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5.Work, Energy, Power and Collision
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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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