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 $l$. 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 $1/4$. Find the displacement of block and plank relative to ground till $t=4\sqrt{\frac{l}{3g}}$ moment.

Answer 1

Here, $f_1=\left(\frac{1}{4}\right)\left(mg\right)$
and $f_2=\left(\frac{1}{2}\right)(m+2m)g=\left(\frac{3}{2}\right)mg$
Retardation of A is $a_1=\frac{f_1}{m}=\frac{g}{4}$
and retardation of B is $a_2=\frac{f_2-f_1}{2m}=\frac{5}{8}g$

Displacement of block $S_A=u_{A}t-\frac{1}{2}{a}_A{t}^2$
or $S_A=4v\sqrt{\frac{l}{3g}}-\frac{1}{2}.\frac{g}{4}.\left(\frac{16l}{3g}\right)(a_A=a_1=\frac{g}{4})$
or $S_A=4v\sqrt{\frac{l}{3g}}-\frac{2}{3}l$
Displacement of plank $S_B=u_{B}t-\frac{1}{2}{a}_B{t}^2$
or $S_B=4v\sqrt{\frac{l}{3g}}-\frac{1}{2}\left(\frac{5}{8}g\right)\left(\frac{16l}{3g}\right)$
or $S_B=4v\sqrt{\frac{l}{3g}}-\frac{5}{3}l$
Note: We can see that $s_A-s_B=l.$ Which is quite obvious because block A has moved a distance $l$ relative to plank.