Question

A bomb of mass $2\text{kg}$ at rest explodes into two chunks of masses $0.5\text{kg}$ and $1.5\text{kg}$. Velocity of $0.5\text{kg}$ chunk is ($2\hat{i}+4\hat{j}+8\hat{k}$) $\text{m/s}$. Velocity of $1.5\text{kg}$ chunk is -

• A. $(\frac{2}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{8}{3}\hat{k})$ $\text{m/s}$
• B. $-(\frac{2}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{8}{3}\hat{k})$ $\text{m/s}$
• C. $(\frac{2}{3}\hat{i}-\frac{4}{3}\hat{j}+\frac{8}{3}\hat{k})$ $\text{m/s}$
• D. $(\frac{2}{3}\hat{i}-\frac{4}{3}\hat{j}-\frac{8}{3}\hat{k})$ $\text{m/s}$

Answer 1

The correct option is B $-(\frac{2}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{8}{3}\hat{k})$ $\text{m/s}$

According to question


We know that in an explosion, internal forces operates and $\sum F_{int}=0$. Also, there is no external forces, $\sum F_{ext}=0$.
On applying law of conservation of momentum, we get
$P_i=P_f$
$\Rightarrow 0=0.5(2\hat{i}+4\hat{j}+8\hat{k})+1.5{\overrightarrow{V}}_B$
$\Rightarrow \frac{3}{2}{\overrightarrow{V}}_B=-(\hat{i}+2\hat{j}+4\hat{k})$
$\Rightarrow {\overrightarrow{V}}_B=\frac{-2}{3}(\hat{i}+2\hat{j}+4\hat{k})\text{m/s}$
Velocity of $1.5\text{kg}$ chunk is $-(\frac{2}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{8}{3}\hat{k})$ $\text{m/s}$