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
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}} $
$\frac{{V_0^2}}{{4g}}$
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
Initial momentum of particle = $m{V_0}$ Final momentum of system (particle + pendulum) $= 2mv$ By the law of conservation of momentum
$⇒$ $m{V_0} = 2mv$
$⇒$ Initial velocity of system $v = \frac{{{V_0}}}{2}$
Initial K.E. of the system = $\frac{1}{2}(2m){v^2}$=$\frac{1}{2}(2m){\left( {\frac{{{V_0}}}{2}} \right)^2}$
If the system rises up to height $h$ then $P.E. =2mgh$
By the law of conservation of energy $\frac{1}{2}(2m){\left( {\frac{{{V_0}}}{2}} \right)^2} = 2mgh$
$⇒$ $h = \frac{{V_0^2}}{{8g}}$
= $\frac{1}{2}(2m){\left( {\frac{{{V_0}}}{2}} \right)^2}$
If the system rises up to height h then $P.E. = 2mgh$
By the law of conservation of energy
$\frac{1}{2}(2m){\left( {\frac{{{V_0}}}{2}} \right)^2} = 2mgh$
$⇒$ $h = \frac{{V_0^2}}{{8g}}$
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