Coefficient of friction between two blocks shown in figure is mu=0.6. The blocks are  given velocities in the directions shown in figure. Find the time when relative motion between them is stopped.

 

 Relative motion of2kg block is towards right. Therefore, maximum friction on this block will act towards left

$12N\stackrel{2kg}{⟵}{a}_2\to 3 \mathrm{m}/\mathrm{s}$

$$\begin{gathered} 18 \mathrm{m}/\mathrm{s}\stackrel{1 \mathrm{kg}}{⟶}{\mathrm{B}}_1 \mathrm{N} \\ f=\mu \mathrm{N}=(0.6)\left(2\right)\left(10\right)=12 \mathrm{N} \\ {a}_2=-\frac{12}{2}=-6 \mathrm{m}/{\mathrm{s}}^2 \\ {a}_1=\frac{12}{1}=12 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 Relative motion between them will stop when,

or

$$\begin{gathered} v_1=v_2\text{ or }{u}_1+a_{1}t=u_2+a_{2}t \\ -18+12t=3-6t \end{gathered}$$                                                                                    ……………. (i)

Solving we get,

$t=\frac{7}{6} \mathrm{s}$