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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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