A ball of mass $10 \mathrm{~g}$ is allowed to fall down from 10 $\mathrm{m}$ height. After collision with…
- $4 \mathrm{~m}$
- $6 \mathrm{~m}$
- $5 \mathrm{~m}$
- $7 \mathrm{~m}$
Solution

$\mathrm{K} . \mathrm{E}^{\prime}=50 \% \mathrm{~K} . \mathrm{E}$ 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 $5 \mathrm{~m}$.
Asked in: AP EAMCET 2023 (17 May Shift 1)