A ball of mass `'m'` drops from a height which sticks to a mass-less hanger after striking it. Neglecting overturning, Find out the maximum extension in rod, assuming that the rod is mass-less.

Explanation

Applying energy conservation
`mg(h+x)=(1)/(2)(k_(1)k_(2))/(k_(1)+k_(2))x^(2)`
where `k_(1)=(A_(1)y_(1))/(l_(1)) k_(2)=(A_(2)y_(2))/(l_(2))`
`& K_(eq)=(A_(1)A_(2)y_(1)y_(2))/(A_(1)y_(1)l_(2)+A_(2)y_(1)l_(1))`
`k_(eq)x^(2)-2mgx-2mgh=0`
`x=(2mg+-sqrt(4m^(2)g^(2)+8mghk_(eq)))/(2k_(eq)) x_(max)=(mg)/(k_(eq))+sqrt((m^(2)g^(2))/(k_(eq)^(2))+(2mgh)/(k_(eq)))`
BY `S.H.M.`
`w=sqrt((k_(eq))/(m)) v=omegasqrt(a^(2)-y^(2))`
`sqrt(2gh)=sqrt((k_(eq))/(m))sqrt(a^(2)-y^(2))rArr sqrt((2mgh)/(k_(eq))+(m^(2)g^(2))/(k_(eq)^(2)))=a`
`max^(m)` extension `a+y=(mg)/(k_(eq))+sqrt((m^(2)g^(2))/(k_(eq))+(2mgh)/(k_(eq)))`