m દળના સાદા લોલક સાથે m દળ અને v_0 વેગથી ગ

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
$\frac{1}{2}(2m){\left( {\frac{{{V_0}}}{2}} \right)^2} = 2mgh$ 

$⇒$ $h = \frac{{V_0^2}}{{8g}}$