Question

A stationary body of mass $3\text{kg}$, explodes into three equal parts. Two of the pieces fly off at right angles to each other, with velocities of $2\hat{i}\text{m/s}$ and $3\hat{j}\text{m/s}$. If the explosion takes place in ${10}^{-3}\text{s}$, find out the average force exerted on the third piece.

• A. $(-2\hat{i}-3\hat{j})\times {10}^3\text{N}$
• B. $(2\hat{i}+3\hat{j})\times {10}^3\text{N}$
• C. $(2\hat{i}-3\hat{j})\times {10}^{-3}\text{N}$
• D. None of these

Answer 1

The correct option is A $(-2\hat{i}-3\hat{j})\times {10}^3\text{N}$

According to conservation of momentum principle,

${\overrightarrow{p}}_i={\overrightarrow{p}}_{1_f}+{\overrightarrow{p}}_{2_f}+{\overrightarrow{p}}_{3_f}$

$0=2\hat{i}+3\hat{j}+{\overrightarrow{p}}_{3_f}$

$\therefore {\overrightarrow{p}}_{3_f}=-2\hat{i}-3\hat{j}$

For the third piece, average force,

$\overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}=\frac{(-2\hat{i}-3\hat{j})-0}{{10}^{-3}}$

$\Rightarrow \overrightarrow{F}=(-2\hat{i}-3\hat{j})\times {10}^3\text{N}$