A small block of mass M move with velocity 5m/s towards another block of same mass M placed at a distance of 2m on a rough horizontal surface. Coefficient of friction between the blocks 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 m//s^(2))

A small block of mass M move with velocity 5m/s towards another block

1.	A small block of mass M move with velocity 5m/s towards another block of same mass M placed at a distance of 2m on a rough horizontal surface. Coefficient of friction between the blocks 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 m//s^(2))
Answer» SOLUTION :Retardation due to FRICTION `= mu g= (0.25) (10) = 2.5M//s^(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 in this manner that second block is PERMANENTLY at rest while only the first block moves. Distance travelled by it will be `s= (v^(2))/(2a)= ((5)^(2))/((2)(2.5))= 5m` `therefore` Final separaton will be `(5-2)= 3m`