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4-1.Newton's Laws of Motion
medium
$m$ દડાને $h_1$ ઊંચાઇ પરથી મુકત કરતાં તે અથડાઇને $h_2$ ઊંચાઇ પર આવતો હોય,તો અથડામણ દરમિયાન વેગમાનમાં કેટલો ફેરફાર થશે?
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
