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Home » A particle of mass at rest suddenly breaks on its own into three fragments. Two fragments of mass each move along mutually perpendicular direction with speed v each. The energy released during the process is, (1) (2) (3) (4)
A particle of mass 5 \mathrm{~m} at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. The energy released during the process is, (1) \frac{4}{3} \mathrm{mv}^{2} (2) \frac{3}{5} \mathrm{mv}^{2} (3) \frac{5}{3} m v^{2} (4) \frac{3}{2} \mathrm{mv}^{2}
Physics | Physics
A particle of mass 5 \mathrm{~m} at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with speed v each. The energy released during the process is, (1) \frac{4}{3} \mathrm{mv}^{2} (2) \frac{3}{5} \mathrm{mv}^{2} (3) \frac{5}{3} m v^{2} (4) \frac{3}{2} \mathrm{mv}^{2}

Answer (1)
Sol. From conservation of linear momentum.
0=m v \hat{j}+m v \hat{i}+3 m v_{1}
\vec{v}_{1}=-\frac{v}{3}(\hat{i}+\hat{j})
v_{1}=\frac{\sqrt{2}}{3} v
K E_{i}=0
K E_{f}=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m)\left(\frac{\sqrt{2}}{3}\right)^{2} v^{2}
=m v^{2}+\frac{m v^{2}}{3}=\frac{4}{3} m v^{2}
\Delta K E=K E_{f}-K E_{i}=\frac{4}{3} m v^{2}

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