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

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))`

and `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)`