Applying conservation of linear momentum in x direction

$$2mv = \left( {2m + m} \right){V_x}$$

$$ \Rightarrow {V_x} = {2 \over 3}v$$

Applying conservation of linear momentum in y direction

$$2mv = \left( {2m + m} \right){V_y}$$

$$ \Rightarrow {V_y} = {2 \over 3}v$$

Final speed of 3m mass, $${V_f}$$ = $$\sqrt {V_x^2 + V_y^2} $$

= $$\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}} $$

= $$\sqrt {{{8{v^2}} \over 9}} $$

Initial kinetic energy

$${E_i} = {1 \over 2}m{\left( {2v} \right)^2} + {1 \over 2}\left( {2m} \right){\left( v \right)^2}$$

$$ = 2m{v^2} + m{v^2}$$

= $$3m{v^2}$$

Final kinetic energy,

$${E_f} = {1 \over 2}\left( {3m} \right)$$$${V_f^2}$$

$$ = {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} \right]$$

$$ = {{4m{v^2}} \over 3}$$

Energy loss = $${E_i} - {E_f}$$

= $$3m{v^2} - {{4m{v^2}} \over 3}$$

= $${{5m{v^2}} \over 3}$$

Percentage loss in the energy during the collision

= $${{{E_i} - {E_f}} \over {{E_i}}}$$$$ \times 100$$

= $${{{{5m{v^2}} \over 3}} \over {3m{v^2}}}$$$$ \times 100$$

= $${5 \over 9} \times 100$$

$$ \simeq 56\% $$