Question

Four identical particles each of mass 1 kg are arranged at the corners of a square of side length 2Ö2 m. If one of the particles is removed, the shift in the centre of mass will be

• A. 1/2m
• B. 3/2m
• C. 2/3 m
• D. 1m

Answer 1

The correct option is C

2/3 m

When all the identical particles are at the vertices of a square, centre of mass will be at the centre of square. If one of the particle is removed as shown below

$x_{cm}=\frac{m\left(0\right)+m\left(a\right)+m\left(0\right)}{3m}$

$x_{cm}=\frac{a}{3}$

$y_{cm}=\frac{m\left(0\right)+m\left(0\right)+m\left(a\right)}{3m}$

$y_{cm}=\frac{a}{3}$

$\therefore Newpositionofcentreofmassfromorigin=c^1=\sqrt{\frac{a^2}{4}+\frac{a^2}{4}}=\frac{\sqrt{2a}}{3}$

$\therefore shift=c-c^1=\frac{a}{\sqrt{2}}-\frac{a\sqrt{2}}{3}=\frac{2\sqrt{2}}{3\sqrt{2}}=\frac{2}{3}m$.