Q125 / 155 · WBJEE 2021 · Physics
Centre of mass (C.M.) of three particles of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}$ and $3 \mathrm{~kg}$ lies at the point $(1,2,3)$ and C.M. of another system of particles of $3 \mathrm{~kg}$ and $2 \mathrm{~kg}$ lies at the point $(-1,3,-2)$. Where should we put a particle of mass $5 \mathrm{~kg}$ so that the C.M. of entire system lies at the C.M. of the first system ?
- Correct answerA$(3,1,8)$
- B$(0,0,0)$
- C$(1,3,2)$
- D$(-1,2,3)$
Solution
$\left.\begin{array}{l}6 \mathrm{~kg}[1,2,3] \\ 5 \mathrm{~kg}[-1,3,-2] \\ 5 \mathrm{~kg}[\mathrm{x}, \mathrm{y}, \mathrm{z}]\end{array}\right) \rightarrow(\overline{\mathrm{x}}, \overline{\mathrm{y}}, \overline{\mathrm{z}})=(1,2,3)$
$\bar{x}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$
$1=\frac{5(-1)+5 \times x}{10}$
$-5+5 x=10 \Rightarrow x=3$
$\bar{y}=2=\frac{5 \times 3+5 y}{10}, 15+5 y=20, y=1$
$\bar{z}=3=\frac{5(-2)+5 . z}{10},-10+5 z=30, z=8$
Hence, the correct answer is option: 'A'.