A ball of mass 0.5 kg is dropped from a height of 40 m…
Question & Answer (English)
- 21 Ns — Correct Answer
- 7 Ns
- 84 Ns
Correct Answer: 21 Ns
Step I: Calculate the downward velocity $v_1$ just before hitting the ground: $v_1 = -\sqrt{2gh_1} = -\sqrt{2 \times 9.8 \times 40} = -\sqrt{784} = -28~m/s$ (taking upward as positive).
Step II: Calculate the upward velocity $v_2$ just after rebounding from the ground: $v_2 = +\sqrt{2gh_2} = +\sqrt{2 \times 9.8 \times 10} = +\sqrt{196} = +14~m/s$.
Step III: Apply the impulse formula: $J = \Delta p = m(v_2 - v_1)$.
Step IV: Substitute the values: $J = 0.5 \times (14 - (-28)) = 0.5 \times (14 + 28) = 0.5 \times 42 = 21~Ns$.
प्रश्न एवं उत्तर (हिंदी)
- 21 Ns — सही उत्तर
- 7 Ns
- 84 Ns
सही उत्तर: 21 Ns
Step I: जमीन से टकराने से ठीक पहले नीचे की ओर वेग $v_1$ की गणना करें: $v_1 = -\sqrt{2gh_1} = -\sqrt{2 \times 9.8 \times 40} = -\sqrt{784} = -28~m/s$ (ऊपर की ओर धनात्मक मानते हुए)।
Step II: जमीन से उछलने के ठीक बाद ऊपर की ओर वेग $v_2$ की गणना करें: $v_2 = +\sqrt{2gh_2} = +\sqrt{2 \times 9.8 \times 10} = +\sqrt{196} = +14~m/s$।
Step III: आवेग सूत्र लागू करें: $J = \Delta p = m(v_2 - v_1)$।
Step IV: मानों को प्रतिस्थापित करें: $J = 0.5 \times (14 - (-28)) = 0.5 \times (14 + 28) = 0.5 \times 42 = 21~Ns$।
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This multiple choice question is from demo. It has 4 options with a detailed explanation of the correct answer and is available in both English and Hindi (द्विभाषी). Practice more MCQs from demo to strengthen your preparation.
