momentum qestion.

no one to answer

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 ?

cn u plz explain the point tht after striking why it will nt hve any velocity perpendicular to incline??plz

it is an inelastic collision ...e = 0 = velocity of seperation velocity of approach

force expression is ,

Ndt= \frac{Mvdx}{L\sqrt{2}}

so N = MvL√2 dxdt =Mv²L√2 = 2MgxL√2 ??

its nt correct an wht abut the impulse...leave the force expression i gt it...u plz solve ffr impulse plz

n e 1 interested here??right ans bata do ?