A) \[\text{(0, 0)}\]
B) \[\text{(0, 1)}\]
C) \[\text{(}\cdot 89\text{, }\cdot \text{28)}\]
D) \[\text{(}\cdot 28\text{, }\cdot \text{89)}\]
Correct Answer: C
Solution :
[c] Conservation of momentum |
\[m{{v}_{0}}+0=m{{v}_{1}}+2m{{v}_{2}}\] |
\[{{v}_{0}}=({{v}_{1}}+2{{v}_{2}}).....(i)\] |
\[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{v}_{0}}}(e=1)\] |
\[{{v}_{2}}-{{v}_{1}}={{v}_{0}}...(ii)\] |
\[{{v}_{0}}-2{{v}_{1}}=2{{v}_{0}}+{{v}_{1}}\] |
\[3{{v}_{1}}=-{{v}_{0}}\Rightarrow {{v}_{1}}=\frac{-{{v}_{0}}}{3}\] |
Fractional loss of its K.E. |
\[=\frac{\frac{1}{2}m{{v}_{0}}^{2}-\frac{1}{2}m{{\left( \frac{{{v}_{0}}}{3} \right)}^{2}}}{\frac{1}{2}m{{v}_{0}}^{2}}=\frac{8}{9}=0.88\approx 0.89\] |
Neutron colliding with carbon |
Conservation of momentum |
\[m{{v}_{0}}=m{{v}_{1}}+12m{{v}_{2}}\] |
\[{{v}_{1}}+12{{v}_{2}}={{v}_{0}}...(i)\] |
\[e=1=\frac{{{V}_{2}}-{{V}_{1}}}{{{V}_{0}}}\] |
\[{{V}_{2}}-{{V}_{1}}={{V}_{0}}...(ii)\] |
\[\Rightarrow {{V}_{1}}+12{{v}_{1}}={{v}_{0}}-12{{v}_{0}}\] |
\[=\frac{-11{{v}_{0}}}{13}\] |
Fractional loss in K.E. |
\[=\frac{\frac{1}{2}m{{v}_{0}}^{2}-\frac{1}{2}m{{\left( \frac{11}{13}{{v}_{0}} \right)}^{2}}}{\frac{1}{2}m{{v}_{0}}^{2}}=1-\frac{121}{169}=\frac{48}{169}\] |
\[\approx 0.28\] |
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