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)] |
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