Two spheres A and B of masses m_1 and m_2 res | vedclass

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Question

$\begin{array}{l}
According\,to\,law\,of\,conservation\,of\
linerar\,momentum\,along\,x - axis,\,we\,get\
{m_1} 	imes 0 + {m_2} 	imes v = {m_1}

Vedclass · Accepted Answer

$	heta = ta{n^{ - 1}}\left( { - \frac{1}{2}} ight)$ to the $x-$axis

Vedclass · Answer

$\begin{array}{l}
According\,to\,law\,of\,conservation\,of\
linerar\,momentum\,along\,x - axis,\,we\,get\
{m_1} 	imes 0 + {m_2} 	imes v = {m_1}v'\cos 	heta \
{m_2}v = {m_1}v'\cos 	heta \
or\,\cos 	heta  = \frac{{{m_2}v}}{{{m_1}v'}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)\
According\,to\,law\,of\,conservation\,of\,\
linear\,momentum\,along\,y - axis,\,we\
get
\end{array}$

$\begin{array}{l}
{m_1} 	imes 0 + {m_2} 	imes 0 = {m_1}v'\sin 	heta  + {m_2}\frac{v}{2} - {m_2}\frac{v}{2}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {m_1}v'\sin 	heta \
\sin 	heta  =  - \frac{{{m_2}v}}{{2{m_1}v'}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)\
Divide\,\left( {ii} ight)\,by\,\left( i ight),\,we\,get\
	an 	heta  =  - \frac{1}{2}\,or\,	heta  = {	an ^{ - 1}}\left( { - \frac{1}{2}} ight)\,to\,the\,x - axis
\end{array}$

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