A small block of mass $M$ moves with velocity 5 $ms^{- 1}$ towards n another block of same mass $M$ placed at a distance of 2 m on a rough horizontal surface. Coefficient of friction between the block and ground is 0.25. Collision between the two blocks is elastic, the separation between the blocks, when both of them come to rest, is (g=10 $ms^{- 2}$)

Question

Dr. Abubacker · Accepted Answer

(a) Retardation due to friction
$a = \mu g = (0.25)(10)$
$= 2.5\ ms^{- 2}$
Collision is elastic, i.e. after collision first block comes to rest and the second block acquires the velocity of first block. Or we can understand it is this manner that second block is permanently at rest while only the first block moves. Distance travelled by it will be
$s = \frac{v^{2}}{2a} = \frac{(5)^{2}}{(2)(2.5)} = 5$m
$	herefore$ Final separation will be ($s - 2$)=3 m

Explanation

Reviewed by Dr. Abubacker

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