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4-1.Newton's Laws of Motion
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A ball of mass m falls vertically to the ground from a height$ h_1$ and rebound to a height ${h_2}$. The change in momentum of the ball on striking the ground is
A
$mg({h_1} - {h_2})$
B
$m(\sqrt {2g{h_1}} + \sqrt {2g{h_2}} )$
C
$m\sqrt {2g({h_1} + {h_2})} $
D
$m\sqrt {2g} ({h_1} + {h_2})$
Solution
(b)When ball falls vertically downward from height ${h_1}$ its velocity ${\overrightarrow v _1} = \sqrt {2g{h_1}} $
and its velocity after collision ${\overrightarrow v _2} = \sqrt {2g{h_2}} $
Change in momentum
$\Delta \vec P = m({\overrightarrow v _2} – {\overrightarrow v _1}) = m(\sqrt {2g{h_1}} + \sqrt {2g{h_2}} )$
(because ${\overrightarrow v _1}$ and ${\overrightarrow v _2}$ are opposite in direction)
Standard 11
Physics
