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) \[m\sqrt{2g}({{h}_{1}}+{{h}_{2}})\]
Correct Answer: B
Solution :
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)
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