What a chain! 2

The weight of the fallen part is $\dfrac{Mgy}{L}$ , dalso as the chain is falling .....consider a length dx in it which is just about to fall

$N \, dt = d(mv)$

$N =$ $\dfrac{v dm}{dt}$

$N =$ $\dfrac{v M}{L}$ $\dfrac{dx}{dt}$

$N =$ $\dfrac{Mv^2}{L}$

$v =$ $(2gy)^.5$

Putting this we get $N =$ $\dfrac{2Mgy}{L}$

This added to the weight of the fallen part give the total normal reaction

Hence net $N =$ $\dfrac{3Mgy}{L}$

I have a few doubts. 1)why have you taken v constant in d(mV) as m is changing v is also changing. 2) how is energy conserved in this case as the locks come to rest by inelastic collision. 3)even if energy conservation is assumed. After y length has fallen, COM moves a distance y/2. ∆PE =√(gy). Please help and sorry if I am wrong.