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A ball of mass $10 \mathrm{~g}$ is allowed to fall down from 10 $\mathrm{m}$ height. After collision with the ground if $50 \%$ of its energy is lost, then the height reached by the ball is
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The correct answer is:
$5 \mathrm{~m}$
Mass of ball, $\mathrm{m}=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}$ Height, $\mathrm{h}=10 \mathrm{~m}$
Apply the conservation of energy potential energy $=$ kinetic energy
$\mathrm{K} . \mathrm{E}=\mathrm{mgh}=10 \mathrm{mg}...(i)$
$\mathrm{K} . \mathrm{E}^{\prime}=50 \% \mathrm{~K} . \mathrm{E}$
Use equation (1), we have $=5 \times 10 \mathrm{mg}$
$\begin{aligned}
& =5 \mathrm{mg} \\
& \mathrm{P} \cdot \mathrm{E}^{\prime}=\mathrm{K} \cdot \mathrm{E}^{\prime}=\mathrm{mgh} \Rightarrow 5 \mathrm{mg}=\mathrm{mgh} \\
& \mathrm{h}=5 \mathrm{~m} .
\end{aligned}$
The height reached by the ball is $5 \mathrm{~m}$.
Apply the conservation of energy potential energy $=$ kinetic energy
$\mathrm{K} . \mathrm{E}=\mathrm{mgh}=10 \mathrm{mg}...(i)$
$\mathrm{K} . \mathrm{E}^{\prime}=50 \% \mathrm{~K} . \mathrm{E}$
Use equation (1), we have $=5 \times 10 \mathrm{mg}$
$\begin{aligned}
& =5 \mathrm{mg} \\
& \mathrm{P} \cdot \mathrm{E}^{\prime}=\mathrm{K} \cdot \mathrm{E}^{\prime}=\mathrm{mgh} \Rightarrow 5 \mathrm{mg}=\mathrm{mgh} \\
& \mathrm{h}=5 \mathrm{~m} .
\end{aligned}$
The height reached by the ball is $5 \mathrm{~m}$.
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