Question 121 of 130

It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of its energy is ${P}_{d}$, while for its similar collision with a carbon nucleus at rest, the fractional loss of energy is ${P}_{c}$. The values of ${P}_{d}$ and ${P}_{c}$ are respectively

A$0,1$B$0.89,0.28$C$0.28,0.89$D$0,0$

Show answer and solution▾

Answer: (B) $0.89,0.28$

Since, the linear momentum is conserved, $u={V}_{1}+2{V}_{2}$.

$e=1=\dfrac{{V}_{2}-{V}_{1}}{u}$ (where $e\Rightarrow$the coefficient of restitution), $u={V}_{2}-{V}_{1}$

${V}_{2}=\dfrac{2u}{3};{V}_{1}=-\dfrac{u}{3}$. Initial energy, $=\dfrac{1}{2}{u}^{2}$. Final energy, $=\dfrac{1}{2}\times 1\times \dfrac{{u}^{2}}{9}=\dfrac{{u}^{2}}{18}$. Fractional change $=\dfrac{\dfrac{{u}^{2}}{2}-\dfrac{{u}^{2}}{18}}{\dfrac{{u}^{2}}{2}}$$=0.88$



$u={V}_{1}+12{V}_{2}$$\Rightarrow 1=\dfrac{{V}_{2}-{V}_{1}}{u}$$\Rightarrow u={V}_{2}-{V}_{1}$
$\Rightarrow {V}_{2}=\dfrac{2u}{13}$$\Rightarrow {V}_{1}=-\dfrac{11u}{13}$.
Change in energy,$={u}^{2}-{\left(\dfrac{11u}{13}\right)}^{2}\times 100$$=0.294$