The Inelastic Chain

Let the length of the falling chain be $x$ at any time. The tension at the top end will be zero as the chain above it is loose. So, the only force acting on the falling chain is the gravitational force $F = \lambda x g$ . Now we can't write $F = ma$ as the mass is a variable, so we write:

$F = \lambda x g = \dfrac{ dp} {dt} = \dfrac{d(\lambda x v)} {dt}$

$\implies xg dt = d(xv)$

$\implies x^2g (vdt) = xvd(xv)$

$\implies x^2g dx = xvd(xv)$

On integrating this with the boundary condition that $v = 0$ when $x = 0$ , we get $v = \sqrt{ \dfrac{2gx} 3 }\;\;\;\ldots \text{ 1 }$

$\implies \dfrac {dx}{dt} = \sqrt{ \dfrac{2gx} 3 }$

Solving this equation with the boundary condition that $x = 0$ when $t = 0$ , we get: $x = \dfrac{gt^2}{6}$ .

Plugging this in equation $\text { 1 }$ , we get:

$v = \dfrac{ gt} { 3 }$

Hence, the answer is $v(2) = 6.533$ .