JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 18)
A spherical body of mass $$100 \mathrm{~g}$$ is dropped from a height of $$10 \mathrm{~m}$$ from the ground. After hitting the ground, the body rebounds to a height of $$5 \mathrm{~m}$$. The impulse of force imparted by the ground to the body is given by : (given, $$\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$$)
$$43.2 \mathrm{~kg} \mathrm{~ms}^{-1}$$
$$2.39 \mathrm{~kg} \mathrm{~ms}^{-1}$$
$$4.32 \mathrm{~kg} \mathrm{~ms}^{-1}$$
$$23.9 \mathrm{~kg} \mathrm{~ms}^{-1}$$
Explanation
$$\begin{aligned}
\vec{I} & =\Delta \vec{P}=\vec{P}_f-\vec{P}_i \\
\mathrm{M} & =0.1 \mathrm{~kg} \\
I & =\Delta P=0.1(\sqrt{2 \times 9.8 \times 5}-(-\sqrt{2 \times 9.8 \times 10})) \\
& =0.1(14+7 \sqrt{2}) \approx 2.39 \mathrm{~kg} \mathrm{~ms}^{-1}
\end{aligned}$$
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