A plank of mass 10kg rests on a smooth horizontal surface. Two blocks A and B of masses `m_(A)=2kg and m_(B)=1kg` lies at a distance of 3m on the plank as shown in the distance of 3m on the plank as shown in the figure. The friction coefficient between the blocks and plank are `mu_(A)=0.3 and mu_(B)=0.1`. Now a force F = 15 N is applied to the plank in horizontal direction. Find the time after which block A collides with B.

Explanation

Maximum friction force between the plank (P) and A is `F_(A)=mu_(A)m_(A)g=6N`
Maximum friction force between the plank (P) and B is `F_(8)=mu_(B)m_(B)g=1N`
Let system moves together with acceleration a then `a=(F)/(m_(A)+m_(B)+M)=(15)/(13)m//s`
Now, friction force on `m_(a), f_(B)=m_(B)a=(15)/(13)N`
Friction force on `m_(A), f_(A)=m_(A)a=(30)/(13)N`
As `f_(B) gt F_(B) rArr` Hence B will not move with plank.
`f_(A) lt F_(A) rArr` Hence A will move with plank.
`a_(B)=(F_(B))/(m_(B))=1m//s^(2)`
`A_(P)=(F-F_(B))/(m_(A)+M)=(7)/(6)m//s^(2)`
Now F.B.D. of the plank and A & B are Acceleration of B w.r.t. the plank

`a_(BP)=1-(7)/(6)= -(1)/(6)m//s^(2)`
Now `L_(BP)=(1)/(2)a_(Bv)t^(2) rArr -3=(1)/(2)xx(-(1)/(6))t^(2)`
Time of collision t = 6 sec
Alternate
In ground frame, at time of collision:
`(1)/(2)A_(p)t^(2)=L+(1)/(2)a_(B)t^(2) rArr (1)/(2) (A_(p)-a_(B))t^(2)=L rArr (1)/(2) ((7)/(6)-1)t^(2)=3 rArr t=6 sec`