Question

Coefficient of friction between two blocks shown in figure is $\mu =\mathrm{0.4.}$ The blocks gives velocities of $2m/s$ and $8m/s$ in the directions shown in figure. Find displacements of $1kg$ and $2kg$ blocks upto the instant when the relative motion between the blocks will stop. (g= $10m/s^2$).

Answer 1

With respect to 2kg block, 1kg block is moving in backward direction at velocity of, $v_r=2-8=-6m/s$

frictional force on 1kg block, in the forward direction is, $f=\mu N=0.4\times 1kg\times 10m/s^2=4N$

So, acceleration of 1kg block in ground's frame is, $a_1=f/1kg=4m{s}^{-2}$, in the forward direction.

Similar frictional force applies on 2kg block in the backward direction.

So, deceleration of 2kg block is, $a_2=f/2kg=2m/s^2$, in the backward direction.

With respect to 2kg block, there is pseudo force of $1kg\times a_2=2N$, on 1kg block, in the forward direction.

So, acceleration of 1kg block with respect to 2kg block is, $a_r=(4N+2N)/1kg=6m/s^2$, in the forward direction.

So time taken to stop relative motion is, $t=v_r/a_r=6m{s}^{-1}/6m{s}^{-2}=1s$

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

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

$s_2=8t-\frac{1}{2}{a}_2{t}^2$

At $t=1s$,

$s_1=4m$

$s_2=7m$