On a smooth horizontal surface,  $n$ identical cubical blocks lie at rest parallel to each other along a line. The separation between the near surfaces of any two adjacent blocks is $L$. The block at one end is given a speed $\upsilon$ towards the next one at time $t=0.$ Given that all collisions are completely inelastic, the last block starts moving at a time  $t=x\left[\frac{n(n-1)L}{\upsilon }\right]$ Find the value of $x$.

Since, collision is perfectly inelastic, so all the block will stick together one by one and move in a form of  combined mass.
Time required to cover distance (d) by first block  $=\frac{L}{\upsilon }.$   Now first and second block will stick together and move with $\text{υ/2}$  velocity (by applying conservation of momentum) and combined system will take  $\frac{L}{\upsilon /2}=\frac{2L}{\upsilon }$ to  reach upto block  third.  Now, these three blocks will  move with velocity  $\text{υ/3}$ and  combined system will take time  $\frac{\text{L}}{\text{υ/3}}=\frac{3L}{\upsilon }$ to reach up to the  fourth block. So, total time $\frac{L}{\upsilon }+\frac{2L}{\upsilon }+\frac{3L}{\upsilon }+\mathrm{.....}$$\frac{\left(n-1\right)L}{\upsilon }=\frac{n\left(n-1\right)L}{2\upsilon }$     Final velocity of the centre of mass of the system will be   $\text{υ/n.}$