A vessel at rest explodes into three pieces. | vedclass

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(A) Let the two pieces have equal mass $m$ and the third piece have a mass of $3m$.According to the law of conservation of linear momentum,since the i

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$10\sqrt{2} \, m/s$ and $135^{\circ}$ from either.

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(A) Let the two pieces have equal mass $m$ and the third piece have a mass of $3m$.According to the law of conservation of linear momentum,since the initial momentum of the system was zero,the final momentum must also be zero.Let the velocities of the two pieces be $\vec{v}_1 = 30\hat{i} \, m/s$ and $\vec{v}_2 = 30\hat{j} \, m/s$.The momentum of these two pieces is $\vec{p}_1 = m(30\hat{i})$ and $\vec{p}_2 = m(30\hat{j})$.The resultant momentum of these two pieces is $\vec{p}_{12} = \vec{p}_1 + \vec{p}_2 = 30m\hat{i} + 30m\hat{j}$.The magnitude of this resultant momentum is $p_{12} = \sqrt{(30m)^2 + (30m)^2} = 30m\sqrt{2}$.For the total momentum to be zero,the momentum of the third piece $\vec{p}_3$ must be equal and opposite to $\vec{p}_{12}$.Thus,$3m \cdot V = 30m\sqrt{2}$,where $V$ is the magnitude of the velocity of the third piece.$V = \frac{30m\sqrt{2}}{3m} = 10\sqrt{2} \, m/s$.The direction of the third piece must be opposite to the resultant of the first two pieces. Since the resultant of the two perpendicular pieces makes an angle of $45^{\circ}$ with each,the third piece must make an angle of $180^{\circ} - 45^{\circ} = 135^{\circ}$ with each of the first two pieces.

$A$ vessel at rest explodes into three pieces. Two pieces having equal masses $m$ fly off perpendicular to one another with the same velocity $30 \, m/s$. The third piece has three times the mass of each of the other pieces. The magnitude and direction of the velocity of the third piece will be:

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