A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed v towards the next one at time t=0. All collisions are completely inelastic, then

Since collision is perfectly inelastic so all the blocks will stick together one by one and move in a form of combined mass

Time required to cover a distance $\text{'}L\text{'}$by first block $=\frac{L}{v}$

Now first and second block will stick together and move with $\mathrm{v}/2$ velocity (by applying conservation of momentum) and combined system will take time $\frac{L}{v/2}=\frac{2L}{v}$to reach up to block third.

So total time 

$\mathrm{t}=\frac{\mathrm{L}}{\mathrm{v}}+\frac{\mathrm{L}}{2\mathrm{v}}+\frac{\mathrm{L}}{3\mathrm{v}}+......\left(\mathrm{n}-1\right)$

$\mathrm{t}=\frac{\mathrm{n}\left(\mathrm{n}-1\right)\mathrm{L}}{2\mathrm{v}}$

As momentum is conserved 

$\mathrm{mv}=\mathrm{nmv}\text{'}$

$\mathrm{v}\text{'}=\mathrm{v}/\mathrm{n}$