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A bomb at rest explodes into three equal fragments

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

(A) 600 N

(B) 60 N

(D) 1200 N

Solution :-

Mass of each fragment = $\frac{12/}{3} = 4 kg$ .

As two fragments fly off at right angles, let one moves along x-axis and another along y-axis. Let the velocity of third fragment be $\displaystyle \overrightarrow{v}$ .

Applying Conservation of Linear Momentum,

p_f = p_i

$\displaystyle \left( 4\times 12\widehat{i} \right)+\left( 4\times 9\widehat{j} \right)+4\overrightarrow{v}=0$

∴ Speed of third particle, $\displaystyle \left| \overrightarrow{v} \right|=\sqrt{{{12}^{2}}+{{9}^{2}}}=15m/s$

Average force acting on the third particle = change in momentum/time

$\displaystyle \Rightarrow F=\frac{4\times 15-0}{0.1}=600N$

A bomb at rest explodes into three equal fragments

A bomb at rest explodes into three equal fragments

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