Two block A and B of equal masses are placed on rough inclined plane as shown in figure. When and where will the two blocks come on the same line on the inclined plane if they are released simultaneously? Initially the block A is `sqrt2` m behind the block B. Co-efficient of kinetic friction for the blocks A and B are 0.2 and 0.3 respectively `(g=10m//s^2)`.

Answer

B, C

Explanation

`a=(mgsintheta-mu_k mg cos theta)/(m)`
`:.` `a_A=g sin theta-mu_(k,A)g cos theta` …(i)
and `a_B= g sin theta -mu_(k,B) g cos theta` …(ii)
Putting values we get
`a_A=4sqrt2m//s^2` and `a_B=3.5sqrt2m//s^2`
Let `a_(AB)` is relative acceleration of A w.r.t. B. Then
`a_(AB)=a_A-a_B`
`L=sqrt2m`
[where L is the relative distance between A and B]
Then `L=(1)/(2)a_(AB^(t^2))`
or `t^2=(2L)/(a_(AB))=(2L)/(a_A-a_B)`
Putting values we get, `t^2=4` or `t=2s`.
Distantce moved by B during that time is given by
`S=(1)/(2)a_(B^(t^2))=1/2xx3.5sqrt2xx4=7sqrt2m`
Similarly for `A=8sqrt2m`.