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

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Solution

## 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: $m{a}_{1}+\frac{\mathrm{\mu }}{2}R=0\phantom{\rule{0ex}{0ex}}⇒m{a}_{1}=-\left(\frac{\mathrm{\mu }}{2}\right)mg=\left(\frac{\mathrm{\mu }}{2}\right)m×10\phantom{\rule{0ex}{0ex}}⇒{a}_{1}=-5\mathrm{\mu }$ 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 $⇒{a}_{2}=\frac{-\mathrm{\mu }mg-2\mu Mg}{2M}$ Substituting the values of a1 and a2 in equation (i), we get: $t=\sqrt{\frac{4Ml}{\left(M+m\right)\mathrm{\mu }g}}$

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