coefficient of friction between two blocks shown in figure is `mu =0.6`. The blocks are given velocities in the direction shown in figure. Find
(a) the time when relative motion between them is stopped.
(b) the common velocity of the two blocks.
(c) the displacement of `1kg` and `2kg` blocks upto that instant. `(Take g =10m//s^(2))`

Explanation

Relstive motion of `2kg` block is toward right. Therefore maximum friction on this block will act towards left

(a)Relstive motionbetween them will slop when,
`u_(1) =u_(2)` or `u_(1) +a_(1)t =u_(2)+a_(2)t`
or `-18+12t =3-6t` ...(i)
Solving we get `t =(7)/(6) s`
(b)Substituting value of 't' in Eq. (i) either on `RHS` or on `LHS` we have,
common velocity `=-4m//s`
`s_(t) =u_(1) t+(1)/(2)a_(1)t^(2)`
`= (-18) ((7)/(6)) +(1)/(2) (12) ((7)/(6))^(2)`
` =-12.83m`
`s_(2) =u_(2) t+(1)/(2)a_(2)t^(2)`
`= (3) ((7)/(6)) +(1)/(2) (-6) ((7)/(6))^(2)`
` =-0.58m`