Centre of mass (C.M.) of three particles of m | vedclass

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(A) Let the first system have a total mass $M_1 = 1 + 2 + 3 = 6 \,kg$ with $C.M.$ at $R_1 = (1, 2, 3)$.Let the second system have a total mass $M_2 =

Vedclass · Accepted Answer

$(3, 1, 8)$

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(A) Let the first system have a total mass $M_1 = 1 + 2 + 3 = 6 \,kg$ with $C.M.$ at $R_1 = (1, 2, 3)$.Let the second system have a total mass $M_2 = 3 + 2 = 5 \,kg$ with $C.M.$ at $R_2 = (-1, 3, -2)$.We add a third particle of mass $M_3 = 5 \,kg$ at position $R_3 = (x, y, z)$.The $C.M.$ of the entire system is given as $R_{cm} = (1, 2, 3)$.The formula for the $C.M.$ of the combined system is $R_{cm} = \frac{M_1 R_1 + M_2 R_2 + M_3 R_3}{M_1 + M_2 + M_3}$.Substituting the values: $(1, 2, 3) = \frac{6(1, 2, 3) + 5(-1, 3, -2) + 5(x, y, z)}{6 + 5 + 5}$.Total mass $M = 16 \,kg$. So,$16(1, 2, 3) = (6, 12, 18) + (-5, 15, -10) + (5x, 5y, 5z)$.$(16, 32, 48) = (1, 27, 8) + (5x, 5y, 5z)$.For $x$-coordinate: $16 = 1 + 5x \Rightarrow 5x = 15 \Rightarrow x = 3$.For $y$-coordinate: $32 = 27 + 5y \Rightarrow 5y = 5 \Rightarrow y = 1$.For $z$-coordinate: $48 = 8 + 5z \Rightarrow 5z = 40 \Rightarrow z = 8$.Thus,the position is $(3, 1, 8)$.

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