Physics-
General
Easy

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 equals 0. All collision are completely elastic. Then

  1. The last block starts moving at time t equals open parentheses n minus 1 close parentheses fraction numerator L over denominator v end fraction    
  2. The last block starts moving at time t equals fraction numerator open parentheses n minus 1 close parentheses L over denominator 2 v end fraction    
  3. The centre of mass of the system will have a final speed v    
  4. The centre of mass of the system will have a final speed v divided by n    

The correct answer is: The centre of mass of the system will have a final speed v divided by n


    Time taken by first block to reach second block equals fraction numerator L over denominator v end fraction
    Since collision is 100% elastic, now first block comes to rest and 2nd block starts moving towards the 3rd block with a velocity v and takes time equals fraction numerator L over denominator v end fraction to reach 3rd block and so on
    therefore Total time equals t plus t plus... left parenthesis n minus 1 right parenthesis time equals open parentheses n minus 1 close parentheses fraction numerator L over denominator v end fraction
    Finally only the last nth block is in motion velocity v, hence final velocity of centre of mass
    v subscript C M end subscript equals fraction numerator m v over denominator n m end fraction equals fraction numerator v over denominator n end fraction

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