Question

Four particles are in $x-y$ plane at :
$\left(1\right)1kg$ at $(0,0)\left(2\right)2kg$ at $(1,0)\left(3\right)3kg$ at $(1,2)\left(4\right)4kg$ at $(2,0)$
The centre of mass is located at:

• A. $(0.3,1.2)$
• B. $(1.3,0.6)$
• C. $(0.5,1.4)$
• D. $(1.2,0.3)$

Answer 1

The correct option is B $(1.3,0.6)$

Centre of mass is given by the relation, $(\sum m_i\times X_i)/\sum m_i$ and $(\sum m_i\times Y_i)/\sum m_i$

Thus we get $(\frac{\left(1\right(0)+2(1)+4(2)+3(1\left)\right)}{(1+2+3+4)},\frac{1\left(0\right)+2\left(0\right)+3\left(2\right)+4\left(0\right)}{1+2+3+4})$

Thus we get the coordinates as $(1.3,0.6)$