JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(2.39 \mathrm{~kg} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(2.39 \mathrm{~kg} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\vec{I} \)\( =\Delta \vec{P}=\vec{P}_f-\vec{P}_i \) \(\mathrm{M} \)\( =0.1 \mathrm{~kg} \) I \( =\Delta P=0.1(\sqrt{2 \times 9.8 \times 5}-(-\sqrt{2 \times 9.8 \times 10})) \) \( =0.1(14+7 \sqrt{2}) \approx 2.39 \mathrm{~kg} \mathrm{~ms}^{-1}\)
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