Question

Figure shows a small block $A$ of mass $m$ kept at the left end of a plank $B$ of mass $M=2m$ and length $\lambda$. The system can slide on a horizontal road. The system is started towards right with the initial velocity $v$. The friction coefficients between the road and the plank is $1/2$ and that between the plank and the block is $1/4$
(b) displacement of block and plank with respect to ground till that moment?

Answer 1

Ref. image I

Let $a_1$ and $a_2$ be acceleration of mass m and 2m also $a_1>a_2$ so mass m moves on 2m.

Let after time t mass m is separated from 2m.

using equation of motion. during this time, mass m covers $vt+\frac{1}{2}{a}_1{t}^2$ & $5m=vt+\frac{1}{2}{a}_2{t}^2$ for mass m to separate from 2m, we have $vt+\frac{1}{2}{a}_1{t}^2=vt+\frac{1}{2}{a}_2{t}^2+1$

Ref. image II

As $\frac{1}{2}{\mu }_1=\frac{1}{4}{\mu }_2$

$\mu =\frac{\mu }{2}$

from free body diagram we have, $m{a}_1+\frac{\mu }{2}R=0$

$m{a}_1=\frac{-\mu }{2}mg$

$=\frac{\mu }{2}g$

$a_1=-5\mu$

again ,

$\left(2m\right)a_2+\mu (2m+m)g-\left(\frac{\mu }{2}\right)mg=0$

$\Rightarrow 2\left(2m\right)a_2+2\mu (2m+m)g-\mu mg=0$

$\Rightarrow 2\left(2m\right)a_2=\mu mg-2\mu mg-2\mu mg$

$a_2=-\mu mg-\frac{2\mu \left(2m\right)g}{2\left(2m\right)}$

Substituting values of $a_1$ & $a_2$ in eq. (1)

$t=\frac{4\left(2m\right)l}{(2m+m)\mu g}$