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A body falling from a height of $10 \mathrm{~m}$ rebounds from hard floor. If it loses $20 \%$ energy in the impact, then coefficient of restitution is
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Verified Answer
The correct answer is:
0.89
As \(20 \%\) energy lost in collision therefore \(\mathrm{mgh}_2=80 \%\) of \(\mathrm{mgh}_1 \Rightarrow \frac{\mathrm{h}_2}{\mathrm{~h}_1}=0.8\)
Here \(h \propto v^2\)
and \(e=\frac{v_f}{v_i}\)
so in such cases:
\(\mathrm{e}=\sqrt{\frac{\mathrm{h}_2}{\mathrm{~h}_1}}=\sqrt{0.8}=0.89\)
Here \(h \propto v^2\)
and \(e=\frac{v_f}{v_i}\)
so in such cases:
\(\mathrm{e}=\sqrt{\frac{\mathrm{h}_2}{\mathrm{~h}_1}}=\sqrt{0.8}=0.89\)
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