From momentum conservation

m(u)$$\widehat i$$ + 3m(0) = mv$$\widehat j$$ + 3m$$\overrightarrow {v'} $$

$$ \Rightarrow $$ 3m$$\overrightarrow {v'} $$ = m(u)$$\widehat i$$ - mv$$\widehat j$$

$$ \Rightarrow $$ $$\overrightarrow {v'} = {{u\widehat i - v\widehat j} \over 3}$$

$$ \therefore $$ $$\left| {\overrightarrow {v'} } \right| = {{\sqrt {{u^2} + {v^2}} } \over 3}$$

$$ \Rightarrow $$ $${\left| {\overrightarrow {v'} } \right|^2} = {{{u^2} + {v^2}} \over 9}$$ ......(1)

As collision is perfectely elastic hence

KE_i = KE_j

$$ \Rightarrow $$ $${1 \over 2}m{u^2} + {1 \over 2}\left( {3m} \right) \times {0^2} = {1 \over 2}m{v^2} + {1 \over 2}3m{\left( {v'} \right)^2}$$

$$ \Rightarrow $$ u² = v² + 3$${\left( {v'} \right)^2}$$

$$ \Rightarrow $$ u² = v² + 3$$\left( {{{{u^2} + {v^2}} \over 9}} \right)$$

$$ \Rightarrow $$ 3u² = 3v² + u² + v²

$$ \Rightarrow $$ 2u² = 4v²

$$ \Rightarrow $$ $$v = {u \over {\sqrt 2 }}$$