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,

Question

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

TestHub · Accepted Answer

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

$e = 1 = \frac{V_{2} - V_{1}}{u}$ (where $e \Rightarrow$ the coefficient of restitution), $u = V_{2} - V_{1}$

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

$u = V_{1} + 12 V_{2}$ $\Rightarrow 1 = \frac{V_{2} - V_{1}}{u}$ $\Rightarrow u = V_{2} - V_{1}$
$\Rightarrow V_{2} = \frac{2 u}{13}$ $\Rightarrow V_{1} = - \frac{11 u}{13}$ .
Change in energy, $= u^{2} - {(\frac{11 u}{13})}^{2} \times 100$ $= 0.294$

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