Answer:
Here \[{{\text{p}}_{\text{1}}}={{\text{p}}_{\text{2}}}\],
i.e., \[{{\text{m}}_{\text{1}}}{{\upsilon
}_{\text{1}}}={{\text{m}}_{\text{2}}}{{\upsilon }_{\text{2}}}\] \[\therefore
\] \[\frac{{{\upsilon }_{1}}}{{{\upsilon
}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\] ...(i)
As \[{{\text{E}}_{\text{2}}}=\frac{1}{2}{{\text{m}}_{\text{2}}}\upsilon
_{2}^{2}\] and \[{{\text{E}}_{1}}=\frac{1}{2}{{\text{m}}_{1}}\upsilon
_{1}^{2}\] \[\therefore \] \[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{\frac{1}{2}{{m}_{2}}\upsilon
_{2}^{2}}{\frac{1}{2}{{m}_{1}}\upsilon _{1}^{2}}=\frac{{{m}_{2}}}{{{m}_{1}}}{{\left(
\frac{{{\upsilon }_{2}}}{{{\upsilon }_{1}}} \right)}^{2}}\]
using (i),
\[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{{{m}_{2}}}{{{m}_{1}}}.{{\left(
\frac{{{m}_{1}}}{{{m}_{2}}} \right)}^{2}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]
If \[{{\text{m}}_{\text{1}}}<{{\text{m}}_{\text{2}}},{{\text{E}}_{\text{1}}}<\text{
}{{\text{E}}_{\text{1}}}\] or \[{{\text{E}}_{\text{1}}}>\text{
}{{\text{E}}_{\text{2}}}\]i.e. lighter body has more K.E.
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