Question

The centre of mass of a system of three particles of masses $2\text{g},3\text{g}$ and $5\text{g}$ is taken as the origin of a co-ordinate system. The position vector of a fourth particle of mass $8\text{g}$, such that the centre of mass of the four particle system lies at the point $(3,4,7)\text{m}$, is $\lambda (3\hat{i}+4\hat{j}+7\hat{k})\text{m}$ where $\lambda$ is a constant. Find the value of $\lambda$.

• A. $2.7$
• B. $5.4$
• C. $1.35$
• D. $2.25$

Answer 1

The correct option is D $2.25$

Centre of mass $2\text{g},3\text{g}$ and $5\text{g}$ is at origin.
$\Rightarrow \frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=0\dots \left(i\right)$

Position vector of $8\text{g}$ particle is $\lambda (3\hat{i}+4\hat{j}+7\hat{k})\text{m}$
$x$-coordinate of centre of mass of system of particles consisting of $2\text{g},3\text{g},5\text{g}$ and $8\text{g}$ particles:-
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
From (i): $m_1{x}_1+m_2{x}_2+m_3{x}_3=0$
$\Rightarrow 3=\frac{0+8\left(3\lambda \right)}{2+3+5+8}$
$\Rightarrow 3=\frac{24\lambda }{18}$
$\Rightarrow \lambda =2.25$
[Note:- Value of $\lambda$ can be found out through equating any one of $x_{com}$,$y_{com}$, or $z_{com}$. In this problem , $\lambda$ is found out by equating $x_{com}$]