Three identical spheres, each of mass m kg, a | vedclass

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(B) Let the position of center $A$ be at the origin, so $x_A = 0$.Since the spheres are identical and touching, the distance between consecutive cente

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$\frac{AB + AC}{3}$

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(B) Let the position of center $A$ be at the origin, so $x_A = 0$.Since the spheres are identical and touching, the distance between consecutive centers is equal to the diameter of the spheres, say $d$.Thus, $x_A = 0$, $x_B = d$, and $x_C = 2d$.Note that $AB = d$ and $AC = 2d$.The center of mass $X_{cm}$ is given by:$X_{cm} = \frac{m(x_A) + m(x_B) + m(x_C)}{m + m + m}$$X_{cm} = \frac{m(0) + m(d) + m(2d)}{3m} = \frac{3md}{3m} = d$.Now, expressing $d$ in terms of the given distances:Since $AB = d$ and $AC = 2d$, we have $d = AB$ and $d = AC/2$.Also, $AB + AC = d + 2d = 3d$, so $d = \frac{AB + AC}{3}$.Therefore, the distance of the center of mass from $A$ is $\frac{AB + AC}{3}$.

Three identical spheres, each of mass $m$ kg, are kept touching each other, with their centers on a straight line. If their centers are marked as $A$, $B$, and $C$ respectively, the distance of the center of mass of the system from $A$ is

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