Coefficient of friction between two block shown in figure ia `mu = 0.4`. The blocks are given velocities of `2 m//s` and `8 m//s` in the directions figure. Find

(a) The time when relative motion between them will stop
(b) the common velocities of blocks upto that instant. (c) Displacement of `1 kg`and `2 kg` block upto that instant `(g = 10 m//s^(2))`.

Explanation

`2 kg` block has relative motion towards right.Therefore , maximum friction will acts on it towards left
`f = muN = (0.4) (1) (10) = 4 N`
`a_(1) = (4)/(1) = 4 m//s^(2)`
`a_(2) = - (4)/(2) = 2 m//s^(2)`

(b) Relative motion between them will stop when ,
`v_(1) = v_(1)` ...(i)
`:. u_(1) + a_(1)t = u_(2) + a_(2)t` ...(ii)
`:. 2 + 4t = 8 - 2t` ...(iv) or `t = 1 s`
(b) Substituting value of 't' in Eq. (ii),either on `RHS` or on `LHS` common velocity `= 6 m//s`
( c) `s_(1) = u_(1)t + (1)/(2) a_(1) t^(2)`
`= (2) (1) + (1)/(2) (4) (1)^(2) = 4 m`
`s_(2) = u_(2)t + (1)/(2) a_(2) t^(2)`
`= (8) (1) + (1)/(2) (-2) (1)^(2) = 7 m`