A small block of mass `m` is projected on a larger block of mass `10 m` and length `l` with a velocity `v` as shown in the figure. The coefficient of friction between the two block is `mu_(2)` while that between the lower block and the ground is `mu_(1)`. Given that `mu_(2) gt 11 mu_(1)`.

(a) Find the minimum value of `v`, such that the mass `m` falls off the block of mass `10 m` .
(b) If `v` has minimum value, find the time taken by block `m` to do so.

Explanation

(a) Force of friction at different contacts are shown in figure
Here `f_(1) = mu_(2) mg`
and `f_(2) = mu_(1) (11 mg)`
Given that `mu_(2) gt 11 mu_(1)`
`:. f_(1) gt f_(2)`
Retardation of upper block
`a_(1) = (f_(1))/(m) = mu_(2) g`
Acceleration of lower block
`a_(2) = (f_(1) - f_(2))/(m) = ((mu_(2) - 11 mu_(1))1 g)/(10)`
Relative retardation of upper block
`a_(r) = a_(1) + a_(2)` or `a_(r) = (11)/(10) (mu_(2) - mu_(1)) g`
Now `0 = nu_(min)^(2) - 2 a_(r) l`
`:. nu_(min) = sqrt(2 2 a_(r) l) = sqrt((22(mu_(2) - mu_(1)) g l)/(10)`
(b) `0 = nu_(min) - a_(r) t`
or `t = (nu_(min))/( a_(r)) = sqrt((20 l)/(11(mu_(2) - mu_(1)) g)`