${x_1},{x_2},{x_n}$ is the distance of the particle from the origin
Let us consider n number of particles, not lying on a straight line.
The position of centre of mass of the system of n particles is defined and located by the coordinates (X, Y) given by
$X = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + ....... + {m_n}{x_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
${m_1},{m_2},{m_n}$ is the mass of the particles along x axis
${x_1},{x_2},{x_n}$ is the distance of the particle from the origin in x axis
$y = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + ....... + {m_n}{y_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
${m_1},{m_2},{m_n}$ is the mass of the particles along y axis
${y_1},{y_2},{y_n}$ is the distance of the particle from the origin in y axis
Form a diagram with given information for better understanding of the question

          FOUR IDENTICAL PARTICLES   

From the diagram,
The points $({x_1},{y_1}),{\text{ }}({x_2},{y_2}),{\text{ }}({x_3},{y_3}),{\text{ }}({x_4},{y_4})$ are $(0,2\sqrt 2 ),{\text{ }}(0,0),{\text{ }}(2\sqrt 2 ,0),{\text{ }}(2\sqrt 2 ,2\sqrt 2 )$ respectively.
The mass of each particle is $1Kg$
Given that the particles are arranged in a square, so the particles are not present in straight line
So, the position of centre of mass of the system of n particles which are not lying on the same line is given by
Along the x axis
$\Rightarrow X = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + ....... + {m_n}{x_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
Substitute the point values
$\Rightarrow X = \dfrac{{(1 \times 0) + (1 \times 0) + (1 \times 2\sqrt 2 ) + (1 \times 2\sqrt 2 )}}{{1 + 1 + 1 + 1}}$
$\Rightarrow X = \dfrac{{4\sqrt 2 }}{4}$
$\Rightarrow X = \sqrt 2 {\text{ }} \to 1$
Along the y axis
$y = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + ....... + {m_n}{y_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
$\Rightarrow y = \dfrac{{(1 \times 2\sqrt 2 ) + (1 \times 0) + (1 \times 0) + (1 \times 2\sqrt 2 )}}{{1 + 1 + 1 + 1}}$

$\Rightarrow y = \dfrac{{4\sqrt 2 }}{4}$
$\Rightarrow y = \sqrt 2 {\text{ }} \to {\text{2}}$
From 1 and 2 the position of centre of mass of the four particles arranged in square C1 is
${C_1} = ({x_1},{y_1}) = (\sqrt 2 ,\sqrt 2 )$
Now, if we remove one of the particle, we get

THREE IDENTICAL PARTICLES
The points $({x_1},{y_1}),{\text{ }}({x_2},{y_2}),{\text{ }}({x_3},{y_3})$ are $(0,2\sqrt 2 ),{\text{ }}(0,0),{\text{ }}(2\sqrt 2 ,0)$ respectively
The position of centre of mass of the system of n particles
Along the x axis
$\Rightarrow X = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + ....... + {m_n}{x_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
$\Rightarrow X = \dfrac{{(1 \times 0) + (1 \times 0) + (1 \times 2\sqrt 2 )}}{{1 + 1 + 1}}$
$\Rightarrow X = \dfrac{{2\sqrt 2 }}{3}{\text{ }} \to {\text{3}}$
Along the y axis
$\Rightarrow y = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + ....... + {m_n}{y_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
$\Rightarrow y = \dfrac{{(1 \times 2\sqrt 2 ) + (1 \times 0) + (1 \times 0)}}{{1 + 1 + 1}}$
$\Rightarrow y = \dfrac{{2\sqrt 2 }}{3}{\text{ }} \to {\text{4}}$
From 3 and 4 the position of centre of mass of the four particles arranged in square C2 is
${C_2} = ({x_2},{y_2}) = (\dfrac{{2\sqrt 2 }}{3},\dfrac{{2\sqrt 2 }}{3})$
The shift in the centre of mass is given by
$shift = \sqrt {({x_1} - {x_2}^2 + {{({y_1} - {y_2})}^2}}$
The points are ${C_1} = ({x_1},{y_1}) = (\sqrt 2 ,\sqrt 2 )$ and ${C_2} = ({x_2},{y_2}) = (\dfrac{{2\sqrt 2 }}{3},\dfrac{{2\sqrt 2 }}{3})$
$\Rightarrow shift = \sqrt {{{\left( {\sqrt 2 - \dfrac{{2\sqrt 2 }}{3}} \right)}^2} + {{\left( {\sqrt 2 - \dfrac{{2\sqrt 2 }}{3}} \right)}^2}}$

$\Rightarrow shift = \sqrt {{{\left( {\dfrac{{3\sqrt 2 - 2\sqrt 2 }}{3}} \right)}^2} + {{\left( {\dfrac{{3\sqrt 2 - 2\sqrt 2 }}{3}} \right)}^2}}$

$\Rightarrow shift = \sqrt {{{\left( {\dfrac{{\sqrt 2 }}{3}} \right)}^2} + {{\left( {\dfrac{{\sqrt 2 }}{3}} \right)}^2}}$
$\Rightarrow shift = \sqrt {\dfrac{2}{9} + \dfrac{2}{9}}$
$\Rightarrow shift = \sqrt {\dfrac{4}{9}}$
$\Rightarrow shift = \dfrac{2}{3}m$
The shift in the centre of mass is $\dfrac{2}{3}m$

Hence the correct answer is option (C) $\dfrac{2}{3}m$

Note It is a simple question. If you know the right formulas to solve this question it will be very easy to solve. Since these points deal with the distance of the shift in centre of mass, its unit is metre.