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 $\mu =\mathrm{0.5.}$
(Take $g=10m/s^2$ and consider $4kg$ block doesn't fall on ground)

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

Answer 1

Answer: 10.80

$a_2=-\frac{20}{4}=-5m/s^2$

$a_1=\frac{20}{2}=+10m/s^2$
Relative motion stopped when $v_1=v_2$, i.e.,
$-6+10t=12-5t$
$t=\frac{18}{15}=1.2s$
Common velocity $=-6+10t$
$=-6+10\times 1.2=6m/s$
Displacement of $4kg$ block
$=\frac{12\times 6}{5}-\frac{1}{2}\times 5\times {\left(\frac{6}{5}\right)}^2$
$=\frac{72}{5}-\frac{36}{10}=10.8m$