A small block of mass M moves with velocity 5 m/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 the blocks, When both of them come to rest, is (g=10 m//s^(2))

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

1.	A small block of mass M moves with velocity 5 m/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 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)=2.5 m//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 separation will be `(5-2)=3m`