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A sphere of mass m moving with velocity v enters, a hanging bag of sand and stops. If the mass of bag is M and it is raised by height h, then the velocity of sphere was

\[(a)\;\frac{M+m}{m} \sqrt {2gh} \quad (b)\;\frac{M}{m} \sqrt {2gh} \quad (c)\;\frac{m}{M+m} \sqrt {2gh} \quad (d)\;\frac{m}{M} \sqrt {2gh}\]
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Answer: $\large\frac{M+m}{m}$$ \sqrt {2gh} $
Per law of conservation of linear momentum $mV=(m+M)v_{\large system}$-----(1)
Per law of conservation of energy $\large\frac{1}{2} $$(m+M)v_{\large system}^2=(m+M)gh$
$\Rightarrow v_{\large system}=\sqrt {2gh}$
From (1) we get Therefore $V=\bigg(\large\frac{m+M}{m}\bigg) $$\sqrt {2gh}$
answered Jul 17, 2013 by meena.p
edited Aug 27, 2014 by balaji.thirumalai

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