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A particle of mass $m$ moving with velocity $V_0$ strikes a simple pendulum of mass $m$ and sticks to it. The maximum height attained by the pendulum will be
$h = \frac{{V_0^2}}{{8g}}$
$\sqrt {{V_0}g} $
$2\sqrt {\frac{{{V_0}}}{g}} $
$\frac{{V_0^2}}{{4g}}$
Solution
Initial momentum of particle $=\mathrm{mV}_{\mathrm{o}}$
Final momentum of system (particle + pendulum)
$=2 \mathrm{mv}$
By the law of conservation of momentum
$\Rightarrow \mathrm{mV}_{0}=2 \mathrm{mv} \Rightarrow$ Initial velocity of system $\mathrm{v}=\frac{\mathrm{V}_{0}}{2}$
Initial $K.E.$ of the system
$=\frac{1}{2}(2 \mathrm{m}) \mathrm{v}^{2}=\frac{1}{2}(2 \mathrm{m})\left(\frac{\mathrm{V}_{0}}{2}\right)^{2}$
If the system rises up to height h then $P . E .=2 \mathrm{mgh}$ By the law of conservation of energy
$\frac{1}{2}(2 m)\left(\frac{V_{0}}{2}\right)^{2}=2 m g h \Rightarrow h=\frac{V_{0}^{2}}{8 g}$
