A) \[{{\left( \frac{1+A}{1-A} \right)}^{2}}E\]
B) \[{{\left( \frac{1-A}{1+A} \right)}^{2}}E\]
C) \[\left( \frac{1+A}{1-A} \right)E\]
D) \[\left( \frac{1-A}{1+A} \right)E\]
Correct Answer: B
Solution :
In elastic collision, kinetic energy and momentum both are conserved. Let\[{{v}_{1}}\]be velocity after collision \[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{1}}+\left( \frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{2}}\] Given, \[{{u}_{1}}=v,\,{{u}_{2}}=0,\,{{m}_{1}}=1,\,{{m}_{2}}=A\] \[\therefore \] \[{{v}_{1}}=\left( \frac{1-A}{1+A} \right)v\] Initial kinetic energy of neutron \[E=\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}\times 1\times {{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\therefore \] \[\frac{E}{E}={{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\Rightarrow \] \[E={{\left( \frac{1-A}{1+A} \right)}^{2}}E\] Final KE of neutron \[E=\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}\times 1\times {{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\therefore \] \[\frac{E}{E}={{\left( \frac{1-A}{1+A} \right)}^{2}}\] \[\Rightarrow \] \[E={{\left( \frac{1-A}{1+A} \right)}^{2}}E\]You need to login to perform this action.
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