A particle of mass m moving with velocity {V_ | vedclass

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Question

(a)Initial momentum of particle = $m{V_0}$ Final momentum of system (particle + pendulum) $= 2mv$ By the law of conservation of momentum 

$&rA

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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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