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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