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  4. Centre of mass (C.M.) of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1,2,3) and C.M. of another system of particles of 3 kg and 2 kg lies at the point (-1,3,-2). Where should we put a particle of mass 5 kg so that the C.M. of entire system lies at the C.M. of the first system ?

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Q. Centre of mass (C.M.) of three particles of masses $1 \,kg$, $2 \,kg$ and $3\, kg$ lies at the point $(1,2,3)$ and C.M. of another system of particles of $3 \,kg$ and $2\, kg$ lies at the point $(-1,3,-2)$. Where should we put a particle of mass $5 \,kg$ so that the C.M. of entire system lies at the C.M. of the first system ?

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Solution:

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$\bar{x}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$
$1=\frac{5(-1)+5 \times x}{10}$
$-5+5 x=10 \Rightarrow x=3$
$\bar{y}=2=\frac{5 \times 3+5 y}{10}, 15+5 y=20, y=1$
$\bar{z}=3=\frac{5 x-2+5 \cdot z}{10},-10+5 z=30, z=8$