Question

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Q. A bullet of mass m passes through a pendulum bob of mass M with velocity v and comes of it with a velocity $\frac{v}{2}$ . The minimum value of v so that the bob completes one revolution in the vertical circle is (l = length of pendulum)


$$\begin{gathered} \left(1\right)\frac{M}{m}\sqrt{2gl}\left(2\right)\frac{2M}{m}\sqrt{2lg} \\ \left(3\right)\frac{2M}{m}\sqrt{5lg}\left(4\right)\frac{M}{m}\sqrt{5lg} \end{gathered}$$

Answer 1

Dear Student,

$$\begin{gathered} energyprovidedtobob \\ E=\frac{m}{2}\left(v^2-\frac{v^2}{4}\right)=\left(\frac{3m{v}^2}{8}\right) \\ theminimumvelocityrequiredattopmostpointbeu \\ \frac{M{u}^2}{l}=Mg \\ u=\sqrt{gl} \\ fromconservationtheorem \\ \left(\frac{3m{v}^2}{8}\right)=2Mgl+\frac{M{u}^2}{2} \\ \left(\frac{3v^2}{8}\right)=\frac{M}{m}\left(2gl+\frac{gl}{2}\right)=\frac{M}{m}\times \frac{5gl}{2} \\ v=2\sqrt{\frac{5Mgl}{3m}} \\ bestofmyknowledge. \\ Regards \end{gathered}$$