consider an element of chain dx at a distance x from the point of contact of incline and chain.
mass of dx = dm =Mdx/L
let v =vel just b4 hitting
now , the tension in the chain will be zero bcz , wen the element of chain hits the incline , then just after striking, it wont hav n e velocity _|_ to the incline ,and thus it wont give n e impulse( along _|_) to another differential element just above it wich will be striking the incline next , with vel v/√2 _|_ to the incline .
thus the chain is under freefall
thus vel of differenetial element wich was at a height x b4 releasing =√2gx
component of v along _|_ to the incline is changing due to inpulse of incline frm v/√2 to 0 .
thus Ndt = dP = dm(v√2 )=\frac{Mvdx}{L\sqrt{2}}
thus total impulse = \int \frac{M\sqrt{2gx}dx}{L\sqrt{2}}=\frac{M\sqrt{g}}{L}\int \sqrt{x}dx=\frac{2M\sqrt{gL}}{3}
is it correct ?