The two blocks A and B of equal mass are initially in contact when released from rest on the inclined plane. The coefficients of friction between the inclined plane and A and B are `mu_(1)` and `mu_(2)` respectiely. Friction is not sufficient to prevent slipping.

Options

A. If `mu_(1) gt m_(2)`, the blocks will always remain in contact
B. If `mu_(1) lt mu_(2)`, the blocks will slide down with different accelerations
C. If `mu_(1) gt mu_(2)`, the blocks will have a common acceleration `(1)/(2) (mu_(1) + mu_(2)) g sin theta`
D. If `mu_(1) gt mu_(2)` the blocks will have a common acceleration `(mu_(2) mu_(2) g)/(mu_(1) + mu_(2)) sin theta`

Answer

A, B, C

Explanation

`mg sin theta + R - N_(1) mu_(1) = ma`
`N_(1) = mg cos theta`
`mg sin theta - R - N_(2) mu_(2) = ma`
`N_(2) = mg cos theta`
`implies 2ma = 2m g sin theta - mu_(2) m g cos theta - mu_(2) m g cos theta`
`implies a = (g[2 sin theta - (mu_(1) + mu_(2)) cos theta])/(2)` and `R = (m g (mu_(1) - mu_(2)) cos theta)/(2)`
`implies a` is common and R is positive if `mu_(1) gt mu_(2)`
and if `mu_(1) lt mu_(2)` then they no longer remain in contact, A and B slides down with acceleration `g (sin theta - mu_(1) cos theta)` and `f (sin theta - mu_(2) co theta)` respectively.