question_answer

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]            

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  \[\frac{v}{n}\]

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 \[\frac{v}{n}\].