Question

Coefficient of friction between two blocks shown in figure is $\mu =0.6$ The blocks are given velocities in the direction shown in figure. Find the displacements of $1kg$ and $2kg$ blocks upto that instant. (Take g= $10m/s^2$)

Answer 1

With respect to 1kg block, 2kg block is moving in backward direction at velocity of, $v_r=3+18=21m/s$, in the forward direction.

frictional force on 2kg block, in the backward direction is, $f=\mu N=0.6\times 2kg\times 10m/s^2=12N$

So, acceleration of 2kg block in ground's frame is, $a_2=f/2kg=6m{s}^{-2}$, in the backward direction.

Similar frictional force applies on 1kg block in the forward direction.

So, deceleration of 1kg block is, $a_1=f/1kg=12m/s^2$, in the forward direction.

With respect to 1kg block, there is pseudo force of $2kg\times a_1=24N$, on 2kg block, in the backward direction.

So, acceleration of 2kg block with respect to 1kg block is, $a_r=(24N+12N)/2kg=18m/s^2$, in the backward direction.

So time taken to stop relative motion is, $t=v_r/a_r=21m{s}^{-1}/18m{s}^{-2}=1.167s$

Now, displacement of 1kg block and 2kg block are given as,

$s_1=-18t+\frac{1}{2}{a}_1{t}^2=-18t+\frac{1}{2}12t^2$

$s_2=3t-\frac{1}{2}{a}_2{t}^2=3t-\frac{1}{2}6t^2$

At $t=1.167s$,

$s_1=-\frac{77}{6}m$

$s_2=-\frac{7}{12}m$