On a frictionless surface, a block of mass M moving at speed V co… - Vidyadip

MCQ

On a frictionless surface, a block of mass $M$ moving at speed $V$ collides elastically with another block of same mass $M$ which is initially at rest. After collision the first block moves at an angle $\theta$ to its initial direction and has a speed $\frac{V}{3}$ The second block's speed after the collision is

✓
$\frac{{2\sqrt 2 }}{3}V$
B
$\frac{{\sqrt 3 }}{2}V$
C
$\frac{3}{4}V$
D
$\frac{3}{{\sqrt 2 }}V$

✓

Answer

Correct option: A.

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

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

$\begin{array}{l}{v^2} = \frac{{{v^2}}}{9} + {v^2}\,or\,v{'^2} = {v^2} - \frac{{{v^2}}}{9}\\ = \frac{{9{v^2} - {v^2}}}{9} = \frac{8}{9}{v^2}\\v' = \sqrt {\frac{8}{9}{v^2}} = \frac{{\sqrt 8 }}{3}v = \frac{{2\sqrt 2 }}{3}v\end{array}$

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