Applying conservation of momentum :

mv + 0 = mv₁ + 2mv₂

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

As collision is elastic,

So, coefficient of restitution, e = 1

$$\therefore\,\,\,$$ e = 1 = $${{velocity\,\,of\,\,separation} \over {Velocity\,\,of\,\,approach}}$$

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

$$ \Rightarrow $$$$\,\,\,$$ v = v₂ $$-$$ v₁ . . . . .(2)

Add (1) and (2),

2v = 3v₂

$$ \Rightarrow $$$$\,\,\,$$ v₂ = $${{2v} \over 3}$$

put value of v₂ in equation (1),

v₁ = v $$-$$ 2v₂

= v $$-$$ $${{4v} \over 3}$$

= $$-$$ $${v \over 3}$$

$$\therefore\,\,\,$$ Fractional loss of energy of neutron.

P_d = $${{{k_i} - {k_f}} \over {{k_i}}}$$

= $${{{1 \over 2}m{v^2} - {1 \over 2}mv_1^2} \over {{1 \over 2}m{v^2}}}$$

= $${{{v^2} - {{{v^2}} \over 9}} \over {{v^2}}}$$

= $${8 \over 9}$$

= 0.89



Applying momentum of conservation,

mv + 0 = mv₁ + 12mv₂

$$ \Rightarrow $$$$\,\,\,$$ v = v₁ + 12v₂ . . . . . (3)

Here also e = 1

$$\therefore\,\,\,$$ e = 1 = $${{{v_2} - v{}_1} \over {v - 0}}$$

$$ \Rightarrow $$$$\,\,\,$$ v = v₂ $$-$$ v₁ . . . . . . (4)

adding (3) and (4), we get

2v = 13v₂

$$ \Rightarrow $$$$\,\,\,$$ v₂ = $${{2v} \over {13}}$$

put this v₂ in equation (3), we get

v₁ = v $$-$$ 12 $$ \times $$ $${{2v} \over {13}}$$

= $$-$$ $${{11v} \over {13}}$$

$$\therefore\,\,\,$$ Frictional loss

p_c = $${{{1 \over 2}m{v^2} - {1 \over 2}m{{\left( {{{11} \over {13}}v} \right)}^2}} \over {{1 \over 2}m{v^2}}}$$

= $${{48} \over {169}}$$

= 0.28