Question

A bomb at rest explodes into three equal fragments. Two fragments fly off at right angles with velocity $12\text{m/s}$ and $9\text{m/s}$. If the explosion process occurs in $0.1\text{s}$, find out the average force acting on the third particle, if total mass of the bomb is $12\text{kg}$.

• A. $600\text{N}$
• B. $60\text{N}$
• C. $15\text{N}$
• D. $1200\text{N}$

Answer 1

The correct option is A $600\text{N}$

Bomb explodes into three equal parts, therefore mass of each fragment will be $\frac{12}{3}=4\text{kg}$.


Let the velocities of the first two fragments be $12\hat{i}$ and $9\hat{j}$. Let the velocity of third component be $\overrightarrow{v}$.

Applying momentum conservation :
Initial momentum $=$ Final monentum
$0=4\times 12\hat{i}+4\times 9\hat{j}+4\overrightarrow{v}$
$\Rightarrow \overrightarrow{v}=-12\hat{i}-9\hat{j}$
Therefore, speed of the third particle $=|\overrightarrow{v}|=\sqrt{{12}^2+9^2}=15\text{m/s}$

Average force acting on the third particle $=\frac{\text{change in momentum}}{\text{time}}$
$=\frac{m\times |\overrightarrow{v}|-0}{t}=\frac{4\times 15}{0.1}=600\text{N}$