JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2018
- Hard
- MCQ
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
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Answer & Solution
Correct Answer
(D) \((0.89,0.28)\)
Step-by-step Solution
Detailed explanation
For collision of a nenutron with deuterium \(:\)
Applying conservation of momentum\(:\)
\(m v+0=m v_{1}+2 m v_{2} \dots(i)\)
\(v_{2}-v_{1}=v \ldots(ii)\)
Therefore, Collision is elastic, \(e=1\)
From equ \((i)\) and equ \((ii)\) \(v_{1}=-v / 3\)
\(P_{d}=\frac{\frac{1}{2} m v^{2}-\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v^{2}}=\frac{8}{9}=0.89\)
Now, for the collision of neutron with carbon nucleus
Applying conservation of momentum
\(m v+0=m v_{1}+12 m v_{2} \ldots\) \((iii)\)
\(v=v_{2}-v_{1}\) \(...(iv)\)
\(v_{1}=-\frac{11}{13} v\)
\(P_{0}=\frac{\frac{1}{2} m v^{2}-\frac{1}{2} m\left(\frac{11}{13} v\right)^{2}}{\frac{1}{2} m v^{2}}=\frac{48}{169} \approx 0.28\)
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