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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

The last block starts moving at time $t = (n - 1) \frac{L}{v}$
The last block starts moving at time $t = \frac{(n - 1) L}{2 v}$
The centre of mass of the system will have a final speed $v$
The centre of mass of the system will have a final speed $v / n$

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 $n$ th 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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