Question

Three identical spheres each of mass $1\text{kg}$ are kept as shown in the figure. They are kept in such a way that they are touching each other with their respective centres on a straight line . If their centres are marked $P,Q,R$ respectively, then the distance of the centre of mass of the system from $P$ is:

• A. $\frac{PQ+QR+PR}{3}$
• B. $\frac{PQ+PR}{3}$
• C. $\frac{PQ+PR}{2}$
• D. $\frac{PQ+QR}{2}$

Answer 1

The correct option is B $\frac{PQ+PR}{3}$


Since the centre of spheres are lying on a straight line, we can consider the coordinates of their centers on x-axis.
Assigning origin i.e $x=0$ at point $P$:
$m_1=1\text{kg}$, $x_1=0$
$m_2=1\text{kg}$, $x_2=PQ$
$m_3=1\text{kg}$, $x_3=PR$

Applying the formula for $x$-coordinate of COM:
$x_{\text{COM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

$=\frac{(1\times 0)+(1\times PQ)+(1\times PR)}{1+1+1}$

$\therefore x_{\text{COM}}=\left(\frac{PQ+PR}{3}\right)$
This is the distance of COM from point $P$.