The centre of mass of three particles of mass | vedclass

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(A) Let the masses be $m_1=1 ~kg, m_2=2 ~kg, m_3=3 ~kg$ and $m_4=4 ~kg$.The initial centre of mass $(X_{CM}, Y_{CM}, Z_{CM})$ of the first three parti

Vedclass · Accepted Answer

$(-3,-3,-3)$

Vedclass · Answer

(A) Let the masses be $m_1=1 ~kg, m_2=2 ~kg, m_3=3 ~kg$ and $m_4=4 ~kg$.The initial centre of mass $(X_{CM}, Y_{CM}, Z_{CM})$ of the first three particles is $(2,2,2)$.The formula for the centre of mass is $X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$.For the first three particles,the sum of masses is $M = 1+2+3 = 6 ~kg$.Thus,$\sum m_i x_i = X_{CM} 	imes M = 2 	imes 6 = 12$. Similarly,$\sum m_i y_i = 12$ and $\sum m_i z_i = 12$.Let the position of the fourth mass $m_4$ be $(x_4, y_4, z_4)$.The new centre of mass is $(0,0,0)$.Using the formula for the new centre of mass: $X_{CM}^{\prime} = \frac{\sum m_i x_i + m_4 x_4}{M + m_4} = 0$.Substituting the values: $0 = \frac{12 + 4x_4}{6 + 4} \implies 12 + 4x_4 = 0 \implies 4x_4 = -12 \implies x_4 = -3$.Similarly,$y_4 = -3$ and $z_4 = -3$.Therefore,the position of the fourth mass is $(-3,-3,-3)$.

The centre of mass of three particles of masses $1 ~kg, 2 ~kg$ and $3 ~kg$ is at $(2,2,2)$. The position of the fourth mass of $4 ~kg$ to be placed in the system so that the new centre of mass is at $(0,0,0)$ is:

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