Question

A uniform heavy chain of length 'a' initially has length 'b' hanging off of a table. The remaining part of chain a - b, is coiled on the table. If the chain is released, the velocity of the chain when the last link leaves the table is

• A. $\sqrt{g\frac{a^3-b^3}{3a^2}}$
• B. $\sqrt{2g\frac{a^3-b^3}{3a^2}}$
• C. $\sqrt{2g\frac{a^4-b^4}{3a^3}}$
• D. $\sqrt{2g\frac{a^3-b^3}{a^2}}$

Answer 1

The correct option is B

$\sqrt{2g\frac{a^3-b^3}{3a^2}}$

This is a variable mass problem, so momentum is constantly changing:
$F_{ext}=m\left(t\right)g=\frac{dP}{dt}=\frac{d\left(m\right(t\left)v\right(t\left)\right)}{dt}=ma+v\dot{m}$
$mg=ma+v\dot{m}$

Simplying the differential equations, we will get the answer