Question

Three particles of masses $2\text{kg},3\text{kg}$ & $4\text{kg}$ are placed at the three vertices of a right angled triangle with sides $3\text{cm},4\text{cm}$ & $5\text{cm}$ as shown in the figure. What will be the coordinates of centre of mass of the system of particles ?

• A. $(\frac{16}{9},2)$
• B. $(1,\frac{9}{16})$
• C. $(\frac{16}{9},1)$
• D. $(1,\frac{16}{9})$

Answer 1

The correct option is C $(\frac{16}{9},1)$

Assigning origin at mass $2\text{kg}$ and writing the $x$ and $y$ coordinates of masses in the table given below:


$\begin{array}{ccc}\text{mass}& x\left(\text{cm}\right)& y\left(\text{cm}\right)\\ m_1=2\text{kg}& 0& 0\\ m_2=3\text{kg}& 0& 3\\ m_3=4\text{kg}& 4& 0\end{array}$

Applying the formula for the coordinates of $\text{CM}$ of system:
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=\frac{(2\times 0)+(3\times 0)+(4\times 4)}{2+3+4}$
$\therefore x_{\text{CM}}=\frac{16}{9}$
Similarly
$y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}=\frac{(2\times 0)+(3\times 3)+(4\times 0)}{2+3+4}=\frac{9}{9}=1$

$\therefore (x_{\text{CM}},y_{\text{CM}})=(\frac{16}{9},1)$