Work Energy And Power Question 78
Question: 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 [IIT 1995]
Options:
A) The last block starts moving at $ t=\frac{(n-1)L}{v} $
B) The last block starts moving at $ t=\frac{n(n-1)L}{2v} $
C) The centre of mass of the system will have a final speed v
D) The centre of mass of the system will have a final speed $ \frac{v}{n} $
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Answer:
Correct Answer: B
Solution:
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} $ .
BETA
