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Question

(B) Let the masses of the three spheres be $m_1 = m_2 = m_3 = 1 \ kg$.Let the positions of the centres $P, Q, R$ be $x_P, x_Q, x_R$ along the $x$-axis

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$\frac{PQ + PR}{3}$

Vedclass · Answer

(B) Let the masses of the three spheres be $m_1 = m_2 = m_3 = 1 \ kg$.Let the positions of the centres $P, Q, R$ be $x_P, x_Q, x_R$ along the $x$-axis.Since $P$ is the origin,$x_P = 0$.The position of the centre of mass $(x_{cm})$ is given by the formula:$x_{cm} = \frac{m_1 x_P + m_2 x_Q + m_3 x_R}{m_1 + m_2 + m_3}$Substituting the values:$x_{cm} = \frac{1 	imes 0 + 1 	imes PQ + 1 	imes PR}{1 + 1 + 1} = \frac{PQ + PR}{3}$Since all centres lie on the $x$-axis,the $y$-coordinate of the centre of mass is $y_{cm} = 0$.Thus,the distance of the centre of mass from $P$ is $\frac{PQ + PR}{3}$.

Column-$I$	Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$	$(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$	$(b)$ Position vector of the center of mass for a system of two equal masses

Three identical spheres,each of mass $1 \ kg$,are kept as shown in the figure,touching each other,with their centres on a straight line. If their centres are marked $P, Q, R$ respectively,the distance of the centre of mass of the system from $P$ is

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Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$ $(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$ $(b)$ Position vector of the center of mass for a system of two equal masses

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