Question

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

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

Answer 1

The correct option is B $-(2\hat{i}+3\hat{j}){10}^5$ N

The situation is shown in the adjacent figure.

As there is no external force in this situation, linear momentum is conserved.

And masses are also equal (that is $1kg$) thus,

the resultant of $\overrightarrow{P_1}$ and $\overrightarrow{P_2}$ must be equal and opposite to $\overrightarrow{P_3}$, i.e. $\overrightarrow{P_3}=-(\overrightarrow{P_1}+\overrightarrow{P_2})$.

The average force acting on it will be $F=\frac{P_3}{t}=-(2\hat{i}+3\hat{j})\times {10}^{5}N$

*$\overrightarrow{P}$ $=$ momentum vector