A) \[{{E}_{1}}<{{E}_{2}}\]
B) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]
C) \[{{E}_{1}}>{{E}_{2}}\]
D) \[{{E}_{1}}={{E}_{2}}\]
Correct Answer: A
Solution :
Kinetic energy is given by \[E=\frac{1}{2}m{{v}^{2}}=\frac{1}{2m}{{(mv)}^{2}}\] but\[mv=\]momentum of the particle\[=p\] \[\therefore \] \[E=\frac{{{p}^{2}}}{2m}\] or \[p=\sqrt{2mE}\] Therefore, \[\frac{{{p}_{1}}}{{{p}_{2}}}=\sqrt{\frac{{{m}_{1}}{{E}_{1}}}{{{m}_{2}}{{E}_{2}}}}\] but it is given that, \[{{p}_{1}}={{p}_{2}}\] \[\therefore \] \[{{m}_{1}}{{E}_{1}}={{m}_{2}}{{E}_{2}}\] Or \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\] ?(i) Now\[{{m}_{1}}>{{m}_{2}}\] Or \[\frac{{{m}_{1}}}{{{m}_{2}}}>1\] ?(ii) Thus, Eqs. (i) and (ii) give \[\frac{{{E}_{1}}}{{{E}_{2}}}>1\] Or \[{{E}_{1}}<{{E}_{2}}\]
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