On a frictionless surface, a block of mass M | vedclass

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Question

$\begin{array}{l}\,\,\,\,\,\,\,\,\,The\,situation\,is\,shown\,in\,the\,figure.\Let\,v'\,be\,speed\,of\,{m{second}}\,block\,after\the\,collisio

Vedclass · Accepted Answer

$\frac{{2\sqrt 2 }}{3}V$

Vedclass · Answer

$\begin{array}{l}\,\,\,\,\,\,\,\,\,The\,situation\,is\,shown\,in\,the\,figure.\Let\,v'\,be\,speed\,of\,{m{second}}\,block\,after\the\,collision.\,As\,the\,collision\,is\,elastic,\so\,kinetic\,energy\,is\,conserved.\According\,to\,conservation\,of\,kinetic\energy,\\frac{1}{2}M{v^2} + 0 = \frac{1}{2}M{\left( {\frac{v}{3}} ight)^2} + \frac{1}{2}Mv{'^2}\end{array}$

$\begin{array}{l}{v^2} = \frac{{{v^2}}}{9} + {v^2}\,or\,v{'^2} = {v^2} - \frac{{{v^2}}}{9}\ = \frac{{9{v^2} - {v^2}}}{9} = \frac{8}{9}{v^2}\v' = \sqrt {\frac{8}{9}{v^2}}  = \frac{{\sqrt 8 }}{3}v = \frac{{2\sqrt 2 }}{3}v\end{array}$

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