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(A) Let the position of the center $P$ be at the origin $(0, 0)$.Since the spheres are identical and touching,the distance between the centers of adja

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$(PQ + PR) / 3$

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(A) Let the position of the center $P$ be at the origin $(0, 0)$.Since the spheres are identical and touching,the distance between the centers of adjacent spheres is equal to the diameter of a sphere,say $d$.Thus,the coordinates of the centers are:$P = (0, 0)$$Q = (d, 0)$$R = (2d, 0)$Note that $PQ = d$ and $PR = 2d$.The $x$-coordinate of the center of mass $(x_{cm})$ is given by:$x_{cm} = \frac{m_P x_P + m_Q x_Q + m_R x_R}{m_P + m_Q + m_R}$Since $m_P = m_Q = m_R = 1 \ kg$:$x_{cm} = \frac{1(0) + 1(d) + 1(2d)}{1 + 1 + 1} = \frac{3d}{3} = d$Since $PQ = d$,the distance of the center of mass from $P$ is $PQ$.However,looking at the options provided,the expression for the center of mass in terms of distances is:$x_{cm} = \frac{0 + PQ + PR}{3} = \frac{PQ + PR}{3}$.

Three identical spheres of mass $1 \ kg$ each are arranged as shown in the figure. The centers of the spheres,which are touching each other,lie on a straight line. If their centers are denoted by $P, Q,$ and $R$,what is the distance of the center of mass of the system from $P$?

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