A) \[m_{1}^{2}{{u}_{1}}+m_{2}^{2}{{u}_{2}}-\varepsilon =m_{1}^{2}{{v}_{1}}+m_{2}^{2}{{v}_{2}}\]
B) \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\varepsilon \]
C) \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}-\varepsilon =\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\]
D) \[\frac{1}{2}m_{1}^{2}u_{1}^{2}+\frac{1}{2}m_{2}^{2}u_{2}^{2}+\varepsilon =\frac{1}{2}m_{1}^{2}v_{1}^{2}+\frac{1}{2}m_{2}^{2}v_{2}^{2}\]
Correct Answer: C
Solution :
Total initial energy \[=\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}\] |
Since, after collision one particle absorb energy \[\varepsilon \] |
\[\therefore \] Total final energy \[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{1}^{2}+\varepsilon \] |
From conservation of energy, |
\[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}+\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}+\varepsilon \] |
\[\Rightarrow \] \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}-\varepsilon \] |
\[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] |
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