Question

Two bodies with masses $m_1=2\text{kg}$ and $m_2=4\text{kg}$ have initial velocities $v_1=v_2=5\text{m/s}$ in the directions shown in the figure. If they collide at the origin elastically, then find the position of the center of mass $3\text{seconds}$ after the collision.

• A. $-4\hat{i}-7\hat{j}$
• B. $-4\hat{i}+7\hat{j}$
• C. $4\hat{i}-7\hat{j}$
• D. $4\hat{i}+7\hat{j}$

Answer 1

The correct option is B $-4\hat{i}+7\hat{j}$


Let us calculate the velocity of centre of mass of the system in the given situation.
Let $(v_{\text{com}}{)}_x$ be the velocity of centre of mass in $x$ direction.
$(v_{\text{com}}{)}_x=\frac{m_1{v}_{1x}+m_2{v}_{2x}}{m_1+m_2}$
Similarly,
$(v_{\text{com}}{)}_y=\frac{m_1{v}_{1y}+m_2{v}_{2y}}{m_1+m_2}$
So,
$(v_{\text{com}}{)}_x=\frac{2(-5\cos {37}^{\circ })+(4\times 0)}{6}=\frac{-4}{3}\text{m/s}$
$(v_{\text{com}}{)}_y=\frac{2(-5\sin {37}^{\circ })+(4\times 5)}{6}=\frac{7}{3}\text{m/s}$

Therefore, velocity of the center of mass of the system
${\overrightarrow{v}}_{com}=(\frac{-4}{3}\hat{i}+\frac{7}{3}\hat{j})\text{m/s}$

Here, net force on the system in both directions $x$ and $y$ is zero. So, ${\overrightarrow{v}}_{com}$ will remain unchanged.
Since the collision occurs at the origin, $\overrightarrow{r_i}=0$. So position vector of center of mass $3\text{seconds}$ after collision is
${\overrightarrow{r}}_{\text{com}}(t=3)=\overrightarrow{r_i}+({\overrightarrow{v}}_{\text{com}}\times t)$
$=0+(\frac{-4}{3}\hat{i}+\frac{7}{3}\hat{j})\times 3=-4\hat{i}+7\hat{j}$
Thus, option (b) is the correct answer.