Question

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $p_d$, while for its similar collision with carbon nucleus at rest, fractional loss of energy is $p_c$
The values of $p_d$ and $p_c$ are respectively :

• A. $(0,0)$
• B. $(0,1)$
• C. $(0.89,0.28)$
• D. $(0.28,0.89)$

Answer 1

The correct option is C $(0.89,0.28)$

Case-i:
Applying conservation of momentum,
$2m{V}_2-{mV}_1=mV$
$2V_2-V_1=V$
We also know,
velocity of seperation = velocity of approach
$V_2+V_1=V$
$3V_2=2V$
$V_2=\frac{2V}{3}$
$V_1=\frac{V}{3}$
$P_d=\frac{\frac{1}{2}{mV}^2-\frac{1}{2}{mV}_1^2}{\frac{1}{2}{mV}^2}=\frac{1-\frac{1}{9}}{1}=\frac{8}{9}=0.89$

Case-ii:
Applying conservation of momentum,
$12m{V}_2-m{V}_1=mV$
We also know,
velocity of seperation = velocity of approach
$V_2+V_1=V$
$13V_2=2V$
$V_2=\frac{2V}{13}$
$V_1=V-\frac{2V}{13}=\frac{11V}{13}$
$\Rightarrow P_c=\frac{\frac{1}{2}{mv}^2-\frac{1}{2}{mv}_1^2}{\frac{1}{2}{mv}^2}=\frac{1-\frac{121}{169}}{1}=\frac{48}{169}=0.28$