Centre Of Mass Ques 67
- 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
(2018 Main)
(a) $(0,1)$
(b) $(.89, .28)$
(c) $(.28, .89)$
(d) $(0,0)$
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Answer:
Correct Answer: 67.(b)
Solution:
- Case I Just before Collision,
Just after collision
From momentum conservation,
$$ 2 v _2-v _1=v $$
From the definition of $e$ ( $=1$ for elastic collision)
$$ \begin{gathered} v _2+v _1=v \quad \Rightarrow \quad 3 v _2=2 v \\ v _2=\frac{2 v}{3} \Rightarrow v _1=\frac{v}{3} \\ p _d=\frac{\frac{1}{2} m v^{2}-\frac{1}{2} m v _1^{2}}{\frac{1}{2} m v^{2}}=\frac{1-\frac{1}{9}}{1}=\frac{8}{9}=0.89 \end{gathered} $$
Case II Just before Collision
Just after Collision,
From momentum conservation,
$$ 12 v _2-v _1=v $$
From the definition of $e(=1$ for elastic collision),
$$ \begin{aligned} v _2+v _1 & =v \\ 13 v _2 & =2 v \\ v _2 & =\frac{2 v}{13} \Rightarrow v _1=v-\frac{2 v}{13}=\frac{11 v}{13} \\ \Rightarrow \quad p _c=\frac{\frac{1}{2} m v^{2}-\frac{1}{2} m v _1^{2}}{\frac{1}{2} m v^{2}} & =\frac{1-\frac{121}{169}}{1}=\frac{48}{169}=0.28 \end{aligned} $$
BETA
