Question

Four bodies $A,B,C$ and $D$ of masses $[M_A=m,M_B=2m,M_C=m,M_D=2m]$ are initially at rest. They start moving under the influence of the internal force of attraction and at a particular instant, their velocities are $\overrightarrow{V_A}=2\hat{i},\overrightarrow{V_B}=-\hat{i}+2\hat{j}$ and $\overrightarrow{V_C}=-2\hat{i}+2\hat{j}$. Find the velocity of body $D\left(\overrightarrow{V_D}\right)$ at that instant.

• A. $2\hat{i}-3\hat{j}$
• B. $\hat{i}-3\hat{j}$
• C. $\hat{i}+3\hat{j}$
• D. $2\hat{i}+3\hat{j}$

Answer 1

The correct option is B $\hat{i}-3\hat{j}$

Bodies are moving under effect of mutual force therefore no net external force on the system.
$\Rightarrow$ No net acceleration of the system.$\Rightarrow a_{com}=0$
Velocity of COM before bodies start moving is zero therefore velocity of COM will remain zero even after they start moving.
${\overrightarrow{V}}_{com}=\frac{m_A\overrightarrow{V_A}+m_B\overrightarrow{V_B}+m_C\overrightarrow{V_C}+m_D\overrightarrow{V_D}}{m_A+m_B+m_C+M_D}$
$0=\frac{m\left(2\hat{i}\right)+2m(-\hat{i}+2\hat{j})+m(-2\hat{i}+2\hat{j})+2m\overrightarrow{V_D}}{m+2m+m+2m}$

$\Rightarrow 2\hat{i}+2(-\hat{i}+2\hat{j})+(-2\hat{i}+2\hat{j})+2\overrightarrow{V_D}=0$
$\Rightarrow 2\overrightarrow{V_D}=2\hat{i}-6\hat{j}$
$\Rightarrow \overrightarrow{V_D}=\hat{i}-3\hat{j}$