QJEEM18Ph7735: JEE Main 2018 Physics - Class 12 JEEM18Ph7735 Solution: It is found that if a neutron suffers an elastic collinear…

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