A long block A is at rest on a smooth horizontal surface. A small block B whose mass is half of mass of A is placed on A at one end and is given an initial velocity u as shown in figure. The coefficient of friction between the blocks is `mu`.

Answer

`5/3 sqrt(6 mu g d)`

Explanation

Since the collision is elastic in nature, applying conservation of linear momentum and conservation of kinetic energy
`mv = (4m) u + mv'`
Where u and are the velocities of the block B and mass m
`implies v' v - 4u " "….(i)`
Also, `1/2 m v^2 = 1/2 (4m) u^2 + 1/2 m v'^2`
`implies v^2 = 4u^2 + v'^2 " "....(ii)`
From (i) to (ii) `v^2 = 4 u^2 + (v - 4u)^2 implies u = (2v)/5`
For block A to topple, B should move a distance 2 d. The retardation produced in B due to friction force between B and the table.
`f = mu N = (4m) a implies mu [ 6 mg] = (4m) a implies a = 1.5 mu g`.
For the motion of B.
`u = (2v)/5`, final velocity = 0, `s = 2d, a = - 1.5 mu g`
Now using , `v^2 - u^2 = 2as implies (0)^2 - ((2v)/(5))^(2) = 2(-1.5 mu g) 2d implies v = 5/2 sqrt(6 mu g d) = v_0`