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QJEEM24Ph33fe: JEE Main 2024 Physics - Class 11 JEEM24Ph33fe Solution: A spherical body of mass is dropped from a height of from the…

A spherical body of mass 100 g100 \mathrm{~g} is dropped from a height of 10 m10 \mathrm{~m} from the ground. After hitting the ground, the body rebounds to a height of 5 m5 \mathrm{~m} . The impulse of force imparted by the ground to the body is given by : (given, g=9.8 m/s2\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2 )

Multiple Choice Options

  • A 43.2 kg ms−143.2 \mathrm{~kg} \mathrm{~ms}^{-1}
  • B 2.39 kg ms−12.39 \mathrm{~kg} \mathrm{~ms}^{-1}
  • C 4.32 kg ms−14.32 \mathrm{~kg} \mathrm{~ms}^{-1}
  • D 23.9 kg ms−123.9 \mathrm{~kg} \mathrm{~ms}^{-1}

Step-by-Step Verified Solution

I⃗amp;=ΔP⃗=P⃗f−P⃗iMamp;=0.1 kgIamp;=ΔP=0.1(2×9.8×5−(−2×9.8×10))amp;=0.1(14+72)≈2.39 kg ms−1\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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