Question

For a system of two particles of mass $2\text{kg}$ & $4\text{kg}$ lying on the same straight line. If mass $4\text{kg}$ is displaced rightwards by $5\text{m}$, then by what distance, $2\text{kg}$ must be moved to maintain the position of centre of mass of the system ?

• A. $10\text{m}$
• B. $15\text{m}$
• C. $20\text{m}$
• D. $25\text{m}$

Answer 1

The correct option is A $10\text{m}$

Since particles are placed along a straight line, $\text{CM}$ of system can be expressed in terms of $x$ coordinates only.

$m_1=2\text{kg}$
$m_2=4\text{kg}$

Initial $\text{COM}$ of system:
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$\Rightarrow x_{\text{CM}}=\frac{2x_1+4x_2}{6}.....\left(i\right)$

Considering rightwards as $+$ve $x$ direction. Let mass $2\text{kg}$ be displaced by $d$ metres rightwards to maintain the position of $\text{CM}$.

$\therefore {x_2}^{\prime }\to (x_2+5)$
${x_1}^{\prime }\to (x_1+d)$

New $\text{COM}$ of system:

${x_{\text{CM}}}^{\prime }=\frac{m_1{x_1}^{\prime }+m_2{x_2}^{\prime }}{m_1+m_2}=\frac{2(x_1+d)+4(x_2+5)}{2+4}.....\left(ii\right)$

Equating both Eq. $\left(i\right)$ and $\left(ii\right)$, because both denote the same position of $\text{CM}$.

$\frac{2x_1+4x_2}{6}=\frac{2x_1+4x_2+2d+20}{6}$
$\Rightarrow 2d+20=0$
$\therefore d=-10\text{m}$
which indicates that mass $2\text{kg}$ should be displaced by $10\text{m}$ leftwards.