As linear momentum is conserved before and after the collision,

$ \therefore $ 4m $ \times $ u = 4m $ \times $ v₁ + 2m $ \times $ v₂

$ \Rightarrow $ 2u = 2v₁ + v₂ ......(1)

As collision is elastic so e = 1

We know, e = $${{{v_2} - {v_1}} \over {{u_1} - {u_2}}}$$

$ \Rightarrow $ 1 = $${{{v_2} - {v_1}} \over {u - 0}}$$

$ \Rightarrow $ u = v₂ - v₁ ....(2)

Subtractiong (2) from (1), we get

u = 3v₁

$ \Rightarrow $ v₁ = ${u \over 3}$

After the collision the fraction of energy lost by the colliding body A is

= $${{{1 \over 2}\left( {4m} \right){u^2} - {1 \over 2}\left( {4m} \right){{\left( {{u \over 3}} \right)}^2}} \over {{1 \over 2}\left( {4m} \right){u^2}}}$$

= $${{1 - {1 \over 9}} \over 1}$$ = ${8 \over 9}$