A stationary body of mass \(3\mathrm{kg}\) explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocity of \(3\mathrm{ms}^{-1}\) and the other with a velocity of \(4\mathrm{ms}^{-1}\) . If the explosion occurs in \(10^{-4}\mathrm{s}\) , the average force acting on the third piece in newton is
Choose Your Option
A. \((3\mathrm{i} + 4\mathrm{j})\times 10^{-4}\)
B. \((3\mathrm{i} - 4\mathrm{j})\times 10^{-4}\)
C. \((3\mathrm{i} - 3\mathrm{j})\times 10^{4}\)
D. \((3\mathrm{i} + 4\mathrm{j})\times 10^{4}\)
Solution & Explanation
No details explanation available for this question. Option D is the verified correct answer.