Gujarati
Hindi
5.Work, Energy, Power and Collision
normal

A particle of mass $4\, m$ which is at rest explodes into three fragments. Two of the  fragments each of mass $m$ are found to move with a speed $v$ each in perpendicular directions. The total energy released in the process will be

A

$3\,mv^2$

B

$\frac{7}{2} \, mv^2$

C

$\frac{3}{2} \, mv^2$

D

$4\,mv^2$

Solution

$\mathrm{p}_{3}=\sqrt{\mathrm{p}_{1}^{2}+\mathrm{p}_{2}^{2}}=\sqrt{2} \mathrm{mv}$

$\Rightarrow 2 \mathrm{mv}_{3}=\sqrt{2} \mathrm{mv}$

$\mathrm{v}_{3}=\frac{\mathrm{V}}{\sqrt{2}}$

$\mathrm{KE}_{f}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}(2 \mathrm{m})\left(\frac{\mathrm{v}}{\sqrt{2}}\right)^{2}$

$=\frac{3}{2} \mathrm{mv}^{2}$

Standard 11
Physics

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