The surface below 2 kg is smooth but there is friction between the blocks. Initially the blocks are static if 6N force is applied on both the blocks, then the direction of friction on

Options

A. 2kg block is left
B. 3kg block is right
C. Kinetic friction is surely acting
D. Data insufficient (`mu` not given)

Answer

A, C

Explanation

If `a_(1)` and `a_(2)` be the acceleration of blocks on incline then,
`a_(1)=gsintheta-mu_(1)g costheta`
`a_(2)=gsintheta-mu_(2)g costheta`
`t=sqrt((2d)/((mu_(2)-mu_(1))gcos theta))`
If the collision is to occur on incline, this 't' must be less than the time that block I would take to slide down the entire incline if given a chance to move alone. As that is the case, velocity `v_(1)` and `v_(2)` of bodies just before collision
`v_(1)=a_(1)t`
`v_(2) =a_(2)t`
If v be the common velocity just after collision
`v=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))`
Loss of kinetic energy = `(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)-(1)/(2)(m_(1)+m_(2))v^(2)`
Putting the value and solving we get
Loss in K.E. = 0.16 Joules.