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  • A shell of mass is at rest initially. It explodes into three fragments having mass in the ratio 2:2:1. If the fragments having equal ma

A shell of mass \mathrm{m} is at rest initially. It explodes into three fragments having mass in the ratio 2:2:1. If the fragments having equal mass fly off along mutually perpendicular directions with speed \mathrm{v}, the speed of the third (lighter) fragment is :

Option: 1

\mathrm{\sqrt{2}v}


Option: 2

\mathrm{2\sqrt{2}v}


Option: 3

\mathrm{3\sqrt{2}v}


Option: 4

\mathrm{v}


Answers (1)

best_answer

The ratio of the masses of the fragments is

\mathrm{m_{1}: m_{2}: m_{3}=2: 2: 1} \\

\mathrm{m_{1}+m_{2}+m_{3}=m} \\

\mathrm{\therefore m_{1}=m_{2}=0.4 \mathrm{~m} }\\ \mathrm{\& \;m_{3}=0.2 \mathrm{~m}}

Since initially the shell is at rest and also external force is zero

\mathrm{\therefore \bar{P_{i}}=\bar{P}_{f}=0}

\mathrm{\bar{P}_{f}=m_{1} v(\hat{i})+m_{2} v(\hat{j})+m_{3} \overline{v_{0}} }\\

\mathrm{\bar{V}_{0}=-(2 v \hat{i}+2 v \hat{j})} \\

\mathrm{V_{0}=\left(\sqrt{(2)^{2}+(2)^{2}}\right)=2 \sqrt{2} V }

The speed of the lighter fragment is 2\sqrt{2}V

Hence the correct option is 2

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