Question

Figure (6−E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is μ and that between the block is μ/2. Find the time elapsed before the smaller blocks separates from the bigger block.

Figure

Answer 1

Let a₁ and a₂ be the accelerations of masses m and M, respectively.
Also, a₁ > a₂ so that mass m moves on mass M.

Let after time t, mass m is separated from mass M.

Using the equation of motion
During this time, mass m covers $vt+\frac{1}{2}{a}_1{t}^2$ and ${\mathrm{s}}_m=vt+\frac{1}{2}{a}_2{t}^2$.

For mass m to separate from mass M, we have:
$vt+\frac{1}{2}{a}_1{t}^2=vt+\frac{1}{2}{a}_2{t}^2+1....\left(\mathrm{ii}\right)$




From the free body diagram, we have:

$$\begin{gathered} m{a}_1+\frac{\mathrm{\mu }}{2}R=0 \\ \Rightarrow m{a}_1=-\left(\frac{\mathrm{\mu }}{2}\right)mg=\left(\frac{\mathrm{\mu }}{2}\right)m\times 10 \\ \Rightarrow a_1=-5\mathrm{\mu } \end{gathered}$$

Again,
$M{a}_2+\mathrm{\mu }\left(M+m\right)g-\left(\frac{\mathrm{\mu }}{2}\right)mg=0$
⇒ 2Ma₂ + 2μ (M + m)g − μmg = 0
⇒ 2Ma₂ = μmg − 2μmg − 2μmg
$\Rightarrow a_2=\frac{-\mathrm{\mu }mg-2\mu Mg}{2M}$

Substituting the values of a₁ and a₂ in equation (i), we get:
$t=\sqrt{\frac{4Ml}{\left(M+m\right)\mathrm{\mu }g}}$