Two blocks `A` and `B` of equal masses are placed on rough inclined plane as shown in Fig. When and where will the two blocks come on the same line on the inclined plane if they are released simultaneously ? Initilly the blocks `A` is `sqrt(2) m` behind the block `B` .Coefficient of kinetic friction for the block `A` and `B` are `0.2` and `0.3` respectively `(g = 10 ms^(-2))`

Explanation

`a = (mg sin theta - mu_(k) mg cos theta)/(m)`
`:. a_(A) = g sin theta - mu_(k.A) g cos theta = 4sqrt(2)ms^(-2)`
and `a_(B) = g sin theta - mu_(k.B) g cos theta = 3.5 sqrt(2)ms^(-2)`
Putting values , we get
`a_(AB)` is relative acceleration of `A` w.r.t
`B = a_(A) - a_(B) = (1)/(sqrt(2))ms^(-2)`
`L= sqrt(2) m,L = (1)/(2) a_(A//B)t^(2) rArr t = 2s`
Distance moved by `B` during that time is given by
`S_(B) = (1)/(2) a_(B)t^(2) = (1)/(2) 3.5 sqrt(2) xx 4 = (2 xx 0.7)/(sqrt(2)) xx 10 = 7sqrt(2)m`
Similarly for `A, S_(A) = 8sqrt(2)m`