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Question

Physics Question on work, energy and power

A particle of mass 5 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 :

A

35mv2\frac{3}{5}mv^{2}

B

53mv2\frac{5}{3}mv^{2}

C

32mv2\frac{3}{2}mv^{2}

D

43mv2\frac{4}{3}mv^{2}

Answer

43mv2\frac{4}{3}mv^{2}

Explanation

Solution

3mv+mvi^+mvj^=03m\vec{v} + mv\hat{i} + mv\hat{j} = 0
v=v3i^v3j^\Rightarrow \vec{v} = -\frac{v}{3} \hat{i} - \frac{v}{3} \hat{j}
v=2v3|\vec{v}| = \sqrt{2} \frac{v}{3}
Energy released
=12mv2+12mv2+12(3m)(2v29)= \frac{1}{2} mv^2 + \frac{1}{2} mv^2 + \frac{1}{2} (3m) (\frac{2v^2}{9})
=43mv2=\frac{4}{3} mv^2