A particle of mass m moving with velocity V_0 | vedclass

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Question

Initial momentum of particle $=\mathrm{mV}_{\mathrm{o}}$

Final momentum of system (particle + pendulum)

$=2 \mathrm{mv}$

By the law of conser

Vedclass · Accepted Answer

$h = \frac{{V_0^2}}{{8g}}$

Vedclass · Answer

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

$\frac{1}{2}(2 m)\left(\frac{V_{0}}{2}\right)^{2}=2 m g h \Rightarrow h=\frac{V_{0}^{2}}{8 g}$

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

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