Question

A body of mass $2$ g, moving along the positive $x-$axis in gravity free space with velocity $20$ cms${}^{-1}$ explodes at $x=1$ m, $t=0$ into two pieces of masses $\frac{2}{3}$ g and $\frac{4}{3}$ g. After $5$s, the lighter piece is at the point $(3$m, $2$m, $-4$m$)$. Then the position of the heavier piece at this moment, in metres is

• A. $(1.5,-1,-2)$
• B. $(1.5,-2,-2)$
• C. $(1.5,-1,-1)$
• D. $None$ $of$ $these$

Answer 1

The correct option is D $None$ $of$ $these$

If the body would not have exploded, the CM after $t$ secs would have been at:
$\overrightarrow{r_{cm}}=\overrightarrow{r_{ini}}+v_{cm}t$;
Putting $t=5$,
Let $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$ be the position of the two bodies explosion.
$\overrightarrow{r_{cm}}=\overrightarrow{r_{ini}}+v_{cm}t$;
$v_{cm}=20cm{s}^{-1}=0.2m{s}^{-1}$;
$\overrightarrow{r_{cm}}=1\hat{i}+0.2\times 5\hat{i}=2\hat{i}$
The centre of mass is given by,

$\overrightarrow{r_{cm}}=\frac{m_1\overrightarrow{r_1}+m_1\overrightarrow{r_2}}{m_1+m_2}$
$\Rightarrow 2\hat{i}=\frac{\frac{2}{3}(3\hat{i}+2\hat{j}-4\hat{k})+\frac{4}{3}\overrightarrow{r_2}}{\frac{2}{3}+\frac{4}{3}}$
$\Rightarrow 4\hat{i}=\frac{2}{3}(3\hat{i}+2\hat{j}-4\hat{k})+\frac{4}{3}\overrightarrow{r_2}$
$\Rightarrow 12\hat{i}=2(3\hat{i}+2\hat{j}-4\hat{k})+4\overrightarrow{r_2}$

$\Rightarrow \overrightarrow{r_2}=1.5\hat{i}-\hat{j}+2\hat{k}$.