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Question

$\begin{array}{l}
Let\,\bar v'\,be\,velocity\,of\,third\,piece\,of\,mass\
2m.\
Initial\,momentum\,,\,{{\bar p}_i} = 0\left( {As\,the\,body\,i

Vedclass · Accepted Answer

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Vedclass · Answer

$\begin{array}{l}
Let\,\bar v'\,be\,velocity\,of\,third\,piece\,of\,mass\
2m.\
Initial\,momentum\,,\,{{\bar p}_i} = 0\left( {As\,the\,body\,is\,at\,rest} ight)\
Final\,momentum,\,{{\bar P}_i} = mv\hat i + mv\hat j + 2m\bar v'\
According\,to\,law\,of\,conservation\,of\
momentum\,\
\,\,\,\,\,\,\,\,\,\,\,\,{{\bar P}_i} = {{\bar P}_f}\
0 = mv\hat i + mv\hat j + 2m\bar v'\
\,\,\,\,\,\,\,\,\,\,\,\,\bar v' =  - \frac{v}{2}\hat i - \frac{v}{2}\,\hat j
\end{array}$

$\begin{array}{l}
The\,magnitude\,of\,v'\,is\
\,\,\,\,\,\,\,\,\,\,\,v' = \sqrt {{{\left( { - \frac{v}{2}} ight)}^2} + {{\left( { - \frac{v}{2}} ight)}^2}}  = \frac{v}{{\sqrt 2 }}\
Total\,kinetic\,energy\,generated\,due\,to\
{m{explosion}}\,\
 = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {2m} ight)v'2\
 = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {2m} ight){\left( { - \frac{v}{{\sqrt 2 }}} ight)^2}\
 = m{v^2} + \frac{{m{v^2}}}{2} = \frac{3}{2}m{v^2}
\end{array}$

A body of mass $(4m)$ is lying in $x-y$ plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass $(m)$ move perpendicular to each other with equal speeds $(v).$ The total kinetic energy generated due to explosion is................. $mv^2$

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