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  • Try this! It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of p

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

  • Option 1)

    (0, 1)

  • Option 2)

    (⋅89, ⋅28)

  • Option 3)

    (⋅28, ⋅89)

  • Option 4)

    (0, 0)

 

Answers (2)

 

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

m_{1},m_{2}:masses

u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}

u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}

 

 

 

 

Before Collision

for elastic collision :

\\*mv=2mv_{2}-mv_{1}\\* or v=2v_{2}-v_{1}\: \: ---(1)\\* Velocity \: o\! f \: sepration=Velocity \: o\! f \:approach\\* v_{2}+v_{1}=v---(2)\\* From (1)\: and\: (2)\Rightarrow 3v_{2}= 2v\: \: or\: \: v_{2}= \frac{2v}{3}\\* and \: 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

For collision with carbon

\\*2v_{2}-v_{1}= v\\*and\: v_{2}+v_{1}= v       \Rightarrow      \Rightarrow \: \: \: \: v_{2}= \frac{2v}{13}\: and\: v_{1}= \frac{11v}{13}

 

Fraction Loss =p_{c}= \frac{\frac{1}{2}mv^{2}-\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv^{2}}= 

= \frac{48}{169}= 0.28


Option 1)

(0, 1)

Option 2)

(⋅89, ⋅28)

Option 3)

(⋅28, ⋅89)

Option 4)

(0, 0)

Posted by

Vakul

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