Question

Four particles \(P,Q,R\) and \(S\) of masses \(m,m,m\) and \(2m\) respectively are kept at the four corners of a square of side \(\sqrt{2}\,\text{m}\). The distance of the centre of mass of the system of particles from the particle \(S\) is

• A. \(1.2\,\text{m}\)
• B. \(0.8\,\text{m}\)
• C. \(0.6\,\text{m}\)
• D. \(0.4\,\text{m}\)

Hint:

For particles, \[ \boxed{ x_{CM}=\frac{\sum m_ix_i}{\sum m_i}, \qquad y_{CM}=\frac{\sum m_iy_i}{\sum m_i}. } \] Choose the heaviest particle as the origin whenever it simplifies the calculations.

Answer 1

The Correct Option is B

Step 1: Assign coordinates to the particles. Let the square have side \[ a=\sqrt2\,\text{m}. \] Take particle \(S\) as the origin. Then, \[ S(0,0),\qquad P(a,0),\qquad Q(a,a),\qquad R(0,a). \] The masses are \[ m,\;m,\;m,\;2m \] at \(P,Q,R,S\) respectively.

Step 2: Find the coordinates of the centre of mass. The total mass is \[ M=5m. \] Hence, \[ x_{CM} = \frac{m(a)+m(a)+m(0)+2m(0)}{5m} = \frac{2a}{5}, \] and \[ y_{CM} = \frac{m(0)+m(a)+m(a)+2m(0)}{5m} = \frac{2a}{5}. \] Thus, \[ \boxed{ \left(\frac{2a}{5},\frac{2a}{5}\right) } \] is the centre of mass.

Step 3: Find the distance from \(S\). The required distance is \[ d = \sqrt{\left(\frac{2a}{5}\right)^2+\left(\frac{2a}{5}\right)^2} = \frac{2a}{5}\sqrt2. \] Since \[ a=\sqrt2, \] \[ d = \frac{2\sqrt2}{5}\times\sqrt2 = \frac45 = 0.8\,\text{m}. \] Hence, \[ \boxed{d=0.8\,\text{m}.} \] Therefore, the correct option is \(\boxed{(B)}\).