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
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
The last block starts moving at $t = \frac{{(n - 1)L}}{v}$
The last block starts moving at $t = \frac{{n(n - 1)L}}{{2v}}$
The centre of mass of the system will have a final speed $v$
None of the above
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
(b,d)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 $‘L’$ by first block $ = \frac{L}{v}$
Now first and second block will stick together and move with $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.
Now these three blocks will move with velocity $v/3 $ and combined system will take time $\frac{L}{{v/3}} = \frac{{3L}}{v}$ to reach upto the block fourth.
So, total time $ = \frac{L}{v} + \frac{{2L}}{v} + \frac{{3L}}{v} + ...\frac{{(n - 1)L}}{v}$$ = \frac{{n(n - 1)L}}{{2v}}$
and velocity of combined system having $n$ blocks as $\frac{v}{n}$.
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