A) \[{{E}_{1}}>{{E}_{2}}\]
B) \[{{E}_{1}}={{E}_{2}}\]
C) \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]
D) \[{{E}_{1}}<{{E}_{2}}\]
Correct Answer: D
Solution :
\[KE=\frac{{{p}^{2}}}{2\,m}=\frac{p_{1}^{2}\,2\,{{m}_{1}}}{p_{2}^{2}\,2\,{{m}_{2}}}\,\,\Rightarrow \,\,\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\][where, p is momentum] As \[{{m}_{1}}>{{m}_{2}}\,\,\,\Rightarrow \,\,\,{{E}_{1}}<{{E}_{2}}\]You need to login to perform this action.
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