Question

A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off at right angles to each other. One with a velocity of 2 $\hat{i}$m/s and the other with a velocity of 3 $\hat{j}$ m/s. If the explosion takes place in 10${}^{-5}$s, the average force acting on the third piece in newtons is :

• A. $(2\hat{i}+3\hat{j})\times {10}^{-5}$
• B. $-(2\hat{i}+3\hat{j})\times {10}^5$
• C. $(3\hat{i}+2\hat{j})\times {10}^5$
• D. $-(2\hat{i}+3\hat{j})\times {10}^{-5}$

Answer 1

Answer: (B)

$-(2\hat{i}+3\hat{j})\times {10}^5$

Mass of each fragment $m=\frac{M}{3}=\frac{3kg}{3}=1kg$

Let the momentum of third piece be $\overrightarrow{P_3}$ and that of other two be $\overrightarrow{P_1}$ and $\overrightarrow{P_2}$

$\therefore$ $\overrightarrow{P_1}=m{v}_1=2\hat{i}$ and $\overrightarrow{P_2}=m{v}_2=3\hat{j}$

Initial momentum of the bomb is zero i.e $\overrightarrow{P_i}=0$

Applying conservation of momentum:

$0=\overrightarrow{P_1}+\overrightarrow{P_2}+\overrightarrow{P_3}$

$0=2\hat{i}+3\hat{j}+\overrightarrow{P_3}$ $⟹\overrightarrow{P_3}=-(2\hat{i}+3\hat{j})$ $kgm{s}^{-1}$

Now average force acting on third piece $\overrightarrow{F}=\frac{\overrightarrow{P_3}}{t}=\frac{-(2\hat{i}+3\hat{j})}{{10}^{-5}}$

$⟹\overrightarrow{F}=-(2\hat{i}+3\hat{j})\times {10}^5$ N