A particle of mass m moving with velocity V_0 | vedclass

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(A) Initial momentum of the particle = $m V_0$. Final momentum of the system (particle + pendulum) = $(m + m)v = 2mv$. By the law of conservation of m

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Initial momentum of the particle = $m V_0$. Final momentum of the system (particle + pendulum) = $(m + m)v = 2mv$. By the law of conservation of momentum: $m V_0 = 2mv$ $v = \frac{V_0}{2}$ Initial kinetic energy of the system = $\frac{1}{2}(2m)v^2 = \frac{1}{2}(2m)\left(\frac{V_0}{2}\right)^2 = \frac{1}{2}(2m)\frac{V_0^2}{4} = \frac{m V_0^2}{4}$. If the system rises to a maximum height $h$,the potential energy gained is $P.E. = (2m)gh$. By the law of conservation of energy,the initial kinetic energy is converted into potential energy at the maximum height: $\frac{m V_0^2}{4} = 2mgh$ $h = \frac{m V_0^2}{4 \cdot 2mg} = \frac{V_0^2}{8g}$.

$A$ 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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