Arrangement in motion

We'll work in S.I. and then finally covert our answer to C.G.S..

Let $m_A=4m_B=4m$ (say). Let tension in the string while it is taut be $T$ From constraint relations $\frac{a_B}{2}=a_A=a$ (say).

Then Newton's second law yields

$T-mg=2ma\;---(i)$

$4mg-2T=4ma\;---(ii)$

Now $2\times(i)+(ii)$ gives

$\displaystyle\,a=\frac{g}{4}$ .

So for motion of $A$ till it hits ground after which string becomes slack( $\rightarrow\,T=0$ ) due to sticking, we have

$\displaystyle\,\frac{1}{5}=\frac{1}{2}\cdot\,a\cdot\,t_1^2\rightarrow\,t_1=\sqrt{\frac{8}{5g}}$ .

During this time $B$ travels a distance $\displaystyle\,d_1=\frac{1}{2}\cdot\,2a\cdot\,t_1^2=\frac{2}{5}$ .

At this point velocity of B $\displaystyle\,v=2a\cdot\,t_1=\sqrt{\frac{2g}{5}}$ . After this $T=0$

$\rightarrow\,d_2=\displaystyle\frac{v^2}{2g}=\frac{1}{5}$ .

So total distance moved by B is $d=d_1+d_2=\frac{3}{5}\,m=\boxed{60\,cm}$ .

This question came in Problems in General Physics by I E Irodov

Q.No: 76 Physical Fundamentals of Mechanics.

$\large \text{Solution}$

$\begin{aligned} 4mg - 2T &= 4ma\\ 2T - 2mg &= 4ma\\ \Rightarrow a = \frac{g}{4}\\\\ \begin{array}{l|c|r}\hline s&a\\ \hline 2h&\dfrac{g}{2}\\\hline\end{array}\\\\ \text{Velocity when}&\text{ the other block touches the floor is}\\ v^2 &= 2gh\\ \text{And further, } h' &= h\\ \Rightarrow \boxed{H = 3h}\end{aligned}$

Found another solution online: (it uses energy conservation) http://irodovsolutionsmechanics.blogspot.in/2007/08/irodov-problem-176.html