Two particles of masses $5 \mathrm{~g}$ and $3 \mathrm{~g}$ are separated by a distance of $40 \mathrm{~cm}$…

Two particles of masses $5 \mathrm{~g}$ and $3 \mathrm{~g}$ are separated by a distance of $40 \mathrm{~cm}$. The centre of mass of the system of these two particles
  1. lies at a distance of $15 \mathrm{~cm}$ from $5 \mathrm{~g}$ particle
  2. lies at a distance of $25 \mathrm{~cm}$ from $5 \mathrm{~g}$ particle
  3. lies at a distance of $10 \mathrm{~cm}$ from $3 \mathrm{~g}$ particle
  4. lies at the mid point of the line joining the two particles

Solution

Suppose centre of mass is $\mathrm{x} \mathrm{cm}$ away from $3 \mathrm{~g}$ and assuming that $3 \mathrm{~g}$ is at $\mathrm{x}=0$
$\begin{aligned} & x_{c m}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\ & =\frac{5 \times 40+3 \times 0}{5+3}=\frac{5 \times 40}{8} \\ & =25 \mathrm{~cm} \end{aligned}$ The centre of mass of the system of these two particles lies at a distance $25 \mathrm{~cm}$ from $5 \mathrm{~g}$ particle.

Asked in: AP EAMCET 2023 (15 May Shift 2)

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