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 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 newton 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)

As the two pieces out of these pieces fly off at right angle. Then, the opposite resultant velocity of these two pieces will be equal to the velocity of the third piece $\therefore$ $v=-(2\hat{i}+3\hat{j})$ Since, 3 kg mass is splitted in to three pieces, then each piece will be of mass 1 kg. Now, average force acting on the third piece will be $F=\frac{mv}{t}$ $=\frac{1\times [-(2\hat{i}+3\hat{j})]}{{10}^{-5}}$ $=-(2\hat{i}+3\hat{j}){10}^{5}N$