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

The time elapsed before the smaller blocks separates is   $\begin{equation}T=\sqrt{\frac{4 m l}{(M+m) \mu g}}$

Explanation:

Step 1:  

Let a 1 and a 2 be the accelerations of ma and M correspondingly.

Here, a1 > a2 so that m moves on M

Suppose, after time  period ‘t’ m separate from M.

In this time, m covers v_t+1/2 a_1 t^2and s_M=v_t+1/2 a_2 t^2

For ‘m’ to m to ‘m’ separate from M.  

Step 2:

Again from free body diagram

$\begin{equation}\begin{aligned}v_{t}+\frac{1}{2} a_{1} t^{2} &=v_{t}+\frac{1}{2} a_{2} t^{2} \\m a_{1}+\frac{\mu}{2} R &=0 \\m a_{1} &=-\frac{\mu}{2} m g\end{aligned}$

g= 10 m/$s^2$

$\begin{equation}\begin{aligned}&m a_{1}=-\frac{\mu}{2} m \times 10\\&a_{1}=-\frac{\mu}{2} \frac{m}{m} \times 10\\&\alpha_{1}=-\frac{\mu}{2} \times 10\\&a_{1}=-5 \mu\end{aligned}$

Step 3:

Again,

$\begin{equation}\begin{aligned}M a_{2}+\mu(M+m) g-\mu_{2} m g &=0 \\2 M a_{2}+\mu m g-2 \mu M g-2 \mu m g &=0 \\a_{2}=\frac{-\mu m g-2 \mu m g}{2 M} &\end{aligned}$

$\begin{equation}T=\sqrt{\frac{4 m l}{(M+m) \mu g}}\end$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

Please refer the above attachment