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A rubber ball is dropped from a height of $5 \mathrm{~m}$ on a planet, where the acceleration due to gravity is not known. On bouncing it rises to $1.8 \mathrm{~m}$. The ball loses its velocity on bouncing by a factor of
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Verified Answer
The correct answer is:
$\frac{2}{5}$
Key Idea According to conservation of energy, potential energy at height $h=$ kinetic energy at
ground. Potential energy = Kinetic energy
ie,
$$
m g h=\frac{1}{2} m v^{2}
$$
$$
\Rightarrow \quad v=\sqrt{2 g h}
$$
If $h_{1}$ and $h_{2}$ are initial and final heights, then
$$
v_{1}=\sqrt{2 g h_{1}}, \quad v_{2}=\sqrt{2 g h_{2}}
$$
Loss in velocity
$$
\Delta v=v_{1}-v_{2}=\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}
$$
$\therefore$ Fractional loss in velocity $=\frac{\Delta v}{v_{1}}$
$$
\begin{array}{l}
=\frac{\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}}{\sqrt{2 g h_{1}}} \\
=1-\sqrt{\frac{h_{2}}{h_{1}}}
\end{array}
$$
Substituting the values, we have
$\therefore$
$$
\frac{\Delta v}{v_{1}}=1-\sqrt{\frac{1.8}{5}}
$$
$$
\begin{array}{l}
=1-\sqrt{0.36}=1-0.6 \\
=0.4=\frac{2}{5}
\end{array}
$$
ground. Potential energy = Kinetic energy
ie,
$$
m g h=\frac{1}{2} m v^{2}
$$
$$
\Rightarrow \quad v=\sqrt{2 g h}
$$
If $h_{1}$ and $h_{2}$ are initial and final heights, then
$$
v_{1}=\sqrt{2 g h_{1}}, \quad v_{2}=\sqrt{2 g h_{2}}
$$
Loss in velocity
$$
\Delta v=v_{1}-v_{2}=\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}
$$
$\therefore$ Fractional loss in velocity $=\frac{\Delta v}{v_{1}}$
$$
\begin{array}{l}
=\frac{\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}}{\sqrt{2 g h_{1}}} \\
=1-\sqrt{\frac{h_{2}}{h_{1}}}
\end{array}
$$
Substituting the values, we have
$\therefore$
$$
\frac{\Delta v}{v_{1}}=1-\sqrt{\frac{1.8}{5}}
$$
$$
\begin{array}{l}
=1-\sqrt{0.36}=1-0.6 \\
=0.4=\frac{2}{5}
\end{array}
$$
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