A spherical body of mass 100 mathrm{~g} is dropped from a height of 10 mathrm{~m} from ground. After

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4-1.Newton's Laws of Motion

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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 $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$ )

A

$4.32 \mathrm{~kg} \mathrm{~ms}^{-1}$

B

$43.2 \mathrm{~kg} \mathrm{~ms}^{-1}$

C

$23.9 \mathrm{~kg} \mathrm{~ms}^{-1}$

D

$2.39 \mathrm{~kg} \mathrm{~ms}^{-1}$

(JEE MAIN-2024)

Solution

$\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}$

Standard 11

Physics

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