SATHEE: Work Energy And Power Question 105

Work Energy And Power Question 105

Question: A ball of mass m falls vertically to the ground from a height h1 and rebound to a height $h_{2}$ . The change in momentum of the ball on striking the ground is [AMU (Engg.) 1999]

Options:

A) $m g (h_{1} - h_{2})$

B) $m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$

C) $m \sqrt{2 g (h_{1} + h_{2})}$

D) $m \sqrt{2 g} (h_{1} + h_{2})$

Show Answer

Answer:

Correct Answer: B

Solution:

When ball falls vertically downward from height

$h_{1}$ its velocity ${\vec{v}}_{1} = \sqrt{2 g h_{1}}$

and its velocity after collision ${\vec{v}}_{2} = \sqrt{2 g h_{2}}$

Change in momentum $Δ \vec{P} = m ({\vec{v}}_{2} - {\vec{v}}_{1}) = m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$ (because ${\vec{v}}_{1}$

and ${\vec{v}}_{2}$ are opposite in direction)

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अगला

Work Energy And Power Question 105

Question: A ball of mass m falls vertically to the ground from a height h1 and rebound to a height $h_{2}$ . The change in momentum of the ball on striking the ground is [AMU (Engg.) 1999]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Work Energy And Power Question 105

Question: A ball of mass m falls vertically to the ground from a height h1 and rebound to a height h2 . The change in momentum of the ball on striking the ground is [AMU (Engg.) 1999]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: A ball of mass m falls vertically to the ground from a height h1 and rebound to a height $h_{2}$ . The change in momentum of the ball on striking the ground is [AMU (Engg.) 1999]