A) \[\sqrt{2({{u}^{2}}-gl)}\]
B) \[\sqrt{{{u}^{2}}-gl}\]
C) \[u-\sqrt{{{u}^{2}}-2gl}\]
D) \[\sqrt{2gl}\]
Correct Answer: A
Solution :
Key Idea: When scone reaches a position where string is horizontal, it attains the energy partially as kinetic and partially as potential. When stone is at its lowest position, it has only kinetic energy, given by \[K=\frac{1}{2}m{{u}^{2}}\] At the horizontal position, it has energy \[E=U+K=\frac{1}{2}mu{{'}^{2}}+mgl\] According to conservation of energy, \[K=E\] \[\therefore \] \[\frac{1}{2}m{{u}^{2}}=\frac{1}{2}mu{{'}^{2}}+mgl\] or \[\frac{1}{2}mu{{'}^{2}}=\frac{1}{2}m{{u}^{2}}-mgl\] or\[u{{'}^{2}}{{u}^{2}}-2gl\] or\[u'=\sqrt{{{u}^{2}}-2gl}\] ...(i) So, the magnitude of change in velocity \[|\Delta \overset{\to }{\mathop{u}}\,|=|\overset{\to }{\mathop{u}}\,|=\sqrt{u{{'}^{2}}+{{u}^{2}}+2u'u\cos {{90}^{0}}}\] \[|\Delta \overset{\to }{\mathop{u}}\,|=\sqrt{u{{'}^{2}}+{{u}^{2}}}\] \[=\sqrt{2(u{{'}^{2}}-gL)}\] [from Eq. (1)]You need to login to perform this action.
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