A small block of mass m is placed at the left end of a larger block...

Let the right direction be positive. Both blocks initially move to the right with velocity $v$.

The frictional force exerted by the road on the larger block $M$ acts towards the left:
$f_1 = \mu(M+m)g$

If both blocks were to move together, their common acceleration would be $-\mu g$. The force required to keep the smaller block $m$ moving with this acceleration is $-\mu mg$. However, the maximum static friction available between the blocks is $\dfrac{\mu}{2}mg$. Since the required force exceeds the available friction, the blocks will slip relative to each other.

The smaller block $m$ experiences kinetic friction from $M$ acting towards the left:
$f_2 = \dfrac{\mu}{2}mg$

By Newton's third law, the smaller block $m$ exerts an equal and opposite frictional force $f_2$ on $M$ towards the right.

The acceleration of the smaller block $m$ is:
$a_m = \dfrac{-f_2}{m} = -\dfrac{\mu g}{2}$

The net force on the larger block $M is the sum of the friction from the road and the friction from$m$:<br>$F_M = f_2 - f_1 = \dfrac{\mu}{2}mg - \mu(M+m)g$<br><br>The acceleration of the larger block$M$is:<br>$a_M = \dfrac{F_M}{M} = \dfrac{\dfrac{\mu}{2}mg - \mu Mg - \mu mg}{M} = -\mu g - \dfrac{\mu mg}{2M}$<br><br>The relative acceleration of$m$with respect to$M$is:<br>$a_{rel} = a_m - a_M = -\dfrac{\mu g}{2} - \left(-\mug - \dfrac{\mu mg}{2M}\right)<br><br>a_{rel} = $\dfrac{\mu g}{2} + \dfrac{\mu mg}{2M} = \dfrac{\mu g(M+m)}{2M}<br><br>The initial relative velocity is$u_{rel} = v - v = 0$. The relative distance the smaller block must cover to separate from the larger block is$s_{rel} = l$.<br><br>Using the kinematic equation$s_{rel} = u_{rel}t + $\dfrac{1}{2}$a_{rel}t^2$:<br>$l = 0 + $\dfrac{1}{2} \left[ \dfrac{\mu g(M+m)}{2M} \right]t^2$

$t^2 = \dfrac{4Ml}{\mu g(M+m)}$

$t = \sqrt{\dfrac{4Ml}{\mu g(M+m)}}$