Question

The blocks are given velocity in the direction shown in the figure. The coefficient of friction between the two blocks shown in the figure is $\mu=0.5 .$
(Take $g =10\, m / s ^{2}$ and consider $4\, kg$ block doesn't fall on ground)

The displacement (in $m$ ) of the $4 \,kg$ block when relative motion stops is

Answer 1

$a_{2}=-\frac{20}{4}=-5\, m / s ^{2}$

$a_{1}=\frac{20}{2}=+10\, m / s ^{2}$
Relative motion stopped when $v_{1}=v_{2}$, i.e.,
$-6+10\, t=12-5\, t$
$t=\frac{18}{15}=1.2 s$
Common velocity $=-6+10 t$
$=-6+10 \times 1.2=6\, m / s$
Displacement of $4 \,kg$ block
$=\frac{12 \times 6}{5}-\frac{1}{2} \times 5 \times\left(\frac{6}{5}\right)^{2}$
$=\frac{72}{5}-\frac{36}{10}=10.8\, m$