Two identical blocks A and B of mass m joined together with a massless spring as shown in figure are placed on a smooth surface. If the block A moves with an acceleration `a_0`, then the acceleration of the block B is

Answer

`6d sqrt(3 mu)`

Explanation

Let velocity of mass m just after collision with the block B is `v_1`.
We know for elastic collision between two bodies
`m (2v_0) = 4 m u + m (-v_1)`
`implies 4u - v_1 = 2v_0 " "….(i)`
`u + v_1 = 2v_0" "…..(ii)`
From (i) and (ii)
`u = 4 v_0//5 and v_1 = (6v_0)/(5) = 6/5 5/2 sqrt(6 m u g d) = 3sqrt(6 m u g d)`
It then follows a projectile motion. Considering the vertical direction motion of this projectile.
`implies d = 1/2 gt^2 implies t = sqrt((2d)/(g))`
The horizontal distance traveled by mass m during this time t
`x = u_x t = 3sqrt(6 mu g d) = sqrt((2d)/(g)) = 6sqrt(3 mu d^2) = 6dsqrt(3mu)`