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\[=p\] |
\[\therefore \] \[E=\frac{p}{2m}\] 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 \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\] |
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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